Prasanna

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.

You can Download Samacheer Kalvi 9th Social Science Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 9th Social Science History Solutions Chapter 6 The Middle Ages

The Middle Ages Textual Exercise

I. Choose the correct answer.

The Middle Ages Class 9 Question 1.
……………….. was the old religion of Japan.
(a) Shinto
(b) Confucianism
(c) Taoism
(d) Animism
Answer:
(a) Shinto

Europe In The Middle Ages 9th Standard Question 2.
…………. means great name/lord.
(a) Daimyo
(b) Shogun
(c) Fujiwara
(d) Tokugawa
Answer:
(a) Daimyo

Samacheer Kalvi Guru 9th Social Science Question 3.
The Arab General who conquered Spain was …………..
(a) Tariq
(b) Alaric
(c) Saladin
(d) Mohammad the Conqueror
Answer:
(a) Tariq

Samacheer Kalvi Guru 9th Social Question 4.
Harun-al-Rashid was the able emperor of ……………
(a) Abbasid dynasty
(b) Umayyad dynasty
(c) Sassanid dynasty
(d) Mongol dynasty
Answer:
(a) Abbasid dynasty

Class 9 History Chapter 6 Question 5.
Feudalism centred around ……………
(a) vassalage
(b) slavery
(c) serfdom
(d) land
Answer:
(a) vassalage

II. Find out the correct statement.

Question 1.
(i) Chengiz Khan was an intolerant person in religion
(ii) Mongols destroyed the city of Jerusalem
(iii) Crusades weakened the Ottoman Empire
(iv) Pope Gregory succeeded in making King Henry IV to abdicate the throne by means of Interdict
(a) (i) is correct
(b) (ii) is correct
(c) (ii) and (iii) are correct
(d) (iv) is correct
Answer:
(d) (iv) is correct

Question 2.
(i) Mangu Khan was the Governor of China.
(ii) Mongol court in China impressed Marco Polo.
(iii) The leader of Red Turbans was Hung Chao.
(iv) Mongols established their rule in China in the name of Yuan dynasty.
(a) (i) is correct
(b) (ii) is correct
(c) (ii) and (iv) are correct
(d) (iv) is correct
Answer:
(c) (ii) and (iv) are correct

Question 3.
(i) Boyang and Changon were built during Sung dynasty.
(ii) Peasant uprisings led to the collapse of Tang dynasty.
(iii) Seljuq Turks were a tribe of Tartars.
(iv) Mongols established their rule in China in the name of Yuan dynasty.
(a) (i) is correct
(b) (ii) is correct
(c) (iii) is correct
(d) (iv) is correct
Answer:
(iii) and (iv) are correct

Question 4.
Assertion (A): Buddhism went to China from India.
Reason (R): The earliest Indian inhabitants in China were the followers of Buddhism.
(a) A is correct; R is wrong
(b) Both A & R are wrong
(c) Both A &, R are correct
(d) A is wrong R is irrelevant to A
Answer:
(a) A is correct; R is wrong

Question 5.
Assertion (A): The fall of Jerusalem into the hands of Seljuk Turks led to the Crusades.
Reason (R): European Christian pilgrims were denied access to Jerusalem. .
(a) A is correct; R is not the correct explanation of A ‘
(b) A and R are correct
(c) A and R are wrong
(d) A is correct, R is the correct explanation of A
Answer:
(d) A is correct, R is the correct explanation of A

III. Fill in the blanks.

1. ……………. were the original inhabitants of Japan.
2. ……………. was the original name of Japan.
3. ………… was the original name of Medina. ,
4. ………….were the barbarians posing a threat to the Chinese in the north.
5. …………… established Ottoman supremacy in the Balkans.
Answer:
1. Ainus
2. Yamato
3. Yethrib
4. The Mongols
5. Mohammed II

IV. Match the following:

Answer:
1. (d)
2. (e)
3. (a)
4. (c)
5. (b)

V. Answer all questions given under each heading.

Question 1.
Shogunate in Japan.
(a) Name the two Daimyo families that fought for power in Japan.
Answer:
The Tara and Minamota

(b) Who emerged successful in the fight?
Answer:
Yoritomo emerged successful in the fight.

(c) What was the title given by the Emperor to the victorious?
Answer:
The high sounding title of Sei-i-tai Shogun (which means the Barbarian-Subduing-Great- General).

(d) Where was the capital of the first Shogunate established?
Answer:
The capital of the first Shogunate established at Kamakura,

Question 2.
Rule of Abbasids.
(a) Who were the Abbasids?
Answer:
The descendants of the Prophet Mohammad’s uncle Abbas and his followers were called Abbasids. ’ ,

(b) What was the title assumed by Abbasid Caliph?
Answer:
“The commander of the faithful” was the title assumed by Abbasid Caliph.

(c) Where did they have their new capital?
Answer:
Baghdad in Iraq

(d) In whose period was the Abbasid Empire at the height of its glory?
Answer:
The Abbasid Empire was at the height of its glory during the reign of Harun-al-Rashid.

