Category: Class 12

Students those who are preparing for the 12th Physics exam can download this Samacheer Kalvi 12th Physics Book Solutions Questions and Answers for Chapter 2 Current Electricity from here free of cost. These Tamilnadu State Board Solutions for Class 12th Physics PDF cover all Chapter 2 Current Electricity Questions and Answers. All these concepts of Samacheer Kalvi 12th Physics Book Solutions Questions and Answers are explained very conceptually as per the board prescribed Syllabus & guidelines.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 2 Current Electricity

Kickstart your preparation by using this Tamilnadu State Board Solutions for 12th Physics Chapter 2 Current Electricity Questions and Answers and get the max score in the exams. You can cover all the topics of Chapter 2 Current Electricity Questions and Answers pdf. Download the Tamilnadu State Board Solutions by accessing the links provided here and ace up your preparation.

Samacheer Kalvi 12th Physics Current Electricity Textual Evaluation Solved

Samacheer Kalvi 12th Physics Current Electricity Multiple Choice Questions   

Current Electricity Class 12 Problems With Solutions State Board Question 1.
The following graph shows current versus voltage values of some unknown conductor. What is the resistance of this conductor?

(a) 2 ohm
(b) 4 ohm
(c) 8 ohm
(d) 1 ohm
Answer:
(a) 2 ohm

Current Electricity Class 12 State Board Question 2.
A wire of resistance 2 ohms per meter is bent to form a circle of radius 1 m. The equivalent resistance between its two diametrically opposite points, A and B as shown in the figure is-

(a) π Ω
(b) \(\frac { π }{ 2 }\) Ω
(c) 2π Ω
(d) \(\frac { π }{ 4 }\) Ω
Answer:
(b) \(\frac { π }{ 2 }\) Ω

12th Physics Chapter 2 Book Back Answers Question 3.
A toaster operating at 240 V has a resistance of 120 Ω. The power is
(a) 400 W
(b) 2 W
(c) 480 W
(d) 240 W
Answer:
(c) 480 W

12th Physics Samacheer Kalvi Question 4.
A carbon resistor of (47 ± 4.7) k Ω to be marked with rings of different colours for its identification. The colour code sequence will be
(a) Yellow – Green – Violet – Gold
(b) Yellow – Violet – Orange – Silver
(c) Violet – Yellow – Orange – Silver
(d) Green – Orange – Violet – Gold
Answer:
(b) Yellow – Violet – Orange – Silver

Samacheer Kalvi 12th Physics Question 5.
What is the value of resistance of the following resistor?

(a) 100 k Ω
(b) 10 k Ω M
(c) 1 k Ω
(d) 1000 k Ω
Answer:
(a) 100 k Ω

Class 12 Physics Samacheer Kalvi Question 6.
Two wires of A and B with circular cross section made up of the same material with equal lengths. Suppose R_A = 3 R_B, then what is the ratio of radius of wire A to that of B?
(a) 3
(b) √3
(c) \(\frac { 1 }{ √3 }\)
(d) \(\frac { 1 }{ 3 }\)
Answer:
(c) \(\frac { 1 }{ √3 }\)

Current Electricity Short Notes Question 7.
A wire connected to a power supply of 230 V has power dissipation P₁ Suppose the wire is cut into two equal pieces and connected parallel to the same power supply. In this case power dissipation is P₂. The ratio \(\frac {{ p }_{2}}{{ p }_{1}}\) is.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Samacheer Kalvi Guru 12th Physics Question 8.
In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60W bulb for use in India is R, the resistance of a 60W bulb for use in USA will be
(a) R
(b) 2R
(c) \(\frac { R }{ 4 }\)
(d) \(\frac { R }{ 2 }\)
Answer:
(c) \(\frac { R }{ 4 }\)

Samacheer Kalvi 12th Physics Solutions Question 9.
In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1kW are connected. The voltage of electric mains is 220 V. The minimum capacity of the main fuse of the building will be (IIT-JEE 2014)
(a) 14 A
(b) 8 A
(c) 10 A
(d) 12 A
Answer:
(d) 12 A

Samacheerkalvi.Guru 12th Physics Question 10.
There is a current of 1.0 A in the circuit shown below. What is the resistance of P ?

(a) 1.5 Ω
(b) 2.5 Ω
(c) 3.5 Ω
(d) 4.5 Ω
Answer:
(c) 3.5 Ω

Samacheer Kalvi Physics Question 11.
What is the current out of the battery?

(а) 1 A
(b) 2 A
(c) 3 A
(d) 4 A
Answer:
(а) 1 A

12th Physics Solutions Samacheer Kalvi Question 12.
The temperature coefficient of resistance of a wire is 0.00125 per °C. At 300 K, its resistance is 1 Ω. The resistance of the wire will be 2 Ω. at
(a) 1154 K
(ft) 1100 K
(c) 1400 K
(d) 1127 K
Answer:
(d) 1127 K

Samacheer Kalvi 12th Physics Solution Book Question 13.
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10Ω is
(a) 0.2 Ω
(b) 0.5 Ω
(c) 0.8 Ω
(d) 1.0 Ω
Answer:
(b) 0.5 Ω

Samacheer Kalvi Guru Physics Question 14.
A piece of copper and another of germanium are cooled from room temperature to 80 K. The resistance of
(a) each of them increases
(b) each of them decreases
(c) copper increases and germanium decreases
(d) copper decreases and germanium increases
Answer:
(d) copper decreases and germanium increases

Physics Class 12 Samacheer Kalvi Question 15.
In Joule’s heating law, when I and t are constant, if the H is taken along the y axis and I² along the x axis, the graph is
(a) straight line
(b) parabola
(c) circle
(d) ellipse
Answer:
(a) straight line

Samacheer Kalvi 12th Physics Current Electricity Short Answer Questions

Samacheer Kalvi.Guru 12th Physics Question 1.
Why current is a scalar?
Answer:
The current I is defined as the scalar product of current density and area vector in which the charges cross.
I = \(\vec { j } \) . \(\vec { A } \)
The dot product of two vector quantity is a scalar form. Hence, current is called as a scalar quantity.

12th Samacheer Physics Solutions Question 2.
Distinguish between drift velocity and mobility.
Answer:

Physics Solution Class 12 Samacheer Kalvi Question 3.
State microscopic form of Ohm’s law.
Answer:
Current density J at a point in a conductor is the amount of current flowing per unit area of the conductor around that point provided the area is held in a direction normal to the current.

Current Electricity Pdf Question 4.
State macroscopic form of Ohm’s law.
Answer:
The macroscopic form of Ohm’s Law relates voltage, current and resistance. Ohm’s Law states that the current through an object is proportional to the voltage across it and inversely proportional to the object’s resistance.
V = IR.

Current Electricity Class 12 Question 5.
What are ohmic and non-ohmic devices?
Answer:
Materials for which the current against voltage graph is a straight line through the origin, are said to obey Ohm’s law and their behaviour is said to be ohmic. Materials or devices that do not follow Ohm’s law are said to be non-ohmic.

Samacheer Kalvi Class 12 Physics Solutions Question 6.
Define electrical resistivity.
Answer:
Electrical resistivity of a material is defined as the resistance offered to current flow by a conductor of unit length having unit area of cross section.

Question 7.
Define temperature coefficient of resistance.
Answer:
It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity at T₀

Question 8.
What is superconductivity?
Answer:
The ability of certain metals, their compounds and alloys to conduct electricity with zero resistance at very low temperatures is called superconductivity.

Question 9.
What is electric power and electric energy?
Answer:
1. Electric power:
It is the rate at which an electric appliance converts electric energy into other forms of energy. Or, it is the rate at which work is done by a source of emf in maintaining an electric current through a circuit.
P = \(\frac { W }{ t }\) = VI = I²R = \(\frac {{ V }_{2}}{ R }\)

2. Electric energy:
It is the total workdone in maintaining an electric current in an electric circuit for a given time.
W = Pt = VIt joule = I²Rt joule.

Question 10.
Define current density.
Answer:
The current density (J) is defined as the current per unit area of cross section of the conductor
J = \(\frac { 1 }{ A }\)
The S.I unit of current density.
\(\frac { A }{{ m }^{2}}\)
Or
Am²

Question 11.
Derive the expression for power P = VI in electrical circuit.
Answer:
The electrical power P is the rate at which the electrical potential energy is delivered,
P = \(\frac { dU }{ dt }\) = \(\frac { d }{ dt }\) (V.dQ) = V\(\frac { dQ }{ dt }\)
Since the electric current I = \(\frac { dQ }{ dt }\)
So the equation can be rewritten as P = VI.

Question 12.
Write down the various forms of expression for power in electrical circuit.
Answer:
The electric power P is the rate at which the electrical potential energy is delivered,
P = \(\frac { dU }{ dt }\) = \(\frac { 1 }{ dt }\) (V.dQ) = V.\(\frac { dQ }{ dt }\)
[dU = V.dQ]
The electric power delivered by the battery to any electrical system.
P = VI
The electric power delivered to the resistance R is expressed in other forms.
P = VI = I(IR) = I²R
P = IV = \((\frac { V }{ R })\) V = \(\frac {{ V }^{2}}{ R }\).

Question 13.
State Kirchhoff’s current rule.
Answer:
It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a statement of conservation of electric charge.

Question 14.
State Kirchhoff’s voltage rule.
Answer:
It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system.

Question 15.
State the principle of potentiometer.
Answer:
The basic principle of a potentiometer is that when a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.

Question 16.
What do you mean by internal resistance of a cell?
Answer:
The resistance offered by the electrolyte of a cell to the flow of current between its electrodes is called internal resistance of the cell. An ideal battery has zero internal resistance and the potential difference across the battery equals to its emf. But a real battery is made of electrodes and electrolyte, there is resistance to the flow of charges within the battery. A freshly prepared cell has low internal resistance and it increases with ageing.

Question 17.
State Joule’s law of heating.
Answer:
It states that the heat developed in an electrical circuit due to the flow of current varies directly as:

1. the square of the current
2. the resistance of the circuit and
3. the time of flow.
   H = I²R?

Question 18.
What is Seebeck effect?
Answer:
Seebeck discovered that in a closed circuit consisting of two dissimilar metals, when the junctions are maintained at different temperatures, an emf (potential difference) is developed.

Question 19.
What is Thomson effect?
Answer:
Thomson showed that if two points in a conductor are at different temperatures, the density of electrons at these points will differ and as a result the potential difference is created between these points. Thomson effect is also reversible.

Question 20.
What is Peltier effect?
Answer:
When an electric current is passed through a circuit of a thermocouple, heat is evolved at one junction and absorbed at the other junction. This is known as Peltier effect.

Question 21.
State the applications of Seebeck effect.
Answer:
Applications of Seebeck effect:

1. Seebeck effect is used in thermoelectric generators (Seebeck generators). These thermoelectric generators are used in power plants to convert waste heat into electricity.
2. This effect is utilised in automobiles as automotive thermoelectric generators for increasing fuel efficiency.
3. Seebeck effect is used in thermocouples and thermopiles to measure the temperature difference between the two objects.

Samacheer Kalvi 12th Physics Current Electricity Long Answer Questions

Question 1.
Describe the microscopic model of current and obtain genera! form of Ohm’s Law.
Answer:
Microscopic model of current: Consider a conductor with area of cross-section A and an electric field E applied from right to left. Suppose there are n electrons per unit volume in the conductor and assume that all the electrons move with the same drift velocity \(\vec { v } \)_d.
The drift velocity of the electrons = v_d
The electrons move through a distance dx within a small interval of dt

v_d = \(\frac { dx }{ dt }\); dx = v_ddt ….. (1)
Since A is the area of cross section of the conductor, the electrons available in the volume or length dx is
= volume x number per unit volume
= A dx × n …… (2)
Substituting for dx from equation (1) in (2)
= (A v_d dt)n
Total charge in volume element dQ = (charge) x (number of electrons in the volume element)
dQ= (e)(A v_d dt)n
Hence the current, I = \(\frac { dQ }{ dt }\) = \(\frac{n e A v_{d} d t}{d t}\)
I = ne A v_d …….. (3)
Current denshy (J):
The current density (J) is defined as the current per unit area of cross section of the conductor
J = \(\frac { I }{ A }\)
The S.I. unit of current density,\(\frac { A }{{ m }^{2}}\) (or) Am^-2
J = \(\frac {{ neA v }_{d}}{ A }\) (from equation 3)
J = nev_d …….. (4)
The above expression is valid only when the direction of the current is perpendicular to the area A. In general, the current density is a vector quantity and it is given by
\(\vec { J } \) = ne\(\vec { v } \)_d
Substituting i from equation \(\vec { v } \)_d = \(\frac { -eτ }{ m }\) \(\vec { E } \)
\(\vec { J } \) = –\(\frac{n \cdot e^{2} \tau}{m}\)\(\vec { E } \) …… (5)
\(\vec { J } \) = -σ\(\vec { E } \)
But conventionally, we take the direction of (conventional) current density as the direction of electric field. So the above equation becomes
\(\vec { J } \) = σ\(\vec { E } \) …….. (6)
where σ = \(\frac{n \cdot e^{2} \tau}{m}\) is called conductivity.
The equation 6 is called microscopic form of ohm’s law.

Question 2.
Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.
Answer:
Ohm’s law: The Ohm’s law can be derived from the equation J = σE. Consider a segment of wire of length l and cross sectional area A.

When a potential difference V is applied across the wire, a net electric field is created in the wire which constitutes the current. For simplicity, we assume that the electric field is uniform in the entire length of the wire, the potential difference (voltage V) can be written as V = El
As we know, the magnitude of current density
J = σE = σ\(\frac { V }{ l }\) ……. (1)
But J = \(\frac { I }{ A }\), so we write the equation as
\(\frac { I }{ A }\) σ\(\frac { V }{ l }\)
By rearranging the above equation, we get
V = I \(\left( \frac { l }{ \sigma A } \right) \) ……… (2)
The quantity \(\frac { l }{ \sigma A } \)is called resistance of the conductor and it is denoted as R. Note that the oA resistance is directly proportional to the length of the conductor and inversely proportional to area of cross section.
Therefore, the macroscopic form of Ohm’s law can be stated as
V = IR.

Question 3.
Explain the equivalent resistance of a series and parallel resistor network.
Answer:
1. Resistors in series:
When two or more resistors are connected end to end, they are said to be in series. The resistors could be simple resistors or bulbs or heating elements or other devices. Fig. (a) shows three resistors R₁, R₂ and R₃ connected in series.

The amount of charge passing through resistor R₁ must also pass through resistors R₂ and R₃ since the charges cannot accumulate anywhere in the circuit. Due to this reason, the current I passing through all the three resistors is the same. According to Ohm’s law, if same current pass through different resistors of different values, then the potential difference across each resistor must be different. Let V₁, V₂ and V₃ be the potential difference (voltage) across each of the resistors R₁, R₂ and R₃ respectively, then we can write V₁ = IR_1, V₂ = IR₂ and V₃= IR₃. But the total voltage V is equal to the sum of voltages across each resistor.
V = V₁ + V₂ + V₃
= IR₁ + IR₂ + IR₃ ….. (1)
V = I(R₁ + R₂ +R₃)
V = I.R_S …… (2)
where R_s = R₁ + R₂ R₃ ……. (3)
When several resistances are connected in series, the total or equivalent resistance is the sum of the individual resistances as shown in fig. (b).

Note:
The value of equivalent resistance in series connection will be greater than each individual resistance.

2. Resistors in parallel:
Resistors are in parallel when they are connected across the same potential difference as shown in figer.

In this case, the total current I that leaves the battery in split into three separate paths. Let I₁, I₂ and I₃ be the current through the resistors R₁, R₂ and R₃ respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.
I = I₁ + I₂ + I₃ ……. (1)
Since the voltage across each resistor is the same, applying Ohm’s law to each resistor, we have
I₁ = \(\frac { V }{{ R }_{1}}\) I₂ = \(\frac { V }{{ R }_{2}}\),I₁ = \(\frac { V }{{ R }_{3}}\)
Substituting these values in equation (1), we get
I = \(\frac { V }{{ R }_{1}}\) + \(\frac { V }{{ R }_{2}}\) +\(\frac { V }{{ R }_{3}}\) = V\(\left[ \frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } \right] \)
I = \(\frac { V }{{ R }_{p}}\)
\(\frac { 1 }{{ R }_{p}}\) = \(\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } \)
Here R_P is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination as shown in the fig. (b).

Note:
The value of equivalent resistance in parallel connection will be lesser than each individual resistance.

Question 4.
Explain the determination of the internal resistance of a cell using voltmeter.
Answer:
Determination of internal resistance:
The emf of cell ξ is measured by connecting a high resistance voltmeter across it without connecting the external resistance R. Since the voltmeter draws very little current for deflection, the circuit may be considered as open. Hence, the voltmeter reading gives the emf of the cell. Then, external resistance R is included in the circuit and current I is established in the circuit. The potential difference across R is equal to the potential difference across the cell (V).

The potential drop across the resistor R is
V= IR …… (1)
Due to internal resistance r of the cell, the voltmeter reads a value V, which is less than the emf of cell ξ. It is because, certain amount of voltage (Ir) has dropped across the internal resistance r.

V = ξ – Ir
Ir = ξ – V …… (2)
Dividing equation (2) by equation (1), we get
\(\frac { Ir }{IR}\) = \(\frac { ξ – V }{V}\)
r = |\(\frac { ξ – V }{V}\)| R …… (3)
Since ξ, V and R are known, internal resistance r can be determined.

Question 5.
State and explain Kirchhoff’s rules.
Answer:
Kirchhoff’s first rule (current rule or junction rule):
Statement: It states that the algebraic sum of the currents at any junction of a circuit is zero. It is a statement of conservation of electric charge.

Explanation:
All charges that enter a given junction in a circuit must leave that junction since charge cannot build up or disappear at a junction. Current entering the junction is taken as positive and current leaving the junction is taken as negative.
Applying this law to the junction A,
I₁ + I₂ – I₃ – I₄ – I₅ = o
Or
I₁ + I₂ = + I₃ I₄ + I₅
Kirchhoff’s second rule (voltage rule or loop rule):
Statement: It states that in a closed circuit the algebraic sum of the products of the current and resistance of each part of the circuit is equal to the total emf included in the circuit. This rule follows from the law of conservation of energy for an isolated system. (The energy supplied by the emf sources is equal to the sum of the energy delivered to all resistors).

Explanation:
The product of current and resistance is taken as positive when the direction of the current is followed. Suppose if the direction of current is opposite to the direction of the loop, then product of current and voltage across the resistor is negative. It is shown in Fig. (a) and (b). The emf is considered positive when proceeding from the negative to the positive terminal of the cell. It is shown in Fig. (c) and (d).

Kirchhoff voltage rule has to be applied only when all currents in the circuit reach a steady state condition (the current in various branches are constant).

Question 6.
Obtain the condition for bridge balance in Wheatstone’s bridge.
Answer:
An important application of Kirchhoff’s rules is the Wheatstone’s bridge. It is used to compare resistances and also helps in determining the unknown resistance in electrical network. The – bridge consists of four resistances P, Q, R and S connected. A galvanometer G is connected between the points B and D. The battery is connected between the points A and C. The current through the galvanometer is I_G and its resistance is G.

Applying KirchhofFs current rule to junction B,
I₁ – I_G – I₃ = 0 …….. (1)
Applying Kirchhoff’s current rule to junction D,
I₂ – I_G – I₄ = 0 …….. (2)
Applying Kirchhoff’s voltage rule to loop ABDA,
I₁P + I_GG – I₂R = 0 …….. (3)
Applying Kirchhoff’s voltage rule to loop ABCDA,
I₁P + I₃Q – I₄S – I₂R = 0 …….. (4)
When the points B and D are at the same potential, the bridge is said to be balanced. As there is no potential difference between B and D, no current flows through galvanometer (I_G = 0). Substituting I_G = 0 in equation, (1), (2) and (3), we get
I₁ = I₃ …….. (5)
I₂ = I₄ …….. (6)
I₁P = I₂R …….. (7)
Substituting the equation (5) and (6) in equation (4)
I₁P + I₁Q – I₂R = 0
I₁(P + Q) = I₂ (R + S) …….. (8)
Dividing equation (8) by equation (7), we get
\(\frac { P + Q }{ P }\) = \(\frac { R + S }{ R }\)
1 + \(\frac { Q }{ P }\) = 1 + \(\frac { S }{ R }\)
⇒ \(\frac { Q }{ P }\) = \(\frac { S }{ R }\)
\(\frac { P }{ Q }\) = \(\frac { R }{ S }\) …….. (9)
This is the bridge balance condition. Only under this condition, galvanometer shows null deflection. Suppose we know the values of two adjacent resistances, the other two resistances can be compared. If three of the resistances are known, the value of unknown resistance (fourth one) can be determined.

Question 7.
Explain the determination of unknown resistance using meter bridge.
Answer:
The meter bridge is another form of Wheatstone’s bridge. It consists of a uniform manganin wire AB of one meter length. This wire is stretched along a meter scale on a wooden board between two copper strips C and D. Between these two copper strips another copper strip E is mounted to enclose two gaps G₁ and G₂ An unknown resistance P is connected in G₁ and a standard resistance Q is connected in G₂.

A jockey (conducting wire) is connected to the terminal E on the central copper strip through a galvanometer (G) and a high resistance (HR). The exact position of jockey on the wire can be read on the scale. A Lechlanche cell and a key (K) are connected across the ends of the bridge wire.

The position of the jockey on the wire is adjusted so that the galvanometer shows zero deflection. Let the point be J. The lengths AJ and JB of the bridge wire now replace the resistance R and S of the Wheatstone’s bridge. Then
\(\frac { P }{ Q }\) = \(\frac { R }{ S }\) = \(\frac { R’.AJ }{ R’.JB }\) …….. (1)
where R’ is the resistance per unit length of wire
\(\frac { P }{ Q }\) = \(\frac { AJ }{ JB }\) = \(\frac {{ l }_{1}}{ { l }_{2} }\) …….. (2)
P = Q \(\frac {{ l }_{1}}{ { l }_{2} }\) ……… (3)
The bridge wire is soldered at the ends of the copper strips. Due to imperfect contact, some resistance might be introduced at the contact. These are called end resistances. This error can be eliminated, if another set of readings are taken with P and Q interchanged and the average value of P is found.
To find the specific resistance of the material of the wire in the coil P, the radius r and length l of the wire is measured. The specific resistance or resistivity r can be calculated using the relation
Resistance = ρ-\(\frac { l }{ A }\)
By rearranging the above equation, we get
ρ = Resistance x \(\frac { A }{ l }\) …….. (4)
If P is the unknown resistance, equation (4) becomes
ρ = P\(\frac {{ πr }^{2}}{ l }\).

Question 8.
How the emf of two cells are compared using potentiometer?
Answer:
Comparison of emf of two cells with a potentiometer:
To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer wire CD is connected to a battery Bt and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emf ξ₁ and ξ₂ to be compared are connected to the terminals M_1, N₁ and M₂, N₂ of the DPDT switch.

The positive terminals of Bt, ξ₁ and ξ₂ should be connected to the same end C. The DPDT switch is pressed towards M₁, N₁ so that cell ξ₁ is included in the secondary circuit and the balancing length l₁ is found by adjusting the jockey for zero deflection, Then the second cells ξ₂ is included in the circuit and the balancing length l₂ is determined. Let r be the resistance per unit length of the potentiometer wire and I be the current flowing through the wire.
we have.
ξ₁ = Irl₁ …… (1)
ξ₂ = Irl₂ ……. (2)
By dividing (1) by (2)
\(\frac {{ ξ }^{1}}{ { ξ }^{2} }\) = \(\frac {{ l }^{1}}{ { l }^{2} }\) ……. (3)
By including a rheostat (Rh) in the primary circuit, the experiment can be repeated several times by changing the current flowing through it.

Samacheer Kalvi 12th Physics Current Electricity Numerical Problems

Question 1.
The following graphs represent the current versus voltage and voltage versus current for the six conductors A,B,C,D,E and F. Which conductor has least resistance and which has maximum resistance?
Solution:

According to ohm’s law, V = IR
Resistance of conductor, R = \(\frac { V }{ I }\)

Graph-I:
Conductor A, I = 4 A and V = 2 V
R = \(\frac { V }{ I }\) = \(\frac { 2 }{ 4 }\) = 0.5 Ω
Conductor B, I = 3 A and V = 4 V
R = \(\frac { V }{ I }\) = \(\frac { 4 }{ 3 }\) = 1.33 Ω
Conductor C, I = 2 A and V = 5 V
R= \(\frac { V }{ I }\) = \(\frac { 5 }{ 2 }\) = 2.5 Q Ω.

Graph-II:
Conductor D, I = 2 A and V = 4 V
R = \(\frac { V }{ I }\) = \(\frac { 4 }{ 2 }\) = 2 Ω d
Conductor E, I = 4A and V = 3 V
R = \(\frac { V }{ I }\) = \(\frac { 3 }{ 4 }\) = 1.75 Ω
Conductor F, I = 5A and V = 2 V
R = \(\frac { V }{ I }\) = \(\frac { 2 }{ 5 }\) = 0.4 Ω
Conductor F has least resistance, R_F = 0.4 Ω,
Conductor C has maximum resistance, R_C = 2.5 G.

Question 2.
Lightning is very good example of natural current. In typical lightning, there is 10⁹ J energy transfer across the potential difference of 5 x 10⁷ V during a time interval of 0.2 s.

Using this information, estimate the following quantities (a) total amount of charge transferred between cloud and ground (b) the current in the lightning bolt (c) the power delivered in 0.2 s.
Solution:
During the lightning energy, E = 10⁹ J
Potential energy, V = 5 x 10⁷ V
Time interval, t = 0.2 s
(a) Amount of charge transferred between cloud and ground,
q = It

(b) Current in the lighting bolt, E = VIt
I = \(\frac { E }{ Vt }\) = \(\frac{10^{9}}{5 \times 10^{7} \times 0.2}\) = 1 × 10⁹ × 1^-7
I = 1 × 10²
I = 100 A
∴ q = It = 100 × 0.2
q = 20 C.

(c) Power delivered, E = VIt
P = VI = 5 × 10⁷ × 100 = 500 × 10⁷
I = 5 × 10⁹ W
P = 5 GW.

Question 3.
A copper wire of 10⁶ m² area of cross section, carries a current of 2 A. If the number of electrons per cubic meter is 8 x 10²⁸, calculate the current density and average drift velocity.
Solution:
Cross-sections area of copper wire, A = 10⁶ m²
I = 2 A
Number of electron, n = 8 x 10²⁸
Current density, J = \(\frac { 1 }{ A }\) = \(\frac { 2 }{{ 10 }^{-6}}\)
J = 2 × 10⁶ Am^-2
Average drift velocity, V_d = \(\frac { 1 }{ neA }\)
e is the charge of electron = 1.6 × 10^-9 C
V_d = \(\frac{2}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6}}\) = \(\frac { 1 }{{ 64. × 10 }{3}}\)
V_d = 0.15625 × 10^-3
V_d = 15.6 × 10^-5 ms^-1

Question 4.
The resistance of a nichrome wire at 0 °C is 10 Ω. If its temperature coefficient of resistance is 0.004/°C, find its resistance at boiling point of water. Comment on the result.
Solution:
Resistance of a nichrome wire at 0°C, R₀ = 10 Ω
Temperature co-efficient of resistance, α = 0.004/°C
Resistance at boiling point of water, RT = ?
Temperature of boiling point of water, T = 100 °C
R_T=R₀ ( 1 + αT) = 10[1 + (0.004 x 100)]
R_T= 10(1 +0.4) = 10 x 1.4
R_T = 14 Ω
As the temperature increases the resistance of the wire also increases.

Question 5.
The rod given in the figure is made up of two different materials.

Both have square cross sections of 3 mm side. The resistivity of the first material is 4 x 10^-3 Ω.m and it is 25 cm long while second material has resistivity of 5 x 10^-3 Ω.m and is of 70 cm long. What is the resistivity of rod between its ends?
Solution:
Square cross section of side, a = 3 mm = 3 x 10^-3 m
Cross section of side, A = a² = 9 x 10⁶ m
First material:
Resistivity of the material, ρ₁ = 4 x 10^-3 Ωm
length, l₁ = 25 cm = 25 x 10^-2 m
Resistance of the lord, R₁ = \(\frac{\rho_{l} l_{l}}{A}\) = \(\frac{4 \times 10^{-3} \times 25 \times 10^{-2}}{9 \times 10^{-6}}\) = \(\frac{100 \times 10^{-5} \times 10^{6}}{9}\)
R₁ = 11.11 x 10¹ Ω

Second material:
Resistivity of the material, ρ₂ = 5 x 10^-3 Ωm
length, l₂ = 70 cm = 70 x 10^-2 m
Resistance of the rod, R₂ = \(\frac{\rho_{2} l_{2}}{A}\) = \(\frac{5 \times 10^{-3} \times 70 \times 10^{-2}}{9 \times 10^{-6}}\) = \(\frac{350 \times 10^{-5} \times 10^{6}}{9}\)
R₂ = 38.88 x 10¹
R₂ = 389 Ω
Total resistance between the ends f the rods
R = R₁ + R₂ = 111 + 389
= 500 Ω

Question 6.
Three identical lamps each having a resistance R are connected to the battery of emf as shown in the figure.

Suddenly the switch S is closed, (a) Calculate the current in the circuit when S is open and closed (b)
What happens to the intensities of the bulbs A,B and C. (c) Calculate the voltage across the three bulbs when S is open and closed (d) Calculate the power delivered to the circuit when S is opened and closed (e) Does the power delivered to the circuit decreases, increases or remain same?
Solution:
Resistance of the identical lamp = R
Emf of the battery = ξ
According to Ohm’s Law, ξ = IR
(a) Current:
When Switch is open— The current in the circuit. Total resistance of the bulb,
R_s = R₁ + R₂ + R₃
R₁ = R₂ = R₃ = R
R_s = R + R + R = 3R
∴ Current, I = \(\frac { ξ }{{ R }_{s}}\)
⇒ I₀ = \(\frac { ξ }{ 3R }\)
Switch is closed— The current in the circuit. Total resistance of the bulb,
R_s = R + R = 2R
Current I = \(\frac { ξ }{{ R }_{s}}\)
I_c = \(\frac { ξ }{ 2R }\).

(b) Intensity:
When switch is open — All the bulbs glow with equal intensity.
When switch is closed — The intensities of the bulbs A and B equally increase. Bulb C will not glow since no current pass through it.

(c) Voltage across three bulbs:
When switch is open — Voltage across bulb A, V_A = I₀ R = \(\frac { ξ }{ 3R }\) x R = \(\frac { ξ }{ 3 }\)
similarly:
Voltage across bulb B, V_B = \(\frac { ξ }{ 3 }\)
Voltage across bulb C, V_C = \(\frac { ξ }{ 3 }\)
When switch is closed— Voltage across bulb A, V_A = I_cR = \(\frac { ξ }{ 2R }\) x \(\frac { ξ }{ 2 }\)
similarly:
Voltage across bulb B, V_B = I_cR \(\frac { ξ }{ 2 }\)
Voltage across bulb C, V_C = 0

(d) Power delivered to the circuit,
When switch is opened — Power P, = VI

When switch is closed — Power P, = VI

(e) Total power delivered to circuit increases.

Question 7.
The current through an element is shown in the figure. Determine the total charge that pass through the element at
(a) t = 0 s
(b) t = 2 s
(c) t = 5 s

Solution :
Rate of flow of charge is called current, I = \(\frac { dq }{ dt }\)
Total charge pass through element, dq = Idt
(a) At t= 0 s, I = 10 A
dq = Idt= 10 x 0 = 0 C.

(b) At t = 2 s, I = 5 A
dq = Idt = 5 x 2 = 10 C.

(c) At t = 5 s, I = 0
dq = Idt = 0 x 5 = 0 C.

Question 8.
An electronics hobbyist is building a radio which requires 150 Ω in her circuit, but she has only 220 Ω, 79 Ω and 92 Ω resistors available. How can she connect the available resistors to get desired value of resistance?
Solution:
Required effective resistance = 150 Ω
Given resistors of resistance, R₁ = 220 Ω, R₂ = 79 Ω, R₃ = 92 Ω
Parallel combination R₁ and R₂
\(\frac { 1 }{{ R }_{p}}\) = \(\frac { 1 }{{ R }_{1}}\) + \(\frac { 1 }{{ R }_{2}}\) = \(\frac { 1 }{ 220 }\) + \(\frac { 1 }{ 79 }\) = \(\frac { 79 + 220 }{ 220 × 79 }\)
R_p = 58 Ω

Parallel combination R_p and R₃
R_s = R_p + R₃ = 58 + 92
R_s = 150 Ω
Parallel combination of 220 Ω and 79 Ω in series with 92 Ω.

Question 9.
A cell supplies a current of 0.9 A through a 2 Ω resistor and a current of 0.3 A through a 7 Ω resistor. Calculate the internal resistance of the cell.
Solution:
Current from the cell, I₁ = 0.9 A
Resistor, R₁ = 2 Ω
Current from the cell, I₂ = 0.3 A
Resistor, R₂ 7 Ω
Internal resistance of the cell, r = ?
Current in the circuit I₁ = \(\frac { ξ }{{ r + R }_{1}}\)
ξ = I₁ (r + R₁) …… (1)
Current in the circuit, I₂ = \(\frac { ξ }{{ r + R }_{2}}\) …… (2)
Equating equation (1) and (2),
I₁r + I₁R₁ = I₂R₂ + I₂r
(I₁ – I₂)r = I₂R₂ – I₁R₁

r = 0.5 Ω.

Question 10.
Calculate the currents in the following circuit.

Solution:
Applying Kirchoff’s 1^st Law at junction B

I₁ – I₁ – I₃ = 0
I₃ = I₁ – I₂ …… (1)
Applying Kirchoff’s II^nd Law at junction in ABEFA
100 I₃ + 100 I₁ = 15
100 (I₃ + I₁) = 15
100 I₁ – 100 I₂ + 100 I₁ =15
200 I₁ – 100 I₂ = 15 …… (2)
Applying Kirchoff’s IInd Law at junction in BCDED
-100I₂ + 100 I₃ = 9
-100 I₂+ 100(I₁ – I₂) = 9
100I₁ – 200 I₂ = 9
Solving equating (2) and (3)

Substitute I₁ values in equ (2)
200(0.07) – 100 I₂ = 15
14 – 100 I₂ = 15
– 100 I₂ = 15 – 14
I₂ = \(\frac { -1 }{ 100 }\)
I₂ = -0.01A
Substitute I₁ and I₂ value in equ (1), we get
I₃ = I₁ – I₂ = 0.07 – (-0.01)
I₃ = 0.08 A.

Question 11
A potentiometer wire has a length of 4 m and resistance of 20 Ω. It is connected in series with resistance of 2980 Ω and a cell of emf 4 V. Calculate the potential along the wire.
Solution:
Length of the potential wire, l = 4 m
Resistance of the wire, r = 20 Ω
Resistance connected series with potentiometer wire, R = 2980 Ω
Emf of the cell, ξ = 4 V
Effective resistance, R_s = r + R = 20 + 2980 = 3000 Ω
Current flowing through the wire, I = \(\frac { ξ }{{ r + R }_{s}}\) = \(\frac { 4 }{ 3000 }\)
I = 1.33 x 10^-3 A
Potential drop across the wire, V = Ir
= 1.33 x 10^-33 x 20
V = 26.6 x 10^-3 V

= 6.65 x 10^-3
Potential garadient = 0.66 x 10^-2 Vm^-1

Question 12.
Determine the current flowing through the galvanometer (G) as shown in the figure.

Solution:
Current flowing through the circuit, I = 2A
Applying Kirchoff’s I^st law at junction P, I = I₁ + I₂
I₂ = I – I₁ …(1)
Applying Kirchoff’s I^nd law at junction PQSP
5 I₁ + 10 I_g – 15 I₂ = 0
5 I₁ + 10 I_g -15(I – I₁) = 0
20 I₁ + 10 I_g = 15 I
20 I₁ + 10 I_g = 15 x 2
÷ by 10 21 I₁ + I_g = 3 … (2)
Applying Kirchoff’s II^nd law at junction QRSQ
10(I₁ – I_g) – 20(I – I₁ – I_g) – 10 I_g = 0
10 I₁ – 10_g – 20(I – I₁ – I_g) – 10 I_g = 0
10 I₁ – 10_g – 20 I + 20 I₁ -20I_g – 10 I_g = 0
30 I ₁ – 40 I_g = 20 I
÷ by 10 ⇒ 3 I₁ – 4 I_g = 20 I
3 I₁ – 4 I_g = 2 I
3 I₁ – 4 I_g = 2 x 2 = 4

Question 13.
Two cells each of 5 V are connected in series across a 8 Ω resistor and three parallel resistors of 4 Ω, 6 SI and 12 Ω. Draw a circuit diagram for the above arrangement. Calculate (i) the current drawn from the cell (ii) current through each resistor.
Solution:
V₁ = 5 V; V₂ = 5 V
R₁ = 8 Ω; R₂ = 4 Ω; R₃ = 6 Ω; R₄ = 12 Ω
Three resistors R₂, R₃ and R₄ are connected parallel combination

Resistors R₁ and R_p are connected in series combination
R_s = R₁ +R_p = 8 + 2 = 10
R_s = 10Ω
Total voltage connected series to the circuit
V = V₁ + V₂
= 5 + 5 = 10
V = 10 V.
(i) Current through the circuit, I = \(\frac { V }{{ R }_{s}}\) = \(\frac { 10 }{ 10 }\)
I = 1 A
Potential drop across the parallel combination,
V’ = IR_p = 1 x 2
V’ 2 V

(ii) Current in 4 Ω resistor, I = \(\frac { V’ }{{ R }_{2}}\) = \(\frac { 2 }{ 4 }\) = 0.5 A
Current in 6 Ω resistor, I = \(\frac { V’ }{{ R }_{3}}\) = \(\frac { 2 }{ 6 }\) = 0.33 A
Current in 12 Ω resistor, I = \(\frac { V’ }{{ R }_{4}}\) = \(\frac { 2 }{ 12 }\) = 0.17 A

Question 14.
Four light bulbs P, Q, R, S are connected in a circuit of unknown arrangement. When each bulb is removed one at a time and replaced, the following behavior is observed.

Solution:

Question 15.
In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35 cm length of the wire. If the cell is replaced by another cell and the balance point shifts to 63 cm, what is the emf of the second cell?
Solution:
Emf of the cell₁, ξ₁ = 1.25 V
Balancing length of the cell, l₁ = 35 cm = 35 x 10^-2 m
Balancing length after interchanged, l₂ = 63 cm = 63 x 10^-2 m
Emf of the cell₁, ξ₂ = ?
The ration of emf’s, \(\frac {{ ξ }_{1}}{{ ξ }_{2}}\) = \(\frac {{ l }_{1}}{{ l }_{2}}\)
The ration of emf’s, ξ₂ = ξ₁ = \(\left( \frac { { l }_{ 2 } }{ { l }_{ 1 } } \right) \)
= 1. 25 x \(\left( \frac { { 63×10 }^{ -2 } }{ { 35×10 }^{ -2 } } \right) \) = 12.5 x 1.8
ξ₂ = 2.25 V.

Samacheer Kalvi 12th Physics Current Electricity Additional Questions Solved

I. Choose the Correct Answer

Question 1.
When current I flows through a wire, the drift velocity of the electrons is v. When current 21 flows through another wire of the same material having double the length and area of cross-section, the drift velocity of the electrons will be-
(a) \(\frac { v }{ 4 }\)
(b) \(\frac { v }{ 2 }\)
(c) v
(d) 2 v
Answer:
(c) v
Hint:
V_d = \(\frac { 1 }{ nAe }\); v’_d = \(\frac { 2I }{ (2A)ne }\) = v_d

Question 2.
A copper wire of length 2 m and area of cross-section 1.7 x 10^-6 m² has a resistance of 2 x 10^-2 Ω. The resistivity of copper is
(a) 1.7 x 10^-8 Ωm
(b) 1.9 x ^-8 Ωm
(c) 2.1 x 10^-7 Ωm
(d) 2.3 x 10^-7 Ωm
Answer:
(a) 1.7 x 10^-8 Ωm
Hint:
Resistivity, ρ = \(\frac { RA }{ l }\) = \(\frac{2 \times 10^{-2} \times 1.7 \times 10^{-6}}{2}\) = 1.7 x 10^-8 Ωm.

Question 3.
If the length of a wire is doubled and its cross-section is also doubled, then its resistance will
(a) become 4 times
(b) become 1 / 4
(c) becomes 2 times
(d) remain unchanged
Answer:
(d) remain unchanged

Question 4.
A 10 m long wire of resistance 20 Ω is connected in series with a battery of emf 3 V and a resistance of 10 Ω. The potential gradient along the wire in volt per meter is
(a) 6.02
(b) 0.1
(c) 0.2
(d) 1.2
Answer:
(c) 0.2
Hint:
Potential difference across the wire = \(\frac { 20 }{ 3 }\) x 3 = 2 V
Potential gradient = \(\frac { v }{ l }\) = \(\frac { 2 }{ 10 }\) = 0.2 V/m

Question 5.
The resistivity of a wire
(a) varies with its length
(b) varies with its mass
(c) varies with its cross-section
(d) does not depend on its length, cross-section and mass.
Answer:
(d) does not depend on its length, cross-section and mass.

Question 6.
The electric intensity E, current density and conductivity a are related as
(a) j = σE
(b) j = \(\frac { E }{ σ }\)
(c) JE = s
(d) j = σ²E
Answer:
(a) j = σE

Question 7.
For which of the following dependences of drift velocity V_d on electric field E, is Ohm’s law obeyed?
(a) v_d ∝ E
(b) v_d ∝ E²
(c) v_d ∝ √E
(d) v_d = constant
Answer:
(a) v_d ∝ E
Hint:
v_d = \(\frac { 1 }{ nAe }\) = \(\frac { j }{ ne }\) = \(\left( \frac { σ }{ ne } \right) \)E ⇒ v_d ∝ E.