VI. Answer the following briefly.

Question 1.
The Great Wall of China.
Answer:
Between 8th and 7th centuries B.C. (BCE), the warring states in China built defensive walls to protect themselves from enemies from the north. During Chin (Qin) Dynasty, the separate walls were connected and consequently the wall stretched from east to west for about 5000 kilometres. This wall, considered to be one of the wonders of the world, served to keep nomadic tribes out. The Wall was further extended and strengthened by the succeeding dynasties. Now ’ it is 6,700 kilometres in length.

Question 2.
Contribution of Arabs to Science and Technology.
Answer:
The Arabs had a scientific spirit of inquiry. In some subjects like medicine and mathematics they learnt much from India. Many Arab students went to Takshashila, which was still a great university for specialized medicine. Indian scholars and mathematicians came in large numbers to Baghdad. Sanskrit books on medicine and other subjects were translated into Arabic. In medicine and surgery, Arab physicians and surgeons earned a great reputation.

Question 3.
Impact of Crusades.
Answer:
Crusades ended the feudal relations. Many of the nobles who went to East to take part in the Crusades either stayed too long a period or did not return. The serfs took advantage of their absence to break away from their bondage to the soil. Increasing demand for products of the East led to expansion of trade. Venice, Genoa and Pisa emerged as important commercial centres in the Mediterranean region.

Constantinople ceased to be the middle man in the trade between the East and the West. The elimination of powerful nobles had its influence in strengthening the monarchy in France and England. One notable outcome of Crusades was the loss of prestige suffered by Pope and Papacy.

Question 4.
How was Feudalism organized in the Middle Ages?
Answer:

Question 5.
Write about the two instruments used by Medieval Pope to assert his authority.
Answer:
New elements were included in Christian theology. They were the theory of priesthood and the theory of sacraments. These two elements increased the power of the clergy. These two elements also helped the Church to extend its authority over all of its lay members. Excommunication and Interdict were the two instruments used against those who defied the Church.

VII. Answer the following in detail.

Question 1.
Discuss the emergence of Japan under the Shogunate.
Answer:
During the two-hundred-year rule of Fujiwaras, a new class of large landholders emerged. These landholders were also military men, called Daimyos (meaning great names-lords). The Daimyos became powerful with their retainers and armies. Involved in personal fights, they ignored the central government in Kyoto. Out of the fight between two chief families, the Tara and the Minamota, Yoritomo emerged successful. In AD (CE) 1192, the emperor gave him the high sounding title of Sei-i-tai-Shogun, which means the Barbarian-subduing-Great-General. The title carried full power to govern hereditarily. The Shogun became the real ruler. In this way began the rule of Shogunate.

Question 2.
Who were the Mongols? How did they rule China?
Answer:

1. Mongols were nomads. They came into Europe from the Steppes of Asiatic Russia.
2. They were herdsmen. ‘
3. The Mongols were experts in warfare and produced a remarkable chief, Chengiz Khan.
4. He was a great military genius.
5. His religion was Shamanism, a worship of the “Everlasting Blue Sky. Mongols’ hold over Russia for about 300 years made Russia technologically backward from the rest of Europe until the end of Middle Ages”.

Rule in China

1. The Mongols established their rule in the name of Yuan dynasty.
2. The Mongols, who overran Persia and the whole of Central Asia, did not spare China either.
3. Mangu Khan became the Great Khan in 1252 who appointed Kublai Khan the Governor of China.
4. The Mongol presence from one end of Eurasia to the other played a key role in spreading Chinese technological advances to the less developed societies in the west.
5. Though the Mongol court in Beijing impressed a foreigner like Marco Polo, the poverty of peasantry continued.
6. There were revolts of religious sects and secret societies.
7. Finally, the leader of “Red Turbans” Chu Yuan Chang took the Mongol capital Beijing and proclaimed himself emperor in 1369.
8. The Ming Empire, which replaced the Mongol empire, consciously discouraged industry and foreign trade in order to concentrate on agriculture.
9. This resulted in China lagging behind in the 16th century. ‘
10. Other parts of Eurasia, building on the techniques of the Chinese, began to march ahead.

Student Activities

Question 1.
In an outline map of Europe, the students are to sketch the extent of Ottoman Empire at the height of its glory
Answer:

Question 2.
Students are to be guided by teachers to look through Google the architectural splendours of Saracenic architecture.
Answer:
You can do this activity under the guidance of your teacher.

Assignment with Teacher’s guidance.

Question 1.
Sketching Ottoman family tree and attempting a biographical account of Saladin of Egypt and Suleiman the Magnificent of Ottoman Empire.
Answer:
The teacher can guide the students to google and find out. Narrate the entire Ottoman family tree.

Question 2.
Attempting an account of the Crusades led by Richard the Lion-Hearted of England and German Emperor Frederick Barbarossa.
Answer:
You can do this activity under the guidance of your teacher.

The Middle Ages Additional Questions

I. Choose the correct answer.

Question 1.
Historians call the period between ………………. and …………… as the Middle Ages.
(a) 470 A.D and 1400 A.D.-(C.E)
(b) 460 A.D (C.E) and 1450 A.D (C.E)
(c) 475 A.D (C.E) and 1453 A.D (C.E)
(d) 476 A.D (C.E) and 1453 A.D (C.E)
Answer:
(d) 476 A.D (C.E) and 1453 A.D (C.E)

Question 2.
The founders of Saracenic Civilization were …………….
(a) Arabs
(b) Jews
(c) Persians
(d) Syrians
Answer:
(a) Arabs

Question 3.
……………. dynasty undertook enormous public works.
(a) Sui
(b) Tang
(c) Sung
(d) Yuan
Answer:
(b) Tang

Question 4.
……………… excelled in Ceramics and Porcelain-making.
(a) Japan
(b) Korea
(c) China
(d) Europe
Answer:
(c) China

Question 5.
Japan’s name was given by a ………….. Emperor.
(a) Mongol
(b) Chinese
(c) Korean
(d) Russian
Answer:
(b) Chinese

Question 6.
…………… established Islam.
(a) Abu Bakr
(b) Abbas
(c) Prophet Mohammed
(d) None of the above
Answer:
(c) Prophet Mohammed

Question 7.
……………. is the holy city of the Christians.
(a) Jerusalem
(b) Baghdad
(c) Venice
(d) Pisa
Answer:
(a) Jerusalem

Question 8.
Chengiz Khan was the remarkable chief of ……………..
(a) Turks
(b) Arabs
(c) Mongols
(d) Chinese
Answer:
(c) Mongols

II. Find out the correct statement.

Question 1.
(i) Tang dynasty undertook enormous public works.
(ii) Land was divided into smdll peasant holdings.
(iii) Now the length of the Great Wall of China is 6,800 kilometres.
(iv) The agricultural surplus went to the Aristocrats as rents.
(a) (i) is correct
(b) (i) and (ii) are correct
(c) (iii) is correct
(d) (iv) is correct
Answer:
(b) (i) and (ii) are correct

Question 2.
(i) The message of equality and brotherhood had great appeal only for Arabs.
(ii) Mohammed and his followers stayed in their birth place.
(iii) The flight of Mohammad from Mecca in 622 A.D is called Hijrat. :
(iv) Mohammad died 20 years after the Hijrat.
(a) (i) is correct
(b) (ii) is correct
(c) (iii) is correct
(d) (iv) is correct
Answer:
(c) (iii) is correct

Question 3.
(i) Traders and artisans were brought under the feudal system.
(ii) The merchants and artisans formed guilds and groups.
(iii) In course of time they obeyed the nobles and kings. ,
(iv) This development continued the Feudal system. –
(a) (i) is correct
(b) (ii) is correct
(c) (iii) is correct
(d) (iv) is correct
Answer:
(b) (ii) is correct

Question 4.
Assertion: Seljuq Turks were a tribe of Tartars from Central Asia.
Reason: They established a powerful empire in Persia.
(a) A is correct R is wrong
(b) Both A and R are wrong
(c) Both A and R are correct
(d) A is correct R is irrelevant to A.
Answer:
(c) Both A and R are correct

Question 5.
Assertion: The Abbasid Empire was at the height of its glory during the reign of Harun-al- Rashid.
Reason: The Arab Empire flourished soon after the death of Harun-al-Rashid.
(a) A is correct; R is not the correct explanation of A
(b) A and R are correct
(c) A and R are wrong
(d) A is correct, R is the correct explanation of A .
Answer:
(a) A is correct; R is not the correct explanation of A

III. Fill in the blanks.

1. …………….. period was also a period of great prosperity to the landowning class, officials and rich merchants. .
2. The original religion of Japan, the Shinto was a mixture of nature and ……………..
3. The first great family that controlled the state was the ……………… family.
4. The first Shogunate is called the ……………. Shogunate.
5. …………… advocated simplicity and equality.
6. The capital of Umayyads was …………..
7. The other name of Baghdad was ……………..
8. …………….. ended the feudal relations.
9. The religion of Chengiz Khan was ……………..
10. …………… meant depriving a person of all the privileges of a Christian.
Answers:
1. Sung
2. ancestor worship
3. Soga
4. Kamakura
5. Islam
6. Damascus
7. the city of Arabian Nights
8. Crusades
9. Shamanism
10. Excommunication

IV. Match the following:

Answer:
1. (g)
2. (a)
3. (b)
4. (f)
5. (c)
6. (d)
7. (e)

V. Answer all questions given under each heading.

Question 1.
Yuan Dynasty
(a) Who overran Persia and China?
Answer:
The Mongols overran Persia and China.

(b) Who was appointed as the Governor of China?
Answer:
Kublai Khan

(c) Mention the foreigner who impressed the Beijing.
Answer:
Marco Polo

(d) Who was the leader of the “Red Turbans”?
Answer:
Chu Yuan Chang

Question 2.
Punic Wars
(a) Who were Carthaginians?
Answer:
The Carthaginians were the descendants of the Phoenicians who excelled in seafaring and trade.