Question 8.
A cell has an emf of 1.5 V. When short circuited, it gives a current of 3A. The internal resistance of the cell is .
(a) 0.5 Ω
(b) 2.0 Ω
(c) 4.5 Ω
(d) \(\frac { 1 }{ 4.5 }\) Ω
Answer:
(a) 0.5 Ω
Hint:
r = \(\frac { ξ }{ I }\) = \(\frac { 1.5 }{ 3 }\) = 0.5 Ω.

Question 9.
The resistance, each of 1 Ω, are joined in parallel. Three such combinations are put in series. The resultant resistance is
(a) 9 Ω
(b) 3 Ω
(c) 1 Ω
(d) \(\frac { 1 }{ 3 }\) Ω
Answer:
(c) 1 Ω
Hint:
R_p = 1 + 1 + 1 = 3 Ω;
\(\frac { 1 }{{ R }_{s}}\) = \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ 3 }\) + \(\frac { 1 }{ 3 }\) = \(\frac { 3 }{ 3 }\) = 1
⇒ R_s = 1 Ω.

Question 10.
Constantan is used for making standard resistance because it has
(a) high resistivity
(b) low resistivity
(c) negligible temperature coefficient of resistance
(d) high melting point
Answer:
(c) negligible temperature coefficient of resistance

Question 11.
Kirchhoff’s two laws for electrical circuits are magnifestations of the conservation of
(a) charge only
(b) both energy and momentum
(c) energy only
(d) both charge and energy
Answer:
(d) both charge and energy

Question 12.
The resistance R₀ and R_t of a metallic wire at temperature 0° C and t° C are related as (a is the temperature co-efficient of resistance).
(a) R_t = R₀(1 + αt)
(b) R_t = R₀(1 – αt)
(c) R_t = R₀(1 + αt)²
(d) R_t = R₀(1 – αt) ²
Answer:
(a) R_t = R₀(1 + αt)

Question 13.
A cell of emf 2 V and internal resistance 0.1 Ω is connected with a resistance of 3.9 Ω. The voltage across the cell terminals will be
(a) 0.5 V
(b) 1.9 V
(c) 1.95 V
(d) 2 V
Answer:
(c) 1.95 V
Hint:
V = \(\frac { ER }{ R + r }\) = \(\frac { 2 x 3.9 }{ 3.9 + 0.1 }\) = 1.95 V.

Question 14.
A flow of 10⁷ electrons per second in a conduction wire constitutes a current of
(a) 1.6 x 10^-26 A
(b) 1.6 x 10¹² A
(c) 1.6 x 10^-12 A
(d) 1.6 x 10²⁶ A
Answer:
(c) 1.6 x 10^-12 A
Hint:
I = \(\frac { Q }{ t }\) = \(\frac{10^{7} \times 1.6 \times 10^{-19}}{1}\) = 1.6 x 10¹² A.

Question 15.
Sensitivity of a potentiometer can be increased by
(a) increasing the emf of the cell
(b) increasing the length of the wire
(c) decreasing the length of the wire
(d) none of the above
Answer:
(b) increasing the length of the wire

Question 16.
Potential gradient is defined as
(a) fall of potential per unit length of the wire.
(b) fall of potential per unit area of the wire.
(c) fall of potential between two ends of the wire.
(d) none of the above.
Answer:
(a) fall of potential per unit length of the wire.

Question 17.
n equal resistors are first connected in series and then in parallel. The ratio of the equivalent resistance in two cases is
(a) n
(b) \(\frac { 1 }{{ n }^{2}}\)
(c) n²
(d) \(\frac { 1 }{ n }\)
Answer:
(c) n²
Hint:
Required ratio = \(\frac{n \mathrm{R}}{\left(\frac{\mathrm{R}}{n}\right)}\) = n(c) n²

Question 18.
A galvanometer is converted into an ammeter when we connect a
(a) high resistance in series
(b) high resistance in parallel
(c) low resistance in series
(d) low resistance in parallel
Answer:
(d) low resistance in parallel

Question 19.
The reciprocal of resistance is
(a) conductance
(b) resistivity
(c) conductivity
(d) none of the above
Answer:
(a) conductance

Question 20.
A student has 10 resistors, each of resistance r. The minimum resistance that can be obtained by him using these resistors is
(a) 10r
(b) \(\frac { r }{ 10 }\)
(c) \(\frac { r }{ 100 }\)
(d) \(\frac { r }{ 5 }\)
Answer:
(b) \(\frac { r }{ 10 }\)

Question 21.
The drift velocity of electrons in a wire of radius r is proportional to
(a) r
(b) r²
(c) r³
(d) none of the above
Answer:
(d) none of the above

Question 22.
Kirchhoff’s first law, i.e. ∑I = 0 at a junction deals with conservation of
(a) charge
(b) energy
(c) momentum
Answer:
(a) charge

Question 23.
The resistance of a material increases with temperature. It is a
(a) metal
(b) insulator
(c) semiconductor
(d) semi-metal
Answer:
(a) metal

Question 24.
Five cells, each of emf E, are joined in parallel. The total emf of the combination is
(a) 5E
(b) \(\frac { E }{ 5 }\)
(c) E
(d) \(\frac { 5E }{ 2 }\)
Answer:
(c) E

Question 25.
A carbon resistance has colour bands in order yellow, brown, red. Its resistance is
(a) 41 Ω
(b) 41 x 10² Ω
(c) 41 x 10³ Ω
(d) 4.2 Ω
Answer:
(b) 41 x 10² Ω

Question 26.
The conductivity of a superconductor is
(a) infinite
(b) very large
(c) very small
(d) zero
Answer:
(a) infinite

Question 27.
The resistance of an ideal voltmeter is
(a) zero
(b) very high
(c) very low
(d) infinite
Answer:
(d) infinite

Question 28.
Carriers of electric current in superconductors are
(a) electrons
(b) photons
(c) holes
Answer:
(c) holes

Question 29.
Potentiometer measures potential more accurately because
(a) It measure potential in the open circuit.
(b) It uses sensitive galvanometer for null detection.
(c) It uses high resistance potentiometer wire.
(d) It measures potential in the closed circuit.
Answer:
(a) It measure potential in the open circuit.

Question 30.
Electromotive force is most closely related to
(a) electric field
(b) magnetic field
(c) potential difference
(d) mechanical force
Answer:
(c) potential difference

Question 31.
The capacitance of a pure capacitor is 1 farad. In DC circuit, the effective resistance will be
(a) zero
(b) infinite
(c) 1 Ω
(d) 0.5 Ω
Answer:
(b) infinite

Question 32.
The resistance of an ideal ammeter is
(a) zero
(b) small
(c) high
(d) infinite
Answer:
(a) zero

Question 33.
A milliammeter of range 10 mA has a coil of resistance 1 Ω. To use it as a voltmeter of range 10 V, the resistance that must be connected in series with it is
(a) 999 Ω
(b) 1000 Ω
(c) 9 Ω
(d) 99 Ω
Answer:
(a) 999 Ω
Hint:
R = \(\frac { V }{{ I }_{g}}\)-R_g = \(\frac { 10 }{{ 10 × 10 }^{-3}}\)-1 = 999 Ω.

Question 34.
A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. The current in the circuit is. 0.5 A. The terminal voltage of the battery when the circuit is closed is
(a) 10 V
(b) zero
(c) 8.5 V
(d) 1.5 V
Answer:
(c) 8.5 V
Hint:
V = ξ – Ir = 10 – (0.5 x 3) = 8.5 V.

Question 35.
Good resistance coils are made of
(a) copper
(b) manganin
(c) iron
(d) aluminium
Answer:
(b) manganin

Question 36.
A wire of resistance R is stretched to three times its original length. The new resistance is
(a) 3R
(b) 9R
(c) R/3
(d) R/9
Answer:
(b) 9R

Question 37.
n resistances, each of r Ω, when connected in parallel give an equivalent resistance of R Ω. If these resistances were connected in series, the combination would have a resistance in horns equal to
(a) n²R
(b) \(\frac { R }{{n}^{ 2 }}\)
(c) \(\frac { R }{ n }\)
(d) nR
Answer:
(a) n²R
Hint:
Resistance in parallel combination, R = \(\frac { r }{ n }\) ⇒ r = Rn
Resistance in series combination, R’ = nr = n²R

Question 38.
When a wire of uniform cross-section, having resistance R, is bent into a complete circle, the resistance between any two of diametrically opposite points will be
(a) \(\frac { R }{ 8 }\)
(b) \(\frac { R }{ 2 }\)
(c) 4R
(d) \(\frac { R }{ 4 }\)
Answer:
(d) \(\frac { R }{ 4 }\)
Hint:
It becomes two resistors each of (d) \(\frac { R }{ 2 }\), connected in parallel.

Question 39.
A steady current is set up in a metallic wire of non uniform cross-section. How is the rate of flow K of electrons related to the area of cross-section A?
(a) K is independent of A
(b) K ∝ A
(c) K ∝ A^-1
(d) K ∝ A²
Answer:
(c) K ∝ A^-1

Question 40.
Ohm’s Law is not obeyed by
(a) electrolytes
(b) discharge tubes
(c) vacuum tubes
(d) all of these
Answer:
(d) all of these

Question 41.
Which of the following has negative temperature coefficient of resistance?
(a) Copper
(b) Aluminium
(c) Germanium
(d) Iron
Answer:
(c) Germanium

II. Fill in the blanks

Question 1.
The material through which electric charge can flow easily is ……………….
Answer:
Copper.

Question 2.
A toaster operating at 240 V has a resistance of 120 Ω. The power is ……………….
Answer:
480 W.

Question 3.
In the case of insulators, as the temperature decreases, resistivity ……………….
Answer:
Increases.

Question 4.
When n resistors of equal resistance (R) are connected in series, the effective resistance is ……………….
Answer:
nR.

Question 5.
The net flow of charge at any point in the conductor is ……………….
Answer:
Zero.

Question 6.
The flow of free electrons in a conductor constitutes ……………….
Answer:
Electric current.

Question 7.
The rate of flow of charge through any wire is called ……………….
Answer:
Current.

Question 8.
The drift velocity acquired per unit electric field is the ……………….
Answer:
Mobility.

Question 9.
The reciprocal of resistance is ……………….
Answer:
Conductance.

Question 10.
The unit of specific resistance is ……………….
Answer:
Ohm meter.

Question 11.
The reciprocal of electrical resistivity is called ……………….
Answer:
Electrical conductivity.

Question 12.
With increase in temperature the resistivity of metals ……………….
Answer:
Increases.

Question 13.
The resistivity of insulators is of the order of ……………….
Answer:
10⁸ 10¹⁴ Ωm.

Question 14.
The resistivity of semiconductors is of the order of ……………….
Answer:
10^-2 -10² Ωm.

Question 15.
The materials which conduct electricity at zero resistance are called ……………….
Answer:
Superconductors.

Question 16.
Conductors turn into superconductors at ……………….
Answer:
Low temperatures.

Question 17.
The resistance of superconductors is ……………….
Answer:
Zero.

Question 18.
The phenomenon of superconductivity was discovered by ……………….
Answer:
Kammerlingh onnes.

Question 19.
Mercury becomes a superconductor at ……………….
Answer:
4.2 K.

Question 20.
With increase of temperature, resistance of conductors ……………….
Answer:
increases

Question 21.
In insulators and semiconductors, as temperature increases, resistance ……………….
Answer:
Decreases.

Question 22.
A material with a negative temperature coefficient is called a ……………….
Answer:
Thermistor.

Question 23.
The temperature coefficient for alloys is ……………….
Answer:
Low.

Question 24.
The electric current in an external circuit flows from the ……………….
Answer:
Positive to negative terminal.

Question 25.
In the electrolyte of the cell, current flows from ……………….
Answer:
Negative to positive terminal.

Question 26.
A freshly prepared cell has ………………. internal resistance.
Answer:
Low.

Question 27.
Kirchhoff’s first law is ……………….
Answer:
Current law.

Question 28.
The current law states that the algebraic sum of the currents meeting at any junction in a circuit is ……………….
Answer:
Zero.

Question 29.
Current law is a consequence of conservation of ……………….
Answer:
Charges.

Question 30.
Kirchhoff’s second law is ……………….
Answer:
Voltage law.

Question 31.
Kirchhoff’s second law is a consequence of conservation of ……………….
Answer:
Energy.

Question 32.
Wheatstone bridge is an application of ……………….
Answer:
Kirchhoff’s Law.

Question 33
………………. is a form of Wheatstone’s bridge.
Answer:
Metre bridge.

Question 34.
The temperature coefficient of manganin wire is ……………….
Answer:
Low.

Question 35
………………. is an instrument to measure potential difference.
Answer:
Potentiometer.

Question 36.
Unit of electrical energy is ……………….
Answer:
Joule.

Question 37.
An instrument to measure electrical power consumed is ……………….
Answer:
Watt meter.

Question 38.
………………. first introduced the electrochemical battery
Answer:
Volta.

Question 39.
Charging is a process of reproducing ……………….
Answer:
Active materials.

III. Match the following

Question 1.

Answer:
(i) → (b)
(ii) → (a)
(iii) → (d)
(iv) →(c)

Question 2.

Answer:
(i) → (b)
(ii) → (c)
(iii) → (d)
(iv) → (a)

Question 3.

Answer:
(i) → (d)
(ii) → (c)
(iii) → (b)
(iv) → (a)

Question 4.

Answer:
(i) → (b)
(ii)→ (d)
(iii) → (a)
(iv) → (c)

IV.Assertion and reason type

(a) If both assertion and reason are true and the reason in the correct explanation of the assertion.
(b) If both assertion and reason are true but the reason is not correct explanation of the assertion.
(c) If assertion is true but reason is false.
(d) If the assertion and reason both are false.
(e) If assertion is false but reason is true.

Question 1.
Assertion: Fuse wire must have high resistance and low melting point.
Reason: Fuse is used for small current flow only
Answer:
(c) If assertion is true but reason is false.

Question 2.
Assertion: In practical application, power rating of resistance is not important.
Reason: Property of resistance remains same even at high temperature
Answer:
(d) If the assertion and reason both are false.

Question 3.
Assertion: Electric appliances with metallic body e.g. heaters, presses, etc, have three pin connections, whereas an electric bulb has two pins.
Reason: Three pin connection reduce heating of connecting cables.
Answer:
(c) If assertion is true but reason is false.

Samacheer Kalvi 12th Physics Current Electricity Short Answer Questions

Question 1.
Define current?
Answer:
Current is defined as a net charge Q passes through any cross section of a conductor in time t
then, I = \(\frac { Q }{ t }\).

Question 2.
Define instantaneous current?
Answer:
The instantaneous current I is defined as the limit of the average current, as ∆t → 0.

Question 3.
What is resistance? Give its unit?
Answer:
The resistance is the ratio of potential difference across the given conductor to the current passing through the conductor V.
R = \(\frac { V }{ I }\).

Question 4.
What is meant by transition temperature?
Answer:
The resistance of certain materials become zero below certain temperature T_c. This temperature is known as critical temperature or transition temperature.

Question 5.
What is Joule’s heating effect?
Answer:
When current flows through a resistor, some of the electrical energy delivered to the resistor is converted into heat energy and it is dissipated. This heating effect of current is known as Joule’s heating effect.

Question 6.
What is meant by thermoelectric effect?
Answer:
Conversion of temperature differences into electrical voltage and vice versa is known as thermoelectric effect.

Question 7.
What is a thermopile? On what principle does it work?
Answer:
Thermopile is a device used to detect thermal radiation. It works on the principle of seebeck effect.

Question 8.
What is a thermistor?
Answer:
A material with a negative temperature coefficient is called a thermistor.
Eg:

1. Insulator
2. Semiconductor.

Question 9.
State principle of potentiometer?
Answer:
The principle of potentiometer states that the emf of the cell is directly proportional to its balancing length.
ξ ∝ l
ξ = Irl.
Samacheer Kalvi 12th Physics Current Electricity Long Answer Questions

Question 1.
Explain the concept of colour code for carbon resistors.
Answer:
Color code for Carbon resistors:
Carbon resistors consists of a ceramic core, on which a thin layer of crystalline carbon is deposited. These resistors are inexpensive, stable and compact in size. Color rings are used to indicate the value of the resistance according to the rules.

Three coloured rings are used to indicate the values of a resistor: the first two rings are significant figures of resistances, the third ring indicates the decimal multiplier after them. The fourth color, silver or gold, shows the tolerance of the resistor at 10% or 5%. If there is no fourth ring, the tolerance is 20%. For the resistor, the first digit = 5 (green), the second digit = 6 (blue), decimal multiplier = 10³ (orange) and tolerance = 5% (gold). The value of resistance = 56 x 10³ Q or 56 kΩ with the tolerance value 5%.

Question 2.
Explain in details of temperature dependence of resistivity.
Answer:
Temperature dependence of resistivity:
The resistivity of a material is dependent on temperature. It is experimentally found that for a wide range of temperatures, the resistivity of a conductor increases with increase in temperature according to the expression,
ρ_T = ρ₀ [1 + α(T -T₀)] ……. (1)
where ρ_T is the resistivity of a conductor at T₀C, ρ₀ is the resistivity of the conductor at some reference temperature To (usually at 20°C) and a is the temperature coefficient of resistivity. It is defined as the ratio of increase in resistivity per degree rise in temperature to its resistivity atT₀.
From the equation (1), we can write
ρ_T – ρ₀ = αρ₀ (T -T₀)
∴ α = \(\frac{\rho_{\mathrm{T}}-\rho_{0}}{\rho_{0}\left(\mathrm{T}-\mathrm{T}_{0}\right)}\) = \(\frac{\Delta p}{\rho_{0} \Delta T}\)
where ∆_ρ = ρ_T – ρ₀ is change in resistivity for a change in temperature ∆T = T – T₀. Its unit is per °C.

1. α of conductors:
For conductors a is positive. If the temperature of a conductor increases, the average kinetic energy of electrons in the conductor increases. This results in more frequent collisions and hence the resistivity increases. Even though, the resistivity of conductors like metals varies linearly for wide range of temperatures, there also exists a nonlinear region at very low temperatures. The resistivity approaches some finite value as the temperature approaches absolute zero. As the resistance is directly proportional to resistivity of the material, we can also write the resistance of a conductor at temperature T °C as
R_T -R₀ = [1 + α(T -T₀)] ……. (2)
The temperature coefficient can be also be obtained from the equation (2), +

where ∆R = R_T -R₀ is change in resistance during the change in temperature ∆T = T – T₀

2. α of semiconductors:
For semiconductors, the resistivity decreases with increase in temperature. As the temperature increases, more electrons will be liberated from their atoms (Refer unit 9 for conduction in semi conductors). Hence the current increases and therefore the resistivity decreases. A semiconductor with a negative temperature coefficient of resistance is called a thermistor.
We can understand the temperature dependence of resistivity in the following way. The electrical conductivity, σ = \(\frac{n e^{2} \tau}{m}\) \(\frac{m}{n e^{2} \tau}\). As the resistivity is inverse of σ, it can be written as,
σ = \(\frac{n e^{2} \tau}{m}\) \(\frac{m}{n e^{2} \tau}\) …… (4)
The resistivity of materials is

1. inversely proportional to the number density (n) of the electrons
2. inversely proportional to the average time between the collisions (τ).

In metals, if the temperature increases, the average time between the collision (τ) decreases and n is independent of temperature. In semiconductors when temperature increases, n increases and τ decreases, but increase in n is dominant than decreasing x, so that overall resistivity decreases.

Question 3.
Explain the effective internal resistance of cells connected in series combination. Compare the results to the external resistance.
Answer:
Cells in series Several cells can be connected to form a battery. In series connection, the negative terminal of one cell is connected to the positive terminal of the second cell, the negative terminal of second cell is connected to the positive terminal of the third cell and so on. The free positive terminal of the first cell and the free negative terminal of the last cell become the terminals of the battery.

Suppose n cells, each of emf ξ volts and internal resistance r ohms are connected in series with an external resistance R.
The total emf of the battery = nξ
The total resistance in the circuit = nr + R
By Ohm’s law, the current in the circuit is

Case (a) If r << R, then,
I = \(\frac { nξ }{ R }\)nI₁ ≈ nI₁ ……. (2)
where, I, is the current due to a single cell \(\left(\mathrm{I}_{1}=\frac{\xi}{\mathrm{R}}\right)\)
Thus, if r is negligible when compared to R the current supplied by the battery is n times that supplied by a single cell.
Case (b) If r >> R, I = \(\frac { nξ }{ nr }\) ≈ \(\frac { ξ }{ R }\) …….. (3)
It is the current due to a single cell. That is, current due to the whole battery is the same as that due to a single cell and hence there is no advantage in connecting several cells. Thus series connection of cells is advantageous only when the effective internal resistance of the cells is negligibly small compared with R.

Question 4.
Explain the effective internal resistance of cells connected in parallel combination. Compare the results to the external resistance.
Answer:
Cells in parallel: In parallel connection all the positive terminals of the cells are connected to one point and all the negative terminals to a second point. These two points form the positive and negative terminals of the battery.
Let n cells be connected in parallel between the points A and B and a resistance R is connected between the points A and B. Let ξ, be the emf and r the internal resistance of each cell.

The equivalent internal resistance of the battery is \(\frac { 1 }{{ r }_{eq}}\) = \(\frac { 1 }{ r }\) + \(\frac { 1 }{ r }\) + ….. \(\frac { 1 }{ r }\) (n terms) = \(\frac { n }{ r }\).
So r_eg = \(\frac { r }{ n }\) and the total resistance in the circuit = R + \(\frac { r }{ n }\). The total emf is the potential difference between the points A and B, which is equal to ξ. The current in the circuit is given by

Case (a) If r << R, then,
I = \(\frac { nξ }{ R }\) = nI₁ …….. (2)
where II is the current due to a single cell and is equal to \(\frac { ξ }{ R }\) when R is negligible. Thus, the
current through the external resistance due to the whole battery is n times the current due to a single cell.

Case (b) If r << R. I = \(\frac { ξ }{ R }\) …… (3)
The above equation implies that current due to the whole battery is the same as that due to a single cell. Hence it is advantageous to connect cells in parallel when the external resistance is very small compared to the internal resistance of the cells.

Samacheer Kalvi 12th Physics Current Electricity Numerical Problems

Question 1.
Show that one ampere is equivalent to a flow of 6.25 x 10¹⁸ elementary charges per second.
Solution:
Here I = 1 A, t= 1s, e = 1.6 x 10^-19 C
As I = \(\frac { q }{ t }\) = \(\frac { ne }{ t }\)
Number of electrons, n = \(\frac { It }{ e }\) = \(\frac { 1 × 1 }{{ 1.6 × 10 }^{19}}\) = 6.25 × 10¹⁸

Question 2.
Calculate the resistivity of a material of a wire 10 m long. 0.4 mm in diameter and having a resistance of 2.0 Ω.
Solution:
Here l = 10 cm, r = 0.2 mm = 0.2 x 10^-3 m, R = 2 Ω
\(\left[ r\quad =\quad \frac { d }{ 2 } \right] \)

= 2.513 x 10^-8 Ω.

Question 3.
A wire of 10 ohm resistance is stretched to thrice its original length. What will be its (i) new resistivity and (ii) new resistance?
Solution:
(i) Resistivity p remains unchanged because it is the property of the material of the wire.
(ii) In both cases, volume of wire is same, so

Question 4.
A copper wire has a resistance of 10 Ω. and an area of cross-section 1 mm². A potential difference of 10 V exists across the wire. Calculate the drift speed of electrons if the number of electrons per cubic metre in copper is 8 x 10²⁸ electrons.
Solution:
Here, R = 10 Ω, A = 1 mm² = 10^-6 m², V = 10 V, n = 8 x 10²⁸ electrons/m³
Now,
I = enAv_d
∴ \(\frac { V }{ R }\) = enAv_d (or) v_d = \(\frac { V }{ enAR }\)
= \(\frac{10}{1.6 \times 10^{-19} \times 8 \times 10^{28} \times 10^{-6} \times 10}\) = 0.078 x 10^-3 ms^-1
= 0.078 x ms^-1

Question 5.
(i) At what temperature would the resistance of a copper conductor be double its resistance at 0°C.
(ii) Does this temperature hold for all copper conductors regardless of shape and size? Given a for Cu = 3.9 x 10^-3 °C^-1.
Solution:

Thus the resistance of copper conductor becomes double at 256 °C.
(ii) Since a does not depend on size and shape of the conductor. So the above result holds for all copper conductors.

Question 6.
Find the value of current I in the circuit shown in figure.
Solution:

In the circuit, the resistance of arm ACB (30 + 30 = 60 Ω) is the parallel with the resistance of arm AB (= 30 Ω).
Hence, the effective resistance of the circuit is
R = \(\frac { 30 × 60 }{ 30 + 60 }\) = 20 Ω
Current, I = \(\frac { V }{ R }\) = \(\frac { 2 }{ 20 }\) = 0.1 A.

Common Errors and Its Rectifications:

Common Errors:

1. Sometimes students think that charge and current are same.
2. In doing calculation part students can’t give the importance to mention the units.
3. They may confuse the parallel and series network of the resistance.

Rectifications:

1. Charge q = ne Current I = q/t
2. Unit is very importance to the every physical quantities.
3. If the resistors are series, their resultant is sum of the all the reciprocal of individual resistance. If the resistors are parallel their resultant is sum of the individual resistance.

We believe that the shared knowledge about Tamilnadu State Board Solutions for 12th Physics Chapter 2 Current Electricity Questions and Answers learning resource will definitely guide you at the time of preparation. For more details visit Tamilnadu State Board Solutions for 12th Physics Solutions and for more information ask us in the below comments and we’ll revert back to you asap.

Are you searching for the Samacheer Kalvi 12th Chemistry Chapter Wise Solutions PDF? Then, get your Samacheer Kalvi 12th Chapter Wise Solutions PDF for free on our website. Students can Download Chemistry Chapter 3 p-Block Elements – II Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 3 p-Block Elements – II

All concepts are explained in an easy way on our website. So, students can easily learn and practice Tamilnadu State Board 12th Chemistry Chapter 3 p-Block Elements – II Question and Answers. You can enjoy the digitized learning with the help of the Tamilnadu State Board Chemistry Online Material.

Samacheer Kalvi 12th Chemistry p-Block Elements – II TextBook Evalution

I. Choose the correct answer.

12th Chemistry Chapter 3 Book Back Answers Question 1.
In which of the following, NH₃ is not used?
(a) Nessler’s reagent
(b) Reagent for the analysis of IV group basic radical
(c) Reagent for the analysis of III group basic radical
(d) Tollen’s reagent
Answer:
(a) Nessler’s reagent

Samacheer Kalvi Guru 12th Chemistry Question 2.
Which is time regarding nitrogen?
(a) least electronegative element
(b) has low ionisation enthalpy than oxygen
(c) d-orbitals available
(d) ability to form pπ – pπ bonds with itself
Answer:
(d) ability to form pπ – pπ bonds with itself

12th Chemistry 3rd Lesson Book Back Answers Question 3.
An element belongs to group 15 and 3 rd period of the periodic table, its electronic configuration would be …………
(a) 1s²2s²2p⁴
(b) 1s²2s²2p³
(c) 1s²2s²2p⁶3s²3p²
(d) 1s²2s²2p⁶3s²3p³
Answer:
(d) 1s²2s²2p⁶3s²3p³

Samacheerkalvi.Guru 12th Chemistry Question 4.
Solid (A) reacts with strong aqueous NaOH liberating a foul smelling gas(B) which spontaneously bum in air giving smoky rings. A and B are respectively …………
(a) P₄(red) and PH₃
(b) P₄(white) and PH₃
(c) S₈ and H₂S
(d) P₄(white) and H₂S
Answer:
(b) P₄(white) and PH₃

Samacheer Kalvi Guru Class 12 Chemistry Question 5.
In the brown ring test, brown colour of the ring is due to …………
(a) a mixture of NO and NO₂
(b) Nitroso ferrous sulphate
(c) Ferrous nitrate
(d) Ferric nitrate
Answer:
(b) Nitroso ferrous sulphate

Samacheer Kalvi 12th Chemistry Question 6.
On hydrolysis, PCl₃ gives …………
(a) H₃PO₃
(b) PH₃
(c) H₃PO₄
(d) POOL
Answer:
(a) H₃PO₃

Chemistry Class 12 Samacheer Kalvi Question 7.
P₄O₆ reacts with cold water to give …………
(a) H₃PO₃
(b) H₄P₂O₇
(c) HPO₃
(d) H₃PO₄
Answer:
(a) H₃PO₃

Samacheer Kalvi Class 12 Chemistry Solutions Question 8.
The basicity of pyrophosphorous acid ( H₄P₂O₅) is …………
(a) 4
(b) 2
(c) 3
(d) 5
Answer:
(b) 2

Samacheer Kalvi Guru Chemistry Question 9.
The molarity of given orthophosphoric acid solution is 2M. its normality is …………
(a) 6N
(b) 4N
(c) 2N
(d) none of these
Answer:
(a) 6N

Chapter 3 Chemistry Class 12 Notes Question 10.
Assertion – bond dissociation energy of fluorine is greater than chlorine gas
Reason – chlorine has more electronic repulsion than fluorine
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(d) Both assertion and reason are false. The converse is true.

Samacheer Kalvi 12th Chemistry Solutions Question 11.
Among the following, which is the strongest oxidizing agent?
(a) Cl₂
(b) F₂
(c) Br₂
(d) I₂
Answer:
(b) F₂

Class 12 Chemistry Samacheer Kalvi Question 12.
The correct order of the thermal stability of hydrogen halide is …………
(a) HI > HBr > HCl > HF
(b) HF > HCl > HBr > HI
(c) HCl > HF > HBr > HI
(d) HI > HCl > HF > HBr
Answer:
(b) HF > HCl > HBr > HI

Samacheer Kalvi 12 Chemistry Solutions Question 13.
Which one of the following compounds is not formed?
(a) XeOF₄
(b) XeO₃
(c) XeF₂
(d) NeF₂
Answer:
(d) NeF₂

Class 12 Chemistry Chapter 3 Notes Pdf Download Question 14.
Most easily liquefiable gas is …………
(a) Ar
(b) Ne
(c) He
(d) Kr
Answer:
(c) He

12th Chemistry Chapter 7 Book Back Answers Question 15.
XeF₆ on complete hydrolysis produces …………
(a) XeOF₄
(b) XeO₂F₄
(c) XeO₃
(d) XeO₂
Answer:
(c) XeO₃

Question 16.
On oxidation with iodine, sulphite ion is transformed to …………
(a) S₄\({ O }_{ 6 }^{ 2- }\)
(b) S₂\({ O }_{ 6 }^{ 2- }\)
(c) S\({ O }_{ 4 }^{ 2- }\)
(d) S\({ O }_{ 3 }^{ 2- }\)
Answer:
(c) S\({ O }_{ 4 }^{ 2- }\)

Question 17.
Which of the following is strongest acid among all?
(a) HI
(b) HF
(c) HBr
(d) HCl
Answer:
(a) HI

Question 18.
Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?
(a) Br₂ > I₂ > F₂ > Cl₂
(b) F₂ > Cl₂ > Br₂ > I₂
(c) I₂ > Br₂ > Cl₂ > F₂
(d) Cl₂ > Br₂ > F₂ > I₂
Answer:
(d) Cl₂ > Br₂ > F₂ > I₂

Question 19.
Among the following the correct order of acidity is …………
(a) HClO₂ < HCIO < HClO₃ < HClO₄
(b) HClO₄ < HClO₂ < HCIO < HClO₃
(c) HClO₃ < HClO₄ < HClO₂ < HCIO
(d) HCIO < HClO₂ < HClO₃ < HClO₄
Answer:
(d) HCIO < HClO₂ < HClO₃ < HClO₄

Question 20.
When copper is heated with cone HNO₃ it produces …………
(a) CU(NO₃)₂ , NO and NO₂
(b) Cu(NO₃)₂ and N₂O
(c) CU(NO₃)₂ and NO₂
(d) Cu(NO₃)₂ and NO
Answer:
(c) CU(NO₃)₂ and NO₂

II. Answer the following questions:

Question 1.
What is inert pair effect?
Answer:
In p-block elements, as we go down the group, two electrons present in the valence s-orbital become inert and are not available for bonding (only p-orbital involves chemical bonding). This is called inert pair effect.

Question 2.
Chalcogens belongs to p-block. Give reason.
Answer:
Chalcogens are ore forming elements. Most of the ores are oxides and sulphides, therefore oxygen, sulphur and other group 16 elements are called Chalcogens. In O, S, Se, Te and Po last electron enters to p-orbital. Therefore Chalcogens belongs to p-block.

Question 3.
Explain why fluorine always exhibit an oxidation state of -1?
Answer:
Fluorine the most electronegative element than other halogens and cannot exhibit any positive oxidation state. Fluorine does not have d-orbital while other halogens have d-orbitals. Therefore fluorine always exhibit an oxidation state of-1 and others in halogen family shows +1, +3, +5 and +7 oxidation states.

Question 4.
Give the oxidation state of halogen in the following.

1. OF₂
2. O₂F₂
3. Cl₂O₃
4. I₂O₄

Answer:
1. OF₂
+2 + 2(x) = 0
+2 = -2x
2x = -2
x =-1

2. O₂F₂
2(+1) + 2x = 0
2x = – 2
x = – 1

3. Cl₂O₃
2(x) + 3(-2) =0
2x = +6
x = +3

4. I₂O₄
2(x) + 4(-2) =0
2x = +8
x = +4

Question 5.
What are interhalogen compounds? Give examples.
Answer:
Each halogen combines with other halogens to form a series of compounds called interhalogen compounds. For example, Fluorine reacts readily with oxygen and forms difluorine oxide (F₂O) and difluorine dioxide (F₂O₂).

Question 6.
Why fluorine is more reactive than other halogens?
Answer:
Fluorine is the most reactive element among halogen. This is due to the minimum value of F – F bond dissociation energy. Hence fluorine is more reactive than other halogens.

Question 7.
Give the uses of helium.
Answer:

1. Helium and oxygen mixture is used by divers in place of air oxygen mixture. This prevents the painful dangerous condition called bends.
2. Helium is used to provide inert atmosphere in electric arc welding of metals
3. Helium has lowest boiling point hence used in cryogenics (low temperature science).
4. It is much less denser than air and hence used for filling air balloons

Question 8.
What is the hybridisation of iodine in IF₇? Give its structure.
Answer:
Hybridisation of iodine in IF₇ is sp³d³ Structure of IF₇ is pentagonal bipyramidal.

Question 9.
Give the balanced equation for the reaction between chlorine with cold NaOH and hot NaOH.
Answer:
1. Reaction between chlorine with cold NaOH:
Cl₂+ H₂O → HCl + HOCl
HCl + NaOH → NaCl + H₂O
HOCl + NaOH → NaOCl + H₂O
Overall reaction

Chlorine reacts with cold NaOH to give sodium chloride and sodium hypochlorite.

2. Reaction between chlorine with hot NaOH:
Cl₂ + H₂O → HCl + HOCl
HCl + NaOH → NaCl + H₂O
HOCl + NaOH → NaOCl + H₂O
3NaOCl → NaClO₃+ 2NaCl
Overall reaction

Chlorine reacts with hot NaOH to give sodium chlorate and sodium chloride.

Question 10.
How will you prepare chlorine in the laboratory?
Answer:
1. Chlorine is prepared by the action of cone, sulphuric acid on chlorides in presence of manganese dioxide.
4NaCl + MnO₂ + 4H₂SO₄ → Cl₂ + MnCl₂ + 4NaHSO₄ + 2H₂O

2. It can also be prepared by oxidising hydrochloric acid using various oxidising agents such as manganese dioxide, lead dioxide, potassium permanganate or dichromate.
PbO₂ + 4HCl → PbCl₂ + 2H₂O + Cl₂
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
2KMnO₄ + 16HCl → 2KCl + 2MnCl + 8H₂O + 5Cl₂
K₂Cr₂O₇ + 14HCl → 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂

3. When bleaching powder is treated with mineral acids chlorine is liberated
CaOCl₂ + 2HCl → CaCl₂ + H₂O + Cl₂
CaOCl₂ + H₂SO₄ → CaSO₄ + H₂O + Cl₂

Question 11.
Give the uses of sulphuric acid.
Answer:

1. Sulphuric acid is used in the manufacture of fertilisers, ammonium sulphate and super phosphates and other chemicals such as hydrochloric acid, nitric acid etc.
2. It is used as a drying agent and also used in the preparation of pigments, explosives etc.

Question 12.
Give a reason to support that sulphuric acid is a dehydrating agent.
Answer:
Sulphuric acid is highly soluble in water and has strong affinity towards water and hence it can be used as a dehydrating agent. When dissolved in water it forms mono ( H₂SO₄. H₂O ) and di ( H₂SO₄. 2H₂O ) hydrates and the reaction is exothermic. The dehydration property can also be illustrated by its reaction with organic compounds such as sugar, oxalic acid and formic acid.

Question 13.
Write the reason for the anamolous behaviour of Nitrogen.
Answer:
1. Due to its small size, high electro negativity, high ionisation enthalpy and absence of d-orbitals.

2. N, has a unique ability to form pπ – pπ multiple bond whereas the heavier members of this group (15) do not form pπ – pπ bond, because their atomic orbitals are so large and diffused that they cannot have effective overlapping.

3. Nitrogen exists a diatomic molecule with triple bond between the two atoms whereas other elements form single bond in the elemental state.

4. N cannot form dπ – pπ bond due to the absence of d – orbitals whereas other elements can.

Question 14.
Write the molecular formula and structural formula for the following molecules.
(a) Nitric acid
(b) dinitrogen pentoxide
(c) phosphoric acid
(d) phosphine
Answer:

Question 15.
Give the uses of argon.
Answer:
Argon prevents the oxidation of hot filament and prolongs the life in filament bulbs.

Question 16.
Write the valence shell electronic configuration of group-15 elements.
Answer:
General electronic configuration of group 15 elements are ns²np³.

1. Nitrogen – [He] 2s² 2p³
2. Phosphorous – [Ne] 3s² 3p³
3. Arsenic – [Ar] 3d¹⁰ 4s² 4p³
4. Antimony – [Kr] 4d¹⁰ 5s² 5p³
5. Bismuth – [Ne] 4f¹⁴ 5s¹⁰ 6s² 6p³

Question 17.
Give two equations to illustrate the chemical behaviour of phosphine.
Answer:
1. Phosphine reacts with halogens to give phosphorous penta halides.
PH₃ + 4Cl₂ → PCl₅ + 3HCl

2. Phosphine forms coordination compound with lewis acids such as boron trichloride.

3. Phosphine precipitates some metal from their salt solutions.
3AgNO₃ + PH₃ → Ag₃P + 3HNO₃

Question 18.
Give a reaction between nitric acid and a basic oxide.
Answer:
Nitric acid reacts with bases and basic oxides to form salts and water.

1. ZnO + 2HNO₃ → Zn(NO₃)₂ + H₂O
2. 3FeO + 10HNO₃ → 3Fe(NO₃)₃ + NO + 5H₂O

Question 19.
What happens when PCl₅ is heated?
Answer:
On heating phosphorous pentachloride, it decomposes into phosphorus trichloride and chlorine.
PCl₅ \(\underrightarrow { \triangle }\) PCl₃ + Cl₂

Question 20.
Suggest a reason why HF is a weak acid, whereas binary acids of the all other halogens are strong acids.
Answer:
The hydrogen halides are extremely soluble in water due to the ionisation.
X + H₂O → H₃O⁺ + X^–
( X = F, Cl, Br or I )

Solutions of hydrogen halides are therefore acidic and known as hydrohalic acids. Hydrochloric, hydrobromic and hydroiodic acids are almost completely ionised and are therefore strong acids but HF is a weak acid. For HF,

• HF + H₂O \(\rightleftharpoons\) H₃O⁺ + F^–
• HF + F^– → \({ HF }_{ 2 }^{ – }\)

At high concentration, the equilibrium involves the removal of fluoride ions is important. Since it affects the dissociation of hydrogen fluoride, therefore it is a weak acid.

Question 21.
Deduce the oxidation number of oxygen in hypofluorous acid – HOF.
Answer:
In case of O – F bond is HOF, fluorine is most electronegative element. So its oxidation number is -1. Thereby oxidation number of O is +1. Similarly in case of O – H bond is HOF. O is highly electronegative than H. So its oxidation number is -1 and oxidation number of H is +1. So, Net oxidation of oxygen is – 1 + 1 = 0.

Question 22.
What type of hybridisation occur in

1. BrF₅
2. BrF₃

Answer:
1. BrF₅
BrF₅ is a AX₅ type. Therefore is has sp³d² hybridisation. Hence, BrF₅ molecule has square pyramidal shape.