(b) Write about the Third Punic War.
Answer:
After the defeat and destruction of the Carthage in the Third Punic War, Rome emerged as an unrivalled power in the western world.

(c) Who united to drive out the Greeks?
Answer:
Rome and Carthage united to drive out the Greeks.

(d) Who was Hannibal?
Answer:
Hannibal was a general who defeated the Roman army and made a great part of Italy a desert.

Question 3.
Silk Route
(a) What is meant by Silk route?
Answer:
The trade route from China to Asia Minor and India, known as the Silk Road or Silk Route.

(b) Which places were linked by this route?
Answer:
It linked China with the West. Goods and ideas between the two great civilizations of Rome and China were exchanged through this route.

(c) Name the good exchanged from East to West and West to East.
Answer:
Silk went westward, and wools, gold, and silver went east.

(d) Name the religion which reached China through this route.
Answer:
China received Buddhism from India via the Silk Road. .

Question 4.
Slave Trade in Rome
(a) Why was the new labour force produced by Rome?
Answer:
Rome produced a new labour force for the rich to exploit. Big landholders bought slaves cheaply and used them to cultivate their estates.

(b) What was the strength of the Slave population in the 1st century B.C.?
Answer:
The strength of the slave population in the 1st century B.C. was 3.25 million.

(c) What was the result of Slave trade?
Answer:
Slave trade led to the impoverishment of free labour. Many poor peasants had to abandon their children who also ended up in the slave markets.

(d) Which place became a great slave market?
Answer:
The island of Delos became a great slave market.

Question 5.
Arabs’ Scholarly Pursuits.
(a) What did the Abbasid Caliphs do?
Answer:
Abbasid Caliphs did not attempt to conquer new lands.

(b) What were their interest?
Answer:
They were more interested in scholarly pursuits.

(c) Name the subjects, they learnt from India.
Answer:
Medicine and Mathematics.

(d) In which field (Arabs) they earned a great reputation? .
Answer:
In medicine and surgery, Arab physicians and surgeons earned a great reputation.

VI. Answer the following briefly.

Question 1.
Write a short account of the public works undertaken by Tang dynasty.
Answer:

• Tang dynasty undertook enormous public works. Two capital cities, Boyang and Chang- on, were built.
• Scholar officials, trained in Confucius Philosophy, were appointed to counterbalance the landowning aristocratic class.
• Land was divided into small peasant holdings.
• As a result, the agricultural surplus went to the state as taxes, not to the aristocrats as rents. State monopoly of salt, and tea added to its- revenues.

Question 2.
What led to the collapse of Chin dynasty?
Answer:

• Shih Huang Ti crushed all local rulers and established a strong central government.
• However, uprisings of the peasantry, unlike in other cultures, occurred again and again in China.
• Such uprisings led to the collapse of Chin dynasty.

Question 3.
What do you know about Sung Dynasty?
Answer:

1. The rebellion of hard-pressed peasantry under the leadership of Hung Ch’ao dealt a death knell to the tottering Tang empire.
2. The empire split into five rival states, until it was reunited under a new dynasty, Sung.
3. Trade and industry flourished during the reign of Sung dynasty.
4. Iron and steel industries became highly organized.

Question 4.
Why did Oligarchy start in Greece?
Answer:

1. When the Greek City-States first emerged, they still carried the legacy of the past.
2. The rulers came from lines of traditional chieftains.
3. Those who grew rich from the expansion of trade, resented the privileges enjoyed by the old ruling families.
4. The outcome was the overthrow of the kings and the establishment of “oligarchies” in many city-states.

Question 5.
Name the Empire which replaced the Mongol Empire. What did they do?
Answer:
The Ming Empire, which replaced the Mongol empire, consciously discouraged industry and foreign trade in order to concentrate on agriculture. This resulted in China lagging behind in the 16th century. Other parts of Eurasia, building on the techniques of the Chinese, began to march ahead.

Question 6.
Write about the reign of Alexander, the Great.
Answer:

1. Under Alexander the Great, the Greeks were able to establish a kingdom in Macedonia.
2. This kingdom succeeded in annexing two historic empires of Egypt and the Middle East.
3. But the entire period of Alexander’s reign was spent on wars.
4. The Greek school of Science, Mathematics and Philosophy reached its peak in the Greek.

Question 7.
What was the message given by Islam?
Answer:
Islam gave a message of brotherhood. Mohammed laid stress on the equality of all those who were Muslims. This message of equality and brotherhood had great appeal not only for the Arabs, who were divided into warring tribes but also for people in other parts of the world.

Question 8.
What do you know about Excommunication and Interdict?
Answer:
Excommunication meant depriving a person of all the privileges of a Christian. He was denied the right to sacraments in Church. His or her body could not be buried in the consecrated ground. Interdict was to deny benefits of religion to a ruler’s subject, intended to kindle their resentment against him.

VII. Answer the following in detail.

Question 1.
Describe Feudalism in detail, with diagram.
Answer:
Despite the hold of powerful religions such as Christianity and Islam, the economic life of . people was governed by feudal relations.
In the prevailing anarchy and violence, the mighty living in strong castles seized whatever they could and the poor peasants and labourers suffered. The latter were not organized to defend their interests. There was no strong central government either to protect them. Out of this chaos and disorder evolved the feudal system.

The king, supposed to represent God on earth, was at the head of the feudal regime. Immediately after him were the great nobles, known as dukes, counts, earls. The relationship was one of a vassal. The nobles in turn had vassals of their own, dividing and distributing their fief to lesser nobles called viscounts or barons. Last in this order were the knights, whose fiefs could not be divided.

At the bottom were the villeins or serfs. In the feudal system which centered around vassalage, there was no idea of equality or freedom.
There were only rights and obligations. The Bishops, Abbots and Cardinals and the Church came under this socio-political structure. The nobility and the clergy did not do any physical work. So the burden of producing the food and other necessities of life fell on the peasants and. Feudalism artisans.

You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1

11th Maths Exercise 4.1 Question 1.

(i) A person went to a restaurant for dinner. In the menu card, the person saw 10 Indian and 7 Chinese food items. In how many ways the person can select either an Indian or a Chinese food?
Solution:
Selecting an Indian food item from the given 10 can be done in 10 ways. Selecting a Chinese food item from the given 7 can be done in 7 ways.
∴ Selecting an Indian or a Chinese food can be done in 10 + 7 = 17 ways.

(ii) There are 3 types of toy car and 2 types of toy train available in a shop. Find the number of ways a baby can buy a toy car and a toy train?
Solution:
Given, Number of toy cars = 3
Number of toy trains = 2
∴ A baby buying a toy car from 3 can be done in 3 ways
∴ A baby buying a toy train from 2 can be done in 2 ways
∴ Buying a toy car and a toy train together can be done in 3 × 2 = 6 ways

(iii) How many of two-digit numbers can be formed using 1, 2, 3, 4, 5 without repetition of digits?
Solution:
The given digits are 1, 2, 3, 4, 5
A two digit number has unit place and 10’s place. We are given 5 digits (1, 2, 3, 4, 5). The unit place can be filled (using the 5 digits) in 5 ways. After filling the unit place since repetition is not allowed one number (filled in the unit place) should be excluded. So the 10’s place can be filled (using the remaining 4 digits) in 4 ways
∴ Unit place and 10’s place together can be filled in 5 × 4 = 20 ways. So the number of two digit numbers = 20

(iv) Three persons enter in to a conference hall in which there are 10 seats. In how many ways they can take their places?
Solution:
Given, Number of the persons = 3 and Number of seats = 10
The first person can take his place (from 10 seats) in 10 ways
The second person can take his place (from the remaining 9 seats) in 9 ways
The third person can take his place (from the remaining 8 seats) in 8 ways
∴ The three persons together can take their places in 10 × 9 × 8 = 720 ways

(v) In how many ways 5 persons can be seated in a row?
Solution:
5 persons can be arranged among themselves in 5! ways
(i.e) 5 × 4 × 3 × 2 × 1 = 120 ways

11th Maths 4th Chapter Solutions Question 2.

(i) A mobile phone has a passcode of 6 distinct digits. What is the maximum number of attempts one makes to retrieve the passcode?
Solution:
Number of digits = 10
∴ Number of attempts made = 10 × 9 × 8 × 7 × 6 × 5 = 151200 ways

(ii) Given four flags of different colours, how many different signals can be generated if each signal requires the use three flags, One below the other?
Solution:
Number of flags given = 4
Number of flag needed (to show a signal) = 3
The first flag can be chosen in 4 ways (from the 4 flags)
The second flag can be chosen (from the remaining 3 flags) in 3 ways
The third flag can be chosen (from the remaining 2 flags) in 2 ways
So the first, second and the third flags together can be chosen in (to generate a signal) 4 × 3 × 2 = 24 ways
(i.e) 24 signals can be generated

11th Maths Exercise 4.1 In Tamil Question 3.
Four children are running a race.

(i) In how many ways can the first two places be filled?
Solution:
Number of children in the running race = 4
The first place can be filled in (from the 4 children) 4 ways
After filling in first place only 3 children are left out
So the second place can be filled in (from the remaining 3 children) 3 ways
So the first and the second places together can be filled in 4 × 3 = 12 ways

(ii) In how many different ways could they finish the race?
Solution:
The first and second places can be filled in 12 ways
The third place can be filled (from the remaining 2 children) in 2 ways and the fourth place can be filled in 1 way
So the race can be finished in 12 × 2 × 1 = 24 ways

11th Maths Exercise 4.1 Solutions Question 4.
Count the number of three-digit numbers which can be formed from the digits 2, 4, 6, 8? if.