2. BrF₃
BrF₃ is a AX₃ type. Therefore it has sp³d hybridisation. Hence, BrF₃ molecule has T-shape.

Question 23.
Complete the following reactions.
Answer:

Samacheer Kalvi 12th Chemistry p-Block Elements – II Evaluate yourself

Question 1.
Write the products formed in the reaction of nitric acid (both dilute and concentrated) with zinc.
Answer:

Samacheer Kalvi 12th Chemistry p-Block Elements – II Additional Questions

Samacheer Kalvi 12th Chemistry p-Block Elements – II 1 Mark Questions and Answers

I. Choose the best answer.

Question 1.
About 78% of earth atmosphere contains,…………
(a) P
(b) As
(c) N
(d) Bi
Answer:
(c) N

Question 2.
Which one of the following is not a pnictogens?
(a) Nitrogen
(b) Oxygen
(c) Phosphorous
(d) Antimony
Answer:
(b) Oxygen

Question 3.
Which one of the following shows isotopes?
(a) Nitrogen
(b) Arsenic
(c) Antimony
(d) Bismuth
Answer:
(a) Nitrogen

Question 4.
Nitrogen gas in atmosphere is separated industrially from liquid air by …………
(a) simple distillation
(b) Fractional distillation
(c) Sublimation
(d) Distillation under reduced pressure
Answer:
(b) Fractional distillation

Question 5.
Bond order for nitrogen molecule is …………
(a) 1
(b) 2
(c) 3
(d) 0
Answer:
(c) 3

Question 6.
Nitrogen gas is …………
(a) Inert
(b) Noble
(c) More reactive
(d) Less reactive
Answer:
(a) Inert

Question 7.
Which one of the following is used in cryosurgery?
(a) Liq N₂
(b) Liq NH₃
(c) Liq Na
(d) Liq H₂
Answer:
(a) Liq N₂

Question 8.
The dielectric constant of ammonia is (K) …………
(a) 10^-30
(b) 10^-14
(c) 10³⁰
(d) 10¹⁴
Answer:
(a) 10^-30

Question 9.
When ammonia reacts with copper sulphate solution to give complex, the colour of complex is …………
(a) violet
(b) deep blue
(c) blue
(d) Red
Answer:
(b) deep blue

Question 10.
H – N – H bond angle in NH₃ is …………
(a) 109° 28’
(b) 107° 28’
(c) 104°
(d) 107°
Answer:
(d) 107°

Question 11.
Shape of ammonia is …………
(a) Planar
(b) Square planar
(c) Pyramidal
(d) Square pyramidal
Answer:
(c) Pyramidal

Question 12.
Nitric acid prepared in large scales using …………
(a) Ostwald’s process
(b) Haber’s process
(c) Contact process
(d) Deacon’s process
Answer:
(a) Ostwald’s process

Question 13.
Benzene undergoes nitration reaction to form nitrobenzene in this reaction takes place due to the formation of …………
(a) Hydronium ion
(b) Hydride ion
(c) Nitronium ion
(d) Nitrasonium ion
Answer:
(c) Nitronium ion

Question 14.
Oxidation state of N m FINO₃ is…………
(a) ±2
(b) +3
(c) +4
(d) +5
Answer:
(d) +5

Question 15.
Compound used in photography is …………
(a) AgNO₃
(b) AgBr
(c) AgCl
(d) AgI
Answer:
(a) AgNO₃

Question 16.
Sodium nitrate
(a) Photography
(b) Firearms
(c) Royal water Sosurgerv …………
(d) Cryosurgery
Answer:
(b) Firearms

Question 17.
Nitrogen sesquoxide colour is …………
(a) colourless
(b) Brown
(c) Blue
(d) Red
Answer:
(c) Blue

Question 18.
White phosphorous is also called as …………
(a) Red phosphorous
(b) Black phosphorous
(c) Scarlet phosphorous
(d) Yellow phosphorous
Answer:
(d) Yellow phosphorous

Question 19.
White (Yellow) phosphorous glows in the dark due to oxidation which is called …………
(a) phosphorescence
(b) phosphorus
(c) Fluorescence
(d) Liminoscence
Answer:
(a) phosphorescence

Question 20.
Yellow phosphorous reacts with alkali on boiling in an inert atmosphere liberates …………
(a) Phosphorous acid
(b) Phosphoric acid
(c) Phosphine
(d) Pyrophosphoric acid
Answer:
(c) Phosphine

Question 21.
Consider the following statements.
(i) phosphine is the most important hydride of phosphorous
(ii) phosphine is a poisonous gas with rotten egg smell.
(iii) phosphine is a powerful reducing agent

Which of the above statement(s) is / are correct?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (ii) only
Ansswer:
(c) (i) and (iii)

Question 22.
When phosphine is heated with air it bums to gives …………
(a) Orthophosphoric acid
(b) Metaphosphoric acid
(c) Pyrophosphoric acid
(d) Phosphoroustrioxide
Answer:
(b) Metaphosphoric acid

Question 23.
Hybridisation of P in phosphine is …………
(a) sp³d
(b) sp³d²
(c) sp3d3
(d) sp3
Answer:
(d) sp3

Question 24.
Compounds used in Holme’s signal are …………
(a) Phosphine + Acetylene
(b) H₃PO₃+H₃PO₃
(c) Calcium carbide + calcium phosphide
(d) Calcium carbonate + calcium phosphate
Answer:
(c) Calcium carbide + calcium phosphide

Question 25.
Chalgogens are also called as …………
(a) Ore forming elements
(b) Group-16 elements
(c) group 17 elements
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 26.
Element present in the volcanic ashes is …………
(a) Oxygen
(b) Sulphur
(c) Selenium
(d) Tellurium
Answer:
(b) Sulphur

Question 27.
The decomposition of potassium chlorate speed up in the presence of …………
(a) MnO₂
(h) Mn₃O₄
(c) MnSO₄
(d) KMnO₄
Answer:
(a) MnO₂

Question 28.
Pure ozone is …………
(a) yellow gas
(b) blue gas
(c) Pale blue gas
(d) bright blue gas
Answer:
(c) Pale blue gas

Question 29.
Shape of ozone …………
(a) V-shape
(b) Linear shape
(c) bent shape
(d) spherical shape
Answer:
(c) bent shape

Question 30.
The rate of decompositon of ozone drops sharply in …………
(a) acidic medium
(b) alkaline medium
(c) neutral medium
(d) Ether medium
Answer:
(b) alkaline medium

Question 31.
Which one of the following used as fuel in rockets?
(a) Liq O₂
(b) Liq CO₂
(c) Liq N₂
(d) Liq He – O₂
Answer:
(a) Liq O₂

Question 32.
Find out crystalline allotrophic form of sulphur?
(a) γ – sulphur
(b) λ – sulphur
(c) α – sulphur
(d) milk of sulphur
Answer:
(c) α – sulphur

Question 33.
Consider the following statements
(i) α – sulphur is the only thermodynamically stable allotrophic form.
(ii) At 140 ° C the mono clinic sulphur melts to form mobile pale yellow liquid called γ – sulphur
(iii) Monoclinic sulphur is stable between 96°C-119°C and slowly changes into λ- sulphur

Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii)
Answer:
(d) (ii) and (iii)

Question 34.
Sulphur di oxide, how many times heavier than air?
(a) 2 times
(b) 2.5 times
(c) 2.2 times
(d) 2.3 times
Answer:
(c) 2.2 times

Question 35.
Which one of following has temporary bleaching action?
(a) Chlorine
(b) SO₃
(c) H₃SO₄
(d) SO₂
Answer:
(d) SO₂

Question 36.
Sulphuric acid can be manufactured by …………
(a) Ostwald’s process
(b) Lead chamber process
(c) Deacon’s process
(d) Haber’s process
Answer:
(b) Lead chamber process

Question 37.
Sulphuric acid is manufactured by contact process, catalyst used in contact process is …………
(a) V₂O₅
(b) TiCl₄
(c) Fe
(d) Mo
Answer:
(c) V₂O₅

Question 38.
Benzene reacts with sulphuric acid to gives …………
(a) sulphate
(b) sulphide
(c) sulphonic acid
(d) sulphite
Answer:
(c) sulphonic acid

Question 39.
Reagent used to detect sulphate ion is …………
(a) BaCl₂
(b) BaSO₃
(c) (CH,COO),Pb
(d) both (a) and (c)
Answwer:
(d) both (a) and (c)

Question 40.
Deacon’s process is used to manufacture …………
(a) Cl₂
(b) F₂
(C) Br
(d) I₂
Answer:
(a) Cl₂

Question 41.
Catalyst used in Deacon’s process is …………
(a) CuCl₂
(b) Cu₂Cl₂
(c) CuBr
(d) Cu₂Br₂
Answer:
(b) Cu₂Cl₂

Question 42.
C₁₀H₁₆ + 8C₁₂ \(\underrightarrow { \triangle }\) A. Identify A?
(a) Methane
(b) Ethane
(c) Carbon
(d) Propane
Answer:
(c) Carbon

Question 43.
Passing chlorine gas through dry slaked lime to produce …………
(a) CaOCl
(b) CaOCl₂
(c) CaO
(d) CaCl₂
Answer:
(b) CaOCl₂

Question 44.
Which one of the following is used for purification of drinking water?
(a) SO₃
(b) SO₂
(c) Br₂ / H₂O
(d) Cl₂
Answer:
(d) Cl₂

Question 45.
Which one of the following is a weak acid?
(a) HF
(b) HCl
(c) HBr
(d) HI
Answer:
(a) HF

Question 46.
Reagent not stored in glass bottles?
(a) HCI
(b) HBr
(c) HF
(d) HI
Answer:
(c) HF

Question 47.
More reactive element is
(a) Fluorine
(b) Chlorine
(c) Bromine
(d) Iodine
Answer:
(a) Fluorine

Question 48.
The correct order of the acidity of hydrohalic acids?
(a) HF > HCI > HBr > HI
(b) HCI >HF >HBr >HI
(c) HBr > HCI >HF > HI
(d) HI > HBr > HCI > HF
Answer:
(d) HI > HBr > HCI > HF

Question 49.
Consider the following statements
(i) In interhalogen compounds, the central atom will be the smaller one.
(ii) It can be formed only between two halogen and not more than two halogens.
(iii) They are strong reducing agents.

Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (iii) only
Answer:
(c) (i) and (iii)

Question 50.
Shape of ClF₃ is …………
(a) Linear
(b) T-shape
(c) Pyrimidal
(d) Square planar
Answer:
(b) T-shape

Question 51.
Which one of the following is more acidic?
(a) HOCl
(b) HCIO₂
(c) HClO₃
(d) HClO₄
Answer:
(d) HClO₄

Question 52.
When XeF₆ reacts with 2.5 M NaOH gives …………
(a) Na₄XeO₆
(b) Na₂XeO₃
(c) XeO₂F₂
(d) XeO₃
Answer:
(a) Na₄XeO₆

Question 53.
Shape of XeF₆ is …………
(a) Octahedron
(b) Distorted octahedron
(c) Pyramidal
(d) Tetrahedron
Answer:
(b) Distorted octahedron

Question 54.
Which one of the following can penetrate through dense fog?
(a) He
(b) Ne
(c) Kr
(d) Rn
Answer:
(c) Kr

Question 55.
Find out radioactive element?
(a) He
(b) Rn
(c) Xe
(d) Ar
Answer:
(b) Rn

II. Fill in the blanks:

1. The 11th most abundant element is ………….
2. …………. is the principle gas of atmosphere.
3. Nitrogen is chemically ………….
4. …………. process for the synthesis of ammonia.
5. …………. is used for the manufacture of calcium cyanamide
6. …………. is a pungent smelling gas.
7. Ammonia acts as a …………. agent.
8. With excess of chlorine, ammonia reacts to give …………. an explosive substance.
9. When excess ammonia is added to aqueous solution of copper sulphate …………. colour compound is formed.
10. Pure nitric acid becomes …………. on standing.
11. …………. is used in gunpower for firearms.
12. The decomposition of ammonium nitrate gives ………….
13. White phosphorous is colourless but becomes pale yellow due to formation of a …………. upon standing.
14. The white phosphorous can be changed into …………. by heated it to 420°C in the absence of air and light.
15. …………. reacts with alkali on boiling in an inert atmosphere liberating phosphine.
16. …………. is used in the match boxes.
17. Phosphine is …………. smelling gas.
18. Phosphine has a …………. shape.
19. …………. is used for producing smoke screen.
20. When phosphorous trichloridie is hydrolysed with cold water it gives ………….
21. Elements belonging group 16 are called ………….
22. Under ordinary condition oxygen exists as a …………. gas.
23. Allotrophic form of oxygen is …………. and ………….
24. Pure ozone is …………. gas.
25. …………. is used in welding purpose.
26. Monoclinic sulphur is stable between 96° and 119°C and slowly changes into ………….
27. …………. gas is found in volcanic eruptions.
28. A large amount of …………. gas is released into atmosphere from plants using coal and oil and copper melting plants.
29. Sulphurdioxide gas has …………. odour.
30. Sulphurdioxide can be used for …………. and …………. in agriculture.
31. In SO₃ S-atom undergoes …………. hybridisation.
32. In SO₃ a double bond arises between S and O is due to …………. overlapping.
33. High boiling point and viscosity of sulphuric acid is due to ………….
34. …………. is used as a drying agent.
35. The main source of fluorine is ………….
36. The main source of chlorine is ………….
37. Chlorine is a …………. gas.
38. Chlorine is soluble in water and its solution is referred as ………….
39. …………. is produced by passing chlorine gas through dry slaked lime.
40. …………. is used in extraction of gold and platinum.
41. …………. is used for extraction of glue from bone.
42. At room temperature, hydrogen halides are gases but …………. can be readily liquefied.
43. Liberation of iodine which gives a …………. colouration with starch.
44. Each halogen combines with other halogens to form a series of compounds called ………….
45. Structure of AX₇ type is ………….
46. Oxidation state of Cl in HClO₄ is ………….
47. Noble gases have the …………. ionisation energy.
48. Xenon reacts with PtF₆ and gave an ………….
49. Kr and fluorine gases are irradiated with SbF it forms ………….
50. Shape of XeOF₄ is ………….
51. Helium used for filling air ………….
52. …………. is used in fluorescent bulbs.
53. …………. is used in high speed electronic flash bulbs.
54. Radon is a source of …………. rays.
55. …………. is formed by the hydrolysis of urea.
56. …………. element subtimes at 889 K.
57. Yellow phosphorus has a characteristics …………. smell.

Answers:

1. phosporous
2. Nitrogen
3. Inert
4. Haber’s
5. Nitrogen
6. Ammonia
7. Reducing
8. Nitrogen trichloride
9. Deep blue
10. Yellow
11. Nitric acid / NaNO
12. Nitrous oxide
13. Layer of red phosphorous
14. Red phosphorous
15. White phosphorous
16. Red phosporous
17. Rotten fish
18. Pyramidal
19. Phosphine
20. phosporous acid
21. Chalgogens
22. diatomic
23. dioxygen and ozone
24. pale blue
25. Oxyacetylene
26. Rhombic sulphur
27. Sulphurdioxide
28. SO₂
29. Suffocating
30. disinfecting crops and plants
31. sp²
32. pπ – dπ
33. Hydrogen bonding
34. Sulphuric acid
35. Fluorite
36. Sodium chloride
37. Green yellow
38. Chlorine water
39. Bleaching powder
40. Chlorine
41. Hydrochloric acid
42. hydrogen fluroide
43. blue-black
44. Inter-halogen compounds
45. pentagonal bipyramidal
46. +7
47. largest
48. orange yellow solid [XePtF₆]
49. KrF₂. 2SbF₃
50. Square pyramidal
51. Balloons
52. Krypton
53. Xenon
54. Gamma
55. Ammonia
56. Arsenic
57. Garlic

II. Match the following:

Question 1.
(i) Haber’s process – (a) HNO₃
(ii) Deacon’s process – (b) Ammonia
(iii) Contact process – (c) Chlorine
(iv) Ostwald’s process – (d) H₂SO₄
Answer:
(i) b
(ii) c
(iii) d
(iv) a

Question 2.
(i) Nitric acid – (a) Purification of bone black
(ii) HCl – (b) Photography
(iii) White (yellow) phosphorous – (c) Rotten fish smell
(iv) Phosphine – (d) Phosphorescence
Answer:
(i) b
(ii) a
(iii) d
(iv) c

Question 3.
(i) Nitrogen sesquoxide – (a) H₂N₂O₂
(ii) Nitrous oxide – (b) H₄N₂O₄
(iii) Hyponitrous acid – (c) N₂O
(iv) Hydronitrous acid – (d) N₂O₃
Answer:
(i) d
(ii) c
(iii) a
(iv) b

Question 4.
(i) N₂O – (a) +5
(ii) N₂O₄ – (b) +3
(iii) N₂O₅ – (c) +1
(iv) N₂O₃ – (d) +4
Answer:
(i) c
(ii) d
(iii) a
(iv) b

Question 5.
(i) White phosphorous – (a) Volcanic eruptions
(ii) Red phosphorous – (b) Yellow phosphorous
(iii) Phosphine – (c) Match boxes
(iv) SO₂ – (d) smoke screen
Answer:
(i) b
(ii) c
(iii) d
(iv) a

Question 6.
(i) ammonia – (a) suffocating odour
(ii) SO₂ – (b) Rotten fish smell
(iii) PH₃ – (c) Greenish yellow gas
(iv) Cl₂ – (d) pungent smelling gas
Answer:
(i) d
(ii) a
(iii) b
(iv) c

Question 7.
(i) XeF₄ – (a) sp³
(ii) XeOF₂ – (b) sp³d²
(iii) XeO₃ – (c) sp³d³
(iv) XeF – (d) sp³d
Answer:
(i) b
(ii) d
(iii) a
(iv) c

Question 8.
(i) Helium – (a) flash bulbs
(ii) Neon – (b) radioactive
(iii) Krypion – (c) air balloons
(iv) Radon – (d) Brilliant red
Answer:
(i) c
(ii) d
(iii) a
(iv) b

IV. Assertion and Reason

Question 1.
Assertion (A) – Xenon is used in high speed electronic flash bulbs used by photographers.
Reason (R) – Xenon emits an intense light in discharge tubes instantly.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct and R explains A

Question 2.
Assertion (A) – Noble gases have the largest ionisation energy compared to any other elements.
Reason (R) – Noble gases have incomplete filled orbital.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(c) A is correct but R is wrong

Question 3.
Assertion (A) – Hydrogen iodide decomposes at 400⁰C while hydrogen fluoride and hydrogen chloride are stable at this temperature.
Reason (R) – Thermal stability of hydrogen halides decreases from fluoride to iodide.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct and R explains A

Question 4.
Assertion (A) – The bleaching of chlorine is temporary.
Reason (R) – Chlorine oxidises ferrous salts to ferric salts.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(d) A is wrong but R is correct

Question 5.
Assertion (A) – Sulphuric acid is highly reactive.
Reason (R) – Sulphuric acid can act as strong acid and an oxidising agent.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct but doesn’t explains A

Question 6.
Assertion (A) – Sulphuric acid is a high boiling point and viscous liquid.
Reason (R) – This is due to the association of molecules together through hydrogen bonding.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct and R explains A

Question 7.
Assertion (A) – Monoclinic sulphur is less stable than rhomobic sulphur.
Reason (R) – Monoclinic sulphur is stable between 96°C – 119°C and slowly changes into rhombic sulphur.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) A and R are correct and R explains A

Question 8.
Assertion (A) – Nitrogen gas is chemically inert.
Reason (R) – Nitrogen has low bonding energy.
(a) A and R are correct and R explains A
(b) A and R are correct but doesn’t explains A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(c) A is correct but R is wrong

V. Find the odd one out
Question 1.
(a) NO
(b) HNO₃
(C) NO₂
(d) N₂O
Answer:
(b) HNO₃
Hint: HNO₃ is acid and other are oxides.

Question 2.
(a) Nitrous acid
(b) Nitric acid
(c) Hyponitrous acid
(d) Pemitrous acid
Answer:
(d) Pemitrous acid
Hint: Pemitrous acid contains peroxide linkage others doesn’t have peroxide linkage.

Question 3.
(a) White phosphorous
(b) Red phosphorous
(c) phosphorous pentaoxide
(d) black phosphorous
Answer:
(c) phosphorous pentaoxide
Hint: P₂O₅ is a compound of phosphorous and others are allotropic form of phosphorous.

Question 4.
(a) PH₃
(b) HPO₃
(c) H₃PO₃
(d) H₃PO₄
Answer:
(a) PH₃
Hint: PH₃ is hydrides of phosphorous and others are oxo acids of phosphorous.

Question 5.
(a) He
(b) Ne
(c) Ar
(d) Xe
Answer:
(d) Xe
Hint: Xe forms several chemical compounds than others.

VI. Find out the correct pair.

Question 1.
(a) Helium – filament bulbs
(b) Krypton – prevent bonds
(c) Xenon – Lasers
(d) Radon – flash bulbs
Answer:
(c) Xenon – Lasers

Question 2.
(a) Ra – gamma rays
(b) Xe – cancer growth
(c) Ne – balloons
(d) Kr – advertisement bulb
Answer:
(a) Ra – gamma rays

Question 3.
(a) ClF₃ – Linear
(b) BrF₅ – T Shaped
(c) IF₄ – square pyrimidal
(d) BrF₅ – square pyrimidal
Answer:
(d) BrF₅ – square pyrimidal

Question 4.
(a) XeOF₂ – sp³
(b) XeF₆ – sp³d³
(c) XeF₄ – sp³
(d) XeOF₄ – sp³d
Answer:
(b) XeF₆ – sp³d³

Question 5.
(a) OF₂ = -1
(b) Cl₄O₄ = -1
(c) I₂O₄ = -1
(d) I₂O₉ = -1
Answer:
(a) OF₂ = -1

Question 6.
(a) Royal water – Dissolving gold
(b) Chlorine – Entraction of glue from bone
(c) HCl- Extraction of gold
(d) Chlorine – Temporary bleaching
Answer:
(a) Royal water – Dissolving gold

Question 7.
(a) O – 2s²2p³
(b) S – 3s²3p⁴
(c) Se – 4d¹⁰5s²5p⁴
(d) Te – 3d¹⁰4s²4p⁴
Answer:
(b) S – 3s²3p⁴

VI.Find out the incorrect pair.

Question 1.
(a) Nitrogen gas – inert
(b) Ammonia – pungent smelling gas
(c) Nitric acid – oxidizing agent
(d) Phosphine – rotten egg smell
Answer:
(d) Phosphine – rotten egg smell

Question 2.
(a) Liquid nitrogen – biological preservation
(b) Nitric acid – photography
(c) white phosphorous – yellow phosphorous
(d) phosphorous – welding
Answer:
(d) phosphorous – welding

Question 3.
(a) N₂O = +1
(b) N₂O = +2
(c) N₂O₃ = +5
(d) NO₂ = +4
Answer:
(c) N₂O₃ = +5

Question 4.
(a) Hvponitrous acid – N₂O
(b) Nitrous acid – HNO₂
(c) pernitric acid – HNO₄
(d) pernitrous acid – HOONO
Answer:
(a) Hvponitrous acid – N₂O

Question 5.
(a) PH₃ – Holme’s signal
(b) O₂ – welding
(c) H₂SO₄ – Disinfecting crops
(d) SO₃ – Bleaching hair
Answer:
(c) H₂SO₄ – Disinfecting crops

Question 6.
(a) H₂SO₄ drying agent
(b) Chlorine – Deacon’s process
(c) HCI – purification of bone black
(d) Helium – flash bulbs
Answer:
(d) Helium – flash bulbs

Question 7.
(a) ICl – Linear
(b) CIF₃ – T shape
(c) IF₅ – pentagonal bipyramidal
(d) IF₇ – pentagonal bipyramidal
Answer:
(c) IF₅ – pentagonal bipyramidal

Question 8.
(a) HOCl = +2
(b) HOCl = +3
(e) HOCI = +5
(d) HOCl₄ = +7
Answer:
(a) HOCl = +2

Question 9.
(a) XeF₄ – sp³d³
(b) XeF₆ – sp³d³
(e) XeOF₄ – sp³d²
(d) XeO₃ – sp³d
Answer:
(d) XeO₃ – sp³d

Question 10.
(a) He – cryogenics
(b) Ne – advertisement
(c) Kr – flurescent bulbs
(d) Ra – Lasers.
Answer:
(d) Ra – Lasers.

Samacheer Kalvi 12th Chemistry p-Block Elements – II 2 Mark Questions and Answers

Question  1.
What are pnictogens?
Answer:
The group – 15 elements like nitrogen, phosphorous Arsenic, Antimony and Bismuth are collectively called as pnictogens. Their general outer electronic configuration is ns²np³.

Question 2.
What happen when sodium azide undergoes thermal decomposition?
Answer:
Pure nitrogen gas can be obtained by the thermal decomposition of sodium azide about 575K
2NaN₃ \(\underrightarrow { 575K }\) 2Na + 3N₂

Question 3.
How will you prepare ammonia from nitrogen? and mention the name of process?
Answer:
Nitrogen directly reacts with hydrogen gives ammonia. This reaction is favoured by high pressures and at optimum temperature in the presence of iron catalyst,
N₂ + 3H₃ \(\rightleftharpoons\) 2NH₃
This process is called as Haber’s process.

Question 4.
Why nitrogen gas is chemically inert?
Answer:
The chemically inert character of nitrogen is largely due to high bonding energy of the molecules 225 cal mol^-1 .Interestingly the triply bonded species is notable for its less reactivity in comparison with other iso-electronic triply bonded systems such as -C \(\equiv\) C – C \(\equiv\) Q, X- C \(\equiv\) N, etc.

Question 5.
Mention the uses of nitrogen?
Answer:

1. Nitrogen is used for the manufacture of ammonia, nitric acid and calcium cyanamide etc.
2. Liquid nitrogen is used for producing low temperature required in cryosurgery, and so in biological preservation

Question 6.
Explain the action of heat on ammonia.
Answer:
Above 500°C ammonia decomposes into its elements. The decomposition may be accelerated by metallic catalysts like Nickel, Iron. Almost complete dissociation occurs on continuous sparking.
2NH₃ \(\underrightarrow { { 500 }^{ 0 }C } \) N₂ + 3H₂

Question 7.
Prove ammonia act as a reducing agent?
Answer:
Ammonia reduces the metal oxides to metal when passed over heated metallic oxide.
3PbO + 2NH₃ → 3Pb + N₂ + 3H₂O

Question 8.
What happen when copper sulphate reacts with ammonia?
Answer:
When excess ammonia is added to aqueous solution copper sulphate a deep blue colour Complex [Cu(NH₃)₄ ]²⁺ is formed.

Question 9.
Pure nitric acid is colourless, on standing it becomes yellow. Justify your answer.
Answer:
Nitric acid decomposes on exposure to sunlight or on being heated, into nitrogen dioxide, water and oxygen.
4HNO₄ → 4NO₂ + 2H₂O + O₂
Due to this reaction pure acid or its concentrated solution becomes yellow on standing.

Question 10.
Write the products formed in the reaction of nitric acid with dilute and concentrated with magnesium.
Answer:
1. Magnesium with cone. HNO₃:
4Mg + 10HNO₃ → 4Mg(NO₃)₂ + NH₄NO₃ + 3H₂O
Magnesium reacts with concentraated nitric acid to gives both magnesium nitrate and ammonium nitrate.

2. Magnesium with dil HNO₃:
Magnesium reacts with dilute nitric acid to gives both magnesium nitrate and nitrous oxide.
4Mg + 10HNO₃ → 4 Mg(NO₃)₂ + N₂O + 5H₂O

Question 11.
Give the uses of nitric acid.
Answer:

1. Nitric acid is used as a oxidising agent and in the preparation of aquaregia.
2. 4. Salts of nitric acid are used in photography (AgNO₃) and gunpowder for firearms. (NaNO₃) .

Question 12.
How will you prepare nitric oxide from sodium nitrite?
Answer:
When sodium nitrite reacts with ferrous sulphate in the presence of sulphuric acid to gives nitric oxide.
2NaNO₂ + 2FeSO₄ + 3H₂S → Fe₂(SO₄)₃ + 2NaHSO₄ + 2H₂O + 2NO

Question 13.
How will you prepare nitrogen pentoxide?
Answer:
Nitric acid reacts with phosphorous pentaoxide to give nitrogen pentoxide.
2HNO₃ + P₂O₅ → N₂O₅ + 2HPO₃

Question 14.
Draw the structure of

1. N₂O
2. N₂O₃

Answer:
1. N₂O (Nitrous oxide)

2. N₂O₃ (Nitrogen sesquoxide)

Question 15.
Draw- the structure of

1. N₂O₄
2. N₂O₃

Answer:
1. N₂O₄ (Nitrogen tetraoxide)

2. N₂O₅

Question 16.
Draw the structure of

1. Hyponitrous acid
2. Hydronitrous acid.

Answer:
(a) Hypon trous acid – H₂N₂O₂
HO – N = H – OH

(b) Hydronitrous acid:

Question 17.
Mention the allotropic forms of phosphorous?
Answer:
The most common allotropic forms of phosphorous

1. white phosphorous
2. Red phosphorous
3. Black phosphorous

Question 18.
Why white phosphorous is also known as yellow phosphorous?
Answer:
The freshly prepared white phosphorus is colourless but becomes pale yellow due to formation of a layer of red phosphorus upon standing. Hence it is also known as yellow phosphorus.

Question 19.
What is phosphorescence?
Answer:
White (yellow) phosphorous glows in the dark due to oxidation which is called phosphorescence.

Question 20.
Why white phosphorous undergoes spontaneous combustion in air?
Answer:
White phosphorous ignition temperature is very low and hence it undergoes spontaneous combustion in air at room temperature and during in the combusition process, white phosphorous produces P₂O₅.

Question 21.
Draw the structure of

1. white phosphorous
2. red phosphorous

Answer:
1. White phosphorous

2. Red phosphorous

Question 22.
How will you convert red phosphorous into P₂O₃ and P₂Q₅?
Answer:
Red phosphorus reacts with oxygen on heating to give phosphorus trioxide or phosphorus pentoxide.
P₄ + 3O₂ \(\underrightarrow { \triangle }\) P₄O₆(phosphorous trioxide)
P₄ + 5O₂ \(\underrightarrow { \triangle }\) P₄Q₁₀ (phosphorous pentaoxide)

Question 23.
How will you prepare orthophosphoric acid from phosphorous?
Answer:
When phosphorous is treated with cone.nitric acid it is oxidised to orthophosphoric acid. This reaction is catalysed by iodine crystals.

Question 24.
Mention the uses of phosphorous?
Answer:

1. The red phosphorus is used in the match boxes.
2. It is also used for the production of certain alloys such as phosphor bronze.

Question 25.
Show that phosphine is weakly basic?
Answer:
Phosphine is weakly basic and forms phosphonium salts with halogen acids.
PH₃ +HI → PH₄I
PH₄I + H₂O \(\underrightarrow { \triangle }\) PH₃+ H₃O^– + I^–

Question 26.
Draw the structure of PCl₃
Answer:

Question 27.
How will you prepare PCl₃ from white phosphorous?

1. When a slow stream of chlorine is passed over white phosphorous, PCl₃ is formed
2. It can also be obtained by treating white phosphorous with thionyl chloride.

Question 28.
What happen when PCl₃ is treated with cold water?
Answer:
When PCl₃ is hydrolysed with cold water it gives phosphorous acid.

Question 29.
How will you prepare PCl₃ ?
Answer:
When PCl₃ is treated with excess chlorine, phosphorous pentachloride is obtained.
PCl₃+ Cl₂ → PCl₅

Question 30.
Mention the uses of PCl₃ and PCl₅?
answer:

1. Phosphorus trichloride is used as a chlorinating agent and for the preparation of H₂PO₃.
2. Phosphorous pentachloride is a chlorinating agent and is useful for replacing hydroxyl groups by chlorine atom.

Question 31.
Draw the structure of H₃PO₂ and H₃PO₃?
Answer:

Question 32.
Draw the structure of H₄P₂O₆ and H₄P₂O₃?
Answwer:
(i) H₄P₂O₆
(ii) H₄P₂O₃

Question 33.
Give the method to prepare hypophosphorous acid and pyrophosphoric acid?
Answer:
1. Hypophosphorous acid – Phosphorous reacts with water to give hypophosphorous acid.

2. Pyrophosphoric acid – Phosphorous acid is heated they produce pyrophosphoric acid.

Question 34.
Complete the reactions

1. HgO \(\underrightarrow { \triangle }\) ?
2. BaO₂ \(\underrightarrow { \triangle }\) ?

Answer:

1. HgO \(\underrightarrow { \triangle }\) 2Hg + O₂
2. 2BaO₂ \(\underrightarrow { \triangle }\) 2BaO + O₂

Question 35.
Give and explain the reaction used to estimation of ozone.
Answer:
Ozone oxidises potassium iodide to iodine.
O₃ + 2KI + H₂O → 2KOH + O₂ + I₂
This reaction is quantitative and can be used for estimation of ozone.

Question 36.
Write a short notes on Rhombic sulphur?
Answer:
Rhombic sulphur also known as a sulphur, is the only thermodynamically stable allotropic form at ordinary temperature and pressure. The crystals have a characteristic yellow colour and composed of S₈ molecules. When heated slowly above 96°C, it converts into monoclinic sulphur. Upon cooling below 96°C the β form converts back to a form.

Question 37.
Write a notes on monoclinic sulphur?
Answer:
Monoclinic sulphur also contains S₈ molecules in addition to small amount of S₆ molecules. It exists as a long needle like prism and is also called as prismatic sulphur. It is stable between 96° – 119°C and slowly changes into rhombic sulphur.

Question 38.
What are λ – sulphur?
Answer:
Sulphur also exists in liquid and gaseous states. At around 140° C the monoclinic sulphur melts to form mobile pale yellow liquid called λ sulphur.

Question 39.
How will you prepare sulphurdioxide by laboratory method?
Answer:
Sulphur dioxide is prepared in the laboratory treating a metal or metal sulphite with sulphuric acid.
Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O
\({ SO }_{ 3 }^{ – }\) + 2H⁺ → H₂O + SO₂

Question 40.
Complete the reactions,
Answer:

1. Zns + O₂ \(\underrightarrow { \triangle }\) ?
2. FeS₂ + O₂ \(\underrightarrow { \triangle }\) ?

Answer:

1. 2Zns + 3O₂ \(\underrightarrow { \triangle }\) 2ZnO + 2SO₂
2. 4FeS₂ + 11O₂ \(\underrightarrow { \triangle }\) 2Fe₂O₃ + 8SO₂

Question 41.
What happen when sulphurdioxdie reacts with sodium hydroxide and sodium carbonate?
Answer:
Sulphur dioxide reacts with sodium hydroxide and sodium carbonate to form sodium bisulphite and sodium sulphite respectively.

Question 42.
Mention the uses of sulphurdioxide?
Answer:

1. Sulphur dioxide is used in bleaching hair, silk, wool etc…
2. It can be used for disinfecting crops and plants in agriculture.

Question 43.
Why sulphuric acid is high boiling and viscous liquid ?
Answer:
Sulphuric acid is high boiling and viscous liquid this is due to the association of molecules together through hydrogen bonding.

Question 44.
Explain the reaction between benzene and sulphuric acid.
Answer:
Sulphuric acid reacts with benzene to give benzene sulphuric acid.

Question 45.
Give the uses of sulphuric acid?
Answer:

1. Sulphuric acid is used in the manufacture of fertilisers, ammonium sulphate and super phosphates and other chemicals such as hydrochloric acid, nitric acid etc…
2. It is used as a drying agent and also used in the preparation of pigments, explosives etc..

Question 46.
Draw the structure of H₂SO₃ and H₂SO₄ ?
Answer:
1. H₂SO₃

2. H₂SO₄

Question 48.
Draw the structure of

1. Pyrosulphuric acid
2. Peroromonosulphuric acid?

Answer:
1. Pyrosulphuric acid H₂S₂O₇

2. Peroromonosulphuric acid H₂SO₅

Question 49.
What happen when chlorine reacts with trupentine?
Answer:
When chlorine burnt with turpentine it forms carbon and hydrochloric acid.

Question 50.
How will you prepare hydrochloric acid by laboratory method?
Answer:
It is prepared by the action of sodium chloride and concentrated sulphuric acid

• NaCl + H₂SO₄ → NaHSO₄ + HCl
• NaHSO₄ + NaCl → Na₂SO₄ + HCl

Dry hydrochloric acid is obtained by passing the gas through conc. sulphuric acid

Question 51.
Mention the uses of hydrochloric acid?
Answer:

1. Hydrochloric acid is used for the manufacture of chlorine, ammonium chloride, glucose from com starch etc.,
2. It is used in the extraction of glue from bone and also for purification of bone black.

Question 52.
Complet the following reaction?
Answer:

Question 53.
Why noble gases have the largest ionisation energy?
Answer:
Noble gases have the largest ionisation energy compared to any other elements in a given row as they have completely filled orbital (ns²np⁶) in their outer most shell. They are extremely stable and have a small tendency to gain or lose electrons. Hence noble gases have the largest ionisation energy.

Question 54.
Mention the uses of Neon?
Answer:
Neon is used in advertisement as neon sign and the brilliant red glow is caused by passing electric current through neon gas under low
pressure.

Question 55.
Give the uses of Krypton?
Answer:

1. Krypton is used in fluorescent bulbs, flash bulbs etc…
2. Lamps filed with krypton are used in airports as approaching lights as they can penetrate through dense fog.

Question 56.
Mention the application of Xenon?
Answer:

1. Xenon is used in fluorescent bulbs, flash bulbs and lasers.
2. Xenon emits an intense light in discharge tubes instantly. Due to this it is used in high speed electronic flash bulbs used by photographers.

Question 57.
Give the uses of Radon?
Answer:

1. Radon is radioactive and used as a source of gamma rays
2. Radon gas is sealed as small capsules and implanted in the body to destroy malignant i.e. cancer growth.

Question 58.
Why are pentahalides more covalent than trihalides?
Answer:
Since elements in the +5 oxidation state have less tendency to lose electrons than in the +3 oxidation state, therefore elements in the +5 oxidation state are more covalent than in the +3 oxidation state. In other words, pentahalides are more covalent than trihalides.

59. Why is N₂ less reactive at room temperature?
Answer:
Due to the presence of triple bond between the two nitrogen atoms, the bond dissociation energy of N₂ (941.4 kJ mol^–) is very high. Therefore, N₂ is less reactive at room temperature.

Question 60.
Why is ICl more reactive than I₂ ?
Answer:
ICl bond is weaker than I – I bond. Therefore ICl can break easily to form halogen atoms which readily brings about the reaction.

Question 61.
Why is helium used in diving apparatus?
Answer:
It is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood.

Question 62.
Give the disproportionation reaction of H₃PO₃?
Answer:
H₃PO₃ on heating undergoes self-oxidation reduction as:

Question 63.
Why is red Phosphorus less reactive than white Phosphorus?
Answer:
White Phosphorus is more reactive than red Phosphorus under normal conditions because of angular strain in the P₄ molecule where the angles are only 60°.

Question 64.
Nitrogen does not form any pentahalide like Phosphorus. Why?
Answer:
Nitrogen does not form pentahalide due to non-availability of the d-orbitals in its valence shell.

Question 65.
Give reason for the following. Among the noble gases only Xenon is well known to form chemical compounds?
Answer:
Xe is largest in size and has the highest polarising power.

Question 66.
Why is hydrogen sulphide, with greater molar mass a gas, while water a liquid at room temperature?
Answer:
H₂O molecules are associated with intermolecular H-bonding, H₂S is not because oxygen is more electronegative and smaller in size than sulphur. That is why H₂O is a liquid and H₂S is a gas.

Question 67.
Noble gases are chemically inert. Give reasons.
Answer:
Noble gases are chemically inert because they have their octet complete except Helium, i.e. they have a stable electronic configurations.

Question 68.
Why do noble gases exist as monoatomic?
Answer:
Noble gases have stable electronic configuration, that is why they have no tendency to lose or gain electrons.Therefore they do not form covalent bond.

Question 69.
Nitrogen exists as diatomic molecule and Phosphrus as P⁴. Why?
Answer:
Nitrogen has a triple bond between its two atoms because of its small size and high electro negativity. Phosphorus P⁴ has single bond, that is why it is tetra-atomic.

Question 70.
Why is H₂S more acidic than water?
Answer:
It is because bond dissociation energy of S – H bond is less than O – H bond due to longer bond length.

Samacheer Kalvi 12th Chemistry p-Block Elements – II 3 Marks Questions and Answers

Question 1.
Complete the following reactions.
(i) 6Li + N₂ → ?
(ii) 3Ca + N₂ → ?
(iii) 2B + N₂ → ?
Answer:

Qustion 2.
How is ammonia prepared?
Answer:
1. Ammonia is formed by the hydrolysis of urea.
NH₂CONH₂ + H₂O → 2NH₃ + CO₂

2. Ammonia is prepared in the laboratory by heating an ammonium salt with a base.
\({ 2NH }_{ 4 }^{ + }\) + OH^– → 2NH₃+ H₂O
2NH₄Cl + CaO → CaCl₂ + 2NH₃ + H₂O

3. It can also be prepared by heating a metal nitrides such as magnesium nitride with water.
Mg₃N₂+ 6H₂O → 3Mg(OH)₂ + 2NH₃

Question 3.
Explain the structure of ammonia.
Answer:
Ammonia molecule is pyramidal in shape N – H bond distance is 1.016 A and H – H bond distance is 1.645 A with a bond angle 107°. The structure of ammonia may be regarded as a tetrahedral with one lone pair of electrons in one tetrahedral position hence it has a pyramidal shape as shown in the figure.

Question 4.
How will you prepare nitric acid?
Answer:
Nitric acid is prepared by heating equal amounts of potassium or sodium nitrate with concentrated sulphuric acid.
KNO₃ + H₂SO₄ → KHSO₄ + HNO₃

The temperature is kept as low as possible to avoid decomposition of nitric acid. The acid condenses to a fuming liquids which is coloured brown by the presence of a little nitrogen dioxide which is formed due to the decomposition of nitric acid.
4HNO₃ → 4NO₂ + 2H₂O + O₂

Question 5.
Discuss the Commercial method to prepare Nitric acid.
(Or)
How will you prepare nitric acid by Ostwald’s process?
Answer:
Nitric acid prepared in large scales using Ostwald’s process. In this method ammonia from Haber’s process is mixed about 10 times of air. This mixture is preheated and passed into the catalyst chamber where they come in contact with platinum gauze. The temperature rises to about 1275 K and the metallic gauze brings about the rapid catalytic oxidation of ammonia resulting in the formation of NO, which then oxidised to nitrogen dioxide.
4NH₃ + 5O₂ → 4NO + 6H₃O + 120kJ
2NO + O₂ → 2NO₃

The nitrogen dioxide produced is passed through a series of adsorption towers. It reacts with water to give nitric acid. Nitric acid formed is bleached by blowing air.
6NO₂ + 3H₂O → 4HNO₃ + 2NO + H₂O

Question 6.
Draw the structure of the following compounds.
(a) Nitrous acid
(b) Nitric acid
(c) Pernitric acid
Answer:

Question 7.
Identify A, B and C from the following reactions.

1. P₄ + Mg → A
2. P₄+ Ca → B
3. P₄ + Na → C

Answer:

1. P₄ + 6Mg → 2Mg₃P₂ (Magnesium phosphide)
2. P₄ + 6Ca → 2Ca₃P₂ (Calcium phosphide)
3. P₄+ 12Na → 2Na₃P (Sodium phosphide)

Question 8.
How will you prepare phosphine and explain the purification of phosphine?
Answer:
Phosphine is prepared by action of sodium hydroxide with white phosphorous in an inert atmosphere of carbon dioxide or hydrogen.