(i) repetitions of digits is allowed
Solution:
Number of digit given = 4 (2,4, 6, 8)
So the unit place can be filled in 4 ways, 10’s place can be filled in 4 ways and 100’s place can be filled in 4 ways
∴ The unit place, 10’s place and 100’s place together can be filled (i.e) So the Number of 3 digit numbers = 4 × 4 × 4 = 64 ways

(ii) repetitions of digits is not allowed.
Solution:
The unit place can be filled (using the 4 digits) in 4 ways after filling the unit place since repetition of digits is not allowed that digit should be excluded.
So the 10’s place can be filled in (4 – 1) 3 ways and the 100’s place can be filled in (3 – 1) 2 ways
So the unit place, 10’s and 100’s places together can be filled in 4 × 3 × 2 = 24 ways
(i.e) The number of 3 digit numbers = 4 × 3 × 2 = 24 ways

11th Maths Exercise 4.1 Solution Question 5.
How many three-digit numbers are there with 3 in the unit place?

(i) with repetition
Solution:
with repetition
The unit place is filled (by 3) in 1 way
The 10’s place can be filled in 10 ways
The 100’s place can be filled in 9 ways (excluding 0)
So the number of 3 digit numbers with 3 unit – place = 9 × 10 × 1 = 90

(ii) without repetition
Solution:
The digits are 0, 1,2, 3, 4, 5, 6, 7, 8, 9
A three digit number has 3 digits l’s, 10’s and 100’s place.
The unit place is (filled by 3) filled in one way.
After filling the unit place since the digit ‘0’ is there, we have to fill the 100’s place. Now to fill the 100’s place we have 8 digits (excluding 0 and 3) So 100’s place can be filled in 8 ways.
Now to fill the 10’s place we have again 8 digits (excluding 3 and any one of the number) So 10’s place can be filled in 8 ways.
∴ Number of 3 digit numbers with ‘3’ in unit place = 8 × 8 × 1 = 64

Combinatorics And Mathematical Induction Question 6.
How many numbers are there between 100 and 500 with the digits 0, 1, 2, 3, 4, 5 if

(i) repetition of digits allowed
Solution:
repetition of digits allowed
The given digits are 0, 1, 2, 3, 4, 5
We have to find numbers between 100 and 500. So the 100’s place can be filled (by the numbers 1, 2, 3, 4) in 4 ways.
The 10’s place can be filled in (using 0, 1, 2, 3, 4, 5) 6 ways
and the unit-place can be filled in (using 0,1, 2, 3, 4, 5) 6 ways
But the number 100 should be excluded
So the number of numbers between 100 and 500 = 4 × 6 × 6 = 144

(ii) the repetition of digits is not allowed
Solution:
The 100’s place can be filled (by using 1, 2, 3, 4) 10’s in 4 ways
The 10’s place can be filled in (6 – 1) 5 ways and the unit place can be filled in (5 – 1) 4 ways
So the number of 3 digit number 4 × 5 × 4 = 80

10th Maths Exercise 4.1 11th Sum Question 7.
How many three-digit odd numbers can be formed by using the digits 0, 1, 2, 3, 4, 5 if

(i) The repetition of digits is not allowed
Solution:
The repetition of digits is not allowed
The given digits are 0, 1, 2, 3, 4, 5. Here the odd number are 1, 3, 5.
So the unit place can be filled in 3 ways (using the 3 odd number)
After filling the unit place since 0 is a given digit be fill the 100’s place which can be
filled in

Then the 10’s place can be filled in (6 – 2) 4 ways.
So the number of 3 digit odd numbers = 3 × 4 × 4 = 48

(ii) The repetition of digits is allowed
Solution:
The unit place can be filled in 3 ways. We are given 6 digits.
So 10’s place can be filled in 6 ways and the 100’s place can be filled in (6 – 1) (excluding zero) 5 ways
So the Number of 3 digit numbers = 3 × 6 × 5 = 90

11th Maths 4.1 Question 8.
Count the numbers between 999 and 10000 subject to the condition that there are

(i) no restriction
Solution:
no restriction
We have to find 4 digit numbers
The 1000’s place can be filled in 9 ways (excluding zero) and the 100’s, 10’s and unit places respectively can be filled in 10, 10, 10 ways (including zero)
So the number of numbers between 999 and 10000 = 9 × 10 × 10 × 10 = 9000

(ii) no digit is repeated
Solution:
Since 0 is given as a digit we have to start filling 1000’s place.
Now 1000’s place can be filled in 9 ways (excluding 0)
Then the 100’s place can be filled in 9 ways (excluding one digit and including 0)
10’s place can be filled in (9 – 1) 8 ways and unit place can be filled in (8 – 1) 7 ways So the number of 4 digit numbers are 9 × 9 × 8 × 7 = 4536 ways

(iii) at least one of the digits is repeated
Solution:
Required number of numbers = 9000 – 4536 = 4464 numbers

11th Maths Chapter 4 Exercise 4.1 Question 9.
How many three-digit numbers, which are divisible by 5, can be formed using the digits 0, 1, 2, 3, 4, 5 if

(i) The repetition of digits are not allowed?
Solution:
The repetition of digits are not allowed.
The given digits are 0, 1, 2, 3, 4, 5. A number will be divisible by 5 if the digit in the unit place is 0 or 5
So the unit place can be filled by 0 or 5

(a) When the unit place is 0 it is filled in 1 way
And so 10’s place can be filled in 5 ways (by using 1, 2, 3, 4, 5) and 100’s place can be filled in (5 – 1) 4 ways
So the number of 3 digit numbers with unit place 0 = 1 × 5 × 4 = 20

(b) When the unit place is 5 it is filled in 1 way
Since 0 is given as a digit to fill 100’s place 0 should be excluded
So 100’s place can be filled in (excluding 0 and 5) 4 ways and 10’s place can be filled in (excluding 5 and one digit and including 0) 4 ways So the number of 3 digit numbers with unit place 5 = 1 × 4 × 4 = 16
∴ Number of 3 digit numbers ÷ by 5 = 20 + 16 = 36

(ii) The repetition of digits are allowed.
Solution:
The digits are
0 1 2 3 4 5
To get a number divisible by 5 we should have the unit place as 5 or 0 So the unit place (using 0 or 5) can be filled in 2 ways.
The 10’s place can be filled (Using 0, 1, 2, 3, 4, 5) in 6 ways and the 100’s place (Using 1, 2, 3, 4, 5) can be filled in 5 ways.
So the number of 3 digit numbers ÷ by 5 (with repetition) = 2 × 6 × 5 = 60

Class 11 Maths Exercise 4.1 Solutions Question 10.
To travel from a place A to place B, there are two different bus routes B₁, B₂ two different train routes T₁, T₂ and one air route A₁. From place B to place C there is one bus route say B₁‘, two different train routes say T₁‘, T₂‘ and one air route A₁‘. Find the number of routes of commuting from place A to place C via place B without using similar mode of transportation.
Solution:

From the above diagram the number of routes from A to C
= (2 × 2 + 2 × 1) + [(2 × 1) + (2 × 1)] + [(1 × 1) + (1 × 2)]
= 4 + 2 + 2 + 2 + 1 + 2 = 13

Class 11 Maths Chapter 4 Exercise 4.1 Solution Question 11.
How many numbers are there between 1 and 1000 (both inclusive) which are divisible neither by 2 nor by 5?
Solution:
From 1 to 1000, the numbers ÷ by 2 = 500
the number ÷ by 5 = 200
and the numbers ÷ by 10 = 100(5 × 2 = 10)
So number ÷ by 2 or 5 = 500 + 200 – 100 = 600
Total numbers from 1 to 1000 = 1000
So the number of numbers which are ÷ neither by 2 nor by 5 = 1000 – 600 = 400

11th Maths Chapter 4 Question 12.
How many strings can be formed using the letters of the word LOTUS if the word

(i) either starts with L or ends with S?
Solution:
either starts with L or ends with S?
To find the number of words starting with L
Number of letters in LOTUS = 5 when the first letter is L it can be filled in 1 way only. So the remaining 4 letters can be arranged in 4! =24 ways = n(A). When the last letter is S it can be filled in the 1 way and the remaining 4 letters can be arranged is 4! = 24 ways = n(B)

(1) (1) 3! = 6 = n(A ∩ B)
Now n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 24 + 24 – 6 = 42
Now, neither words starts with L nor ends with S = 42

(ii) neither starts with L nor ends with S?
Solution:
Number of letters of the word LOTUS = 5.
They can be arranged in 5 ! = 120 ways
Number of words starting with L and ending with S = 42
So the number of words neither starts with L nor ends with S = 120 – 42 = 78

Ex 4.1 Class 11 Pdf Question 13.
(i) Count the total number of ways of answering 6 objective type questions, each question having 4 choices.
Solution:
Count the total number of ways of answering 6 objective type questions, each question having 4 choices.
One question can be answered in 4 ways
Two questions can be answered in 4 × 4 = 4² ways
∴ Six questions can be answered in 46 ways

(ii) In how many ways 10 pigeons can be placed in 3 different pigeon holes ?
Solution:
First pigeons can be placed in pigeon hole in 3 ways (selecting 1 from 3 holes)
Second pigeons can be placed in pigeon hole in 3 ways Tenth pigeons can be placed in pigeon hole in 3 ways
So total number of ways in which all the number 10 place can be sent = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 = 310 ways

(iii) Find the number of ways of distributing 12 distinct prizes to 10 students?
Solution:
To give the first prize we have to select, from the 10 students which can be done in 10 ways.
To give the second prize we have to select one from the 10 students which can be done is 10 ways.
To give the 12th prize we have to select one from 10 students which can be done in 10 ways.
So all the 12 prizes can be given in (10 × 10 × 10 …. 12 times) = 1012 ways.

Samacheerkalvi.Guru 11th Maths Question 14.
Find the value of

(i) 6!
Solution:
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

(ii) 4! + 5!
Solution:
4! + 5! = (4 × 3 × 2 × 1) + (5 × 4 × 3 × 2 × 1)
= 24 + 120 = 144

(iii) 3! – 2!
Solution:
3! – 2! = (3 × 2 × 1) – (2 × 1)
= 6 – 2 = 4

(iv) 3! × 4!
Solution:
3! × 4! = (3 × 2 × 1) × (4 × 3 × 2 × 1) = 6 × 24 = 144
12!