Phosphine is freed from phosphine dihydride (P₂H₄) by passing through a freezing mixture. The dihydride condenses while phosphine does not. Phosphine can also prepared by the hydrolysis of metallic phosphides with water or dilute mineral acids.

Phosphine is prepared in pure form by heating phosphorous acid.

A pure sample of phosphine is prepared by heating phosphonium iodide with caustic soda solution.

Question 9.
What happens when PH₃ reacts with oxygen or air?
Answer:
When phosphine air or oxygen it bums to give meta phosphoric acid.

Question 10.
Explain the structure of phosphine.
Answer:
In phosphine, phosphorus shows sp₃ hybridisation. Three orbitals are occupied by bond pair and fourth comer is occupied by lone pair of electrons. Hence, bond angle is reduced to 94°. Phosphine has a pyramidal shape.

Question 11.
Discuss the uses of phosphine.
Answer:
Phosphine is used for producing smoke screen as it gives large smoke. In a ship, a pierced container with a mixture of calcium carbide and calcium phosphide, liberates phosphine and acetylene when thrown into sea. The liberated phosphine catches fire and ignites acetylene. These burning gases serves as a signal to the approaching ships. This is known as Holmes signal.

Question 12.
How does PCl₃ reacts with the following reagents?

1. C₂H₅OH
2. C₂H₅COOH

Answer:

1. C₂H₅OH + PCl₃ → C₂H₅Cl + H₃PO₃
2. C₂H₅COOH + PCl₃ → C₂H₅COCl + H₃PO₃

Question 13.
Explain the reaction between PCl₅ and water.
Answer:
Phosphorous pentachloride reacts with water to give phosphoryl chloride and orthophosphoric acid.

Question 14.
Explain the structure of phosphorous trioxide (P₂O₃).
Answer:
In phosphorous trioxide four phosphorous atoms lie at the comers of a tetrahedron and six oxygen atoms along the edges. The P – O bond distance is 165.6 pm which is shorter than the single bond distance of P – O (184 pm) due to pπ – dπ bonding and results in considerable double bond character.

Question 15.
Discuss the structure of phosphorous pentaoxide (P₂O₃).
Answer:
In P₄O₁₀ each P atoms form three bonds to oxygen atom and also an additional coordinate bond with an oxygen atom. Terminal coordinate P – O bond length is 143 pm, which is less than the expected single bond distance. This may be due to lateral overlap of filled p orbitals of an oxygen atom with empty d-orbital on phosphorous.

Question 16.
Mention the uses of oxygen.
Answer:

1. Oxygen is one of the essential component for the survival of living organisms.
2. It is used in welding (oxyacetylene welding)
3. Liquid oxygen is used as fuel in rockets etc.

Question 17.
Give and explain reducing behaviour of sulphur dioxide.
Answer:
As Sulphur dioxide can readily be oxidised, it acts as a reducing agent. It reduces chlorine into hydrochloric acid.
SO₂ + 2H₂O + Cl₂ → H₂SO₄ + 2HCl
It also reduces potassium permanganate and dichromate to Mn²⁺ and Cr³⁺ respectively.

• 2KMnO₄ + 5SO₂ + 2H₂O → K₂SO₄ + 2MnSO₄ + 2H₂SO₄
• K₂Cr₂O₇ + 3SO₂ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O

Question 18.
Why bleaching action of sulphur dioxide is temporary?
Answer:
In presence of water, sulphur dioxide bleaches coloured wool, silk, sponges and straw into colourless due to its reducing property.
SO₂ + 2H₂O → 2H₂SO₄ + 2(H)

However, the bleached product (colourless) is allowed to stand in air, it is reoxidised by atmospheric oxygen to its original colour. Hence bleaching action of sulphur dioxide is temporary.

Question 19.
Explain the structure of sulphur dioxide.
Answer:
In sulphur dioxide, sulphur atom undergoes sp² hybridisation. A double bond arises between S and O is due to pπ- dπ overlapping.

Question 20.
How will you manufacture sulphuric acid by contact process?
Answer:
Manufacture of sulphuric acid by contact process involves the following steps:
1. Initially sulphur dioxide is produced by burning sulphur or iron pyrites in oxygen/air.
S + O₂ → SO₂
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

2. Sulphur dioxide formed is oxidised to sulphur trioxide by air in the presence of a catalyst such as V₂O₅ or platinised asbestos.

3. The sulphur trioxide is absorbed in concentrated sulphuric acid and produces oleum (H₂S₂O₇). The oleum is converted into sulphuric acid by diluting it with water.

To maximise the yield the plant is operated at 2 bar pressure and 720 K. The sulphuric acid obtained in this process is over 96 % pure.

Question 21.
Prove that H₂SO₄ is a strong dibasic acid.
Answer:
Sulphuric acid forms two types of salts namely sulphates and bisulphates.

Question 22.
Draw the structure of
(a) Marshall’s acid
(b) polythionic acid
(c) Dithionic acid
Answer:

Question 23.
What happens when chlorine reacts with ammonia?
Answer:
1. If chlorine reacts with excess ammonia, it gives nitrogen and ammonium chloride.
8NH₃ + 3Cl₂ → N₂ + 6NH₄Cl

2. If ammonia reacts with excess chlorine, it gives nitrogen trichloride and ammonium chloride.
4NH₂ + 3Cl₂ → NCl₂ + 3NH₄Cl

Question 24.
Bleaching action of chlorine is permanent. Justify this statement.
Answer:
Chlorine is a strong oxidising and bleaching agent because of the nascent oxygen.

Colouring matter + Nascent oxygen → Colourless oxidation product
Therefore, bleaching of chlorine is permanent. It oxidises ferrous salts to ferric, sulphites to sulphates and hydrogen sulphide to sulphur.

Question 25.
Give the uses of chlorine.
Answer:

1. Purification of drinking water
2. Bleaching of cotton textiles, paper and rayon
3. It is used in extraction of gold and platinum

Question 26.
What is Royal water?
Answer:
When three parts of concentrated hydrochloric acid and one part of concentrated nitric acid are mixed, aquaregia is obtained. This is also known as Royal water. This is used for dissolving gold, platinum etc.
Au + 4H⁺ + \({ 4NO }_{ 3 }^{ – }\) + 4 Cl^– → \({ AuCl }_{ 4 }^{ – }\) + NO + 2H₂O
3Pt + 16H⁺ + \({ 4NO }_{ 3 }^{ – }\) + 18Cl^– → \({ 3PtCl }_{ 3 }^{ 2- }\) + NO + 8H₂O

Question 27.
Why HF is not stored in glass bottles?
Answer:
HF rapidly reacts with silica and glass to form soluble salt and that process leads to break the glass bottles. Hence HF is not stored in glass bottles.
SiO₂ + 4HF → SiF₄ + 2H₂O
Na₂SiO₃ + 6HF → Na₂SiF₆ + 3H₂O

Question 28.
Give the properties of inter halogen compounds.
Answer:
Properties of inter halogen compounds:

1. The central atom will be the larger one.
2. It can be formed only between two halogen and not more than two halogens.
3. Fluorine can’t act as a central metal atom being the smallest one.
4. Due to high electronegativity with small size fluorine helps the central atom to attain high coordination number
5. They can undergo the auto ionization.
6. They are strong oxidizing agents.

Question 29.
How will you prepare Xenon fluoride?
Answer:
Xenon fluorides are prepare by direct reaction of xenon and fluorine under different conditions as shown below.

Question 30.
Complete the following reaction.
(i) XeOF₄ + SiO₂ → A + SiF₆
(ii) A + SiO₂ → B + SiF₆
Answer:

Question 31.
What happens when XeF₆ reacts 2.5 M solution of NaOH?
Answer:
When XeF₆ reacts with 2.5 M of NaOH, sodium per xenate is obtained.

Question 32.
Give two examples to show the anomalous behaviour of fluorine.
Answer:
The anomalous behaviour of fluorine is due to its:

1. Small size
2. highest electronegativity
3. low F-F bond dissociation enethalpy
4. non-availability of d-orbitals in its valence shell.

The two examples are:
1. Due to non-availability of d-orbitals in its valence shell, fluorine cannot expand its octet, therefore, shows only -1 oxidation state while all other halogens due to the presence of d-orbitals shows positive oxidation states of +1, +3, +5 and +7 besides oxidation state of -1.

2. Due to its small size, the three lone pair of electrons on each F atom in F – F molecule, repel the bond pair. As a result, F – F bond dissociation energy is lower than that of Cl – Cl bond.

Question 33.
Why does the reactivity of nitrogen differ from Phosphorus?
Answer:
Nitrogen exists as a diatomic molecule (N \(\equiv\) N). Due to the presence of a triple bond between the two N – atoms the bond dissociation energy is large (941.4 kJ mol^-1). As a result nitrogen is said to be chemically inert in its elemental state. In contrast, P – P single bond is much weaker (213 kJ mol^-1) than N \(\equiv\) N triple bond Therefore, phosphorus is much more reactive than nitrogen.

Question 34.
Why does NH₃ form hydrogen bond but PH₃ does not?
Answer:
The electronegativity of N (3.0) is much higher than that of FI (2.1). As a result, N – H bond is quite polar and hence NH₃ undergoes intermolecular H-bonding. In contrast, both P and H have an electronegativity of 2.1. Therefore P – H bond is non-polar and hence PH₃ does not undergo H – bonding.

Question 35.
Can PCl₅ act as an oxidising as well as a reducing agent? Justify.
Answer:
Oxidation state of P in PCl₅ is +5. Since P has five valence electrons in its valence shell, therefore it cannot increase its oxidation state beyond +5 by donating its electrons, therefore PCl₅ cannot act as a reducing agent. However, it can decrease its oxidation number from +5 to +3 or some lower value, therefore PCl₅ acts as an oxidising agent. For example, it oxidises Ag to AgCl.

Samacheer Kalvi 12th Chemistry p-Block Elements – II 5 Mark Questions and Answers

Question 1.
How does ammonia react with

1. Excess Cl₂
2. Na
3. CuSO₄
4. O₂ / ∆

Answer:
1. Reaction with Excess Cl₂

2. Reaction with Na:

3. Reaction with CuSO₄

4. Reaction with O₂

Question 2.
Explain the reaction of metals with nitric acid.
Answer:
The reactions of metals with nitric acid are explained in 3 steps as follows:
Primary reaction:
Metal nitrate is formed with the release of nascent hydrogen
M + HNO₃ → MNO₃ + (H)

Secondary reaction:
Nascent hydrogen produces the reduction products of nitric acid.

Tertiary reaction:
The secondary products either decompose or react to give final products

For examples:
Copper reacts with nitric acid in the following manner
3Cu + 6HNO₃ → 3Cu(NO₃)₂ + 6(H)
6(H) + 3HNO₃ → 3HNO₂ + 3H₂O
3HNO₂ → HNO₃ + 2NO + H₂O

Overall reaction
3Cu + 8HNO₃ → 3CU(NO₃)₂ + 2NO + 4H₂O

The concentrated acid has a tendency to form nitrogen dioxide
Cu + 4HNO₃ → 3CU(NO₃)₂ + 2NO₂ + 2H₂O

Question 3.
How will you prepare ozone by laboratory method? Explain the structure of ozone.
Answer:
In the laboratory ozone is prepared by passing electrical discharge through oxygen. At a potential of 20,000 V about 10% of oxygen is converted into ozone it gives a mixture known as ozonised oxygen. Pure ozone is obtained as a pale blue gas by the fractional distillation of liquefied ozonised oxygen.


Structure of ozone:
The ozone molecule have a bent shape and symmetrical with delocalised bonding between the oxygen atoms.

Question 4.
A is a king of acid. A reacts with HBr to give B and Bromine. A reacts with Na₂CO₃ to give C and carbon dioxide. Identify A, B and C. Give the reaction.
Answer:
1. King of acid is sulphuric acid (A).

2. Sulphuric acid (A) reacts with HBr to give SO₂ (B) and Bromine.

3. Sulphuric acid (A) reacts with Na2CO₃ to give sodium sulphate (C) and CO₂.

Question 5.
How does Sulphuric acid react with the following:
(a) Al
(b) KNO₃
(c) NaBr
(d) C₆H₆
Answer:

Question 6.

1. Explain the test for sulphate (or) sulphuric acid.
2. What happens when sulphuric acid reacts with oxalic acid?

Answer:
1. Dilute solution of sulphuric acid or aqueous solution of sulphates gives white precipitate with barium chloride solution. It can also be detected using lead acetate solution. Here a white precipitate of lead sulphate is obtained.

2. Sulphuric acid reacts with oxalic acid to give CO and CO₂

Question 7.
Discuss the manufacture of chlorine.
Answer:
Electrolytic process:
When a solution of brine (NaCl) is electrolysed, Na+ and CP ions are formed. Na⁺ ion reacts with OH^– ions of water and forms sodium hydroxide. Hydrogen and chlorine are liberated as gases.

Deacons process:
In this process a mixture of air and hydrochloric acid is passed up a chamber containing a number of shelves, pumice stones soaked in cuprous chloride are placed. Hot gases at about 723 K are passed through a jacket that surrounds the chamber.

The chlorine obtained by this method is dilute and is employed for the manufacture of bleaching powder. The catalysed reaction is given below,

Question 8.

1. How is bleaching powder prepared?
2. What happens when benzene reacts with chlorine?

Answer:
1. Bleaching powder is prepared by passing chlorine gas through dry slaked lime (Calcium hydroxide).

2. Benzene reacts with chlorine in the presence of ferric chloride to give chlorobenzene.

Question 9.
Cone. H₂SO₄ is added followed by heating to each of the following test tubes labelled (I) to (IV). Identify in which of the above test tube the following change will be observed. Support your answer with the help of chemical equation:
(a) formation of black substance
(b) evolution of brown gas
(c) evolution of colourless gas
(d) formation of a brown substance which on dilution becomes blue.
(e) disappearance of yellow powder along with evolution of colourless gas.
Answer:

Question 10.
Give reasons for each of the following:

1. Bleaching of flowers by Cl₂ is permanent while by SO₂ is temporary.
2. Molten aluminium bromide is a poor conductor of electricity.
3. Nitric oxide becomes brown when released in air.
4. PCl₅ is ionic in nature in the solid state.
5. Ammonia is a good complexing agent.

Answer:
1. Cl₂ bleaches by oxidation, while SO₂ does it by reduction. The reduced product gets oxidise again and the colour is regained back.

2. Aluminium bromide exists as a dimer, Al₂Br₆. In this structure, each aluminium atom forms one coordinate bond by accepting a lone pair of electrons from the bromine atom of another aluminium bromide molecule and thus complete the octet of electrons. Due to lack of free electrons, molten aluminium bromide is a poor conductor of electricity.

3. Nitric oxide reacts with air and oxidised into NO₂ which is brown in colour.
2NO + O₂ → 2NO₂

4. In solid state PCl₅ exists as [PCl₄]⁺ [PCl₆]^– and hence it is ionic in nature. Due to its ionic nature, it conducts current on fusion.

5. N atom in ammonia has lone pair of electrons which can coordinate with other atoms or cations required for the stability of electron pair.

Question 11.
How will you prepare the following compounds.
(a) Hyponitrous acid
(b) Nitrous acid
(c) Pernitrous acid
(d) Pernitric acid
Answer:

Common Errors And Its Rectificaations:

Common Errors:

1. Oxidation number rules may be confused.
2. Oxidation number of oxygen may get confused.
3. Molecular formula and compound name may get confused.

Rectifications:

1. Oxidation number of fluorine is always -1.
2. Always oxygen is -2 but in peroxide it is -1. When it comes as a first element its oxidation number is positive..
3. Student should memorise the formula using short cut method, e.g. Phosphorous acid – H₃PO₃ Phosphoric acid – H₃PO₄ e.g. Phosphorous acid – H₃PO₃ By pronunciation, ‘O’ indicates oxygen is more.

Hope you love the Samacheer Kalvi 12th Chemistry Chapter Wise Material. Clearly understand the deep concept of Chemistry learning with the help of Tamilnadu State Board 12th Chemistry Chapter 3 p-Block Elements – II Questions and AnswersPDF. Refer your friends to and bookmark our website for instant updates. Also, keep in touch with us using the comment section.

Students those who are preparing for the 12th Physics exam can download this Samacheer Kalvi 12th Physics Book Solutions Questions and Answers for Chapter 4 Electromagnetic Induction and Alternating Current from here free of cost. These Tamilnadu State Board Solutions for Class 12th Physics PDF cover all Chapter 4 Electromagnetic Induction and Alternating Current Questions and Answers. All these concepts of Samacheer Kalvi 12th Physics Book Solutions Questions and Answers are explained very conceptually as per the board prescribed Syllabus & guidelines.

Tamilnadu Samacheer Kalvi 12th Physics Solutions Chapter 4 Electromagnetic Induction and Alternating Current

Kickstart your preparation by using this Tamilnadu State Board Solutions for 12th Physics Chapter 4 Electromagnetic Induction and Alternating Current Questions and Answers and get the max score in the exams. You can cover all the topics of Chapter 4 Electromagnetic Induction and Alternating Current Questions and Answers pdf. Download the Tamilnadu State Board Solutions by accessing the links provided here and ace up your preparation.

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Textual Evaluation Solved

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Multiple Choice Questions

12th Physics Chapter 4 Book Back Answers Question 1.
An electron moves on a straight, line path XY as shown in the figure. The coil abed is adjacent to the path of the electron. What will be the direction of current, if any, induced in the coil? (NEET – 2015)

(а) The current will reverse its direction as the electron goes past the coil
(b) No current will be induced Electron
(c) abcd
(d) adcb
Answer:
(a) The current will reverse its direction as the electron goes past the coil

12th Physics Lesson 4 Book Back Answers Question 2.
A thin semi-circular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure. The potential difference developed across the ring when its speed v, is- (NEET 2014)

(a) Zero
(b) \(\frac {{ Bvπr }^{2}}{ 2 }\)
(c) πrBv and R is at higher potential
(d) 2rBv and R is at higher potential
Answer:
(d) 2rBv and R is at higher potential

Samacheerkalvi.Guru 12th Physics Question 3.
The flux linked with a coil at any instant t is given by Φ_B = 10t² – 50t + 250. The induced emf at t = 3s is-
(a) -190 V
(b) -10 V
(c) 10 V
(d) 190 V
Answer:
(b) -10 V

Samacheer Kalvi 12th Physics Solutions Question 4.
When the current changes from +2A to -2A in 0.05 s, an emf of 8 V is induced in a coil. The co-efficient of self-induction of the coil is-
(a) 0.2 H
(b) 0.4 H
(c) 0.8 H
(d) 0.1 H
Answer:
(d) 0.1 H

Samacheer Kalvi 12th Physics Solution Book Question 5.
The current i flowing in a coil varies with time as shown in the figure. The variation of induced emf with time would be- (NEET-2011)


Answer:

Tn 12th Physics Solution Question 6.
A circular coil with a cross-sectional area of 4 cm² has 10 turns. It is placed at the centre of a long solenoid that has 15 turns/cm and a cross-sectional area of 10 cm². The axis of the coil coincides with the axis of the solenoid. What is their mutual inductance?
(a) 7.54 μH
(b) 8.54 μH
(c) 9.54 μH
(d) 10.54 μH
Answer:
(a) 7.54 μH

Samacheer Kalvi Guru 12th Physics Question 7.
In a transformer, the number of turns in the primary and the secondary are 410 and 1230, respectively. If the current in primary is 6A, then that in the secondary coil is-
(a) 2 A
(b) 18 A
(c) 12 A
(d) 1 A
Answer:
(a) 2 A

Physics Class 12 Chapter 4 Notes Question 8.
A step-down transformer reduces the supply voltage from 220 V to 11 V and increase the current from 6 A to 100 A. Then its efficiency is-
(a) 1.2
(b) 0.83
(c) 0.12
(d) 0.9
Answer:
(b) 0.83

Alternating Current Class 12 Notes Pdf Download Question 9.
In an electrical circuit, R, L, C and AC voltage source are all connected in series. When L is removed from the circuit, the phase difference between the voltage and current in the circuit is \(\frac { 1 }{ 2 }\). Instead, if C is removed from the circuit, the phase difference is again \(\frac { π }{ 3 }\). The power factor of the circuit is- (NEET 2012)
(a) \(\frac { 1 }{ 2 }\)
(b) \(\frac { 1 }{ √ 2 }\)
(c) 1
(d) \(\frac { √ 3 }{ 2 }\)
Answer:
(c) 1

Physics Solution Class 12 Samacheer Kalvi Question 10.
In a series RL circuit, the resistance and inductive reactance are the same. Then the phase difference between the voltage and current in the circuit is-
(a) \(\frac { π }{ 4 }\)
(b) \(\frac { π }{ 2 }\)
(c) \(\frac { π }{ 6 }\)
(d) zero
Answer:
(a) \(\frac { π }{ 4 }\)

Samacheer 12th Physics Solutions Question 11.
In a series resonant RLC circuit, the voltage across 100 Ω resistor is 40 V. The resonant frequency co is 250 rad/s. If the value of C is 4 μF, then the voltage across L is-
(a) 600 V
(b) 4000 V
(c) 400 V
(d) 1 V
Answer:
(c) 400 V

Question 12.
An inductor 20 mH, a capacitor 50 μF and a resistor 40 Ω are connected in series across a source of emf v = 10 sin 340 t. The power loss in AC circuit is-
(a) 0.76 W
(b) 0.89 W
(c) 0.46 W
(d) 0.67 W
Answer:
(c) 0.46 W

Questions 13.
The instantaneous values of alternating current and voltage in a circuit are i = \(\frac { 1 }{ √ 2 }\) = sin(100πt) A and v = \(\frac { 1 }{ √ 2 }\) sin \(\left(100 \pi t+\frac{\pi}{3}\right)\) V. The average power in watts consumed in the circuit is-
(IIT Main 2012)
(a) \(\frac { 1 }{ 4 }\)
(b) \(\frac { √3 }{ 4 }\)
(c) \(\frac { 1 }{ 2 }\)
(d) \(\frac { 1 }{ 8 }\)
Answer:
(d) \(\frac { 1 }{ 8 }\)

Question 14.
In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic fields is-
(a) \(\frac { Q }{ 2 }\)
(b) \(\frac { Q }{ √3 }\)
(c) \(\frac { Q }{ √2 }\)
(d) \(\frac { Q }{ 2 }\)
Answer:
(c) \(\frac { Q }{ √2 }\)

Question 15.
\(\frac { 20 }{{ π }^{2}}\) H inductor is connected to a capacitor of capacitance C. The value of C in order to impart maximum power at 50 Hz is-
(a) 50 μF
(b) 0.5 μF
(c) 500 μF
(d) 5 μF
Answer:
(d) 5 μF

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Short Answer Questions

Question 1.
What is meant by electromagnetic induction?
Answer:
Whenever the magnetic flux linked with a closed coil changes, an emf (electromotive force) is induced and hence an electric current flows in the circuit.

Question 2.
State Faraday’s laws of electromagnetic induction.
Answer:
First law:
Whenever magnetic flux linked with a closed circuit changes, an emf is induced in the circuit.

Second law:
The magnitude of induced emf in a closed circuit is equal to the time rate of change of magnetic flux linked with the circuit.

Question 3.
State Lenz’s law.
Answer:
State Lenz’s law:
Lenz’s law states that the direction of the induced current is such that it always opposes the cause responsible for its production.

Question 4.
State Fleming’s right hand rule.
Answer:
The thumb, index finger and middle finger of right hand are stretched out in mutually perpendicular directions. If the index finger points the direction of the magnetic field and the Electromagnetic Induction and Alternating Current thumb indicates the direction of motion of the conductor, then the middle finger will indicate the direction of the induced current.

Question 5.
How is Eddy current produced? How do they flow in a conductor?
Answer:
Even for a conductor in the form of a sheet or plate, an emf is induced when magnetic flux linked with it changes. But the difference is that there is no definite loop or path for induced current to flow away. As a result, the induced currents flow in concentric circular paths. As these electric currents resemble eddies of water, these are known as Eddy currents. They are also called Foucault currents.

Question 6.
Mention the ways of producing induced emf.
Answer:
Induced emf can be produced by changing magnetic flux in any of the following ways:

1. By changing the magnetic field B
2. By changing the area A of the coil and
3. By changing the relative orientation 0 of the coil with magnetic field

Question 7.
What for an inductor is used? Give some examples.
Answer:
Inductor is a device used to store energy in a magnetic field when an electric current flows through it. The typical examples are coils, solenoids and toroids.

Question 8.
What do you mean by self-induction?
Answer:
If the magnetic flux is changed by changing the current, an emf is induced in that same coil. This phenomenon is known as self-induction.

Question 9.
What is meant by mutual induction?
Answer:
When an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil. This phenomenon is known as mutual induction.

Question 10.
Give the principle of AC generator.
Answer:
Alternators work on the principle of electromagnetic induction. The relative motion between a conductor and a magnetic field changes the magnetic flux linked with the conductor which in turn, induces an emf. The magnitude of the induced emf is given by Faraday’s law of electromagnetic induction and its direction by Fleming’s right hand rule.

Question 11.
List out the advantages of stationary armature-rotating field system of AC generator.
Answer:

1. The current is drawn directly from fixed terminals on the stator without the use of brush contacts.
2. The insulation of stationary armature winding is easier.
3. The number of sliding contacts (slip rings) is reduced. Moreover, the sliding contacts are used for low-voltage DC Source.
4. Armature windings can be constructed more rigidly to prevent deformation due to any mechanical stress.

Question 12.
What are step-up and step-down transformers?
Answer:
If the transformer converts an alternating current with low voltage into an alternating current with high voltage, it is called step-up transformer. On the contrary, if the transformer converts alternating current with high voltage into an alternating current with low voltage, then it is called step-down transformer.

Question 13.
Define average value of an alternating current.
Answer:
The average value of alternating current is defined as the average of all values of current over a positive half-cycle or negative half-cycle.

Question 14.
How will you define RMS value of an alternating current?
Answer:
RMS value of alternating current is defined as that value of the steady current which when flowing through a given circuit for a given time produces the same amount of heat as produced by the alternating current when flowing through the same circuit for the same time.

Question 15.
What are phasors?
Answer:
A sinusoidal alternating voltage (or current) can be represented by a vector which rotates about the origin in anti-clockwise direction at a constant angular velocity ω. Such a rotating vector is called a phasor.

Question 16.
Define electric resonance.
Answer:
When the frequency of the applied alternating source is equal to the natural frequency of the RLC circuit, the current in the circuit reaches its maximum value. Then the circuit is said to be in electrical resonance.

Question 17.
What do you mean by resonant frequency?
Answer:
When the frequency of the applied alternating source (ω_r) is equal to the natural frequency \(\left[\frac{1}{\sqrt{L C}}\right]\) of the RLC circuit, the current in the circuit reaches its maximum value. Then the circuit is said to be in electrical resonance. The frequency at which resonance takes place is called resonant frequency. Resonant angular frequency, ω_r = \(\frac { 1 }{ \sqrt { LC } } \)

Question 18.
How will you define Q-factor?
Answer:
It is defined as the ratio of voltage across L or C to the applied voltage.

Question 19.
What is meant by wattles current?
Answer:
The component of current (I_RMS sin φ), which has a phase angle of \(\frac { π }{ 2 }\) with the voltage is called reactive component. The power consumed is zero. So that it is also known as ‘Wattless’ current.

Question 20.
Give any one definition of power factor.
Answer:
The power factor is defined as the ratio of true power to the apparent power of an a.c. circuit. It is equal to the cosine of the phase angle between current and voltage in the a.c. circuit.

Question 21.
What are LC oscillations?
Answer:
Whenever energy is given to a LC circuit, the electrical oscillations of definite frequency are generated. These oscillations are called LC oscillations. During LC oscillations, the total energy remains constant. It means that LC oscillations take place in accordance with the law of conservation of energy.

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Long Answer Questions

Question 1.
Establish the fact that the relative motion between the coil and the magnet induces an emf in the coil of a closed circuit.
Answer:
Whenever the magnetic flux linked with a closed coil changes, an emf (electromotive force) is induced and hence an electric current flows in the circuit.
The relative motion between the coil and the magnet induces:
In the first experiment, when a bar magnet is placed close to a coil, some of the magnetic field lines of the bar magnet pass through the coil i.e., the magnetic flux is linked with the coil. When the bar magnet and the coil approach each other, the magnetic flux linked with the coil increases. So this increase in magnetic flux induces an emf and hence a transient electric current flows in the circuit in one direction (Figure(a)).

At the same time, when they recede away from one another, the magnetic flux linked with the coil decreases. The decrease in magnetic flux again induces an emf in opposite direction and hence an electric current flows in opposite direction (Figure (b)). So there is deflection in the galvanometer when there is a relative motion between the coil and the magnet.

In the second experiment, when the primary coil P carries an electric current, a magnetic field is established around it. The magnetic lines of this field pass through itself and the neighbouring secondary coil S.

When the primary circuit is open, no electric current flows in it and hence the magnetic flux linked with the secondary coil is zero (Figure(a)).
However, when the primary circuit is closed, the increasing current builds up a magnetic field around the primary coil. Therefore, the magnetic flux linked with the secondary coil increases. This increasing flux linked induces a transient electric current in the secondary coil (Figure(b)).

When the electric current in the primary coil reaches a steady value, the magnetic flux linked with the secondary coil does not change and the electric current in the secondary coil will disappear. Similarly, when the primary circuit is broken, the decreasing primary current induces an electric current in the secondary coil, but in the opposite direction (Figure (c)). So there is deflection in the galvanometer whenever there is a change in the primary current

Question 2.
Give an illustration of determining direction of induced current by using Lenz’s law.
Answer:
Illustration 1:
Consider a uniform magnetic field, with its field lines perpendicular to the plane of the paper and pointing inwards. These field lines are represented by crosses (x) as shown in figure (a). A rectangular metallic frame ABCD is placed in this magnetic field, with its plane perpendicular to the field. The arm AB is movable so that it can slide towards right or left.

If the arm AB slides to our right side, the number of field lines (magnetic flux) passing through the frame ABCD increases and a current is induced. As suggested by Lenz’s law, the induced current opposes this flux increase and it tries to reduce it by producing another magnetic field pointing outwards i.e., opposite to the existing magnetic field.

The magnetic lines of this induced field are represented by circles in the figure (b). From the direction of the magnetic field thus produced, the direction of the induced current is found to be anti-clockwise by using right-hand thumb rule.

The leftward motion of arm AB decreases magnetic flux. The induced current, this time, produces a magnetic field in the inward direction i.e., in the direction of the existing magnetic field (figure (c)). Therefore, the flux decrease is opposed by the flow of induced current. From this, it is found that induced current flows in clockwise direction.

Illustration 2:
Let us move a bar magnet towards the solenoid, with its north pole pointing the solenoid as shown in figure (b). This motion increases the magnetic flux of the coil which in turn, induces an electric current. Due to the flow of induced current, the coil becomes a magnetic dipole whose two magnetic poles are on either end of the coil.

In this case, the cause producing the induced current is the movement of the magnet. According to Lenz’s law, the induced current should flow in such a way that it opposes the movement of the north pole towards coil. It is possible if the end nearer to the magnet becomes north pole (figure (b)).

Then it repels the north pole of the bar magnet and opposes the movement of the magnet. Once pole ends are known, the direction of the induced current could be found by using right hand thumb rule. When the bar magnet is withdrawn, the nearer end becomes south pole which attracts north pole of the bar magnet, opposing the receding motion of the magnet (figure (c)). Thus the direction of the induced current can be found from Lenz’s law.

Question 3.
Show that Lenz’s law is in accordance with the law of conservation of energy.
Answer:
Conservation of energy:
The truth of Lenz’s law can be established on the basis of the law of conservation of energy. According to Lenz’s law, when a magnet is moved either towards or away from a coil, the induced current produced opposes its motion. As a result, there will always be a resisting force on the moving magnet.

Work has to be done by some external agency to move the magnet against this resisting force. Here the mechanical energy of the moving magnet is converted into the electrical energy which in turn, gets converted into Joule heat in the coil i.e., energy is converted from one form to another.

Question 4.
Obtain an expression for motional emf from Lorentz force.
Answer:
Motional emf from Lorentz force:
Consider a straight conducting rod AB of length l in a uniform magnetic field \(\vec { B } \) which is directed perpendicularly into the plane of the paper. The length of the rod is normal to the magnetic field. Let the rod move with a constant velocity \(\vec { v } \) towards right side.
When the rod moves, the free electrons present in it also move with same velocity \(\vec { v } \) in \(\vec { B } \). As a result, the Lorentz force acts on free electrons in the direction from B to A and is given by the relation
\(\vec { F } \)_B = -e(\(\vec { v } \) x \(\vec { B } \) ) ……. (1)
The action of this Lorentz force is to accumulate the free electrons at the end A. This accumulation of free electrons produces a potential difference across the rod which in turn establishes an electric field E directed along BA. Due to the electric field E, the coulomb force starts acting on the free electrons along AB and is given by
\(\vec { F } \)_E = -e\(\vec { E } \) ……. (2)
The magnitude of the electric field \(\vec { E } \) keeps on increasing as long as accumulation of electrons at the end A continues. The force \(\vec { F } \)_E also increases until equilibrium is reached. At equilibrium, the magnetic Lorentz force \(\vec { F } \)_B and the coulomb force \(\vec { F } \)_E balance each other and no further accumulation of free electrons at the end A takes place, i.e.,
\(\left| \vec { { F }_{ B } } \right| \) = \(\left| \vec { { F }_{ E } } \right| \)

vB sin 90° = E
vB = E ……. (3)
The potential difference between two ends of the rod is

Figure: Motional emf from Lorentz force
V = El
V = vBl
Thus the Lorentz force on the free electrons is responsible to maintain this . potential difference and hence produces an emf
ε = Blv ….. (4)
As this emf is produced due to the movement of the rod, it is often called as motional emf.

Question 5.
Using Faraday’s law of electromagnetic induction, derive an equation for motional emf.
Answer:
Motional emf from Faraday’s law:
Let us consider a rectangular conducting loop of width l in a uniform magnetic field \(\vec { B } \) which is perpendicular to the plane of the loop and is directed inwards. A part of the loop is in the magnetic field while the remaining part is outside the field.

Figure: Motional emf from Faraday’s law

When the loop is pulled with a constant velocity \(\vec { v } \) to the right, the area of the portion of the loop within the magnetic field will decrease. Thus, the flux linked with the loop will also decrease. According to Faraday’s law, an electric current is induced in the loop which flow’s in a direction so as to oppose the pull of the loop.
Let x be the length of the loop which is still within the magnetic field, then its area is lx. The magnetic flux linked with the loop is

As this magnetic flux decreases due to the movement of the loop, the magnitude of the induced emf is given by
ε = \(\frac {{ dΦ }_{B}}{ dt }\) = \(\frac { d }{ dt }\) (Blx)
Here, both B and l are constants. Therefore,
ε = Bl \(\frac { dx}{ dl }\) = Blv …… (2)
where v = \(\frac { dx}{ dt }\) is the velocity of the loop. This emf is known as motional emf since it is produced due to the movement of the loop in the magnetic field.

Question 6.
Give the uses of Foucault current.
Answer:
Though the production of eddy current is undesirable in some cases, it is useful in some other cases. A few of them are

1. Induction stove
2. Eddy current brake
3. Eddy current testing
4. Electromagnetic damping

1. Induction stove:
Induction stove is used to cook the food quickly and safely with less energy consumption. Below the cooking zone, there is a tightly wound coil of insulated wire. The cooking pan made of suitable material, is placed over the cooking zone.

When the stove is switched on, an alternating current flowing in the coil produces high frequency alternating magnetic field which induces very strong eddy currents in the cooking pan. The eddy currents in the pan produce so much of heat due to Joule heating which is used to cook the food.

2. Eddy current brake:
This eddy current braking system is generally used in high speed trains and roller coasters. Strong electromagnets are fixed just above the rails. To stop the train, electromagnets are switched on. The magnetic field of these magnets induces eddy currents in the rails which oppose or resist the movement of the train. This is Eddy current linear brake.

In some cases, the circular disc, connected to the wheel of the train through a common shaft, is made to rotate in between the poles of an electromagnet. When there is a relative motion between the disc and the magnet, eddy currents are induced in the disc which stop the train. This is Eddy current circular brake.

3. Eddy current testing:
It is one of the simple non-destructive testing methods to find defects like surface cracks and air bubbles present in a specimen. A coil of insulated wire is given an alternating electric current so that it produces an alternating magnetic field.

When this coil is brought near the test surface, eddy current is induced in the test surface. The presence of defects causes the change in phase and amplitude of the eddy current that can be detected by some other means. In this way, the defects present in the specimen are identified.

4. Electro magnetic damping:
The armature of the galvanometer coil is wound on a soft iron cylinder. Once the armature is deflected, the relative motion between the soft iron cylinder and the radial magnetic field induces eddy current in the cylinder. The damping force due to the flow of eddy current brings the armature to rest immediately and then galvanometer shows a steady deflection. This is called electromagnetic damping.

Question 7.
Define self-inductance of a coil interns of (i) magnetic flux and (ii) induced emf.
Answer:
Self-inductance or simply inductance of a coil is defined as the flux linkage of the coil when 1A current flows through it.
When the current i changes with time, an emf is induced in it. From Faraday’s law of electromagnetic induction, this self-induced emf is given by
ε = –\(\frac{d\left(\mathrm{N} \Phi_{\mathrm{B}}\right)}{d t}\) = –\(\frac { d(Li)}{ dt }\)
∴ ε = -L\(\frac { di}{ dt }\) or L = \(\frac { -ε}{ di/dt }\)
The negative sign in the above equation means that the self-induced emf always opposes the change in current with respect to time. If \(\frac { di}{ dt }\) = 1 As^-1, then L= -ε. Inductance of a coil is also defined as the opposing emf induced in the coil when the rate of change of current through the coil is 1 A s^-1.

Question 8.
How will you define the unit of inductance?
Answer:
Unit of inductance: Inductance is a scalar and its unit is Wb A^-1 or V s A^-1. It is also measured in henry (H).
1 H = 1 Wb A^-1 = 1 V s A^-1
The dimensional formula of inductance is M L² T^-2A^-2.
If i = 1 A and NΦ_B = 1 Wb turns, then L = 1 H.
Therefore, the inductance of the coil is said to be one henry if a current of 1 A produces unit flux linkage in the coil.
If \(\frac { di}{ dt }\) = 1 As^-1 and ε = -1 V, then L = 1 H.
Therefore, the inductance of the coil is one henry if a current changing at the rate of 1 A s^-1 induces an opposing emf of 1 V in it.

Question 9.
What do you understand by self-inductance of a coil? Give its physical significance.
Answer:
Self-inductance or simply inductance of a coil is defined as the flux linkage of the coil when 1A current flows through it.
When the current i changes with time, an emf is induced in it. From Faraday’s law of electromagnetic induction, this self-induced emf is given by

Physical significance of inductance:
When a circuit is switched on, the increasing current induces an emf which opposes the growth of current in a circuit. Likewise, when circuit is broken, the decreasing current induces an emf in the reverse direction. This emf now opposes the decay of current.

Figure: Induced emf ε opposes the changing current i

Thus, inductance of the coil opposes any change in current and tries to maintain the original state.

Question 10.
Assuming that the length of the solenoid is large when compared to its diameter, find the equation for its inductance.
Answer:
Self-inductance of a long solenoid:
Consider a long solenoid of length l and cross-sectional area A. Let n be the number of turns per unit length (or turn density) of the solenoid. When an electric current i is passed through the solenoid, a magnetic field is produced by it which is almost uniform and is directed along the axis of the solenoid. The magnetic field at any point inside the solenoid is given by
B = μ₀ni
As this magnetic field passes through the solenoid, the windings of the solenoid are linked by the field lines. The magnetic flux passing through each turn is

The total magnetic flux linked or flux linkage of the solenoid with N turns (the total number of turns N is given by N = nl) is

NΦ_B = n (nl) (μ₀ni)A
NΦ_B = (μ₀n²Al)i ….. (1)
From the self induction
NΦ_B = LI ….. (2)
Comparing equations (1) and (2), we have L = μ₀n²Al
From the above equation, it is clear that inductance depends on the geometry of the solenoid (turn density n, cross-sectional area A, length l) and the medium present inside the solenoid. If the solenoid is filled with a dielectric medium of relative permeability μ_r, then
L = μ₀
L = μn₀μ_rn²Al

Question 11.
An inductor of inductance L carries an electric current i. How much energy is stored while establishing the current in it?
Answer:
Energy stored in an inductor:
Whenever a current is established in the circuit, the inductance opposes the growth of the current. In order to establish a current in the circuit, work is done against this opposition by some external agency. This work done is stored as magnetic potential energy.

Let us assume that electrical resistance of the inductor is negligible and inductor effect alone is considered. The induced emf e at any instant t is
ε = -L\(\frac { di}{ dt }\) …… (1)
Let dW be work done in moving a charge dq in a time dt against the opposition, then
dW = -εdq = -εidi = εidi           [∵dq = idt]
Substituting for s value from equation (1)
= – \(\left(-\mathrm{L} \frac{d i}{d t}\right)\) idt
dW = Lid …… (2)
Total work done in establishing the current i is

This work done is stored as magnetic potential energy.
U_B = \(\frac { 1 }{ 2 }\) Li² …….. (4)

Question 12.
Show that the mutual inductance between a pair of coils is same (M₁₂ = M₂₁).
Answer:
Mutual induction:
When an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil. This phenomenon is known as mutual induction and the emf is called mutually induced emf.

Consider two coils which are placed close to each other. If an electric current i₁ is sent through coil 1, the magnetic field produced by it is also linked with coil 2. Let Φ₂₁ be the magnetic flux linked with each turn of the coil 2 of N₂ turns due to coil 1, then the total flux linked with coil 2 (N₂Φ₂₁) is proportional to the current i₁ in the coil 1.

The constant of proportionality M₂₁ is the mutual inductance of the coil 2 with respect to coil 1. It is also called as coefficient of mutual induction. If i₁ = 1A, then M₂₁ = N₂Φ₂₁.
Therefore, the mutual inductance M₂₁ is defined as the flux linkage of the coil 2 when 1A current flows through coil 1. When the current changes with time, an emf ε₂ is induced in coil 2. From Faraday’s law of electromagnetic induction, this mutually induced emf ε₂ is given by

The negative sign in the above equation shows that the mutually induced emf always opposes the change in current i, with respect to time. If \(\frac { di }{ dt }\) = 1 As^-1, then M₂₁ = -ε₂. Mutual inductance M₂₁, is also defined as the opposing emf induced in the coil 2 when the rate of change of current through the coil 1 is 1 As^-1. Similarly, if an electric current i₂ through coil 2 changes with time, then emf ε₁ is induced in coil 1. Therefore,

where M₁₂ is the mutual inductance of the coil 1 with respect to coil 2. It can be shown that for a given pair of coils, the mutual inductance is same, i.e., M₂₁ = M₁₂ = M.
In general, the mutual induction between two coils depends on size, shape, the number of turns of the coils, their relative orientation and permeability of the medium.

Question 13.
How will you induce an emf by changing the area enclosed by the coil?
Answer:
Induction of emf by changing the area of the coil:
Consider a conducting rod of length 1 moving with a velocity v towards left on a rectangular metallic framework. The whole arrangement is placed in a uniform magnetic field \(\vec { B } \) whose magnetic lines are perpendicularly directed into the plane of the paper. As the rod moves from AB to DC in a time dt, the area enclosed by the loop and hence the magnetic flux through the loop decreases.

The change in magnetic flux in time dt is
dΦ_B = B x change in area
B x Area ABCD
= Blvdt since Area ABCD = l(vdt)
or \(\frac {{ dΦ }_{B}}{ dt }\) = Blv
As a result of change in flux, an emf is generated in the loop. The magnitude of the induced emf is
ε = \(\frac {{ dΦ }_{B}}{ dt }\) = Blv
This emf is called motional emf. The direction of induced current is found to be clockwise from Fleming’s right hand rule.

Question 14.
Show mathematically that the rotation of a coil in a magnetic field over one rotation induces an alternating emf of one cycle.
Answer:
Induction of emf by changing relative orientation of the coil with the magnetic field:
Consider a rectangular coil of N turns kept in a uniform magnetic field \(\vec { B } \) figure (a). The coil rotates in anti-clockwise direction with an angular velocity ω about an axis, perpendicular to the field. At time = 0, the plane of the coil is perpendicular to the field and the flux linked with the coil has its maximum value Φ_m = BA (where A is the area of the coil).

In a time t seconds, the coil is rotated through an angle θ (= ωt) in anti-clockwise direction. In this position, the flux linked is Φ_m cos ωt, a component of Φ_m normal to the plane of the coil (figure (b)). The component parallel to the plane (Φ_m sin ωt) has no role in electromagnetic induction. Therefore, the flux linkage at this deflected position is NΦ_B = NΦ_m cos ωt. According to Faraday’s law, the emf induced at that instant is

ε= \(\frac { d }{ dt }\) (NΦ_B ) = \(\frac { d }{ dt }\) (NΦ_m cos ωt)
= -NΦ_m (-sin ωt)ω = NΦ_m ω sin ωt
When the coil is rotated through 90° from initial position, sin ωt = 1, Then the maximum value of induced emf is
ε_m = NΦ_mω = NBAω since Φ_m = BA
Therefore, the value of induced emf at that instant is then given by
ε = ε_m sin ωt
It is seen that the induced emf varies as sine function of the time angle ωt. The graph between – induced emf and time angle for one rotation of coil will be a sine curve and the emf varying in this manner is called sinusoidal emf or alternating emf.

Question 15.
Elaborate the standard construction details of AC generator.
Answer:
Construction:
lternator consists of two major parts, namely stator and rotor. As their names suggest, stator is stationary while rotor rotates inside the stator. In any standard construction of commercial alternators, the armature winding is mounted on stator and the field magnet on rotor. The construction details of stator, rotor and various other components involved in them are given below.

(i) Stator:
The stationary part which has armature windings mounted in it is called stator. It has three components, namely stator frame, stator core and armature winding.

Stator frame:
This is the outer frame used for holding stator core and armature windings in proper position. Stator frame provides best ventilation with the help of holes provided in the frame itself.

Stator core:
Stator core or armature core is made up of iron or steel alloy. It is a hollow cylinder and is laminated to minimize eddy current loss. The slots are cut on inner surface of the core to accommodate armature windings.

Armature winding:
Armature winding is the coil, wound on slots provided in the armature core. One or more than one coil may be employed, depending on the type of alternator. Two types of windings are commonly used. They are (i) single-layer winding and (ii) double-layer winding. In single-layer winding, a slot is occupied by a coil as a single layer. But in double-layer winding, the coils are split into two layers such as top and bottom layers.

(ii) Rotor:
Rotor contains magnetic field windings. The magnetic poles are magnetized by DC source. The ends of field windings are connected to a pair of slip rings, attached to a common shaft about which rotor rotates. Slip rings rotate along with rotor. To maintain connection between the DC source and field windings, two brushes are used which . continuously slide over the slip rings.

There are 2 types of rotors used in alternators:

1. salient pole rotor
2. cylindrical pole rotor.

1. Salient pole rotor:
The word salient means projecting. This rotor has a number of projecting poles having their bases riveted to the rotor. It is mainly used in low-speed alternators.

2. Cylindrical pole rotor:
This rotor consists of a smooth solid cylinder. The slots are cut on the outer surface of the cylinder along its length. It is suitable for very high speed alternators.

The frequency of alternating emf induced is directly proportional to the rotor speed. In order to maintain the frequency constant, the rotor must run at a constant speed. These are standard construction details of alternators.

Question 16.
Explain the working of a single-phase AC generator with necessary diagram.
Answer:
Single phase AC generator: In a single phase AC generator, the armature conductors are connected in series so as to form a single circuit which generates a single-phase alternating emf and hence it is called single-phase alternator.

The simplified version of a AC generator is discussed here. Consider a stator core consisting of 2 slots in which 2 armature conductors PQ and RS are mounted to form single-turn rectangular loop PQRS. Rotor has 2 salient poles with field windings which can be magnetized by means of DC source.

Working:
The loop PQRS is stationary and is perpendicular to the plane of the paper. When field windings are excited, magnetic field is produced around it. The direction of magnetic field passing through the armature core. Let the field magnet be rotated in clockwise direction by the prime mover. The axis of rotation is perpendicular to the plane of the paper.

Assume that initial position of the field magnet is horizontal. At that instant, the direction of magnetic field is perpendicular to the plane of the loop PQRS. The induced emf is zero. This is represented by origin O in the graph between induced emf and time angle.

When field magnet rotates through 90°, magnetic field becomes parallel to PQRS. The induced cmfs across PQ and RS would become maximum. Since they are connected in series, emfs are added up and the direction of total induced emf is given by Fleming’s right hand rule. Care has to be taken while applying this rule; the thumb indicates the direction of the motion of the conductor with respect to field.

For clockwise rotating poles, the conductor appears to be rotating anti-clockwise. Hence, thumb should point to the left. The direction of the induced emf is at right angles to the plane of the paper. For PQ, it is downwards B and for RS upwards. Therefore, the current flows along PQRS. The point A in the graph represents this maximum emf.

For the rotation of 180° from the initial position, the field is again perpendicular to PQRS and the induced emf becomes zero. This is represented by point B. The field magnet becomes again parallel to PQRS for 270° rotation of field magnet. The induced emf is maximum but the direction is reversed. Thus the current flows along SRQP This is represented by point C.

On completion of 360°, the induced emf becomes zero and is represented by the point D. From the graph, it is clear that emf induced in PQRS is alternating in nature. Therefore, when field magnet completes one rotation, induced emf in PQRS finishes one cycle. For this construction, the frequency of the induced emf depends on the speed at which the field magnet rotates.

Question 17.
How are the three different emfs generated in a three-phase AC generator? Show the graphical representation of these three emfs.
Answer:
Three-phase AC generator:
Some AC generators may have more than one coil in the armature core and each coil produces an alternating emf. In these generators, more than one emf is produced. Thus they are called poly-phase generators. If there are two alternating emfs produced in a generator, it is called two-phase generator. In some AC generators, there are three separate coils, which would give three separate emfs. Hence they are called three-phase AC generators.

In the simplified construction of three-phase AC generator, the armature core has 6 slots, cut on its inner rim. Each slot is 60° away from one another. Six armature conductors are mounted in these slots. The conductors 1 and 4 are joined in series to form coil 1. The conductors 3 and 6 form coil 2 while the conductors 5 and 2 form coil 3. So, these coils are rectangular in shape and are 120° apart from one another.

The initial position of the field magnet is horizontal and field direction is perpendicular to the plane of the coil 1. As it is seen in single phase AC generator, when field magnet is rotated from that position in clockwise direction, alternating emf ε₁ in coil 1 begins a cycle from origin O.

The corresponding cycle for alternating emf ε₂ in coil 2 starts at point A after field magnet has rotated through 120°. Therefore, the phase difference between ε₁ and ε₂ is 120°. Similarly, emf ε₃ in coil 3 would begin its cycle at point B after 240° rotation of field magnet from initial position. Thus these emfs produced in the three phase AC generator have 120° phase difference between one another.

Question 18.
Explain the construction and working of transformer.
Answer:
Construction and working of transformer:
Principle:
The principle of transformer is the mutual induction between two coils. That is, when an electric current passing through a coil changes with time, an emf is induced in the neighbouring coil.

Construction:
In the simple construction of transformers, there are two coils of high mutual inductance wound over the same transformer core. The core is generally laminated and is made up of a good magnetic material like silicon steel. Coils are electrically insulated but magnetically linked via transformer core.

The coil across which alternating voltage is applied is called primary coil P and the coil from which output power is drawn out is called secondary coil S. The assembled core and coils are kept in a container which is filled with suitable medium for better insulation and cooling purpose.

Working:
If the primary’ coil is connected to a source of alternating voltage, an alternating magnetic flux is set up in the laminated core. If there is no magnetic flux leakage, then whole of magnetic flux linked with primary coil is also linked with secondary coif This means that rate at which magnetic flux changes through each turn is same for both primary and secondary coils. As a result of flux change, emf is induced in both primary and secondary coils. The emf induced in the primary coil ε_p is almost equal and opposite to the applied voltage υ_p and is given by
υ_p = ε_p = -N_p \(\frac {{ dΦ }_{B}}{ dt }\) …….. (1)
The frequency of alternating magnetic flux in the core is same as the frequency of the applied voltage. Therefore, induced emf in secondary will also have same frequency as that of applied voltage. The emf induced in the secondary coil eg is given by
ε_s = -N_s \(\frac {{ dΦ }_{B}}{ dt }\)
where N_p and N_s are the number of turns in the primary and secondary coil, respectively. If the secondary circuit is open, then ε_s = υ_s where υ_s is the voltage across secondary coil.
υ_s ε_s = -N_s \(\frac {{ dΦ }_{B}}{ dt }\) ……… (2)
From equation (1) and (2),
\(\frac {{ υ }_{s}}{{ ε }{s}}\) = \(\frac {{ N }_{s}}{{ N }{p}}\) = K …….. (3)
This constant K is known as voltage transformation ratio. For an ideal transformer,
Input power υ_p i_p = Output power υ_si_s
where iυ_p and i_s are the currents in the primary and secondary coil respectively. Therefore,
\(\frac {{ υ }_{s}}{{ υ }{p}}\) = \(\frac {{ N }_{s}}{{ N }{p}}\) = \(\frac {{ i }_{p}}{{ i }{s}}\)
Equation (4) is written in terms of amplitude of corresponding quantities,
\(\frac {{ V }_{s}}{{ V }{p}}\) = \(\frac {{ N }_{s}}{{ N }{p}}\) = \(\frac {{ I }_{p}}{{ I }{s}}\) = K ……. (4)

(i) If N_s > N_p ( or K > 1)
∴ V_s > V_p and I_s < I_p.
This is sthe case of step-up transformer in which voltage is decreased and the corresponding current is decreased.

(ii) If N_s < N_p (or K < 1)
∴ V_s < V_p and I_s > I_p
This is step-down transformer where voltage is decreased and the current is increased.

Question 19.
Mention the various energy losses in a transformer.
Answer:
Energy losses in a transformer: Transformers do not have any moving parts so that its efficiency is much higher than that of rotating machines like generators and motors. But there are many factors which lead to energy loss in a transformer.

(i) Core loss or Iron loss:
This loss takes place in transformer core. Hysteresis loss and eddy current loss are known as core loss or Iron loss. When transformer core is magnetized and demagnetized repeatedly by the alternating voltage applied across primary coil, hysteresis takes place due to which some energy is lost in the form of heat.

Hysteresis loss is minimized by using steel of high silicon content in making transformer core. Alternating magnetic flux in the core induces eddy currents in it. Therefore there is energy loss due to the flow of eddy current, called eddy current loss which is minimized by using very thin laminations of transformer core.

(ii) Copper loss:
Transformer windings have electrical resistance. When an electric current flows through them, some amount of energy is dissipated due to Joule heating. This energy loss is called copper loss which is minimized by using wires of larger diameter.

(iii) Flux leakage:
Flux leakage happens when the magnetic lines of primary coil arc not completely linked with secondary coil. Energy loss due to this flux leakage is minimized by winding coils one over the other.

Question 20.
Give the advantage of AC in long distance power transmission with an example.
Answer:
Advantages of AC in long distance power transmission:
Electric power is produced in a t large scale at electric power stations with the help of AC generators. These power stations are classified based on the type of fuel used as thermal, hydro electric and nuclear power stations. Most of these stations are located at remote places.

Hence the electric power generated is transmitted over long distances through transmission lines to reach towns or cities where it is actually consumed. This process is called power transmission. But there is a difficulty during power transmission. A sizable fraction of electric power is lost due to Joule heating (i²R) in the transmission lines which are hundreds of kilometer long.

This power loss can be tackled either by reducing current i or by reducing resistance R of the transmission lines. The resistance R can be reduced with thick wires of copper or aluminium. But this increases the cost of production of transmission lines and other related expenses. So this way of reducing power loss is not economically viable.

Since power produced is alternating in nature, there is a way out. The most important property of alternating voltage is that it can be stepped up and stepped down by using transformers could be exploited in reducing current and thereby reducing power losses to a greater extent.

At the transmitting point, the voltage is increased and the corresponding current is decreased by using step-up transformer. Then it is transmitted through transmission lines. This reduced current at high voltage reaches the destination without any appreciable loss.

At the receiving point, the voltage is decreased and the current is increased to appropriate values by using step-down transformer and then it is given to consumers. Thus power transmission is done efficiently and economically.

Illustration:
An electric power of 2 MW is transmitted to a place through transmission lines of total resistance, say R = 40 Ω, at two different voltages. One is lower voltage (10 kV) and the other is higher (100 kV). Let us now calculate and compare power losses in these two cases.

Case I:
P = 2 MW; R == 40 Ω; V = 10 kV Power,
Power, P = VI
∴ Current, I = \(\frac { P }{ V }\) = \(\frac {{ 2 × 10 }^{6}}{{ 10 × 10 }^{3}}\) 200 A
Power loss = Heat produced = I²R = (200)² × 40 = 1.6 × 10⁶ W
% of power loss =\(\frac {{ 1.6 × 10 }^{6}}{{ 2 × 10 }^{6}}\) × 100% = 0.8 × 100% = 80%

Case II:
P = 2 MW; R == 40 Ω; V = 100 kV
∴ Current, I = \(\frac { P }{ V }\) = \(\frac {{ 2 × 10 }^{6}}{{ 100 × 10 }^{3}}\) 20 A
Power loss = Heat produced = I²R = (20)² × 40 = 0.016 × 10⁶ W
% of power loss =\(\frac {{ 0.6 × 10 }^{6}}{{ 2 × 10 }^{6}}\) × 100% = 0.008 × 100% = 0.8%

Question 21.
Find out the phase relationship between voltage and current in a pure inductive circuit.
Answer:
AC circuit containing only an inductor:
Consider a circuit containing a pure inductor of inductance L connected across an alternating voltage source. The alternating voltage is given by the equation.
υ = V_m sin ωt …(1)
The alternating current flowing through the inductor induces a self-induced emf or back emf in the circuit. The back emf is given by
Back emf, ε -L\(\frac { di }{ dt }\)
By applying Kirchoff’s loop rule to the purely inductive circuit, we get

υ + ε = 0
V_m sin ωt = L\(\frac { di }{ dt }\)
di = L\(\frac {{ V }_{m}}{ L }\) sin ωt dt
i = \(\frac {{ V }_{m}}{ L }\) \(\int { sin } \) ωt dt = \(\frac{{ V }_{m}}{ Lω }\) (-cos ωt) + constant
The integration constant in the above equation is independent of time. Since the voltage in the circuit has only time dependent part, we can set the time independent part in the current (integration constant) into zero.

where \(\frac{{ V }_{m}}{ Lω }\) = I_m, the peak value of the alternating current in the circuit. From equation (1) and (2), it is evident that current lags behind the applied voltage by \(\frac{π}{ 2 }\) in an inductive circuit.
This fact is depicted in the phasor diagram. In the wave diagram also, it is seen that current lags the voltage by 90°.

Inductive reactance X_L:
The peak value of current I_m is given by I_m = \(\frac{{ V }_{m}}{ Lω }\) . Let us compare this equation with I_m = \(\frac{{ V }_{m}}{ R }\) from resistive circuit. The equantity ω_L Plays the same role as the resistance in resistive circuit. This is the resistance offered by the inductor, called inductive reactance (X_L). It is measured in ohm.
X_L = ω_L
The inductive reactance (X_L) varies directly as the frequency.
X_L = 2πfL …….. (3)
where ƒ is the frequency of the alternating current. For a steady current, ƒ= 0. Therefore, X_L = 0. Thus an ideal inductor offers no resistance to steady DC current.

Question 22.
Derive an expression for phase angle between the applied voltage and current in a series RLC circuit.
Answer:
AC circuit containing a resistor, an inductor and a capacitor in series – Series RLC
circuit:
Consider a circuit containing a resistor of resistance R, a inductor of inductance L and a capacitor of capacitance C connected across an alternating voltage source. The applied alternating voltage is given by the equation.
υ = V_m sin ωt …… (1)

Let i be the resulting circuit current in the circuit at that instant. As a result, the voltage is developed across R, L and C.
We know that voltage across R (V_R) is in phase with i, voltage across L (V_L) leads i by π/2 and voltage across C (V_C) lags i by π/2.
The phasor diagram is drawn with current as the reference phasor. The current is represented by the phasor
\(\vec { OI } \), V_R by \(\vec { OA } \) ; V_L by \(\vec { OB } \) and V_C by \(\vec { OC } \).
The length of these phasors are
OI = I_m, OA = I_mR, OB = I_m,X_L; OC = I_mX_c
The circuit is cither effectively inductive or capacitive or resistive that depends on the value of V₁ or V_c Let us assume that V_L > V_C. so that net voltage drop across L – C combination is V_L < V_C which is represented by a phasor \(\vec { AD } \).
By parallelogram law, the diagonal \(\vec { OE } \) gives the resultant voltage u of VR and (V_L – V_C ) and its length OE is equal to V_m. Therefore,

Z is called impedance of the circuit which refers to the effective opposition to the circuit current by the series RLC circuit. The voltage triangle and impedance triangle are given in the graphical figure.

From phasor diagram, the phase angle between n and i is found out from the following relation

Special cases Figure: Phasor diagram for a series
(i) If X_L > X_C, (X_L – X_C) is positive and phase angle φ
is also positive. It means that the applied voltage leads the current by φ (or current lags behind voltage by φ). The circuit is inductive.
∴ υ = V_m sin ωt; i = I_m sin(ωt + φ)

(ii) If X_L < X_C, (X_L – X_C) is negative and φ is also negative. Therefore current leads voltage by φ and the circuit is capacitive.
∴ υ = Vm sin ωt; i = Im sin(ωt + φ)

(iii) If X_L = X_C, φ is zero. Therefore current and voltage are in the same phase and the circuit is resistive.
∴ υ = V_m sin ωt; i = I_m sin ωt

Question 23.
Define inductive and capacitive reactance. Give their units.
Answer:
Inductive reactance X_L:
The peak value of current I_m is given by I_m = \(\frac{{ V }_{m}}{ Lω }\). Let us compare
this equation with I_m \(\frac{{ V }_{m}}{ R }\) from resistive circuit. The quantity ωL plays the same role as the resistance R in resistive circuit. This is the resistance offered by the capacitor, called capacitive reactance (X_L). It measured in ohm.
X_L = ωL

Capacitive reactance X_C:
The peak value of current I is given by I_m = \(\frac{\mathrm{v}_{\mathrm{m}}}{1 / \mathrm{c} \omega}\). Let us compare this equation with I_m = \(\frac{{ V }_{m}}{ R }\) from resistive circuit. The quantity \(\frac { 1 }{ ωC }\) plays the same role as the resistance R in resistive circuit. This is the resistance offered by the capacitor, called capacitive reactance (X_C). It measured in ohm.
X_C = \(\frac{ 1 }{ ωC }\).

Question 24.
Obtain an expression for average power of AC over a cycle. Discuss its special cases. Power of a circuit is defined as the rate of consumption of electric energy in that circuit.
Answer:
It is given by the product of the voltage and current. In an AC circuit, the voltage and current vary continuously with time. Let us first calculate the power at an instant and then it is averaged over a complete cycle.
The alternating voltage and alternating current in the series RLC circuit at an instant are given by
υ = V_m sin ωt and i = I_m sin (ωt + 4>)
where φ is the phase angle between υ and i. The instantaneous power is then written as
P = υi = V_m I_m sin ωt sin(ωt + φ)
= V_m I_m sin ωt (sin ωt cos φ – cos ωt sin φ)
P = V_m I_m (cos φ sin² ωt – sin ωt cos ωt sin φ) …… (1)
Here the average of sin² ωt over a cycle is \(\frac { 1 }{ 2 }\) and that of sin ωt cos ωt is zero. Substituting these values, we obtain average power over a cycle.
P_av = V_m I_m cos φ x \(\frac { 1 }{ 2 }\) = \(\frac {{ V }_{m}}{ √2 }\) \(\frac {{ I }_{m}}{ √2 }\) cos φ
P_av = V_RMS I_RMS cos φ …… (2)
where V_RMS I_RMS is called apparent power and cos φ is power factor. The average power of an AC circuit is also known as the true power of the circuit.
Special Cases:
(i) For a purely resistive circuit, the phase angle between voltage and current is zero and cos
φ = 1.
∴ P_av = V_RMS I_RMS
(ii) For a purely inductive or capacitive circuit, the phase angle is ± \(\frac { π }{ 2 }\) and cos \(\left(\pm \frac{\pi}{2}\right)\) = 0
∴ P_av = 0
(iii) For series RLC circuit, the phase angle φ = tan^-1 \(\left(\frac{\mathrm{x}_{\mathrm{L}}-\mathrm{x}_{\mathrm{c}}}{\mathrm{R}}\right)\)
∴ P_av = V_RMS I_RMS cos φ
(iv) For series RLC circuit at resonance, the phase angle is zero and cos φ = 1.
∴ P_av = V_RMS I_RMS

Question 25.
Show that the total energy is conserved during LC oscillations.
Answer:
Conservation of energy in LC oscillations: During LC oscillations in LC circuits, the energy of the system oscillates between the electric field of the capacitor and the magnetic field of the inductor. Although, these two forms of energy vary with time, the total energy remains constant. It means that LC oscillations take place in accordance with the law of conservation of energy.
Total energy,

Let us consider 3 different stages of LC oscillations and calculate the total energy of the system.

Case I:
When the charge in the capacitor, q = Q_m and the current through the inductor, i = 0, the total energy is given by

The total energy is wholly electrical.

Case II:
When charge = 0; current = I_m, the total energy is

The total energy is wholly electrical.

Case III:
When charge = q; current = i, the total energy is

Since q = Q_m cos ωt, i = \(\frac { bq }{ dt }\) = Q_mω sin ωt. The negative sign in current indicates that the charge in the capacitor in the capacitor decreases with time.

From above three cases, it is clear that the total energy of the system remains constant.

Question 26.
Prove that energy is conserved during electromagnetic induction.
Answer:
The mechanical energy of the spring-mass system is given by

The energy E remains constant for varying values of x and v. Differentiating E with respect to time, we get

This is the differential equation of the oscillations of the spring-mass system. The general solution of equation (2) is of the form
x(t) = X_m cos (ωt + φ) …… (3)
where X_m is the maximum value of x(t), ω, the angular frequency and φ, the phase constant. Similarly, the electromagnetic energy of the LC system is given by

Differentiating U with respect to time, we get

Equation (2) and (5) are proved the energy of electromagnetic induction is conserved.

Question 27.
Compare the electromagnetic oscillations of LC circuit with the mechanical oscillations of block spring system to find the expression for angular frequency of LC oscillators mathematically.
Answer:
The mechanical energy of the spring-mass system is given by

The energy E remains constant for varying values of x and v. Differentiating E with respect to time, we get

This is the differential equation of the oscillations of the spring-mass system. The general solution of equation (2) is of the form
x(t) = X_m cos (ωt + φ) …… (3)
where X_m is the maximum value of x(t), ω, the angular frequency and φ, the phase constant. Similarly, the electromagnetic energy of the LC system is given by

Differentiating U with respect to time, we get

Equation (2) and (5) are proved the energy of electromagnetic induction is conserved.
q(t) = Q_m cos (ωt + φ) …… (6)
where Q_m is the maximum value of q(t), ω, the angular frequency and φ, the phase constant.

Current in the LC circuit:
The current flowing in the LC circuit is obtained by differentiating q(t) with respect to time.
i(t) = \(\frac { dq }{ dt }\) = \(\frac { d }{ dt }\) [Q_m cos (ωt + φ)] = Q_m ω sin (ωt + φ) since I_m = Q_mω
(or)
i(t) -I_m sin (ωt + φ) ……. (7)
The equation (7) clearly shows that current varies as a function of time t. In fact, it is a sinusoidally varying alternating current with angular frequency ω.

Angular frequency of LC oscillations:
By differentiating equation (6) twice, we get
\(\frac { { d }^{ 2 }q }{ dt } \) = -Q_mω² cos (ωt + φ) …….. (8)
Substituting equations (6) and (8) in equation (5),
we obtain L[-Q_mω² cos (ωt + φ)] + \(\frac { 1 }{ C }\) Q_m cos (ωt + φ) = 0
Rearranging the terms, the angular frequency of LC oscillations is given by
ω = \(\frac { 1 }{ \sqrt { LC } } \) …… (9)
This equation is the same as that obtained from qualitative analogy.

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Numerical Problems

Question 1.
A square coil of side 30 cm with 500 turns is kept in a uniform magnetic field of 0.4 T. The plane of the coil is inclined at an angle of 30° to the field. Calculate the magnetic flux through the coil.
Solution:
Square coil of side (a) = 30 cm = 30 × 10^-2m
Area of square coil (A) = a² = (30 × 10^-2)2 = 9 × 10^-2 m²
Number of turns (N) = 500
Magnetic field (B) = 0.4 T
Angular between the field and coil (θ) = 90 – 30 = 60°
Magnetic flux (Φ) = NBA cos 0 = 500 × 0.4 × 9 × 10^-2 × cos 60° = 18 × \(\frac { 1 }{ 2 }\)
Φ = 9 W b

Question 2.
A straight metal wire crosses a magnetic field of flux 4 mWb in a time 0.4 s. Find the magnitude of the emf induced in the wire.
Solution:
Magnetic flux (Φ) = 4 m Wb = 4 × 10^-3 Wb
time (t) = 0.4 s
The magnitude of induced emf (e) = \(\frac { dΦ }{ dt }\) = \(\frac {{ 4 × 10 }^{-3}}{ 0.4 }\) 10^-2
e = 10 mV

Question 3.
The magnetic flux passing through a coil perpendicular to its plane is a function of time and is given by OB = (2t³ + 4t² + 8t + 8) Wb. If the resistance of the coil is 5 Ω, determine the induced current through the coil at a time t = 3 second.
Solution:
Magnetic flux (Φ_B) = (2t³ + 8t² + 8t + 8)Wb
Resistance of the coil (R) = 5 Ω
time (t) = 3 second
Induced current through the coil, I = \(\frac { e }{ R }\)
Induced emf, e = \(\frac {{ dΦ }_{B}}{ dt }\) = \(\frac { d }{ dt }\) ((2t³ + 4t² + 8t + 8) = 6t² + 8t + 8
Here time (t) = 3 second
e = 6(3)² + 8 × 3 + 8 = 54 + 24 + 8 = 86 V
∴ Induced current through the coil, I = \(\frac { e }{ R }\) = \(\frac { 86 }{ 5 }\) = 17.2 A

Question 4.
A closely wound coil of radius 0.02 m is placed perpendicular to the magnetic field. When the magnetic field is changed from 8000 T to 2000 T in 6 s, an emf of 44 V is induced. Calculate the number of turns in the coil.
Solution:
Radius of the coil (r) = 0.02 m
Area of the coil (A) = πr² = 3.14 × (0.02)²= 1.256 × 10^-3 m²
Change in magnetic field, dB = 8000 – 2000 = 6000 T
Time, dt = 6 second
Induced emf, e = 44 V
θ = 0°
Induced emf in the coil, e = NA \(\frac { d }{ dt }\) cos θ . dt
44 = N × 1.256 × 10^-3 × \(\frac { 600 }{ 6 }\) × Cos 0°

Number of turns N = 35 turns

Question 5.
A rectangular coil of area 6 cm² having 3500 turns is kept in a uniform magnetic field of 0.4 T, Initially, the plane of the coil is perpendicular to the field and is then rotated through an angle of 180°. If the resistance of the coil is 35 Ω, find the amount of charge flowing through the coil.
Solution:
Rectangular coil of their area, A = 6 cm² = 6 x 10^-4 m²
Number of turns N = 3500 turns
Magnetic field, B = 0.4 T
Resistance of the coil, R= 35 Ω
Induced emf (e) = change in flux per second = Φ₂ – Φ₁
e = NAB cos 180° – NBA cos 0° = -NBA – NBA = – 2 NBA
= – 2 x 3500 x 0.4 x 6 x 10^-4 – 16800 x 10^-4 = – 1.68 V
Current flowing the coil, I = \(\frac { e }{ R }\) = \(\frac { -1.68 }{ 35 }\) = 0.048
Magnitude of the current, I = 48 x 10^-3 A
Amount of charge flowing through the coil, q = It = 48 x 10^-3 x 1 = 48 x 10^-3 C

Question 6.
An induced current of 2.5 mA flows through a single conductor of resistance 100 Ω. Find out the rate at which the magnetic flux is cut by the conductor.
Solution:
Induced current, I = 2.5 mA
Resistance of conductor, R = 100 Ω
∴ The rate of change of flux, \(\frac {{ dΦ }_{B}}{ dt }\) = e
\(\frac {{ dΦ }_{B}}{ dt }\) = e = IR = 2.5 x 10-3 x 100 = 250 x 10^-3 dt
\(\frac {{ dΦ }_{B}}{ dt }\) = 250 mWb s^-1

Question 7.
A fan of metal blades of length 0.4 m rotates normal to a magnetic field of 4 x 10^-3 T. If the induced emf between the centre and edge of the blade is 0.02 V, determine the rate of rotation of the blade.
Solution:
Length of the metal blade, l = 0.4 m
Magnetic field, B = 4 x 10^-3 T
Induced emf, e = 0.02 V
Rotational area of the blade, A = πr² = 3.14 x (0.4)² = 0.5024 m²
Induced emf in rotational of the coil, e = NBA ω sin θ

= 9.95222 x 10^-3 x 10³
= 9.95 revolutions/second
Rate of rotational of the blade, ω = 9.95 revolutions/second

Question 8.
A bicycle wheel with metal spokes of 1 m long rotates in Earth’s magnetic field. The plane of the wheel is perpendicular to the horizontal component of Earth’s field of 4 x 10^-5 T. If the emf induced across the spokes is 31.4 mV, calculate the rate of revolution of the wheel.
Solution:
Length of the metal spokes, l = 1 m
Rotational area of the spokes, A = π² = 3.14 x (1)² = 3.14 m²
Horizontal component of Earth’s field, B = 4 x 10^-5 T
Induced emf, e = 31.4 mV
The rate of revolution of wheel,

ω = 250 revolutions / second

Question 9.
Determine the self-inductance of 4000 turn air-core solenoid of length 2m and diameter 0. 04 m.
Solution:
Length of the air core solenoid, l = 2 m
Diameter, d = 0.04 m
Radius, r = \(\frac { d }{ 2 }\) = 0.02 m
Area of the air core solenoid, A = π² = 3.14 x (0.02)² = 1.256 x 10^-3 m²
Number of Turns, N = 4000 turns
Self inductance, L = µ₀n² Al

Question 10.
A coil of 200 turns carries a current of 4 A. If the magnetic flux through the coil is 6 x 10^-5 Wb, find the magnetic energy stored in the medium surrounding the coil.
Solution:
Number of turns of the coil, N = 200
Current, I = 4 A
Magnetic flux through the coil, Φ = 6 x 10^-5 Wb
Energy stored in the coil, U = \(\frac { 1 }{ 2 }\) LI² = \(\frac { 1 }{ I2}\)
Self inductance of the coil, L = \(\frac { NΦ }{ I }\)
U =\(\frac { 1 }{ 2 }\) \(\frac { NΦ }{ I }\) x I² = \(\frac { 1 }{ 2}\) NΦI = \(\frac { 1 }{ 2}\) x 200 x 6 x 10^-5 x 4
U = 2400 x 10^-5 = 0.024 J (or) joules.

Question 11.
A 50 cm long solenoid has 400 turns per cm. The diameter of the solenoid is 0.04 m. Find the magnetic flux of a turn when it carries a current of 1 A.
Solution:
Length of the solenoid, l = 50 cm = 50 x 10^-2 m
Number of turns per cm, N = 400
Number of turns in 50 cm, N = 400 x 50 = 20000
Diameter of the solenoid, d = 0.04 m
Radius of the solenoid, r = \(\frac { d }{ 2}\) = 0.02 m
Area of the solenoid, A = π² = 3.14 x (0.02)² = 1.256 x 10^-3 m²
Current passing through the solenoid, I = 1 A
Magnetic fluex,

Question 12.
A coil of 200 turns carries a current of 0.4 A. If the magnetic flux of 4 mWb is linked with the coil, find the inductance of the coil.
Solution:
Number of turns, N = 200; Current, I = 0.4 A
Magnetic flux linked with coil, Φ = 4 mWb = 4 x 10^-3 Wb
Induction of the coil, L = \(\frac { NΦ }{ I }\) = \(\frac {{ 200 × 4 × 10 }^{-3}}{ 0.4 }\) = \(\frac {{ 800 × 10 }^{-3}}{ 0.4 }\) 2 H

Question 13.
Two air core solenoids have the same length of 80 cm and same cross-sectional area 5 cm². Find the mutual inductance between them if the number of turns in the first coil is 1200 turns and that in the second coil is 400 turns.
Solution:
Length of the solenoids, l = 80 cm = 8 x 10^-2 m
Cross sectional area of the solenoid, A = 5 cm² = 5 x 10^-4 m²
Number of turns in the I^st coil, N₁ = 1200
Number of turns in the IInd coil, N₂ = 400
Mutual inductance between the two coils,

Question 14.
A long solenoid having 400 turns per cm carries a current 2A. A 100 turn coil of cross sectional area 4 cm² is placed co-axially inside the solenoid so that the coil is in the field produced by the solenoid. Find the emf induced in the coil if the current through the solenoid reverses its direction in 0.04 sec.
Solution:
Number of turns of long solenoid per cm =\(\frac { 400 }{{10}^{ -2 }}\); N₂ = 400 x 10²
Number of turns inside the solenoid, N₂ = 100
Cross-sectional area of the coil, A = 4 cm² = 4 x 10^-4 m²
Current through the solenoid, I = 2A; time, t = 0.04 s
Induced emf of the coil, e = -M \(\frac { dI }{ dt }\)
Mutual inductance of the coil,

Induced emf of the coil,

The current through the solenoid reverse its direction if the induced emf, e = -0.2 V

Question 15.
A 200 turn coil of radius 2 cm is placed co-axially within a long solenoid of 3 cm radius. If the turn density of the solenoid is 90 turns per cm, then calculate mutual inductance of the coil.
Solution:
Number of turns of the solenoid, N₂ = 200
Radius of the solenoid, r = 2cm = 2 x 10² m
Area of the solenoid, A = πr²= 3.14 x (2 x 10^-2)² = 1.256 x 10^-3 m²
Turn density of long solenoid per cm, N₁ = 90 x 10²
Mutual inductance of the coil,

= 283956.48 x 10^-8 ⇒ M = 2.84 mH

Question 16.
The solenoids S₁ and S₂ are wound on an iron-core of relative permeability 900. The area of their cross-section and their length are the same and are 4 cm² and 0.04 m, respectively. If the number of turns in S₁ is 200 and that in S₂ is 800, calculate the mutual inductance between the coils. The current in solenoid 1 is increased from 2A to 8A in 0.04 second. Calculate the induced emf in solenoid 2.
Solution:
Relative permeability of iron core, μ_r = 900
Number of turns of solenoid S₁, N₁ = 200
Number of turns of solenoid S₂, N₂ = ‘800
Area of cross section, A = 4 cm² = 4 x 10^-4 m²
Length of the solenoid S₁, l₁ = 0.04 m
current, I =I₂ – I₁ = 8 – 2 = 6A
time taken, t = 0.04 second
emf induced in solenoid S₂ e = -M \(\frac { dI }{ dt }\)
Mutual inductance between the two coils, M = \(\frac{\mu_{0} \mu_{r} N_{1} N_{2} A}{l}\)

M = 180864 x 10^-5 = 1.81 H
Emf induced in solenoid S₂, e = -M\(\frac { dI }{ dt }\) = -1.81 x \(\frac { 6 }{ 0.04 }\)
Magnitude of emf, e = 271.5 V

Question 17.
A step-down transformer connected to main supply of 220 V is made to operate 11 V, 88 W lamp. Calculate (i) Transformation ratio and (ii) Current in the primary.
Solution:
Voltage in primary coil, V_p = 220 V
Voltage in secondary coil, V_s = 11 V
Output power = 88 W
(i) To find transformation ratio, k = \(\frac {{ V }_{ s }}{{ V }_{ p }}\) = \(\frac { 11 }{ 220 }\) = \(\frac { 1 }{ 20 }\)
(ii) Current in primary, I_p = \(\frac {{ V }_{ s }}{{ V }_{ p }}\) x I_s
So, I_s = ?
Outputpower = V_s I_s
⇒ 88 = 11 x I_s
I_s = \(\frac { 88 }{ 11 }\) = 8A
Therefore, I_p = \(\frac {{ V }_{ s }}{{ V }_{ p }}\) x I_s = \(\frac { 11 }{ 220 }\) x 8 = 0.4 A

Question 18.
A 200V/120V step-down transformer of 90% efficiency is connected to an induction stove of resistance 40 Ω. Find the current drawn by the primary of the transformer.
Solution:
Primary voltage, V_p = 200 V
Secondary voltage, V_s = 120 V
Efficiency, η = 90%
Secondary resistance, R_s = 40 Ω
Current drawn by the primary of the transformc, I_p = \(\frac {{ V }_{ s }}{{ R }_{ s }}\) x I_s
Output power = V_s I_s

Question 19.
The 300 turn primary of a transformer has resistance 0.82 Ω and the resistance of its secondary of 1200 turns is 6.2 Ω. Find the voltage across the primary if the power output from the secondary at 1600V is 32 kW. Calculate the power losses in both coils when the transformer efficiency is 80%.
Solution:
Efficiency, η = 80% = \(\frac { 80 }{ 100 }\)
Number of turns in primary, N_p = 300
Number of turns in secondary, N_s = 1200
Resistance in primary, R_p = 0.82 Ω
Resistance in secondary, R_s = 6.2 Ω
Secondary voltage, V_s = 1600 V
Output power = 32 kW
Output power = V_s I_s

Power loss in primary = \({ { I }_{ p }^{ 2 }{ R }_{ p } }\) = (100)² x 0.82 = 8200 = 8.2 kW
Power loss in secondary = \({ { I }_{ s }^{ 2 }{ R }_{ s } }\) = (20)² x 6.2 = 2480 = 2.48 kW

Question 20.
Calculate the instantaneous value at 60°, average value and RMS value of an alternating current whose peak value is 20 A.
Solution:
Peak value of current, I_m = 20 A
Angle, θ = 60° [θ = ωt]
(i) Instantaneous value of current,
i = I_m sin ωt = I_m sin θ
= 20 sin 60° = 20 x \(\frac { √3 }{ 2 }\) = 10√3 = 10 x 1.732
i = 17.32 A

(ii) Average value of current,
I_av = \(\frac {{ 2I }_{m}}{ π }\) = \(\frac { 2 × 20 }{ 3.14 }\)
I_av = 12.74 A

(iii) RMS value of current,
I_RMS = 0.707 I_m
or \(\frac{\mathrm{I}_{\mathrm{m}}}{\sqrt{2}}\) = 0.707 x 20
I_RMS = 14. 14 A

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Conceptual Questions

Question 1.
A graph between the magnitude of the magnetic flux linked with a closed loop and time is given in the figure. Arrange the regions of the graph in ascending order of the magnitude of induced emf in the loop.
Answer:
According to electromagnetic induction, induced emf,

e = \(\frac { dΦ }{ dt }\)
Ascending order of induced emf from the graphical representation is b < c < d < a.

Question 2.
Using Lenz’s law, predict the direction of induced current in conducting rings 1 and 2 when current in the wire is steadily decreasing.
Answer:

According to Lenz’s law, a current will be induced in the coil which will produce a flux in the opposite direction.

If the current decreases in the wire, the induced current flows in ring 1 in clockwise direction, the induced current flows in ring 2 in anti-clockwise direction.

Question 3.
A flexible metallic loop abed in the shape of a square is kept in a magnetic field with its plane perpendicular to the field. The magnetic field is directed into the paper normally. Find the direction of the induced current when the square loop is crushed into an irregular shape as shown in the figure.
Answer:

The magnetic flux linked with the wire decreases due to decrease in area of the loop. The induced emf will cause current to flow in the direction. So that the wire is pulled out ward direction from all sides. According to Fleming’s left hand rule, force on wire will act outward i direction from all sides.

Question 4.
Predict the polarity of the capacitor in a closed circular loop when two bar magnets are moved as shown in the figure.
Answer:

When magnet 1 is moved with its South pole towards to the coil, emf is induced in the coil as the magnetic flux through the coil changes. When seeing from the left hand side the direction of induced current appears to be clockwise. When seeing from the right hand side the direction of induced current appears to be anti-clockwise. In capacitor, plate A has positive polarity and plate B has negative polarity.

Question 5.
In series LC circuit, the voltages across L and C are 180° out of phase. Is it correct? Explain.
Answer:
In series LC circuit, the voltage across the capacitance lag the current by 90° while the voltage across the inductance lead the current by 90°. This makes the inductance and capacitance voltages 180° out of phase.

Question 6.
When does power factor of a series RLC circuit become maximum?
Answer:
For a series LCR circuit, power factor is

For purely resistive, Φ = 0°, cos 0° = 1
Thus the power factor assumes the maximum value for a purely resistive circuit.

Question 7.
Draw graphs showing the distribution of charge in a capacitor and current through an inductor during LC oscillations with respect to time. Assume that the charge in the capacitor is maximum initially.
Answer:
For a capacitor, the graph between charge and time.

The charge decays exponentially decreases with time.

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Additional Questions solved

I Choose The Correct Answer

Question 1.
A coil of area of cross section 0.5 m² with 10 turns is in a plane which is perpendicular to an uniform magnetic field of 0.2 Wb/m². The flux through the coil is –
(a) 100 Wb
(b) 10 Wb
(c) 1 Wb
(d) zero
Answer:
(c) 1 Wb
Hint:
Φ = NBA cos θ
= 10 x 0.2 x 0.5 x cos 0° = 1 Wb

Question 2.
A rectangular coil of 100 turns and size 0.1 m x 0.05 m is placed perpendicular to a magnetic field of 0.1 T. If the field drops to 0.05 T in 0.05 s, the magnitude of the emf induced in the coil is-
(a) 0.5 V
(b) 0.75 V
(c) 1.0 V
(d) 1.5 V
Answer:
(a) 0.5 V
Hint:

ε = 0.5 V

Question 3.
A wire of length 1 m moves with a speed of 10 ms^-1 perpendicular to a magnetic field. If the emf induced in the wire is 1 V, the magnitude of the field is-
(a) 0.01 T
(b) 0.1 T
(c) 0.2 T
(d) 0.02 T
Answer:
(b) 0.1 T
Hint:
ε = Blv
⇒ B = \(\frac { ε }{ lv }\) = \(\frac { 1 }{ 1 × 10 }\) = 0.02 T

Question 4.
A coil of area 10 cm², 10 ms^-1 turns and resistance 20 Ω is placed in a magnetic field directed perpendicular to the plane of the coil and changing at the rate of 10⁸ gauss/second. The induced current in the coil will be-
(a) 5 A
(b) 0.5 A
(c) 0.05 A
(d) 50 A
Answer:
(a) 5 A
Hint:

Question 5.
A coil of cross sectional area 400 cm² having 30 turns is making 1800 rev/min in a magnetic field of IT. The peak value of the induced emf is-
(a) 113 V
(b) 226 V
(c) 339 V
(d) 452 V
Answer:
(b) 226 V
Hint:
ε_m = NBA ω = 30 x 1 x 400 x 10^-4 x 30 x 2π
= 226 V

Question 6.
Eddy currents are produced in a material when it is-
(a) heated
(b) placed in a time varying magnetic field
(c) placed in an electric field
(d) placed a uniform magnetic field
Answer:
(b) placed in a time varying magnetic field

Question 7.
An emf of 5 V is induced in an inductance when the current in it changes at a steady rate from 3 A to 2 A in 1 millisecond. The value of inductance is-
(a) 5 mH
(b) 5 H
(c) 5000 H
(d) zero
Answer:
(a) 5 mH

Question 8.
Faraday’s law of electromagnetic induction is related to the-
(a) Law of conservation of charge
(b) Law of conservation of energy
(c) Third law of motion
(d) Law of conservation of angular momentum
Answer:
(b) Law of conservation of energy

Question 9.
The inductance of a coil is proportional to-
(a) its length
(b) the number of turns
(c) the resistance of the coil
(d) square of the number of turns
Answer:
(d) square of the number of turns

Question 10.
When a direct current ‘i’ is passed through an inductance L, the energy stored is-
(a) Zero
(b) Li
(c) \(\frac { 1 }{ 2 }\) Li²
(d) \(\frac {{ L }^{ 2 }}{2i}\)
Answer:
(c) \(\frac { 1 }{ 2 }\) Li²

Question 11.
A coil of area 80 cm² and 50 turns is rotating with 2000 revolutions per minute about an axis perpendicular to a magnetic field of 0.05 T. The maximum value of the emf developed in it is-
(a) 2000 πV
(b) \(\frac { 10π }{ 3 }\) V
(c) \(\frac { 4π }{ 3 }\)V
(d) \(\frac { 2 }{ 3 }\) V
Answer:
(c) \(\frac { 4π }{ 3 }\)V
Hint:
ε = NBA ω = 50 x 0.05 x 80 x 10^-4 x \(\frac { 2π × 2000 }{ 60 }\) = \(\frac { 4π }{ 3 }\)V

Question 12.
The direction of induced current during electro magnetic induction is given by-
(a) Faraday’s law
(b) Lenz’s law
(c) Maxwell’s law
(d) Ampere’s law
Answer:
(b) Lenz’s law

Question 13.
AC power is transmitted from a power house at a high voltage as-
(a) the rate of transmission is faster at high voltages
(b) it is more economical due to less power loss
(c) power cannot be transmitted at low voltages
(d) a precaution against theft of transmission lines
Answer:
(b) it is more economical due to less power loss

Question 14.
In a step-down transformer the input voltage is 22 kV and the output voltage is 550 V. The ratio of the number of turns in the secondary to that in the primary is-
(a) 1 : 20
(b) 20 : 1
(c) 1 : 40
(d) 40 : 1
Answer:
(c) 1 : 40
Hint:
\(\frac {{ N }_{ s }}{{ N }_{ p }}\) = \(\frac {{ V }_{ s }}{{ V }_{ p }}\) = \(\frac { 550 }{ 22000 }\) = \(\frac { 1 }{ 40 }\)

Question 15.
The self-inductance of a coil is 5 H. A current of 1 A changes to 2 A within 5 s through the coil. The value of induced emf will be-
(a) 10 V
(b) 0.1 V
(c) 1.0 V
(d) 100 V
Answer:
(c) 1.0 V
Hint:

Question 16.
The low-loss transformer has 230 V applied to the primary and gives 4.6 V in the secondary. The secondary is connected to a load which draws 5 amperes of current. The current (in amperes) in the primary is-
(a) 0.1 A
(b) 1.0 A
(c) 10 A
(d) 250 A
Answer:
(a) 0.1 A
Hint:
I_p = \(\frac {{ V }_{ s }{ I }_{ s }}{{ V }_{ p }}\) = \(\frac { 4.6 × 5 }{ 230 }\) = 0.1A

Question 17.
A coil is wound on a frame of rectangular cross-section. If all the linear dimensions of the frame are increased by a factor 2 and the number of turns per unit length of the coil remains the same. Self-inductance of the coil increases by a factor of-
(a) 4
(b) 8
(c) 12
(d) 16
Answer:
(b) 8
Hint:
If all the linear dimensions are doubled, the cross-sectional are a becomes eight times. Therefore, the flux produced by a given current will become eight times. Hence, the selfinductance increases by a factor of 8.

Question 18.
If N is the number of turns in a coil, the value of self-inductance varies as-
(a) N°
(b) N
(c) N²
(d) N^-2
Answer:
(c) N²
Hint:
According to self inductance of long solenoid
L = \(\frac{\mu_{0} \mathrm{N}^{2} \mathrm{A}}{l}\)
⇒ L ∝ N²

Question 19.
A magnetic field 2 x 10^-2 T acts at right angles to a coil of area 100 cm² with 50 turns. The average emf induced in the coil will be 0.1 V if it is removed from the field in time.
(a) 0.01 s
(b) 0.1 s
(c) 1 s
(d) 10 s
Answer:
(b) 0.1 s
Hint:

Question 20.
Number of turns in a coil is increased from 10 to 100. Its inductance becomes-
(a) 10 times
(b) 100 times
(c) 1/10 times
(d) 25 times
Answer:
(a) 10 times

Question 21.
The north pole of a magnet is brought near a metallic ring as shown in the figure. The direction of induced current in the ring, as seen by the magnet is-

(a) anti-clockwise
(b) first anti-clockwise and then clockwise
(c) clockwise
(d) first clockwise and then anti-clockwise
Answer:
(a) anti-clockwise

Question 22.
Quantity that remains unchanged in a transformer is-
(a) voltage
(b) current
(c) frequency
(d) none of these
Answer:
(c) frequency

Question 23.
The core of a transformer is laminated to reduce.
(a) Copper loss
(b) Magnetic loss
(c) Eddy current loss
(d) Hysteresis loss
Answer:
(c) Eddy current loss

Question 24.
Which of the following has the dimension of time?
(a) LC
(b) \(\frac { R }{ L }\)
(c) \(\frac { L }{ R }\)
(d) \(\frac { C }{ L }\)
Answer:
(c) \(\frac { L }{ R }\)

Question 25.
A coil has a self-inductance of 0.04 H. The energy required to establish a steady-state current of 5 A in it is-
(a) 0.5 J
(b) 1.0 J
(c) 0.8 J
(d) 0.2 J
Answer:
(a) 0.5 J

Question 26.
Alternating current can be measured by
(a) moving coil galvanometer
(b) hot wire ammeter
(c) tangent galvanometer
(d) none of the above
Answer:
(b) hot wire ammeter

Question 27.
In an LCR circuit, the energy is dissipated in-
(a) R only
(b) R and L only
(c) R and C only
(d) R, L and C
Answer:
(a) R only

Question 28.
A 40 Ω electric heater is connected to 200 V, 50 Hz main supply. The peak value of the electric current flowing in the circuit is approximately-
(a) 2.5 A
(b) 5 A
(c) 7 A
(d) 10 A
Answer:
(c) 7 A
Hint:
I₀ = \(\frac {{ V }_{0}}{ R }\) = \(\frac{200 \sqrt{2}}{40}\) 5 √2 ≈ 7A

Question 29.
The rms value of an alternating current, which when passed through a resistor produces heat three times of that produced by a direct current of 2 A in the same resistor, is-
(a) 6 A
(b) 3 A
(c) 2 A
(d) 2√3 A
Answer:
(d) 2√3 A
Hint:
\({ I }_{ rms }^{ 2 }\)R = 3(2²R) (or) I_rms = 2√3 A

Question 30.
An inductance, a capacitance and a resistance are connected in series across a source of alternating voltages. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of-
(a) \(\frac { π }{ 4 }\)
(b) zero
(c) π
(d) \(\frac { π }{ 2 }\)
Answer:
(b) zero

Question 31.
In an AC circuit, the rms value of the current I_rms, is related to the peak current I₀ as-
(a) I_rms = \(\frac {{I}_{0}}{ π }\)
(b) I_rms = \(\frac {{I}_{0}}{ √2 }\)
(c) I_rms = √2 I₀
(d) I_rms = πI₀
Answer:
(b) I_rms = \(\frac {{I}_{0}}{ √2 }\)

Question 32.
The impedance of a circuit consists of 3 Ω resistance and 4 Ω resistance. The power factor of the circuit is
(a) 0.4
(b) 0.6
(c) 0.8
(d) 1.0
Answer:
(b) 0.6
Hint:
tan Φ = \(\frac { 4 }{ 3 }\).
Power factor = cos Φ = \(\frac { 3 }{ 5 }\) = 0.06

Question 33.
The reactance of a capacitance at 50 Hz is 5 Ω. Its reactance at 100 Hz will be-
(a) 5 Ω
(b) 10 Ω
(c) 20 Ω
(d) 2.5 Ω
Answer:
(d) 2.5 Ω.

Question 34.
In a LCR AC circuit off resonance, the current-
(a) is always in phase with the voltage
(b) always lags behind the voltage
(c) always leads the voltage
(d) may lead or lag behind the voltage
Answer:
(d) may lead or lag behind the voltage

Question 35.
The average power dissipation in a pure inductance L, through which a current I₀ sin ωt is flowing is-
(a) \(\frac { 1 }{ 2 }\) L\({ I }_{ 0 }^{ 2 }\)
(b) L\({ I }_{ 0 }^{ 2 }\)
(c) 2 L\({ I }_{ 0 }^{ 2 }\)
(d) zero
Answer:
(d) zero

Question 36.
The power in an AC circuit is given by P = V_rms I_rms cos Φ. The value of the power factor cos Φ in series LCR circuit at resonance is-
(a) zero
(b) 1
(c) \(\frac { 1 }{ 2 }\)
(d) \(\frac { 1 }{ √2 }\)
Answer:
(b) 1
Hint:

Question 37.
In an AC circuit with voltage V and current I, the power dissipated is-
(a) VI
(b) \(\frac { 1 }{ 2 }\) VI
(c) \(\frac { 1 }{ √2 }\) VI
(d) depends on the phase difference between I and V
Answer:
(d) depends on the phase difference between I and V

Question 38.
In an AC circuit containing only capacitance, the current-
(a) leads the voltage by 180°
(b) remains in phase with the voltage
(c) leads the voltage by 90°
(d) lags the voltage by 90°
Answer:
(c) leads the voltage by 90°

Question 39.
In a series LCR circuit R = 10 Ω and the impedance Z = 20 Ω. Then the phase difference between the current and the voltage is-
(a) 60°
(b) 30°
(c) 45°
(d) 90°
Answer:
(c) 60°
Hint:
cos Φ = \(\frac { R }{ Z }\) = \(\frac { 10 }{ 20 }\) = \(\frac { 1 }{ 2 }\)
⇒ Φ = 60°

Question 40.
What is the value of inductance L for which the current is maximum in a series LCR circuit with C = 10 μF and ω = 1000 s^-1?
(a) 1 mH
(b) 10 mH
(c) 100 mH
(d) Cannot be calculated unless R is known
Answer:
(c) 100 mH
Hint:
L = \(\frac { 1 }{ { \omega }^{ 2 }C } \) = \(\frac{1}{(1000)^{2} \times 10 \times 10^{-6}}\) = 0.1 H = 100 mH

II Fill in the Blanks

Question 1.
Electromagnetic induction is used in …………….
Answer:
transformer and AC generator

Question 2.
Lenz’s Law is in accordance with the law of …………….
Answer:
conservation of energy.

Question 3.
The self-inductance of a straight conductor is …………….
Answer:
zero

Question 4.
Transformer works on …………….
Answer:
AC only

Question 5.
The power loss is less in transmission lines when …………….
Answer:
voltage is more but current is less

Question 6.
The law that gives the direction of the induced current produced in a circuit is …………….
Answer:
Lenz’s law

Question 7.
Fleming’s right hand rule is otherwise called …………….
Answer:
generator rule

Question 8.
Unit of self-inductance is …………….
Answer:
Henry

Question 9.
The mutual induction is very large, if the two coils are wound on …………….
Answer:
soft iron core

Question 10.
When the coil is in vertical position, the angle between the normal to the plane of the coil and magnetic field is …………….
Answer:
zero

Question 11.
The emf induced by changing the orientation of the coil is ……………. in nature.
Answer:
sinusoidal

Question 12.
In a three phase AC generator, the three coils are inclined at an angle of …………….
Answer:
120°

Question 13.
The emf induced in each of the coils differ in phase by …………….
Answer:
120°

Question 14.
A device which converts high alternating voltage into low alternating voltage and vice versa is …………….
Answer:
transformer

Question 15.
For an ideal transformer efficiency η is …………….
Answer:
1

Question 16.
The alternating emf induced in the coil varies …………….
Answer:
periodically in both magnitude and direction

Question 17.
For direct current, inductive reactance is …………….
Answer:
zero

Question 18.
In an inductive circuit the average power of sinusoidal quantity of double the frequency over a complete cycle is …………….
zero

Question 19.
For direct current, the resistance offered by a capacitor is …………….
Answer:
infinity

Question 20.
In a capacitive circuit, power over a complete cycle is …………….
Answer:
zero

Question 21.
Q-factor measures the …………….in resonant circuit
Answer:
selectivity

Question 22.
Voltage drop across inductor and capacitor differ in phase by …………….
Answer:
180°

Question 23.
Angular resonant frequency (co) is …………….
Answer:
\(\frac { 1 }{ \sqrt { LC } } \)

Question 24.
A circuit will have flat resonance if its Q-value is …………….
Answer:
low

Question 25.
The average power consumed by the choke coil over a complete cycle is …………….
Answer:
zero

III Match the following

Question 1.

Answer:
(i) → (c)
(ii) → (a)
(iii) → (d)
(iv) → (b)

Question 2.

Answer:
(i) → (c)
(ii) → (a)
(iii) → (d)
(iv) → (b)

Question 3.

Answer:
(i) → (c)
(ii) → (d)
(iii) → (b)
(iv) → (a)

Question 4.
Type of impedance Phase between voltage and current

Answer:
(i) → (c)
(ii) → (d)
(iii) → (a)
(iv) → (b)

Question 5.
Energy in two oscillatory systems: (LC oscillator and spring mass system)

Answer:
(i) → (c)
(ii) → (d)
(iii) → (a)
(iv) → (b)

IV Assertion and reason

(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.
(c) If assertion is true but reason is false.
(d) If the assertion and reason both are false.
(e) If assertion is false but reason is true.

Question 1.
Assertion: Eddy currents is produced in any metallic conductor when flux is changed around it.
Reason: Electric potential determines the flow of charge.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.
Solution:
When a metallic conductor is moved in a magnetic field, magnetic flux is “varied. It disturbs the free electrons of the metal and set up an induced emf in it. As there are no free ends of the metal i.e., it will be closed in itself so there will be induced current.

Question 2.
Assertion: Faraday’s laws are consequences of conservation of energy.
Reason: In a purely resistive AC circuit, the current lags behind the emf in phase.
Answer:
(c) If assertion is true but reason is false.
Solution:
According to Faraday’s law, the conversion of mechanical energy into electrical energy is in accordance with the law of conservation of energy. It is also clearly known that in pure resistance, the emf is in phase with the current.

Question 3.
Assertion: Inductance coil are made of copper.
Reason: Induced current is more in wire having less resistance.
Answer:
(a) If both assertion and reason are true and the reason in the correct explanation of the assertion.
Solution:
Inductance coils made of copper will have very small ohmic resistance.

Question 4.
Assertion: An aircraft flies along the meridian, the potential at the ends of its wings will be the same.
Reason: Whenever there is a change in the magnetic flux, and emf is induced.
Answer:
(e) If assertion is false but reason is true.
Solution:
As the aircraft flies magnetic flux change through its wings due to the vertical component of the Earth’s magnetic field. Due to this, induced emf is produced across the wings of the aircraft. Therefore, the wings of the aircraft will not be at the same potential.

Question 5.
Assertion: In series LCR circuit resonance can take place.
Reason: Resonance takes place if inductance and capacitive reactances are equal and opposite.
Answer:
(a) If both assertion and reason are hue and the reason is the correct explanation of the assertion.
Solution:
At resonant frequency X_L = X_C
So, Impedance, Z = R (minimum)
Therefore, the current in the circuit is maximum.

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Short Answer Questions

Question 1.
Define magnetic flux (Φ_B).
The magnetic flux through an area A in a magnetic field is defined as the number of magnetic field lines passing through that area normally and is given by the equation,

Question 2.
Write down the drawbacks of Eddy currents.
Answer:
When eddy currents flow in the conductor, a large amount of energy is dissipated in the form of heat. The energy loss due to the flow of eddy current is inevitable but it can be reduced to a greater extent with suitable measures. The design of transformer core and electric motor armature is crucial in order to minimise the eddy current loss.

To reduce these losses, the core of the transformer is made up of thin laminas insulated from one another while for electric motor the winding is made up of a group of wires insulated from one another. The insulation used does not allow huge eddy currents to flow and hence losses are minimized.

Question 3.
Define the unit of self-inductance.
Answer:
The unit of self-inductance is henry. One henry is defined as the self-inductance of a coil in which a change in current of one ampere per second produces an opposing emf of one volt.

Question 4.
Define mutual inductance in terms of flux and current.
Answer:
The mutual inductance M₂₁ is defined as the flux linkage of the coil 2 when 1A current flows through coil 1.
M₂₁ = \(\frac{\mathrm{N}_{2} \mathrm{\phi}_{21}}{i_{1}}\)

Question 5.
Define mutual inductance in terms of emf and current.
Answer:
Mutual inductance M₂₁ is also defined as the opposing emf induced is the coil 2 when the rate of change of current through the coil 1 is 1 As^-1.
M₁₂ = \(\frac{-\varepsilon_{1}}{d i_{2} / d t}\)

Question 6.
List out the advantages of three phase alternator.
Answer:
Three-phase system has many advantages over single-phase system, which is as follows:
(i) For a given dimension of the generator, three-phase machine produces higher power output than a single-phase machine.
(ii) For the same capacity, three-phase alternator is smaller in size when compared to single phase alternator.
(iii) Three-phase transmission system is cheaper. A relatively thinner wire is sufficient for transmission of three-phase power.

Question 7.
Mentions the differences betw een a step up and step down transformer.
Answer:

Question 8.
Define efficiency of transformer.
Answer:
The efficiency p of a transformer is defined as the ratio of the useful output power to the input power. Thus

Transformers are highly efficient devices having their efficiency in the range of 96 – 99%. Various energy losses in a transformer will not allow them to be 100% efficient.

Question 9.
What is meant by sinusoidal alternating voltage?
Answer:
If the waveform of alternating voltage is a sine wave, then it is known as sinusoidal alternating voltage, which is given by the relation.
υ = V_m sin ωt

Question 10.
An inductor blocks AC but it allows DC. Why? and How?
Answer:
An inductor L is a closely wound helical coil. The steady DC current flowing through L produces uniform magnetic field around it and the magnetic flux linked remains constant. Therefore there is no self-induction and self-induced emf (back emf). Since inductor behaves like a resistor, DC flows through an inductor.

The AC flowing through L produces time-varying magnetic field which in turn induces self- induced emf (back emf). This back emf, according to Lenz’s law, opposes any change in the current. Since AC varies both in magnitude and direction, its flow is opposed in L. For an ideal inductor of zero ohmic resistance, the back emf is equal and opposite to the applied emf. Therefore L blocks AC.

Question 11.
A capacitor blocks DC but allows AC. Explain.
Answer:
Capacitive reactance, X_C = \(\frac { 1 }{ ωC }\) = \(\frac { 1 }{ 2πƒc }\)
where, ƒ = 0, X_C = ∞
where, ƒ is the frequency of the ac supply. In a dc circuit ƒ = 0. Hence the capacitive reactance has infinite value for dc and a finite value for ac. In other words, a capacitor serves as a block for dc and offers an easy path to ac.

Question 12.
Why dc ammeter cannot read ac?
Answer:
A dc ammeter cannot read ac because, the average value of ac is zero over a complete cycle.

Question 13.
Write down the applications of series RLC resonant circuit.
Answer:
RLC circuits have many applications like filter circuits, oscillators and voltage multipliers, etc. An important use of series RLC resonant circuits is in the tuning circuits of radio and TV systems. The signals from many broadcasting stations at different frequencies are available in the air. To receive the signal of a particular station, tuning is done.

Question 14.
What is meant by ‘Wattful current’?
Answer:
The component of current (I_rms cos Φ) which is in phase with the voltage is called active component. The power consumed by this current = V_rms I_rms cos Φ. So that it is also known as ‘Wattful’ current.

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Long Answer Questions

Question 1.
Derive an expression for Mutual Inductance between two long co-axial solenoids.
Answer:
Mutual inductance between two long co-axial solenoids:
Consider two long co-axial solenoids of same length 1. The length of these solenoids is large when compared to their radii so that the magnetic field produced inside the solenoids is uniform and the fringing effect at the ends may be ignored. Let A₁ and A₂ be the area of cross section of the solenoids with A₁ being greater than A₂. The turn density of these solenoids are n₁ and n₂ respectively.

Let i₁ be the current flowing through solenoid 1, then the magnetic field produced inside it is
B₁ = μ₀n₁i₁.
As the field lines of \(\vec {{ B }_{1}} \) are passing through the area bounded by solenoid 2, the magnetic flux is linked with each turn of solenoid 2 due to solenoid 1 and is given by

since θ = 0°
The flux linkage of solenoid 2 with total turns N₂ is
N₂Φ₂₁ = (n₂l)(μ₀ n₁ i₁)
since N₂ = n₂l
N₂Φ₂₁ = (μ₀ n₁ n₂ A₂l)i₁ ….. (1)
From equation of mutual induction
N₂Φ₂₁ = M₂₁ i₁ …… (2)
Comparing the equations (1) and (2),
M₂₁ = μ₀ n₁ n₂ A₂l ….. (3)
This gives the expression for mutual inductance M₂₁ of the solenoid 2 with respect to solenoid 1. Similarly, we can find mutual inductance M₂₁ of solenoid 1 with respect to solenoid 2 as given below.
The magnetic field produced by the solenoid 2 when carrying a current i₂ is
B₂ = μ₀ n₂ i₂
This magnetic field B₂ is uniform inside the solenoid 2 but outside the solenoid 2, it is almost zero. Therefore for solenoid 1, the area A₂ is the effective area over which the magnetic field B₂ is present; not area A₂ Then the magnetic flux Φ₁₂ linked with each turn of solenoid 1 due to solenoid 2 is

The flux linkage of solenoid 1 with total turns N₁ is
[Since N₁ = n₁l]
[Since N₁ Φ₁₂ = M₁₂i₂]
N₁ Φ₁₂ = (n₁l) (μ₀ n₂ i₂) A₂
N₁ Φ₁₂ = (μ₀ n₁ n₂ A₂l) i₂
M₁₂i₂ = (μ₀ n₁ n₂ A₂l) i₂
Therefore, we get
∴ M₁₂ = μ₀ n₁ n₂ A₂l ……. (4)
From equation (3) and (4), we can write
M₁₂ = M₂₁ = M ……. (5)
In general, the mutual inductance between two long co-axial solenoids is given by
M= μ₀ n₁ n₂ A₂l ……. (6)
If a dielectric medium of relative permeability’ pr is present inside the solenoids, then
M = μn₁ n₂ A₂l
or M = μ₀ μ_r n₁ n₂ A₂l

Question 2.
How will you define the unit of mutual-inductance?
Answer:
Unit of mutual inductance:
The unit of mutual inductance is also henry (H).
If i_A= 1 A and N₂ Φ₂₁ = 1 Wb turns, then M₂₁ = 1 H.
Therefore, the mutual inductance between two coils is said to be one henry if a current of 1A in coil 1 produces unit flux linkage in coil 2.
If \(\frac {{ di }_{1}}{ 2 }\) = 1 As^-1 and ε₂ = -1V, theen M₂₁ = 1H.
Therefore, the mutual inductance between two coils is one henry if a current changing at the rate of lAs-1 in coil 1 induces an opposing emf of IV in coil 2.

Question 3.
Find out the phase relationship between voltage and current in a pure resister circuit.
Answer:
AC circuit containing pure resistor:
Consider a circuit containing a pure resistor of resistance R connected across an alternating voltage source. The instantaneous value of the alternating voltage is given by

υ = V_m sin ωt ….. (1)
An alternating current i flowing in the circuit due to this voltage, develops a potential drop across R and is given by
V_R = iR ……. (2)
Kirchoff’s loop rule states that the algebraic sum of potential differences in a closed circuit is zero. For this resistive circuit,
υ – V_R = 0
From equation (1) and (2),
V_m sin ωt = iR
⇒ i = \(\frac {{ V }_{m}}{ R }\) sin ωt
i = I_m sin ωt …… (3)

where V\(\frac {{ V }_{m}}{ R }\) = I_m the peak value of alternating current in the circuit. From equations (1) and and (3), it is clear that the applied voltage and the current are in phase with each other in a resistive circuit. It means that they reach their maxima and minima simultaneously. This is indicated in the phasor diagram. The wave diagram also depicts that current is in phase with the applied voltage.

Question 4.
Find out the phase relationship between voltage and current in a pure capacitor circuit.
Answer:
AC circuit containing only a capacitor:
Consider a circuit containing a capacitor of capacitance C connected across an alternating voltage source.
The alternating voltage is given by

V_m sin ωt …… (1)
Let q be the instantaneous charge on the capacitor. The emf across the capacitor at that instant is \(\frac { q }{ C }\). According to Kirchoff’s loop rule,.
υ = \(\frac { q }{ C }\) = 0
⇒ CV_m sin ωt
By the definition of current,
i = \(\frac { dq }{ dt }\) = \(\frac { d }{ dt }\) CV_m \(\frac { d }{ dt }\) (sin ωt)
= CV_m sin ωt
or i = \(\frac{\frac{\mathrm{V}_{m}}{1}}{\frac{\mathrm{l}}{\mathrm{C} \omega}}\) sin \(\left(\omega t+\frac{\pi}{2}\right)\)
i = I_m sin \(\left(\omega t+\frac{\pi}{2}\right)\) ….. (2)
where \(\frac{\frac{\mathrm{V}_{m}}{1}}{\frac{\mathrm{l}}{\mathrm{C} \omega}}\) = I_m, the peak value of the alternating current. From equation (1) and (2), it is clear that current leads the applied voltage by π/2 in a capacitive circuit. The wave diagram for a capacitive circuit also shows that the current leads the applied voltage by 90°.

Question 5.
What are LC oscillation? and explain the generation of LC oscillation.
Answer:
Whenever energy is given to a circuit containing a pure inductor of inductance L and a capacitor of capacitance C, the energy oscillates back and forth between the magnetic field of the inductor and the electric field of the capacitor. Thus the electrical oscillations of definite frequency are generated. These oscillations are called LC oscillations.

Generation of LC oscillations:
Let us assume that the capacitor is fully charged with maximum charge Q_m at the initial stage. So that the energy stored in the capacitor is maximum and is given by U_Em = \(\frac{\mathrm{Q}_{\mathrm{m}}^{2}}{2 \mathrm{C}}\) As there is no current in the inductor, the energy stored in it is zero i.e., UB = 0. Therefore, the total energy is wholly electrical.

The capacitor now begins to discharge through the inductor that establishes current i in clockwise direction. This current produces a magnetic field around the inductor and the energy stored in the inductor is given by U_B = \(\frac {{ Li }_{ 2 }}{ 2 }\). As the charge in the capacitor decreases, the energy stored in it also decreases and is given by U_E = \(\frac {{ q }_{ 2 }}{ 2C }\). Thus there is a transfer of some part of energy from the capacitor to the inductor. At that instant, the total energy is the sum of electrical and magnetic energies.

When the charges in the capacitor are exhausted, its energy becomes zero i.e., UE = 0. The energy is fully transferred to the magnetic field of the inductor and its energy is maximum. This maximum energy is given by UB = \(\frac{\mathrm{LI}_{\mathrm{m}}^{2}}{2}\) where I_m is the maximum current flowing in the circuit. The total energy is wholly magnetic.

Even though the charge in the capacitor is zero, the current will continue to flow in the same direction because the inductor will not allow it to stop immediately. The current is made to flow with decreasing magnitude by the collapsing magnetic field of the inductor. As a result of this, the capacitor begins to charge in the opposite direction. A part of the energy is transferred from the inductor back to the capacitor. The total energy is the sum of the electrical and magnetic energies.

When the current in the circuit reduces to zero, the capacitor becomes frilly charged in the opposite direction. The energy stored in the capacitor becomes maximum. Since the current is zero, the energy stored in the inductor is zero. The total energy is wholly electrical. The state of the circuit is similar to the initial state but the difference is that the capacitor is charged in opposite direction. The capacitor then starts to discharge through the inductor with anti-clockwise current. The total energy is the sum of the electrical and magnetic energies.

As already explained, the processes are repeated in opposite direction. Finally, the circuit returns to the initial state. Thus, when the circuit goes through these stages, an alternating current flows in the circuit. As this process is repeated again and again, the electrical oscillations of definite frequency are generated. These are known as LC oscillations. In the ideal LC circuit, there is no loss of energy. Therefore, the oscillations will continue indefinitely. Such oscillations are called undamped oscillations.

Samacheer Kalvi 12th Physics Electromagnetic Induction and Alternating Current Numerical Problems

Question 1.
A coil has 2000 turns and area 70 cm². The magnetic field perpendicular to the plane of the coil is 0.3 Wb/m². The coil takes 0.1 s to rotate through 180°. Then what is the value of induced emf?
Solution:
Magnitude of change in flux,
|∆Φ | = |NBA (cos 180° – cos 0°
= |NBA(-1 – 1)| = |-2 NBA| = |2 NBA|
Where,
N = 2000
B = 0.3 Wb/m²
A = 70 x 10^-4 m²
t = 0.1 sec
Induced emf, ε = \(\frac { \left| \Delta \phi \right| }{ \Delta t } \) = \(\frac { 2NBA }{ ∆t }\) = \(\frac {{ 2 × 2000 × 0.3 × 70 × 10 }^{-4}}{ 0.1 }\)
ε = 84 V

Question 2.
A rectangular loop of sides 8 cm and 2 cm is lying in a uniform magnetic field of magnitude 0.5 T with its plane normal to the field. The field is now gradually reduced at the rate of 0.02 T/s. If the resistance of the loop is 1.6 Ω, then find the power dissipated by the loop as heat.
Solution:
Induced emf, |ε| = \(\frac { dΦ }{ dt } \) = A \(\frac { dB }{ dt } \) = 8 × 2 × 10^-4 × 0.02
ε = 3.2 × 10^-5 V
Induced current, I = \(\frac { ε }{ R } \) = 2 × 10^-5 A
Power loss = I²R = 4 × 10^-10 × 1.6 = 6.4 × 10^-10 W

Question 3.
A current of 2 A flowing through a coil of 100 turns gives rise to a magnetic flux of 5 x 10^-5 Wb per turn. What is the magnetic energy associated with the coil?
Solution:
Self inductance of coil, L = \(\frac { NΦ }{ I } \) = \(\frac {{ 100 × 5 × 10 }^{-3}}{ 2 } \)
= 2.5 × 10^-3 H
Magnetic energy associated with inductance,
U = \(\frac { 1 }{ 2 }\) LI² = \(\frac { 1 }{ 2 }\) × 2.5 × 10^-3 × (2)²
= \(\frac { 1 }{ 2 }\) × 2.5 × 10^-3 × 4 = 5 × 10^-3 J

Question 4.
A transformer is used to light a 140 W, 24 V bulb from a 240 V AC mains. The current in the main cable is 0.7 A. Find the efficiency of the transformer.
Solution:

η = \(\frac { 140 }{ 240 × 0.7 }\) × 100 = 83.3%

Question 5.
In an ideal step up transformer the turns ratio is 1 : 10. A resistance of 200 ohm connected across the secondary is drawing a current of 0.5 A. What are the primary voltage and current?
Solution:

Primary current, I_p = 5 A
Promary voltage, E_p = 10 V

Question 6.
A capacitor of capacitance 2 μF is connected in a tank circuit oscillating with a frequency of 1 kHz. If the current flowing in the circuit is 2 mA, then find the voltage across the capacitor.
Solution:

Question 7.
An ideal inductor takes a current of 10 A when connected to a 125 V, 50 Hz AC supply. A pure resistor across the same source takes 12.5 A. If the two are connected in series across a 100 √2 V, 40 Hz supply, then calculate the current through the circuit.
Solution:

Question 8.
An LCR series circuit containing a resistance of 120 Ω. has angular resonance frequency 4 x 10⁵ rad s^-1. At resonance the voltages across resistance and inductance are 60 V and 40 V, respectively. Find the values of L and C.
Solution:

Question 9.
A coil of inductive reactance 31 Ω has a resistance of 8 Ω. It is placed in series with a capacitor of capacitance reactance 25 Ω. The combination is connected to an ac source of 110 volt. Find the power factor of the circuit.
Solution:

Power faactor = 0.8

Question 10.
The power factor of an RL circuit is \(\frac { 1 }{ √2 }\). If the frequency of AC is doubled, what will be the power factor?
Solution:

Question 11.
The instantaneous value of alternating current and voltage are given as i = \(\frac { 1 }{ √2 }\) sin (100 πt) A and e = \(\frac { 1 }{ √2 }\) sin(100 πt + \(\frac { π }{ 3 }\)) volt. Find the average power in watts consumed in the circuit.
Solution:

Common Errors and its Rectifications:

Common Errors:

1. Students sometimes may confuse the peak current and instantaneous value of current and emf.
2. They may confuse in the area of R, L and C with AC. The relation between current and induced emf.

Rectifications:

1. Instantaneous current, i = I₀ sin tot Peak current, I₀ = √2 I_rms Instantaneous emf, e = E₀ sin cor Peak emf, E₀ = √2 E_rms
2. In Inductor: current is \(\frac { π }{ 2 }\) rad less than that of emf.
   In Resistor: current and emf are same phase.
   In Capacitor: current is \(\frac { π }{ 2 }\) rad greater than that of emf.

We believe that the shared knowledge about Tamilnadu State Board Solutions for 12th Physics Chapter 4 Electromagnetic Induction and Alternating Current Questions and Answers learning resource will definitely guide you at the time of preparation. For more details visit Tamilnadu State Board Solutions for 12th Physics Solutions and for more information ask us in the below comments and we’ll revert back to you asap.

Are you searching for the Samacheer Kalvi 12th Chemistry Chapter Wise Solutions PDF? Then, get your Samacheer Kalvi 12th Chapter Wise Solutions PDF for free on our website. Students can Download Chemistry Chapter 7 Chemical Kinetics Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Chemistry Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 12th Chemistry Solutions Chapter 7 Chemical Kinetics

All concepts are explained in an easy way on our website. So, students can easily learn and practice Tamilnadu State Board 12th Chemistry Chapter 7 Chemical Kinetics Question and Answers. You can enjoy the digitized learning with the help of the Tamilnadu State Board Chemistry Online Material.

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements Text Book Evalution

I. Choose the correct answer.

12th Chemistry Chapter 7 Book Back Answers Question 1.
For a first order reaction A → B the rate constant is x min^-1. If the initial concentration of A is 0.01 M, the concentration of A after one hour is given by the expression.
(a) 0.01 e^-x
(b) 1 x 10^-2 (1 – e^-60x)
(c) (1 x 10^-2) e^-60x
(d) none of these
Answer:
(c) (1 x 10^-2) e^-60x
Answer:
Solutions:

In this case
k = x min^-1 and [A₀] = 0.01 M
= 1 x 10^-2 M
t = 1 hour = 60 min
[A] = 1 x 10^-2(e^-60x)

12th Chemistry 7th Lesson Book Back Answers Question 2.
A zero order reaction X → Product, with an initial concentration 0.02M has a half life of 10 min. If one starts with concentration 0.04M, then the half life is …………….
(a) 10 s
(b) 5 min
(c) 20 min
(d) cannot be predicted using the given information
Answer:
(c) 20 min
Solutions:

Given,
[A₀] = 0.02 M ; t_1/2 = 10 min
[A₀] = 0.04 M ; t_1/2 = ?
Substitute in (1)
10 min ? 0.02 M ……………………..(2)
t_1/2 ∝ 0.04 M ……………………..(3)
Dividing Eq.(3) by Eq. (2) we get,
\(\frac { { t }^{ 1/2 } }{ 10min }\) = \(\frac { 0.04M }{ 0.02M }\)
t_1/2 = 2 x 10 min = 20 min

Chemical Kinetics Book Back Answers Question 3.
Among the following graphs showing variation of rate constant with temperature (T) for a reaction, the one that exhibits Arrhenius behavior over the entire temperature range is ……………

Answer:

Solution:

In k = In A – \(\left( \frac { { E }_{ a } }{ R } \right)\) \((\frac { 1 }{ T })\)
this equation is in the form of a straight
line equation y = c + m x
a plot of ink vs \((\frac { 1 }{ T })\) is a straight line with negative slope.

Samacheer Kalvi Guru 12th Chemistry Question 4.
For a first order react ion A → product with initial concentration x mol L^-1, has a half life period of 2.5 hours. For the same reaction with initial concentration mol L^-1 the half life is
(a) (2.5 x 2) hours
(b) \((\frac { 2.5 }{ 2 })\) hours
(c) 2.5 hours
(d) Without knowing the rate constant, t_1/2 cannot be determined from the given data
Answer:
(d) Without knowing the rate constant, t_1/2 cannot be determined from the given data.
Solutions:
For a first order reaction
t_1/2 = \(\frac { 0.693 }{ k }\) t_1/2 does not depend on the initial concentration and it remains constant (whatever may be the initial concentration)
t_1/2 = 2.5 hrs .

Samacheer 12 Chemistry Solutions Question 5.
For the reaction, 2NH₃ → N₂ + 3H₂, if

then the relation between
k₁, k₂ and k₃ is
(a) k₁ = k₂ = k₃
(b) k₁ = 3 k₂ = 2 k₃
(c) 1.5k₁ = 3 k₂ = k₃
(d) 2k₁ = k₂ = 3 k₃
Answer:
(c) 1.5k₁ = 3 k₂ = k₃
Solution:

Samacheer Kalvi 12th Chemistry Question 6.
The decomposition of phosphine (PH₃) on tungsten at low pressure is a first order reaction. It is because the …………….
(a) rate is proportional to the surface coverage
(b) rate is inversely proportional to the surface coverage
(c) rate is independent of the surface coverage
(d) rate of decomposition is slow
Answer:
(c) rate is independent of the surface coverage
Solution:
Given:
At low pressure the reaction follows first order, therefore Rate ∝ [reactant]¹ Rate ∝ (surface area) At high pressure due to the complete coverage of surface area, the reaction follows zero order. Rate ∝ [reactant]°. Therefore the rate is independent of surface area.

Chemical Kinetics Solutions Question 7.
For a reaction Rate = k [acetone]^3/2 then unit of rate constant and rate of reaction respectively is …………..
(a) (mol L^-1 s^-1), (mol^-1/2 L^1/2 s^-1)
(b) (mol^-1/2 L^1/2 s^-1), (mol L^-1 s^-1)
(c) (mol^1/2 L^1/2 s^-1), (mol L^-1 s^-1)
(d) (mol L s^-1), (mol^1/2 L^1/2 s)
Answer:
(b) (mol^1/2 L^1/2 s^-1), (mol L^-1 s^-1)
Solution:
Rate = k [A]ⁿ
Rate = \(\frac { -d[A] }{ dt } \)
unit of rate = \(\frac { mol{ L }^{ -1 } }{ s }\) = mol L^-1 s^-1
unit of rate constant = \(\frac { (mol{ L }^{ -1 }{ S }^{ -1 }) }{ { (mol{ L }^{ -1 }) }^{ n } }\)
= mol^1-n L^n-1 S^-1
in this case, rate k [Acetone]^3/2
n = 3/2
mol^1-(3/2) L^(3/2)-1 s^-1
mol^-(1/2) L^(1/2) s^-1

Samacheer Kalvi Class 12 Chemistry Solutions Question 8.
The addition of a catalyst during a chemical reaction alters which of the following quantities?
(a) Enthalpy
(b) Activation energy
(c) Entropy
(d) Internal energy
Answer:
(b) Activation energy
Solution:
A catalyst provides a new path to the reaction with low activation energy. i.e., it lowers the activation energy.

Chemical Kinetics In Tamil Question 9.
Consider the following statements:
(i) increase in concentration of the reactant increases the rate of a zero order reaction.
(ii) rate constant k is equal to collision frequency A if E_a = o
(iii) rate constant k is equal to collision frequency A if E_a = o
(iv) a plot of ln (k) vs T is a straight line.
(v) a plot of In (k) vs \((\frac { 1 }{ T })\) is a straight line with a positive slope.

Correct statements are
(a) (ii) only
(b) (ii) and (iv)
(c) (ii) and (v)
(d) (i), (ii) and (v)
Answer:
(a) (ii) only
Solutions:
In zero order reactions, increase in the concentration of reactant does not alter the rate, So statement (i) is wrong.

if E_a = O (so, statement (ii) is correct, and statement (iii) is wrong)
k = A e°
k = A
in k = A – \(\left( \frac { { E }_{ a } }{ R } \right)\) \(\frac { 1 }{ T }\)
this equation is in the form of a straight line equation yc + m x. a plot of Ink vs \(\frac { 1 }{ T }\) is a straight line with negative slope so statements (iv) and (v) are wrong.

Samacheer Kalvi Guru Class 12 Chemistry Question 10.
In a reversible reaction, the enthalpy change and the activation energy in the forward direction are respectively – x kJ mol^-1 and y kJ mol^-1. Therefore, the energy of activation in the backward direction is ………..
(a) (v – x)kJ mol^-1
(b) (x + y) J mol^-1
(c) (x – y) kJ mol^-1
(d) (x + y) x 10³ J mol^-1
Answer:
(d) (x + y) x 10³ J mol^-1
Solution:

12th Chemistry Samacheer Kalvi Question 11.
What is the activation energy for a reaction if its rate doubles when the temperature is raised from 200K to 400K? (R 8.314 JK^-1 mol^-1)
(a) 234.65 kJ mol^-1 K^-1
(b) 434.65 kJ mol^-1 K^-1
(c) 434.65 J mol^-1 K^-1
(d) 334.65 J mol^-1 K^-1
Answer:
(c)434.65 J mol^-1 K^-1
Solutions:

Question 12.

This reaction follows first order kinetics. The rate constant at particular temperature is 2.303 x 10² hourd. The initial concentration of cyclopropane is 0.25 M. What will be the concentration of cyclopropane after 1806 minutes? (Log 2 = 0.30 10)
(a) 0.125 M
(b) 0.215 M
(c) 0.25 x 2.303 M
(d) 0.05 M
Answer:
(b) 0.2 15 M
Solution:

Question 13.
For a first order reaction, the rate constant is 6.909 min^-1.The time taken for 75% conversion in minutes is …………
(a) \((\frac { 3 }{ 2 })\) log 2
(b) \((\frac { 3 }{ 2 })\) log 2
(c) \((\frac { 3 }{ 2 })\) log \((\frac { 3 }{ 4 })\)
(d) \((\frac { 2 }{ 3 })\) log \((\frac { 4 }{ 3 })\)
Answer:
(b) \((\frac { 3 }{ 2 })\) log 2
Solution:
k = \((\frac { 2.303 }{ t })\) log \(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
[A₀] = 100
[A] = 25
[A₀]= 100; [A]=25
6.909 = \((\frac { 2.303 }{ t })\) log \((\frac { 100 }{ 25 })\)
t = \((\frac { 2.303 }{ 6.909 })\) log (4) ⇒ t = \((\frac { 1 }{ 3 })\) log 2²
t = \((\frac { 2 }{ 3 })\) log 2

Question 14.
In a first order reaction x → y; if k is the rate constant and the initial concentration of the reactant x is 0.1 M, then, the half life is ……..
(a) \((\frac { log2 }{ k })\)
(b) \((\frac { 0.693 }{ (0.1)k })\)
(c) \((\frac { In2 }{ k })\)
(d) none of these
Answer:
(c) \((\frac { In2 }{ k })\)
Solution:
k = \((\frac { 1 }{ t })\) In \(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
[A₀] = 0.1
[A] = 0.05
k = \(\left( \frac { 1 }{ { t }_{ 1/2 } } \right)\) In \((\frac { 0.1 }{ 0.05 })\)
k = \(\left( \frac { 1 }{ { t }_{ 1/2 } } \right)\) In (2) ⇒ t_1/2 = \((\frac { In(2) }{ k })\)

Question 15.
Predict the rate law of the following reaction based on the data given below:
2A + B → C + 3D

(a) rate = k [A]² [B]
(b) rate = k [A][B]²
(c) rate = k [A][B]
(d) rate = k [A]^1/2 [B]^3/2
Answer:
(b) rate = k [A][B]²
Solution:
rate₁ = k [0.1]ⁿ [0.1]^m ……………(1)
rate₂ = k [0.2]ⁿ [0.1]^m …………(2)
Dividing Eq.(2) by Eq.(1)

\(\frac { 2x }{ x }\) = 2ⁿ
∴ n = 1
rate₃ = k [0.1]ⁿ [0.2]^m …………..(3)
rate₄ = k [0.2]ⁿ [0.2]^m …………..(4)
Dividing Eq.(4) by Eq.(2)

\(\frac { 8 }{ 2 } \) = 2^m
∴m = 1
∴ rate = k [A]¹ [B]²

Question 16.
Assertion: rate of reaction doubles when the concentration of the reactant is doubles if it is a first order reaction.
Reason: rate constant also doubles
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Both assertion and reason are false.
Answer:
(c) Assertion is true but reason is false.
Solution:
For a first reaction, when the concentration of reactant is doubled, then the rate of reaction also doubled. Rate constant is independent of concentration and is a constant at a constant temperature, i.e., it depends on the temperature and hence, it will not be doubled and when the concentration of the reactant is doubled.

Question 17.
The rate constant of a reaction is 5.8 x 102 s¹. The order of the reaction is ………….
(a) First order
(b) zero order
(c) Second order
(a) Third order
Answer:
(a) First order
Solution:
The unit of rate constant is s^-1 and it indicates that the reaction is first order.

Question 18.
For the reaction N₂ O_5(g) → 2NO_2(g) +\(\frac { 1 }{ 2 }\) – O_2(g) the value of rate of disappearance of N₂O₅ is given as 6.5 x 10^-2 mol L^-1s^-1 The rate of formation of NO₂ and O₂ is given respectively as …………….
(a) (3.25 x 10^-2 mol L^-1s^-1) and (1.3 x 10^-2 mol L^-1s^-1)
(b) (1.3 x 10^-2 mol L^-1s^-1) and (3.25 x 10² mol L^-1s^-1)
(c) (1.3 x 10^-1 mol L^-1s^-1) and (3.25 x 10^-2 mol L^-1s^-1)
(d) None of these
Answer:
(c) (1.3 x 10^-1 mol L^-1s^-1) and (3.25 x 10^-2 mol L^-1s^-1)

Question 19.
During the decomposition of H₂O₂ to give dioxygen, 48g O₂ is formed per minute at certain point of time. The rate of formation of water at this point is …………….
(a) 0.75 mol min^-1
(b) 1.5 mol min^-1
(c) 2.25 mol min^-1
(d) 3.0 mol min^-1
Answer:
(d) 3.0 mol min^-1
Solution:
H₂O₂ → H₂O + \(\frac { 1 }{ 2 }\)O₂
Rate =
No. of moles of oxygen = \((\frac { 48 }{ 32 })\) = 1.5 mol
Rate of formation of oxygen = 2 x 1.5
= 3 mol min^-1

Question 20.
If the initial concentration of the reactant is doubled, the time for half reaction is also doubled. Then the order of the reaction is …………
(a) Zero
(b) one
(c) Fraction
(d) none
Answer:
(a) Zero
Solution:
For a first order reaction t^1/2 is independent of initial concentration .i.e., n \(\neq\) 1 for such cases

Question 21.
In a homogeneous reaction A ? B + C + D, the initial pressure was P₀ and after time t it was P. Expression for rate constant in terms of P₀, P and t will be ……….

Answer:

Solution:

Question 22.
If 75% of a first order reaction was completed in 60 minutes, 50% of the same reaction under the same conditions would be completed in ………
(a) 20 minutes
(b) 30 minutes
(c) 35 minutes
(d) 75 minutes
Answer:
(b) 30 minutes
Solution:

Question 23.
The half life period of a radioactive element is 140 days. After 560 days, 1 g of element will be reduced to
(a) \(\frac { 1 }{ 2 }\) g
(b) \(\frac { 1 }{ 4 }\) g
(c) \(\frac { 1 }{ 8 }\) g
(d) \(\frac { 1 }{ 16 }\) g
Answer:
(d) \(\frac { 1 }{ 16 }\) g
Solution:
in 140 days ⇒ initial concentration reduced to \(\frac { 1 }{ 2 }\) g
in 280 days ⇒ initial concentration reduced to \(\frac { 1 }{ 4 }\) g
in 420 days ⇒ initial concentration reduced to \(\frac { 1 }{ 8 }\) g
in 560 days ⇒ initial concentration reduced to \(\frac { 1 }{ 8 }\) g

Question 24.
The correct difference between first and second order reactions is that …………
(a) A first order reaction can be catalysed a second order reaction cannot be catalysed.
(b) The half life of a first order reaction does not depend on [A₀] the half life of a second order reaction does depend on [A₀].
(c) The rate of a first order reaction does not depend on reactant concentrations; the rate of a second order reaction does depend on reactant concentrations.
(d) The rate of a first order reaction does depend on reactant concentrations; the rate of a second order reaction does not depend on reactant concentrations,
Answer:
(b) The half life of a first order reaction does not depend on [A₀]; the half life of a second order reaction does depend on [A₀].
Solution:
For a first order reaction
t_1/2 = \(\frac { 0.6932 }{ k }\)
For a second order reaction

Question 25.
After 2 hours, a radioactive substance becomes \((\frac { 1 }{ 16 })\)^th of original amount. Then the half life (in mm) is ………………
(a) 60 minutes
(b) 120 minutes
(c) 30 minutes
(d) 15 minutes
Answer:
(c) 30 minutes
Solution:

II . Answer the following questions:

Question 1.
Define average rate and instantaneous rate.
Answer:
1. Average rate:
The average rate of a reaction is defined as the rate of change of concentration of a reactant (or of a product) over a specified measurable period of time.

2. insantaneous rate:
Instantaneous rate of reaction gives the tendency of the reaction at a particular point of time during its course (or) The time derivative of the concentration of a reactant (or product) converted to a positive number is called the instantaneous rate of reaction.

Question 2.
Define rate law and rate constant.
1. Rate law:
The expression in which reaction rate is given in terms of molar concentration of the reactants with each term raised to some power, which may or may not be same as the Stoichiometric coefficient of the reacting species in a balanced chemical equation.
x A + y B → products
Rate = k [A]^m [B]^m
k = Rate constant

2. Rate constant:
For a reaction involving the reactants A and B, Reaction rate = k [A]^m [B]^m The constant k is called rate constant of the reaction. If [A] = 1 M and [B] = 1 M; Reaction rate = k Thus, the rate constant (k) of a reaction is equal to the rate of reaction when the concentration of each reactant is equal to 1 mol L^-1. The change in the concentration of reactant or product per unit time under the condition of unit concentration of all the reactant.

Question 3.
Derive integrated rate law for a zero order reaction A product. A reaction in which the rate is independent of the concentration of the reactant over a wide range of concentrations is called as zero order reactions. Such reactions are rare. Let us consider the following hypothetical zero order reaction.
A → Product
The rate law can be written
Rate = k [A]°
(∴[A]° = 1)
– d [A] k (l)
\(\frac { -d[A] }{ dt }\) = k(1)
-d[A] = k dt
Integrate the above equation between the limits of [A₀] at zero time and [A] at some later time ‘t’,

Question 4.
Define half life of a reaction. Show that for a first order reaction half life is independent of Initial concentration.
Answer:
Half life of a reaction is defined as the time required for the reactant concentration to reach one half of its initial value. For a first order reaction, the half life is a constant i.e., it does not depend on the initial concentration. The rate constant for a first order reaction is given by,

Question 5.
What is an elementary reaction? Give the differences between order and molecularity of a reaction.
Answer:
Elementary reaction – Each and every single step in a reaction mechanism is called an elementary reaction. Differences between order and molecularity:
Order of a reaction:

1. It is the sum of the powers of concentration terms involved in the experimentally determined rate law.
2. It can be zero (or) fractional (or) integer.
3. It is assigned for a overall reaction.

Molecularity of a reaction:

1. It is the total number of reactant species that are involved in an elementary step.
2. It is always a whole number, cannot be zero or a fractional number.
3. It is assigned for each elementary step of mechanism.

Question 6.
Explain the rate determining step with an example.
Answer:
1. Most of the chemical reactions occur by multistep reactions. In the sequence of steps it is found that one of the steps is considerably slower than the others. The overall rate of the reaction cannot be lower in value than the rate of the slowest step.

2. Thus in a multistep reaction the experimentally determined rate corresponds to the rate of the slowest step. The step which has the lowest rate value among the other steps of the reaction is called as the rate determining step (or) rate limiting step.

3. Consider the reaction,
2A + B → C + D
going by two steps as follows,

Here the overall rate of the reaction corresponds to the rate of the first step which is the slow step and thus the first step is called as the rate determining step of the reaction. In the above equation, the rate of the reaction depends upon the rate constant k ( only. The rate of second step dosn’t contribute experimentally determined overall rate of the reaction.
For example,
NO_2(g) + CO_2(g) → NO_(g) + CO_2(g)
Which occurs in two elementary steps:

• NO₂ + NO₂ → NO + NO₃ (Slow)
• NO₃+ CO → NO₂ + CO₂ (fast)

Because the first step is the lowest step, the overall reaction cannot proceed any faster than the rate of the first elementary step. The first elementary step in this example is therefore the rate determining step.

The rate equation for this reaction is equal to the rate is constant of step-1 multiplied by the reactants of that first step. If the rate constant of step-1 is denoted as k₁ then the rate of the first step in the reaction (and the total reaction) will be,
Rate = k, [NO₂] [NO₂]
= k₁ [NO₂]₂

Question 7.
Describe the graphical representation of first order reaction.
Answer:
Rate constant for first order reaction is,
kt = ln\(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
kt = In [A₀] – In [A]
In[A] = In [A₀] – kty = c + mx
If we follow the reaction by measuring the concentration of the reactants at regular time interval ‘t’, a plot of ln[A] against ‘t’ yields a straight line with a negative slope. From this, the rate constant is calculated.

Question 8.
Write the rate law for the following reactions.

1. A reaction that is 3/2 order in x and zero order in y.
2. A reaction that is second order in NO and first order in Br₂.

Answer:
1. \(\frac { 3 }{ 2 }\) x + y (excess) → products
– \(\frac { 3 }{ 2 }\) \(\frac { d[x] }{ dt }\) = k [x]^3/2

2. 2NO + Br₂ → products
– \(\frac { 1 }{ 2 }\) \(\frac { d[NO] }{ dt }\) = k [NO]² [Br₂]

Question 9.
Explain the effect of catalyst on reaction rate with an example.
Answer:

1. Significant changes in the reaction can be brought out by the addition of a substance called catalyst.
2. A catalyst is substance which alters the rate of a reaction without itself undergoing any permanent chemical change.
3. They may participate in the reaction, but again regenerated and the end of the reaction.
4. In the presence of a catalyst, the energy of activation is lowered and hence, greater number of molecules can cross the energy barrier and change over to products, thereby increasing the rate of the reaction.
5. For example, decomposition of potassium chlorate is enhanced by addition of MnO₂.

2KClO₃ \(\frac{\mathrm{MnO}_{4}}{\Delta} 2 \mathrm{KCl}\) + 3O₂ (MnO₂ – Catalyst)

Question 10.
The rate law for a reaction of A, B and C has been found to be rate = k[A]² [B][L]^3/2. How would the rate of reaction change when

1. Concentration of [L] is quadrupled
2. Concentration of both [A] and [B] are doubled
3. Concentration of [A] is halved
4. Concentration of [A] is reduced to(1/3) and concentration of [L] is quadrupled.

Solution:
Rate = k [A]² [B] [L]^3/2 ………….(1)
1. when [L] = [4L]
Rate = k [A]² [B] [4L]^3/2
Rate = 8 (k[A]² [B] [L]^3/2) …………………..(2)
Comparing (1) and (3) rate is increased by 8 times.

2. when [A] = [2A] and [B] = [2B]
Rate = k[2A]² [2B ] [L]^3/2
Rate = 8 (k[A]² [B] [L]^3/2 …………….(3)
Comparing (1) and (3); rate is increased by 8 times.

3. when [A] = \([\frac { A }{ 2 }]\)
Rate = k \([\frac { A }{ 2 }]\)² [L]\(\frac { 3 }{ 2 }\)
Rate = \(\frac { 1 }{ 4 }\) (k[A]² [B] [L]^3/2) ……………..(4)
Comparing (1) and ( 4); rate is reduced to \(\frac { 1 }{ 4 }\) times.

4. when [A] = \([\frac { A }{ 3 }]\) and [L] = [4L]
Rate k\(\frac { A }{ 3 }\)² [B] [4L]^3/2
Rate = \([\frac { 8 }{ 9 }]\) (k[A]² [B] [L]^3/2) ……………….(5)
Comparing (1) and (5); rate is reduced to 8/9 times.

Question 11.
The rate of formation of a dimer in a second order reaction is 7.5 x 10^-3 mol L^-1s^-1 at 0.05 mol L^-1 monomer concentration. Calculate the rate constant.
Solution:
Let us consider the dimensation of a monomer M
2M → (M)₂
Rate = k [M]ⁿ
Given that n =2 and [M] = 0.05 mol L^-1
Rate = 7.5 x 10^-3 mol L^-1s^-1
Rate 7.5 x 10³ mol L^-1 s^-1
k = \(\frac { Rate }{ { \left[ M \right] }^{ n } }\)
k= =\(\frac { 7.5\times { 10 }^{ -3 } }{ { \left( 0.05 \right) }^{ 2 } }\) = 3 mol^-1 Ls^-1

Question 12.
For a reaction x +y + z → products, the rate law is given by rate = k [x]^3/2 [y]^1/2 what is the overall order of the reaction and what is the order of the reaction with respect to z.
Solution:
Rate = k [x]^3/2 [y]^1/2
overall order = \(\left( \frac { 3 }{ 2 } +\frac { 1 }{ 2 } \right)\) = 2
i.e., second order reaction.
Since the rate expression does not contain the concentration of Z , the reaction is zero order with respect to Z.

Question 13.
Explain briefly the collision theory of bimolecular reactions.
Answer:
Collision theory is based on the kinetic theory of gases. According to this theory, chemical reactions occur as a result of collisions between the reacting molecules. Let us understand this theory by considering the following reaction.
A_2(g) + B_2(g) → 2AB_(g)

If we consider that, the reaction between A₂ and B₂ molecules proceeds through collisions between them, then the rate would be proportional to the number of collisions per second. Rate ix Number of molecules colliding per litre per second (or) Rate ∝ Collision rate. The number of collisions is directly proportional to the concentration of both A₂ and B₂.
Collision rate ∝ [A₂] [B₂] …………………(1)
Collision rate = Z [A₂] [B₂] ……………………..(2)
Where, Z is a constant.

The collision rate ¡n gases can be calculated from kinetic theory of gases. For a gas at room temperature (298K) and 1 atm pressure, each molecule undergoes approximately 10⁹ collisions per second, i.e., I collision in 10⁹ second. Thus, if every collision resulted in reaction, the reaction would be complete in 10⁹ second.

In actual practice this does not happen. It implies that all collisions are not effective to lead to the reaction. In order to react, the colliding molecules must possess a minimum energy called activation energy. The molecules that collide with less energy than activation energy will remain intact and no reaction occurs.

Fraction of effective collisions (f) is given by the following expression, \({ e }^{ \frac { { -E }_{ a } }{ RT } }\)
Fraction of collisions is further reduced due to orientation factor i.e., even if the reactant collide with sufficient energy, they will not react unless the orientation of the reactant molecules is suitable for the formation of the transition state. The fraction of effective collisions (f) having proper orientation is given by the steric factor P.

Rate = P x f x collision rate
Rate= P x \({ e }^{ \frac { { -E }_{ a } }{ RT } }\) x Z [A₂] [B₂] ……..(1)
As per the rate law, Rate = k [A₂] [B₂] ………………….(2)
Where k is the rate constant
On comparing equation (1) and (2), the rate constant k is,
k = p Z \({ e }^{ \frac { { -E }_{ a } }{ RT } }\)

Question 14.
Write Arrhenius equation and explains the terms involved.
Answer:
Arrhenius equation:
k = A\({ e }^{ \frac { { -E }_{ a } }{ RT } }\)
A = Arrhenius factor (frequency factor)
R = Gas constant
k = Rate constant
E_a = Activation energy
T = Absolute temperature (in K)

Question.15.
The decomposition of Cl₂O₇ at 500K in the gas phase to Cl₂ and O₂ is a first order reaction. After 1 minute at 500K, the pressure of Cl₂O₇ falls from 0.08 to 0.04 atm. Calculate the rate constant in s^-1.
Answer:
Solution:
k = \(\frac { 2.303 }{ t }\) log \(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)
k = \(\frac { 2.303 }{ 1 min }\) log \(\frac { [0.08] }{ [0.04] }\)
k = 2.303 log 2
k = 2.303 x 0.3010
k = 0.693 2 min^-1
k = \((\frac { 0.6932 }{ 60 })\) s-1
k = 1.153 x 10^-2 s^-1

Question 16.
Give the examples for a zero order reaction.
Answer:
Examples for a zero order reaction:
1. Photochemical reaction between H₂ and Cl
H_2(g) + Cl_2(g) \(\underrightarrow { h\nu }\) 2HCI_(g)

2. Decomposition of N₂O on hot platinum surface
N₂ O_(g) \(\rightleftharpoons\) N_2(g) + \(\frac { 1 }{ 2 }\) O_2(g)

3. lodination of acetone in acid medium is zero order with respect to iodine.
CH₃COCH₃ + I₂ \(\underrightarrow { { H }^{ + } } \) ICH₂COCH₃ + HI
Rate k [CH₃COCI₃] [H⁺]

Question 17.
Explain pseudo first order reaction with an example.
Answer:
A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, such reaction is called pseudo first order reaction. Let us consider the acid hydrolysis of an ester,
CH₃COOCH_3(aq) +H₂ O₍₁₎ \(\underrightarrow { { H }^{ + } } \) CH₃COOH_(aq) + CH₃OH_(aq)
Rate = k [CH₃COOCH₃] [H₂O]

If the reaction is carried out with the large excess of water, there is no significant change in the concentration of water during hydrolysis. i.e., concentration of water remains almost a constant. Now we can define k [H₂O] = k
∴ The above rate equation becomes
Rate k [CHCOOCH] Thus it follows first order kinetics.

Question 18.
Identify the order for the following reactions

1. Rusting of Iron
2. Radioactive disintegration of ₉₂U²³
3. 2 A+ B → products; rate = k [A]^1/2 [B]²

Answer:
1.
Theoritically order value may be more than one but practically one.

2. All radioactive disintegrations are first order reactions

3. 2A + 3B → products:
rate = k[A]^1/2 [B]²
Order = \(\frac { 1 }{ 2 }\) + 2 = \(\frac { 5 }{ 2 }\) = 2.5

Question 19.
A gas phase reaction has energy of activation 200 kJ mol^-1. If the frequency factor of the reaction is 1.6 x 10¹³ s^-1. Calculate the rate constant at 600 K. (e^-40.09 = 3.8 x I0^-18 )
Solution:

Question 20.
For the reaction 2x +y → L find the rate law from the following data.

Answer:
Rate = k [x]ⁿ [y]^m
0.15 = k [0.2]ⁿ [0.02]^m ……………..(1)
0.30 = k [0.4]ⁿ [0.02]^m ……………… (2)
1.20 = k [0.4]ⁿ [0.08]^m ……………… (3)
Dividing Eq(3) by Eq (2) we get

Question 21.
How do concentrations of the reactant influence the rate of reaction?
Answer:
The rate of a reaction increases with the increase in the concentration of the reactants. The effect of concentration is explained on the basis of collision theory of reaction rates.

According to this theory, the rate of a reaction depends upon the number of collisions between the reacting molecules. Higher the concentration, greater is the possibility for collision and hence the rate.

Question 22.
How do nature of the reactant influence rate of reaction?
Answer:
Nature and state of the reactant:
We know that a chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product.

The net energy involved in this process is dependent on the nature of the reactant and hence the rates arc different for different reactants. Let us compare the following two reactions that we carried out in volumetric analysis.

1. Redox reaction between ferrous ammonium sulphate (FAS) and KMnO₄
2. Redox reaction between oxalic acid and KMnO₄

The oxidation of oxalate ion by KMnO₄ is relatively slow compared to the reaction between KMnO₄ and Fe . In fact heating is required for the reaction between KMnO₄ and Oxalate ion and is carried out at around 60°C. The physical state of the reactant also plays an important role to influence the rate of reactions. Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants.

For example, reaction of sodium metal with iodine vapours is faster than the reaction between solid sodium and solid iodine. Let us consider another example that we carried out in inorganic qualitative analysis of lead salts.

If we mix the aqueous solution of colorless potassium iodide with the colorless solution of lead nitrate, precipitation of yellow lead iodide take place instantancously, whereas if we mix the solid lead nitrate with solid potassium iodide, yellow coloration will appear slowly.

Question 23.
The rate constant for a first order reaction is 1.54 x 10 s^-1. Calculate its half life time.
Answer:
We know that, t, 0.693 k
t_1/2 = 0.693/1.54 x 1o^-3 = 450 s

Question 24.
The half life of the homogeneous gaseous reaction SO₂CI₂ → SO₂ + Cl₂ which obeys first order kinetics Is 8.0 minutes. How long will it take for the concentration of SO₂Cl₂ to be reduced to 1% of the initial value?
Answer:
We know that, k = 0.693/ t_1/2
k = 0.693/8.0 minutes = 0.087 minutes ^-1
For a first order reaction,
k = \(\frac { 2.303 }{ k }\) log \(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
t = \(\frac { 2.303 }{ 0.087{ min }^{ -1 } }\) log\(\frac { 100 }{ 1 }\)
t = 52.93 mm

Question 25.
The time for half change in a first order decomposition of a substance A is 60 seconds. Calculate the rate constant. How much of A will be left after 180 seconds?
Answer:
1. Order of a reaction = 1
t_1/2 = 60
seconds, k = ?
k = \(\frac { 2.303 }{ 60 }\)
We know that, k = \(\frac { 2.303 }{ { t }_{ 1/2 } }\)
k = \(\frac { 2.303 }{ 60 }\) = 0.01155 s^-1

2. [A₀] = 100%
t = 180 s
k = 0.01155 seconds^-1
[A] = ?
For the first order reaction k = \(\frac { 2.303 }{ 60 }\) log \(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\)
0.9207 = log 100 – log [A]
log [A] = log 100 – 0.9207
log [A] = 2 – 0.9207
log[A] = 1.0973
[A] = antilog of (1.0973)
[A] = l2.5%

Question 26.
A zero order reaction is 20% complete in 20 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?
Answer:
1. A = 100%, x = 20%, Therefore, a – x =100 – 20 = 80
For the zero order reaction k= \((\frac { x }{ t })\) ⇒
k = \((\frac { 20 }{ 20 })\) = 1
Rate constant for a reaction = 1

2. To calculate the time for 80% of completion
k = 1, a = l00, x = 80%, t = ?
Therefore, t = \((\frac { x }{ k })\) = \((\frac { 80 }{ 1 })\) = 80 min

Question 27.
The activation energy of a reaction is 225 k cal mol^-1 and the value of rate constant at 40°C is 1.8 x 10^-5 s^-1. Calculate the frequency factor, A. Here, we arc given that
Answer:
E_a = 22.5 kcal mol^-1 = 22500 cal mol^-1
T = 40°C = 40 + 273 = 313 K
k = 1.8 x 10^-5 sec^-1
Substituting the values in the equation
log A = log k + \(\left( \frac { { E }_{ a } }{ 2.303RT } \right)\)
log A = log (l .8 x 10^-5) + \(\left( \frac { 22500 }{ 2.303\times 1.987\times 313 } \right)\)
log A = log (l.8) – 5 + (15.7089)
log A = (10.9642)
A = antilog ( 10.9642)
A = 9.208 x 10¹⁰ collisions s^-1

Question 28.
Benzene diazonium chloride in aqueous solution decomposes according to the equation
C₆H₅N₂CI C₆H₅CI + N₂. Starting with an initial concentration of 10 g L^-1 volume of N₂. gas obtained at 50°C at different intervals of time was found to be as under:

Show that the above reaction follows the first order kinetics. What is the value of the rate constant ?
Solution:
For a first order reaction
k = \(\frac { 2.303 }{ t }\) log \(\frac { a }{ (a-x) }\)
k = \(\frac { 2.303 }{ t }\) log \(\frac { { V }_{ \infty } }{ { V }_{ \infty }-{ V }_{ t } }\)
In this case, V_∞ = 58.3 ml
The value of k at different time can be calculated as follows:

Since the value of k comes out to be nearly constant, the given reaction is of the first order. The mean value of k = 0.0674 min^-1

Question 29.
From the following data, show that the decomposition of hydrogen peroxide is a reaction of the first order:

Where t is the volume of standard KMnO₄ solution required for titrating the same volume of the reaction mixture.
Solution:
Volume of KMnO₄ solution used Amount of H₂O₂ present. Hence if the given reaction is of the first order, it must obey the equation
k = \(\frac { 2.303 }{ t }\) log \(\frac { a }{ (a-x) }\)
k = \(\frac { 2.303 }{ t }\) log \(\frac { { V }_{ 0 } }{ { V }_{ t } }\)
In this case,V₀ = 46.1 ml
The value of k at each instant can be calculated as follows:

Thus, the value of k comes out to be nearly constant. Hence it is a reaction of the first order.

Question 30.
A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. in what time will the reaction be 80% complete?
Answer:
1. For the first order reaction k = \(\frac { 2.303 }{ t }\) log \(\frac { a }{ (a-x) }\)
Assume, a = 100 %, x = 40%, t = 50 minutes
Therefore, a – x = 100 – 40 = 60
k = (2.303/50) log (100/60)
k = 0.010216 min^-1
Hence the value of the rate constant is 0.010216 min^-1

2. t = ?, when x = 8O%
Therefore, a – x = 100 – 80 = 20
From above, k = 0.0102 16 min^-1
t = (2.303 / 0.010216) log (100 / 20)
t = 157.58 min
The time at which the reaction will be 80% complete is 157.58 min.

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements Evaluate yourself

Question 1.
Write the rate expression for the following reactions, assuming them as elementary reactions.

1. 3A + 5B₂ → 4CD
2. X₂ + Y₂ → 2XY

Answer:
1. 3A + 5B₂ → 4CD
Rate = – \(\frac { 1 }{ 3 }\) \(\frac { \triangle [A] }{ dt }\)
= – \(\frac { 1 }{ 5 }\) \(\frac { \triangle [{ B }_{ 2 }] }{ dt }\)
= + \(\frac { 1 }{ 4 }\) \(\frac { \triangle [CD] }{ dt }\)

2. X₂ + Y₂ → 2XY
Rate = – \(\frac { \triangle [{ X }_{ 2 }] }{ dt }\)
= + \(\frac { 1 }{ 2 }\) \( [latex]\frac { \triangle [{ XY }_{ 2 }] }{ dt }\)

Question 2.
Consider the decomposition of N₂O_5(g) to form NO_2(g) and O_2(g). At a particular instant N₂O₅ disappears at a rate of 2.5 x 10^-2 mol dm^-3 s^-1. At what rates are NO₂ and O₂ formed? What is the rate of the reaction?
Solution:
2N₂O_5(g) → 4NO_2(g) + O_2(g)
from the stoichiometry of the reaction.
– \(\frac { 1 }{ 2 }\) \(\frac { d[{ N }_{ 2 }{ O }_{ 5 }] }{ dt }\)
= \(\frac { 1 }{ 4 }\) \(\frac { d[{ N }{ O }_{ 2 }] }{ dt }\)
= –\(\frac { d[{ N }{ O }_{ 2 }] }{ dt }\)
= 2 –\(\frac { d[{ N }_{ 2 }{ O }_{ 5 }] }{ dt }\)
Rate of disappearance of N₂O₅ is 2.5 x 10^-2 mol dm^-3 s^-1
∴ The rate of formation of NO2 at this temperature is 2 x 2.5 x 10^-2 = 5 x 10^-2 mol dm^-3 s^-1.
– \(\frac { 1 }{ 2 }\) \(\frac { d[{ N }_{ 2 }{ O }_{ 5 }] }{ dt }\)
= – \(\frac { d[{ O }_{ 2 }] }{ dt }\)
∴ \(\frac { d[{ O }_{ 2 }] }{ dt }\) = \(\frac { 1 }{ 2 }\) x 2.5 x 10^-2 mol dm^-3 s^-1
= 1.25 x 10^-2 mol dm^-3 s^-1

Question 3.
For a reaction, X + Y → Product quadrupling [x], increases the rate by a factor of 8. Quailrupling both [x] and [y] increases the rate by a factor of 16. Find the order of the reaction with respect to x and y. what is the overall order of the reaction?
Solution:

z = k[x]^m [y]ⁿ …………(1)
8z = k[x]^m [y]ⁿ …………(2)
16z = k[x]^m [y]ⁿ ……………(3)
Dividing Eq (2) by Eq (1) we get,

8 = 4^m ⇒2³ = (2²)m ⇒ 2 = 2^2m
2m = 3
m = 3/2
1.5 order with respect to x.
Dividing Eq (3) by Eq (1) we get,

16 = 4^m. 4ⁿ
16 = 4². 4ⁿ
\(\frac { 16 }{ 16 }\) = 4ⁿ
1 = 4n
∴ n = 0 [Zero order with respect to y]
Overall order of the reaction.
k [x]^m [y]ⁿ
k [x]^1.5 [y]⁰
Order (1.5+0) = 1.5

Question 4.
Find the individual and overall order of the following reaction using the given data.
2NO(g) + Cl₂ (g) → 2NOCI(g)

Solution:
Rate = k [NO]^m [CI₂]
For experiment 1, the rate law is,
Rate₁ = k [NO]^m [CI₂]ⁿ
7.8 x 10^-5 k[0.1]^m [0.1]ⁿ ………………(1)

For experiment 2, the rate law is.
Rate₂ = k [NO]^m [CI₂]ⁿ
3.12 x 10^-4 = k[O.2]^m [0.1]ⁿ ……………….(2)

For experiment 3, the rate law is,
Rate₃ = k [NO]^m [CI₂]ⁿ
9.36 x 10^-4 = k [O.2]^m [0.3]^m ……………(3)
Dividing Eq (2) by Eq (l) we get,

4 = \(\frac { 0.2 }{ 0.1 }\)^m
⇒ 2² = 2^m
∴m = 2
Therefore the reaction is secondary order with respect to NO.
Dividing Eq (3) by Eq (2) we get,

Therefore the reaction is first order with respect to Cl₂
The rate law is, Rate = k [NO]₂ [Cl₂]¹
The overall order of the reaction (2 +1) = 3.

Question 5.
In a first order reaction A → products, 60% of the given sample of A decomposes in 40 min. what is the half life of the reaction?
Solution:
k = \(\frac { 2.303 }{ t }\) log \(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)
k = \(\frac { 2.303 }{ 40min }\) log \(\frac { 100 }{ (100-60) }\)
k = 0.0575 (0.3979) ⇒ k = 0.02287 min^-1
t_1/2 = \(\frac { 0.6932 }{ k }\) log \(\frac { 0.6932 }{ 0.02287 }\)
t_1/2 = 30.31 min.

Question 6.
The rate constant for a first order reaction is 2.3 x 10^-4 s^-1. If the initial concentration of the reactant is 0.01 M. what concentration will remain after 1 hour?
Solution:
Rate constant of a first order reaction k = 2.3 x 10^-4 s^-1
Initial concentration of the reactant [A₀] = 0.01 M
Initial concentration ot the reactant [A₀] = 0.01 M
Concentration will remain after 1 hour [A] =7
k = \(\frac { 2.303 }{ t }\) log \(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)
2.3 x 10^-4 = \(\frac { 2.303 }{ 1 hour }\) log \(\frac { [0.01] }{ [A] }\)
\(\frac { 2.3\times { 10 }^{ -4 }\times 1 }{ 2.303 }\) = log [0.01] – log[A]
9.986 x 10^-5 = – 2 – log [A]
11.986 x 10^-5 = – log [A]
[A] = Antilog (-11.986 x 10^-5)
[A] = 0.997 M

Question 7.
Hydrolysis of an ester in an aqueous solution was studied by titrating the liberated carboxylic acid against sodium hydroxide solution. The concentrations of the ester at different time intervals are given below.

Show that, the reaction follows first order kinetics.
Solution:
The value of k at different time can be calculated as follows:

This value shows that reaction follows first order kinetics.

Question 8.
For a first order reaction the rate constant at 500K is 8 x 10^-4 s^-1. Calculate the frequency factor, if the energy of activation for the reaction is 190 kJ mol^-1.
k = 8 x 10^-4s
T = 500K
E_a = 190 kJ mol^-1 A = ?
According to Arrhenius equation,

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements Text book Example problems

Question 1.
The oxidation of nitric oxide (NO)
2NO_(g) + O_2(g) → NO_2(g)
Series of experiments are conducted by keeping the concentration of one of the reactants constant and the changing the concentration of the others.

Find out the individual overall order of the reaction.
Solution:
Rate = k [NO]^m [O₂]ⁿ
For experiment 1, the rate law is
Rate₂ = k [NO]^m [O₂]ⁿ
19.26 x 10^-2 = k[1.3]^m [1.1]ⁿ ………………(1)
Similarly for experiment 2
Rate₂ = k [No]^m [O₂]ⁿ
38.40 x 10^-2 = k [1.3]^m [2.2]ⁿ …………………..(2)
For experiment 3
Rate₃ = k [NO]^m [O₂]ⁿ
76.8 x 10^-2 = k [2.6]^m [1.1]ⁿ ……….(3)
Dividing Eq (2) by Eq (1)
\(\frac { 38.40\times { 10 }^{ -2 } }{ 19.26\times { 10 }^{ -2 } }\) = \(\frac { k{ \left[ 1.3 \right] }^{ m }{ \left[ 2.2 \right] }^{ n } }{ k{ \left[ 1.3 \right] }^{ m }{ \left[ 1.1 \right] }^{ n } }\)
2 = \({ \left( \frac { 2.2 }{ 1.1 } \right) }^{ n }\)
2 = 2ⁿ
i.e., n= 1
Therefore the reaction is first order with respect to O₂
Dividing Eq (3) byEq (1)
\(\frac { 76.8\times { 10 }^{ -2 } }{ 19.26\times { 10 }^{ -2 } }\) = \(\frac { k{ \left[ 2.6 \right] }^{ m }{ \left[ 1.1 \right] }^{ n } }{ k{ \left[ 1.3 \right] }^{ m }{ \left[ 1.1 \right] }^{ n } }\)
4 = \({ \left( \frac { 2.6 }{ 1.3 } \right) }^{ m }\)
4 = 2^m
i.e., m = 2
Therefore the reaction is second order with respect to NO
The rate law is Rate₁ = k [NO]² [O₂]¹
The overall order of the reaction = (2 + 1) = 3

Question 2.
Consider the oxidation of nitric oxide to form NO₂
2NO_(g) + O_2(g) → 2NO_2(g)

• Express the rate of the reaction in terms of changes in the concentration of NO, O₂ and NO₂.
• At a particular instant, when [O₂] is decreasing at 0.2 mol L^-1 s^-1 at what rate is
  [NO₂] increasing at that instant?

Solution:

Question 3.
What is the order with respect to each of the reactant and overall order of the following reactions?
1.  5Br^–_(aq) + BrO₃^– _(aq)+ 6H⁺_(aq)  → 3Br₂₍₁₎ +3H₂O₍₁₎
The experimental rate law is
Rate = k [Br^–] [BrO₃^–][H⁺]²

2.  CH₃ CHO_(g) \(\underrightarrow { \triangle }\) CH_4(g) + CO_(g)
the experimental rate law is
Rate = k [CH₃CHO]^3/2
Solution:
1. First order with respect to Br^–, first order with respect to BrO₃^– and second order with respect to H⁺. Hence the overall order of the reaction is equal to 1+1+2=4

2. Order of the reaction with respect to acetaldehyde is \(\frac { 3 }{ 2 }\)and overall order is also \(\frac { 3 }{ 2 }\)

Question 4.
The rate of the reaction x + 2y → product is 4 x 10^-3 mol L^-1 s^-1, if [x] = [y] = 0.2 M and rate constant at 400K is 2 x 10^-2 s^-1. what is the overall order of the reaction.
Solution:

Comparing the powers on both sides, the overall order of the reaction n + m = 1

Question 5.
A first order reaction takes 8 hours for 90% completion. Calculate the time required for 80% completion. (log 5 = 0.6989; log10 = 1)
Solution:
For a first order reaction,

Let [A₀] = 100M
When
t = t_90%; [A] = 10M (given that t_90% = 8 hours)
t = t_80%; [A] = 20M

Find the value of k using the given data
k = \(\frac{2.303}{t_{90 \%}}\) log \((\frac { 100 }{ 10 })\)
k = \(\frac { 2.303 }{ 8 hours }\) log 10
k = \(\frac { 2.303 }{ 8 hours }\) (1)
Substitute the value of k in equation (2)
t_80% = \(\frac { 2.303 }{ 2.303/8 hours }\) log (5)
t_80% = 8 hours x 0.6989
t_80% = 5.59 hours

Question 6.
The half life of a first order reaction x → products is 6.932 x 10⁴ s at 500K . What percentage of x would be decomposed on heating at 500K for 100 min. (e^0.06 = 1.06)
Solution:
Given t_1/2 = 0.6932 x 10⁴ s
To solve: when t = 100 min

We know that for a first order reaction,t_1/2 = \(\frac { 0.6932 }{ k }\)

Question 7.
Show that in case of first order reaction, the time required for 99.9% completion is nearly ten times the time required for half completion of the reaction.
Solution:
Let [A₀] = 100
When t = t_99.9%; [A] = (100 – 99.9) = 0.1

Question 8.
The rate constant of a reaction at 400 and 200K are 0.04 and 0.02 s^-1 respectively. Calculate the value of activation energy.
Answer:
According to Arrhenius equation.

Question 9.
Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation. log k = log A \(\frac { { E }_{ a } }{ 2.303R }\) \(\frac { 1 }{ T }\).

Where E is the activation energy. When a graph is plotted for log k Vs \(\frac { 1 }{ T }\) a straight line with a slope of 4000K is obtained. Calculate the activation energy.
Solution:

E_a = – 2.303 Rm
E_a = – 2.303 x 8.314 JK^-1 mol^-1 x (-4000k)
E_a = 76,589 J mol^-1
E_a = 76,589 kJ mol^-1

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements Additional Questions

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements 1 Mark Questions and Answers

I. Choose the correct answer.

Question 1.
Which one of the following is a slow reaction?
(a) Rusting of iron
(b) Combustion of carbon
(c) Reaction between BaCl₂ and dil. H₂SO₄
(d) Reaction between acidified K₂Cr₂O₇ with NaCl.
Answer:
(a) Rusting of iron

Question 2.
Which one of the following is the unit of rate of reaction?
(a) s^-1
(b) mol s^-1
(c) mol L^-1 s^-1
(d) mol L s
Answer:
(c) mol L^-1 s^-1

Question 3.
For a gas phase reaction, the unit of reaction rate is ……………
(a) s^-1
(b) atm s^-1
(c) mol L^-1 s^-1
(d) mol^-1 L^-1 s^-1
Answer:
(b) atm s^-1

Question 4.
For the reaction A → 2B, the rate of the reaction is ………….
(a) +\(\frac { d[B] }{ dt }\) = 2 – \(\frac { d[A] }{ dt }\)
(b) +\(\frac { d[A] }{ dt }\) = \(\frac { 1 }{ 2 }\) \(\frac { d[B] }{ dt }\)
(c) Rate = \(\frac { 1 }{ 2 }\) = \(\frac { d[A] }{ dt }\)
(d) Rate = 2\(\frac { d[B] }{ dt }\)
Answer:
(a) +\(\frac { d[B] }{ dt }\) = 2 – \(\frac { d[A] }{ dt }\)

Question 5.
Consider the following statement.
(i) In ionisation of cyclopropane, if the concentration of cyclopropane is reduced half, the rate increases twice.
(ii) The rate of the reaction depends upon the concentration of the reactant.
(iii) Order values must be determined experimentally.

Which of the above statement (s) is / are not correct?
(a) (i) only
(b) (ii) and (iii)
(c) (iii) only
(d) (ii) only
Answer:
(a) (i) only

Question 6.
In the reaction 2NO_(g) + O_2(g) → 2NO_2(g) the order of the reaction with respect to NO is…………
(a) first order
(b) second order
(c) third order
(d) zero order
Answer:
(b) second order

Question 7.
In the reaction 2NO_(g) + O_2(g) → 2NO_2(g). the order of the reaction with respect to O₂is …….
(a) zero order
(b) first order
(c) second order
(d) third order
Answer:
(b) first order

Question 8.
The overall order of the reaction 2NO_(g) + O_2(g) → 2NO_2(g) is …………….
(a) 2
(b) 1
(c) 3
(d) 0
Answer:
(c) 3

Question 9.
Consider the following statements.
(i) Rate of the reaction does not depend on the initial concentration of the reactants.
(ii) Rate constant of the reaction depends on the initial concentration of reactants.
(iii) Rate constant of the reaction is equal to the rate of the reaction, when the concentration of each of the reactants is unity.
Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) only
(c) (i) and (ii)
(d) (iii) only
Answer:
(a) (iii) only

Question 10.
The overall molecularity of the reaction 2H₂ O_2(aq) \(\underrightarrow { { I }^{ – } }\) 2H₂O₁ + O_2(g) is …………
(a) unimolecular
(b) bimolecular
(c) termolecular
(d) pentamolecular
Answer:
(b) bimolecular

Question 11.
Which of the following is the order of decomposition of hydrogen peroxide catalysed by I^– ………….
(a) First order
(b) Second order
(c) Zero order
(a) Third order
Answer:
(a) First order

Question 12.
Consider the following statements.
(i) order cannot be zero.
(ii) Molecularity can be zero (or) fractional (or) integer.
(iii) order can be determined only by experiment.

Which of the above statement(s) is / are not correct?
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i) and (ii)
Answer:
(c) (iii) only

Question 13.
The overall order of the reaction 5Br^– + BrO₃^– + 6H⁺ is ……..
(a) 4
(b) 3/2
(c) 12
(d) 1
Answer:
(a) 4

Question 14.
Which one of the following reaction is a fractional order reaction?
(a) 2NO +O₂ → 2NO₂
(b) CH₃CHO_(g) → CH_4(g) + CO_(g)
(c) 2H₂O₂ → 2H₂)O₍₁₎ + O_2(g)
(d) H₂ + Br₂ → 2HBr
Answer:
(b) CH₃CHO_(g) → CH_4(g) + CO_(g)

Question 15.
The order of decomposition of acetaldehyde is …………….
(a) 1
(b) 1.5
(c) 2
(d) 5/2
Answer:
(b) 1.5

Question 16.
Which one of the following is the unit of rate constant for a first order reaction?
(a) mol^-1 L s^-1
(b) mol L^-1 s^-1
(c) s^-1
(d) mol L S
Answer:
(c) s^-1

Question 17.
Which one of the following is an example for first order reaction?
(a) 2NO_(g)+ O_2(g) → 2NO_2(g)
(b) CH₃CHO_(g) → CH_4(g) + CO_(g)
(c) SO₂ Cl₂₍₁₎ → SO_2(g) + Cl_2(g)
(d) 2HBr → H₂ + Br₂
Answer:
(c) SO₂ Cl₂₍₁₎ → SO_2(g) + Cl_2(g)

Question 18.
Which one of the following is not an example for first order reaction?
(a) N₂O_5(g) → 2NO_2(g) \(\frac { 1 }{ 2 }\) O_2(g)
(b) SO₂Cl₂₍₁₎ → SO_2(g) + Cl_2(g)
(e) H₂O_2(aq) → H₂O₁\(\frac { 1 }{ 2 }\)O_2(g)
(d) CH₃CHO_(g) → CH_4(g) + CO_(g)
Answer:
(d) CH₃CHO_(g) → CH_4(g) + CO_(g)

Question 19.
What is the order of isomerisation of cyclopropane to propene?
(a) 1.5
(b) 3/2
(c) 5/2
(d) 1
Answer:
(d) 1

Question 20.
Which one of the following is an example of pseudo first order reaction?
(a) CH₃CHO_4(g) → CH_4(g) + CO_(g)
(b) 2H₂O_2(aq) → H₂O₍₁₎ +O_2(g)
(c) CH₃COOCH_3(aq) + H₂O₍₁₎ \(\underrightarrow { { H }^{ + } }\) CH₃COOH_(aq) + CH₃OH_(aq)
(d) Isomerisation of cyclo propane to propene
Answer:
(c) CH₃COOCH_3(aq) + H₂O₍₁₎ \(\underrightarrow { { H }^{ + } }\) CH₃COOH_(aq) + CH₃OH_(aq)

Question 21.
Which one of the following is called pseudo first order reaction?
(a) Decomposition of acetaldehyde
(b) Acid hydrolysis of an ester
(c) Isomerisation of cyclopropane to propene
(d) Decomposition of hydrogen peroxide
Answer:
(b) Acid hydrolysis of an ester

Question 22.
Which of the following is an example of zero order reaction?
(a) lodination of acetone in acid medium
(b) Hydrolysis of an ester in acid medium
(c) Decomposition of acetaldehyde
(d) Isomerisation of cyclopropane to propene
Answer:
(a) lodination of acetone in acid medium

Question 23.
Which one of the follow is not zero order reaction?
(a) H_2(g) + Cl_2(g) \(\underrightarrow { h\nu }\)  2HCI_(g)
(b) N₂O_(g) \(\rightleftharpoons\) N_2(g) + \(\frac { 1 }{ 2 }\) O_2(g)
(c) CH₃CHO_(g) → CH_4(g)+ CO_(g)
(d) CH₃COCH₃ + I₂ \(\underrightarrow { { H }^{ + } }\) CH₂COCH₃ + HI
Answer:
(c) CH₃CHO_(g) → CH_4(g)+ CO_(g)

Question 24.
Consider the following statements.
(i) For a first order reaction, half life period is independent of initial concentration.
(ii) Photo chemical reaction between H₂ and Cl₂ is a zero order reaction
(iii) Acid hydrolysis of an ester is a second order reaction

Which of the above statement is/are correct?
(a) (i) only
(b) (iii) only
(c) (i) & (ii)
(d) (ii) & (iii)
Answer:
(c) (i) & (ii)

Question 25.
The formula of half life for an nth order reaction involving reactant A and n \(\neq\) 1 is

Answer:

Question 26.
The half life period of first order reaction is 10 seconds. What is the time required for 99.9% completion of that reaction?
(a) 20 seconds
(b) 1000 seconds
(c) 100 seconds
(d) 999 seconds
Answer:
(c) loo seconds
Hint:
10 x t_1/2 = t_99.9%
∴ t_99.9% = 10 x 10 sec = 100 sec

Question 27.
Which one of the following is known as arrhenius equation?

Answer:

Question 28.
Which one of the following does not affect the rate of the reaction?
(a) Nature of the reactant
(b) Concentration of the reactants
(c) Surface area and temperature
(d) pressure
Answer:
(d) pressure

Question 29.
Consider the following statements.
(i) Higher the concentration, slower is the possibility for collision and rate also slower
(ii) Increase in surface area of reactant leads to more collisions per litre per second
(iii) Gas phas reactions are slower as compared to solid or liquid reactants

Which of the above statement is/are not correct?
(a) (ii)
(b) (i) & (iii)
(c) (ii) & (iii)
(d) (i) & (ii)
Answer:
(b) (i) & (iii)

Question 30.
Which of the following reaction take place at a faster rate?

Answer:

Question 31.
Which one of the following graph is not correct ………..

Answer:

Question 32.
The half life of paracetamol with in the body is ………
(a) 2 hours
(b) 2.5 hours
(c) 6 hours
(d) 10 hours
Answer:
(b) 2.5 hours

Question 33.
What is the order of radioactive decay?
(a) first order
(b) zero order
(c) second order
(d) third order
Answer:
(a) first order

Question 34.
t_1/2 of the reaction increases with increase in initial concentration of the reaction means the order of the reaction will be …………
(a) first order
(b) zero order
(c) second order
(d) third order
Answer:
(b) zero order

Question 35.
The reaction rate that does not decrease with time is …………
(a) pseudo first order reaction
(b) first order reaction
(c) zero order reaction
(d) second order reaction
Answer:
(c) zero order reaction

Question 36.
The rate of the reaction X → Y becomes 8 times when the concentration of the reactant ‘X’ is doubled. The rate law of the reaction is ………
(a) – \(\frac { d[x] }{ dt }\) = k[X]²
(b) – \(\frac { d[x] }{ dt }\) = k[X]³
(c) – \(\frac { d[x] }{ dt }\) = k[X]⁴
(d) – \(\frac { d[x] }{ dt }\) = k[X]⁸
Answer:
(b) – \(\frac { d[x] }{ dt }\) = k[X]³

Question 37.
The decomposition of ammonia gas on platinum surface has a rate constant k = 2.5 x 10^-4 mol L^-1 s^-1 What is the order of the reaction?
(a) first order
(b) second order
(c) third order
(d) zero order
Answer:
(d) zero order

Question 38.
A reaction is 50% completed in 2 hours and 75% completed in 4 hours. Then the order of the reaction is ………….
(a) first order
(b) zero order
(c) second order
(d) third order
Answer:
(a) first order
Answer:
(a) first order

Question 39.
What is the rate equation for the reaction A + B → C has zero order?
(a) Rate = k
(b) Rate = k [A]
(c) Rate = k [A]. [B]
(a) Rate = k. \(\frac { 1 }{ [c] }\)
Answer:
(c) Rate = k [A]. [B]

Question 40.
How does the value of rate constant vary with reactant concentration?

Answer:

Question 41.
Identify the reaction order if the unit of rate constant is s^-1 ……….
(a) zero order reaction
(b) second order reaction
(c) first order reaction
(d) third order reaction
Answer:
(c) first order reaction

Question 42.
What is unit of zero order reaction?
(a) s^-1
(b) mol^-1 L^-1 s^-1
(c) mol L^-1 s^-1
(d) mol L s^-1
Answer:
(c) mol L^-1 s^-1

Question 43.
Which of the following factor affect the rate of the reaction’?
(a) volume
(b) pressure
(c) cone
(d) all the above
Answer:
(c) cone

Question 44.
Acid hydrolysis of an ester is an example of ………
(a) zero order reaction
(b) Pseudo first order reaction
(c) second order reaction
(d) first order reaction
Answer:
(b) Pseudo first order reaction

Question 45.
Polymerisation reactions follows ………………. order kinetics.
(a) fractional
(b) first
(c) zero
(d) Pseudo first
Answer:
(a) fractional

Question 46.
Activation energy of a chemical reaction can be determined by ……….
(a) changing concentration of the reactants
(b) Evaluating rate constants at standard temperature
(c) Evaluating rate constants at two different temperature
(d) Evaluating reIocities of reaction at two different temperature
Answer:
(c) Evaluating rate constants at two different temperature

Question 47.
Which of the following is the fastest reaction?

Answer:

Question 48.
Half life period of a reaction is found to be inversely proportional to the cube of its initial concentration. The order of the reaction is ………
(a) 2
(b) 5
(c) 3
(d) 4
Answer:
(d) 4

Question 49.
A large increase in the rate of a reaction for a rise in temperature is due to ………
(a) the decrease in the number of collisions
(b) increase in the number of activated molecules
(c) the shortening of mean free path
(d) the lowering of activation energy
Answer:
(b) increase in the number of activated molecules

Question 50.
The minimum energy of a molecule would possess in order to enter into a fruitful collision is known as …………..
(a) Reaction energy
(b) collision energy
(c) Activation energy
(d) Threshold energy
Answer:
(a) Threshold energy

II. Fill in the blanks.

1. The unit of the rate of a reaction is ………..
2. For a ……….. reaction, the unit of the reaction rate is atm s
3. An elementary step is characterised by its ………..
4. The total number of reactant species involved in an elementary step is called ………..
5. The sum of powers of concentration terms involved in the experimentally determined rate law is called ………..
6. The overall order of decomposition of acetaldehyde to methane and carbon monoxide is ………..
7. A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, it is called ………..
8. A reaction in which rate is independent of the concentration of the reactant over a wide range of concentration is called ………..
9. All radioactive disentegration reactions follow ……….. kinetics.
10. For a first order reaction, half life does not depend on ………..
11. Half life period of zero order reaction is ……….. proportional to initial concentration of the reactant.
12. Half life period ……….. reaction is directly proportional to initial concentration of the reactant.
13.  ……….. was proposed by Max Trautz and William lewis.
14. Collision theory was proposed by ……….. in 1916 and in ……….. 1918.
15. For a gas at room temperature (298 K) and I atm, each molecule undergoes approximately ……….. per second.
16. In order to react, the collidng molecules must possess a minimum energy called ………..
17. Generally the reaction rate tends to double when the temperature is increased by ………..
18. The number of collisions of reactant molecules per second is known as ………..
19. Heating is required for the reaction between KMnO₄ and oxalate ion and is carried out at around ………..
20. ……….. reactions are faster as compared to reactions involving solid or liquid reactants.
21. The rate of the reaction ……….. with the increase in the concentration of the reactants.
22. Higher the concentration of reactants greater is the possibility of and hence the ………..
23. In the presence of catalyst the energy of activation is ……….. and hence greater number of molecules change over to products there by increasing the rate of the reaction.
24. Bio availability of drugs within the body and this branch of study is called ………..
25. ……….. has a half life of 2.5 hours within the body.
26. The change in concentration of species per unit time gives the ……….. of the reaction.
27. The rate constant is equal to the rate of the reaction when concentration of reactants is ………..
28. Increase in surface area of reactant leads to more collisions per litre per second and hence the rate of the reaction is ………..
29. Acid hydroJysis of an ester is an example of ………..
30. Molecularity of a chemical reaction will never be equal to ………..

Answer:

1. moI L^-1 s^-1
2. gas phase
3. molecularity
4. Molecularity
5. order
6. 3/2 or 1.5
7. Pseudo first order reaction
8. Zero order reaction
9. First order
10. initial concentration
11. directly
12. zero order
13. Collision theory
14. Max Trautz, William lewis
15. 10 collisions
16. Activation energy
17. 10°C
18. Frequency factor (A)
19. 60°C
20. Gas phase
21. increases
22. collisions, rate
23. lowered
24. Pharmaco kinetics
25. Paracetamol
26. rate
27. unity
28. increased
29. Pseudo first order reaction
30. zero

III. Match the following

Match the list I and II using the code given below the list.

Quetion 1.

Answer:
(a) 3, 4, 1, 2

Question 2.

Answer:
(a) 3, 4, 1, 2

Question 3.

Answer:
(a) 3, 1, 4, 2

Question 4.

Answer:
(a) 3, 4, 2, 1

IV. Assertion and Reason

Question 1.
Assertion (A): Decomposition of hydrogen peroxide catalysed by I^– is a bimolecular first order reaction.
Reason (R): The above reaction take place in two steps, step 1 involves both H₂O₂ and I and so it is bimolecular but order is determined experimentally as 1.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R arc correct and R is the correct explanation of A

Question 2.
Assertion (A):  The overall order of the reaction is equal to 4.
Reason (R): The experimental rate law is.
Rate = K [Br^–] [BrO₃^–] [H⁺]².
So 1 + 1 + 2 = 4 order value is 4.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are wrong
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation olA.

Question 3.
Assertion (A): The rate of a reaction increases with the increase in the concentration of the reactants.
Reason (R): The rate of the reaction depends upon the number of collisions between the reacting molecules. Higher the concentration, greater is the possibility for collision and hence the rate.
(a) Both A and R are correct and R is the correct explanation of A,
(b) Both A and R are correct but R is not the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 4.
Assertion (A): Powdered calcium carbonate reacts much faster with dilute
HCL than with the same mass of CaCO₃ as marble.
Reason (R): For a given mass of a reactant, when the particle size decreases, surface area increases. Increase in surface area of the reactant leads to more collisions per litre per second and hence the rate of the reaction also increases.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R arc correct but R is not correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 5.
Assertion (A): Catalyst presence increases the rate of the reaction
Reason (R): In the presence of a catalyst, energy of activation is lowered and hence greater number of molecules can across the energy harrier and change over to products thereby increasing the rate of the reaction.
(a) Both A and R are correct but R is not correct explanation of A
(b) Both A and R are correct and R is the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(b) Both A and R are correct and R is the correct explanation of A

Question 6.
Assertion (A): Doctors adviced to take paracetamol once in 6 hours during fever and body pain
Reason (R): Paracetarnol has a half life of 2.5 hours within the body. After 10 hours (4 half lives) only 6.25% of drug remains. Based on this, doctors adviced to take it once in 6 hours.
(a) Both A and R are wrong
(b) A is correct but R is wrong
(c) A and R are correct and R is the correct explanation of A
(d) A and R are correct but R is not correct explanation of A
Answer:
(a) Both A and R are correct and R is the correct explanation of A.

Question 7.
Assertion (A): Order of the reaction can be zero or fractional
Reason (R): We cannot determine order from balanced chemical equation
(a) Both A and R are correct but R is not correct explanation of A.
(b) Both A and R are correct and R is the correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(a) Both A and R are correct and R is not correct explanation of A

Question 8.
Assertion (A): If the activation enery of a reaction is zero, temperature will have no effect on the rate constant
Reason (R): Lower the activation energy, faster is the reaction.
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not correct explanation of A
(c) A is correct but R is wrong
(d) A is wrong but R is correct
Answer:
(b) Both A and R are correct but R is not correct explanation of A

V. Find the odd one out

Question 1.

Answer:

Hint: It is a fractional order reaction with order value 3/2 where as others are first order reaction.

Question 2.

Answer:

Hint: It is first order reaction whereas others are zero order reaction.

Question 3.
(a) Decomposition of dinitrogen pentoxide
(b) Iodination of acetone in acid medium
(c) Decomposition of N20 on hot Pt surface
(d) photochemical reaction between H2 and CI2
Answer:
(a) Decomposition of dinitrogen pentoxide
Hint: It is a first order reaction whereas others are zero order reactions

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements 2 Mark Questions and Answers

Question 1.
Define rate of the reaction. Give the unit of rate.
Answer:
1. In a chemical reaction, the change in the concentration of the species involved in a chemical reaction per unit time gives the rate of a reaction.
A → B
Rate = –\(\frac { \triangle [A] }{ dt }\) (or) \(\frac { \triangle [B] }{ dt }\)

2.

= mol L^-1 s^-1

Question 2.
Define molecularity of a reaction.
Answer:
Molecularity of a reaction is the total number of reactant species that are involved in an elementary step.

Question 3.
Define order of a chemical reaction.
Answer:
Order of a chemical reaction is the sum of powers of concentration terms involved in the experimentally determined rate law.

Question 4.
Define Half life period.
Answer:
The half life ola reaction is defined as the time required for the reactant concentration to reach one half its initial value.

Question 5.
Mention the factors affecting the reaction rate.
Answer:
The rate of the reaction is affected by the following factors.

1. Nature and state of the reactant
2. Concentration of the reactant
3. Surface area of the reactant
4. Temperature of the reaction
5. Presence of a catalyst

Question 6.
How is surface area of the reactant affect the rate of the reaction?
Answer:

1. In heterogeneous reactions, the surface area of the solid reactants play an important role in deciding the rate.
2. For a given mass of a reactant, when the particle size decreases surface area increases. Increase in surface area of reactant leads to more collisions per litre per second and hence the rate of reaction is increased.
3. For example, powdered calcium carbonate reacts much faster with dilute HCl than with the same mass of CaCO₃ as marble.

Question 7.
Paracetamol is prescribed to take once in 6 hours. Justify this statement.
Answer:
1. Paracetamol is a well known antipyretic and analgesic that is prescribed in cases of fever and body pain.

2. Paracetamol has a half life of 2.5 hours within the body. (Le) the plasma concentration of the drug is halved after 2.5 hours. So after 10 hours (4 half lives), only 6.25% of drug remains. Based on this, the dosage and frequency will be decided.

3. In the case of paracetamol, it is usually prescribed to take once in 6 hours.

Question 8.
For a reaction, A + B → product; the rate law is given by r = k[A]^1/2 [B]². What is the order of the reaction?
Answer:
Order of the reaction = \(\frac { 1 }{ 2 }\) + 2 = 2 \(\frac { 1 }{ 2 }\) or 0.5

Question 9.
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will ¡t affect the rate of formation of Y?
Answer:
For the reaction, X → Y as it follows second order kinetics, therefore the rate law equation will be
Rate = k[X]² = ka²
if [X] = a mol^-1
if the concentration of X is increased three times, then
[X] = 3a mol L^-1
∴ Rate = k (3a)² = 9 ka²
Thus, the rate of the reaction will become 9 times. Hence the rate of formation of Y will increase by 9 times.

Question 10.
Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Answer:
For a first order reaction,

Question 11.
What will be the effect of temperature on rate constant?
Answer:
Rate constant of a reaction is nearly doubled with rise in temperature by 10°C. The exact dependence of the rate constant on temperature is given by Arrhenius equation:
Rate constant,

Question 12.
A reaction is first order in A and second order in B.

1. Write the differential rate equation.
2. How is the rate affected on increasing the concentration of B three times?
3. How is the rate affected when the concentrations of both A and B arc doubled?

Answer:

1. \(\frac { dx }{ dt }\) = k [A]¹ [B]²
2. If concentration of ‘B’ is tripled, then the rate will become 9 times.
3. When concentration of both A and B are doubled, then the rate will become 8 times.

Question 13.
Define zero order reaction. Give the unit for its rate constant(k).
Answer:
Zero Order Reaction. The reaction in which the rate of reaction is independent of the concentration of the reactants is called zero order reaction.
Rate = k [A]⁰ ⇒ k
Where k is the rate constant. Its unit is mol L^-1 s^-1

Question 14.
Write units of rate constant k for zero order, first order, second order and n order reaction.
Answer:
Order of Reaction

1. Zero order reaction
2. First order reaction
3. Second order reaction
4. nth order reaction

Unit of k:

1. mol L^-1 s^-1
2. s^-1
3. mol L s^-1
4. (mol/ L)^1-n s^-1

Question 15.
What is the effect of catalyst on the activation energy? Why?
Answer:
A Catalyst lower down the activation energy. It provides an alternate path to the reaction. It forms an unstable intermediate which readily changes into products.

Question 16.
Give two differences between zero order and first order reaction.
Answer:
Zero Order:

1. Its ‘k’ has unit = mol L^-1 s^-1
2. Its t 1/2 is directly proportional to initial conc. of reactant

First order:

1. Its ‘k’ has unit = time^-1 = sec^-1
2. Its half life is independent of the initial conc. of the reactant.

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements 3 Mark Questions and Answers

Question 1.
Write the differences between the rate and rate constant of the reaction.
Answer:
Rate of a reaction:

1. It represents the speed at which the reactants are converted into products at any instant
2. It is measured as decrease in the concentration of the reactants (or) increase in the concentration of products
3. It depends on the initial concentration of reactants

Rate constant of a reaction:

1. It is a proportionality constant
2. It is equal to the rate of the reaction, when the concentration of each of the reactants is unity
3. It does not depend on the initial concentration of the reactants

Question 2.
What are the examples of first order reaction?
Answer:

1. Decompostion of dinitrogen pentoxide
   N₂O_2(g) → 2NO_2(g) + \(\frac { 1 }{ 2 }\) O_2(g)
2. Decomposition of thionylchloride
   SO₂Cl_2(g) → SO_2(g) + CI_2(g)
3. Decomposition of H₂O₂ in aqueous solution
   H₂O_2(aq) → H₂O₍₁₎ + \(\frac { 1 }{ 2 }\) O_2(g)
4. Isomerisation of cyclopropane to propene

Question 3.
For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Answer:

Question 4.
In a reaction, 2 A → products. the concentration of A decreases from 0.5 moI L^-1 to 0.4 mol L^-1 in 10 minutes. Calculate the rate during this interval.
Answer:
Average rate

Question 5.
A first order reaction has a rate constant, 1.15 x 10^-3 s^-1. How long will 5 g of this reactant take to reduce to 3 g?
Answer:
Here,
[R]₀ = 5g
[R] = 3 g
k = 1.15 x 10^-3 s^-1 As the reaction is of first order,

Question 6.
Time required to decompose SO₂CI₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction. For a first order reaction,
Answer:

Question 7.
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is-

1. doubled
2. reduced to half.

Answer:
1. Reaction is second order with respect to the reactant
∴ Rate = k[A]² = ka².
when [A] = 2a
Rate k (2a)²
= 4ka²
= 4 times
Therefore, when concentration of the reactant is doubled the rate will become 4 times

2. when [A] = \(\frac { 1 }{ 2 }\) a
Rate = k \({ \left( \frac { 1 }{ 2 } a \right) }^{ 2 }\) = \(\frac { 1 }{ 4 }\) ka² = \(\frac { 1 }{ 4 }\) k
Therefore, rate will be reduced to one-fourth of the initial rate.

Question 8.
The rate constant for a first order reaction is 60 s¹. How much time will It take to reduce the initial concentration of the reactant to its 1/6^th value?
Answer:

Question 9.
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer:
For a first order reaction, t = k = \(\frac { 2.303 }{ t }\) log \(\frac { a }{ a-x }\)
99% completion means that
x = 99% of a =0.99 a
t_99% = \(\frac { 2303 }{ k }\) log \(\frac { a }{ a-0.99a }\) = \(\frac { 2.303 }{ k }\) log 10² = 2 x \(\frac { 2303 }{ k }\)
90% completion means that
x = 90% of a = 0.90 a
t_99% = \(\frac { 2303 }{ k }\) log \(\frac { a }{ a-0.99a }\) = \(\frac { 2.303 }{ k }\) log 10 = \(\frac { 2.303 }{ k }\)
\(\frac{t_{99 \%}}{t_{90 \%}}=\left(\frac{2 \times 2303}{k}\right) / \frac{2.303}{k}=2\)
or
t_99% = 2 x t_90%

Question 10.
Calculate the half life of a first order reaction whose rate constant is 200 s^-1
Answer:
Here rate constant
k= 200 s^-1
∴ Half – life of a first order reaction is
t_1/2 = \(\frac { 0.693 }{ k }\) = \(\frac { 0.693 }{ 200 }\) = 3.46 x 10^-3 sec

Question 11.
The decomposition of dinitrogen pentoxide (N₂O) follows the first order rate law. Calculate the rate constant from the given data.
t = 800 sec, [N₂O₅] = 1.45 moI L^-1 = [A₂]
t = 1600 sec
[N₂O₃] = 0.88 moI L^-1 = [A₂]
Answer:
Applying the formula,
k = \(\frac { 2.303 }{ \left( { t }_{ 2 }-{ t }_{ 1 } \right) }\) log 10 \(\frac { \left[ { A }_{ 1 } \right] }{ \left[ { A }_{ 2 } \right] }\)
= \(\frac { 2.303 }{ (1600 – 800) }\) log 10 \(\frac { 1.45 }{ 0.88 }\) = \(\frac { 2.303 }{ 800 }\) x 0.2169
= 6.24 x 10^-4 sec^-1

Question 12.
A second order reaction in which both the reactants have same concentration, is 20% completed in 500 seconds. How much time it will take for 60% completion?
Answer:
The second order equation when both the reactants have same concentration is
k = \(\frac { 1 }{ t }\). \(\frac { x }{ a(a – x) }\)
If a = 100, x = 20, 1= 500 seconds
Then k = \(\frac { 1 }{ 500 }\) x \(\frac { 20 }{ 100  x  (100-20) }\)
When
a = 100
x = 60
t = ?
t = \(\frac { 1 }{ k }\) \(\frac { 60 }{ 100  x  40 }\)
Substituting the value of k,
t = \(\frac { 500  x  100  x  80 }{ 20 }\) x \(\frac { 60 }{ 100  x  40 }\)
or
t = 3000 seconds

Question 13.
A first order reaction is 20% completed in 10 minutes. Calculate the time taken for the reaction to go to 80% completion.
Answer:
Applying the first order equation,
k = \(\frac { 2303 }{ t }\) l0g \(\frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] }\)
At t = 10 min
R = 100 – 20
k = \(\frac { 2303 }{ t }\) log 10 \(\frac { 100 }{ (100 – 20) }\)
t = \(\frac { 2303 }{ 10 }\) log 10 \(\frac { 100 }{ 80 }\)
= 0.0223 min^-1

Question 14.
For a reaction: 2NH_3(g)  \(\underrightarrow { Pt }\)  N_2(g)+ 3H_2(g) Rate = K

1. Write the order and molecularity of this reaction.
2. Write the unit of K.

Answer:

1. Order of reaction Zero order. Molecularity = 2
2. Unit of K = mol L^-1 sec^-1

Samacheer Kalvi 12th Chemistry Chemical Kinetics Elements 5 Marks Questions and Answers

Question 1.
How would you calculate the order of the reaction 2NO + O_2(g) → 2NO_2(g) by an experiment?
(or)
prove that 2NO + O₂ → 2 NO₂ is a third order reaction.
Answer:
2NO_(g) + O_2(g) → 2NO_2(g)
Series of experiments are conducted by keeping the concentration of one of the reactants as constant and changing the concentration of the others.

Rate = k [NO]^m[O₂]ⁿ
For experiment 1, the rate law is
Rate₁ = k [NO]^m [O₂]ⁿ
19.26 x 10^-2 = k [1.3]^m [1.1]ⁿ ………………….(1)
For experiment 2
Rate₂ = k [NO]^m [O₂]ⁿ
38.40 x 10^-2 = k [1.3]^m [2.2]ⁿ ……………….(2)
For experiment 3
Rate₃ = k [NO]^m [O₂]ⁿ
76.8 x 10^-2 = k [2.6]^m [1.1]ⁿ ………………(3)

2 = \({ \left( \frac { 2.2 }{ 1.1 } \right) }^{ n }\)
4 = 2^m
⇒ n = 1
Therefore the reaction is first order with respect to O₂

2 = \({ \left( \frac { 2.6 }{ 1.3 } \right) }^{ m }\)
4 = 2^m
⇒ m = 1
Therefore the reaction is second order with respect to NO
The rate law is Rate_– = k [NO]² [O₂]¹
The overall order of the reaction = 2 + 1 = 3

Question 2.
Derive the integrated rate law for a first order reaction?
Answer:
A reaction whose rate depends on the reactant concentration raised to the first power is called a first order reaction. First order reaction is A → product. Rate law can be expressed as, Rate = k [A]¹. Where, k is the first order rate constant
\(\frac { -d[A] }{ dt }\) = k[A]¹
\(\frac { -d[A] }{ [A] }\) = k.dt ……………………(1)
Integrate the above equation (I) between the limits of time t = 0 and time equal to t, while the concentration varies from initial concentration [A₀] to [A] at the later time.

This equation (2) is in natural logarithm. To convert it into usual logarithm with base 10, we have to multiply the term by 2.303

Question 3.
Explain the effect of temperature on reaction rate based on Arrhenius theory.
Answer:
1. Generally, the rate of a reaction increases with increasing temperature. However, there are very few exceptions.

2. As a rough rule, for many reactions near room temperature, reaction rate tends to double when the temperature is increased by 10°C.

3. A large number of reactions are known which do not take place at room temperature but occur readily at higher temperature. Example – Reaction between H₂ and O₂ to form H₂O takes place only when an electric spark is passed.

4. Arrhenius suggested that the rates of most reactions vary with temperature in such a way that the rate constant is directly proportional to
and he proposed a relation between the rate constant and temperature.

where
k = frequency factor
R = gas constant ,
E_a = Activation energy
T = Absolute temperature (in kelvin)
The factor A does not vary significantly with temperature and hence it may be taken as a constant.

5. Taking logarithm on both side of the equation (1)

6. The plot of Ink vs \((\frac { 1 }{ T })\) is a straight line with negative slope\(\frac { { E }_{ a } }{ RT }\). If the rate constant for a reaction at two different temperatures is known, we can calculate the activation energy.

This equation can be used to calculate E from rate E_a constants k₁ & k₂ at temperature T₁ and T₂

Question 4.
Explain about the factors that affecting the reaction rate.
Answer:
The rate of a reaction is affected by the following factors.
1. Nature and state of the reactant
(a) A chemical reaction involves breaking of certain existing bonds of the reactant and forming new bonds which lead to the product. The net energy involved in this process is dependent on the nature of the reactant and hence the rates are different for different reactants.

(b) Gas phase reactions are faster as compared to the reactions involving solid or liquid reactants. For example, reaction of sodium metal with iodine vapours is faster than the reaction between solid sodium and solid iodine.

2. Concentration of the reactant
The rate of the reaction increases with the increase in the concentration of the reactants. According to collision theory, the rate of the reaction depends upon the number of collisions between the reacting molecules. Higher the concentration, greater is the possibility for collision and hence the rate.

3. Effect of surface area of the reactant:
In heterogeneous reactions, the surface areas of the solid reactants plays an important role in deciding the rate. For a given mass of a reactant, when the particle size decreases surface area increases.

Increase in surface area of reactant leads to more collisions per litre per second and hence the rate of reaction is increased. For example, powdered calcium carbonate reacts much faster with dilute HCI than with the same mass of CaCOl as marble

4. Temperature:
For many reactions near room temperature, the reaction rate tends to double when the temperature is increased by 10°C . For eg, Reaction between H₂ and O₂ to form H₂O take place only when an electric spark is passed. So when the temperature increases, the rate of the reaction also increases.

5. Effect of presence of catalyst
(a) A catalyst is substance which alters the rate of a reaction without itself undergoing any permanent chemical change. They may participate in the reaction, but again regenerated and the end of the reaction.

(b) In the presence of a catalyst, the energy of activation is lowered and hence greater number of molecules can cross the energy barrier and change over to products,thereby increasing the rate of the reaction.

Question 5.
The decomposition of A into product has value of k as 4.5 x 10³ s^-1 at 10°C and energy of activation 60 kJ mol^-1. At what temperature would k be 1.5 x 10⁴ s^-1 ?
Answer:
k₁ = 4.5 x 10³ s^-1
T₁ = 10 + 273 K = 283 K
k₂ = 1.5 x 10⁴ s^-1
T₂ = ?
E_a = 6o kJ mol^-1
According to Arrhenius equation

Question 6.
For a decomposition reaction the values of rate constant k at two different temperatures are given below:
k₁ = 2.15 x 10 L mol^-1 s^-1 at 650 K
k₂ = 2.39 x 10 L mol^-1 s^-1 at 700 K
Calculate the value of activation energy for this reaction. (R = 8.314 J K^-1 mol^-1 )
Answer:
Here
k₁ = 2.15 x 10 L mol^-1 s^-1 at 650 K
T₁ = 650 K
T₂ = 700K and
k₂ = 2.39 x 10 L mol^-1 s^-1 at 700 K
R = 8.314 J K^-1 mol^-1
Using the formula

Question 7.
For a certain chemical reaction variation In concentration, In IRI Vs time (s) plot is given below.
For this reaction write/draw:

1. What is the order of the reaction?
2. What is the units of rate constant (k)?
3. Give the relationship between k and t_1/2 (half-life period).
4. What does the slope of above line indicate?
5. Draw the plot of log [R₀]/[R] vs time (s)

Answer:
1. First order
2. time^-1 (s^-1)
3.
4.  Rate constat (k) of reaction
5.

Question 8.
A substance reacts according to the first order rate law and the specific reaction rate for the reaction ¡s 1 x 10^-2 s^-1. If the initial concentration is 1.0 M.

1. What is the initial rate?
2. What ¡s the reaction rate after 1 minute?

Answer:

1. Initial rate of a first order reaction
= k C
= l x 10^-2 x 1.0
= l x 10^-2 mol L^-1 s^-1

2. Concentration after 60 seconds is calculated by applying the first order kinetic equation,

Rate of reaction after 1 minute
= k x C
= l x 10^-2 x 0.5489
5.489 x 10^-3 mol L^-1 s^-1

Question 9.
A first order reaction is 50% completed in 30 minutes at 27°C and in 10 minutes at 47°C. Calculate the reaction rate constant at 27°C and the energy of activation of the reaction in kJ mol^-1
Answer:
For a first order reaction

Now applying the following equation:

Question 10.
In Arrhenius equation for a certain reaction, the values of A and E (activation energy) are 4 x 10¹³ sec^-1 and 98.6 KJ mol^-1 respectively. If the reaction is of first order, at what temperature will its half life period be 10 minutes?
Answer:
According to the Arrhenius equation.

For a first order reaction
t_1/2 = \(\frac { 0.693 }{ k }\) \(\frac { 0.693 }{ 600 }\)
So,
k = \(\frac { 0.693 }{ 600 }\) sec^-1 (t_1/2 = 10 min = 600 sec)
= 1.1 x 10^-3 sec^-1
Hence, log (1.1 x 10^-3)

Common Errors

Common Errors:

1. Order and molecularity may get confused
2. Unit of first order Rate constant and zero order may get con fused.
3. t_1/2 – Half liefe period may be difficult to remember

Rectifications:

1. Order and molecularity are same for the single step process. But for reactions of more than one step, they may be different.
2. First order sec^-1, Zero order – mol litre^-1 sec^-1
3. t_1/2 = 0.693 / k₁ for first order reaction.

Hope you love the Samacheer Kalvi 12th Chemistry Chapter Wise Material. Clearly understand the deep concept of Chemistry learning with the help of Tamilnadu State Board 12th Chemistry Chapter 7 Chemical Kinetics Questions and AnswersPDF. Refer your friends to and bookmark our website for instant updates. Also, keep in touch with us using the comment section.