(v) \(\frac{12 !}{9 ! \times 3 !}\)
Solution:

(vi) \(\frac{(n+3) !}{(n+1) !}\)
Solution:

Exercise 4.1 Class 11 Question 15.
Evaluate \(\frac{n !}{r !(n-r) !}\) when

(i) n = 6,
r = 2
Solution:

(ii) n = 10,
r = 3
Solution:

(iii) For any n with r = 2
Solution:

Chapter 4 Maths Class 11 Question 16.
Find the value of n if

(i) (n + 1)! = 20(n – 1)!
Solution:

(ii) \(\frac{1}{8 !}+\frac{1}{9 !}=\frac{n}{10 !}\)
Solution:

Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.1 Additional Questions Solved

Samacheer Kalvi 11th Maths Example Sums Question 1.
If the letter of the word ‘RACHIT’ are arranged in all possible ways as listed in dictionary, then what is the rank of the word ‘RACHIT’?
Solution:
The alphabetical order of RACHIT is A, C, H, I, R and T
Number of words beginning with A = 5!
Number of words beginning with C = 5!
Number of words beginning with H = 5!
Number of words beginning with 1 = 5!
and Number of words beginning with R (i.e) RACHIT = 1
∴ The rank of the word ’RACHIT’ in the dictionary = 5! + 5! + 5! + 5! + 1 = 4 × 5! + 1
= 4 × 5 . 4 . 3 . 2 . 1 + 1 = 4 × 120 + 1 = 480 + 1 = 481

Samacheer Kalvi Guru 11th Maths Question 2.
Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.
Solution:
Any number divisible by 5, its unit place must have 0 or 5. We have to find 4-digit number greater than 6000 and less than 7000.
So, the unit place can be filled with 2 ways (0 or 5) since, repetition is not allowed.
∴ Tens place can be filled with 7 ways and hundreds place can be filled with 8 ways.
But the required number is greater than 6000 and less than 7000. So, thousand place can be filled with 1 digits (i.e) 6.

So, the total number of integers =1 × 8 × 7 × 2 = 112
Hence, the required number of integers = 112

Ex 4.1 Class 11 Question 3.
Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9 where no digits are repeated.
Solution:
Given that all the 5 digit numbers are greater than 7000.
So, the ways of forming 5-digit numbers = 5 × 4 × 3 × 2 × 1 = 120
Now, all the four digit number greater than 7000 can be formed as follows.
Thousand place can be filled with 3 ways
Hundred place can be filled with 4 ways
Tenths place can be filled with 3 ways
Units place can be filled with 2 ways
So, the total number of 4-digits numbers = 3 × 4 × 3 × 2 = 72
∴ Total number of integers = 120 + 72 = 192
Hence, the required number of integers = 192

Ch 4 Maths Class 11 Question 4.
How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if

Solution:

(c) All letters are used but the first is a vowel = 2 × 5! = 2 × 120 = 240
Hence, the required matching is
(a) ↔ (iii), (b) ↔ (i), (c) ↔ (ii)

Chapter 4 Class 11 Maths Question 5.
Five boys and 5 girls form a line. Find the number of ways of making the seating arrangement under the following condition.

Solution:
(a) Total number of arrangement when boys and girls alternate : = (5!)² + (5!)²
(b) No two girls sit together = 5! 6!
(c) All the girls sit together = 2! 5! 5!
(d) All the girls sit never together = 10! – 5! 6!
Hence, the required matching is (a) ↔ (iii), (b) ↔ (i), (c) ↔ (iv), (d) ↔ (ii)

Class 11 Chapter 4 Maths Question 6.
How many automobile license plates can be made, if each plate contains two different letters followed by three different digits?
Solution:
We have 26 English alphabet and 10 digits (0 to 9)
Since, it is given that each plate contains 2 different letters followed by 3 different digits.
∴ Number of arrangement of 26 letter taken 2 at a time

Three digit number can be formed out of 10 digit = ¹⁰P₃

Total number of license plates = 650 × 720 = 468000
Hence, the required number of plates = 468000.

Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Tamil Book Solutions Guide Pdf Chapter 1.5 ஏழுத்துகளின் பிறப்பு Text Book Back Questions and Answers, Summary, Notes.

Tamilnadu Samacheer Kalvi 8th Tamil Solutions Chapter 1.5 ஏழுத்துகளின் பிறப்பு

கற்பவை கற்றபின்

Question 1.
‘ஆய்தம்’ – இச்சொல்லில் உள்ள ஒவ்வோர் எழுத்தின் வகையையும், அது பிறக்கும் இடம் கழுத்து கழுத்து மார்பு, கழுத்து றுக்கு
Answer:

பாடநூல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
இதழ்களைக் குவிப்பதால் பிறக்கும் எழுத்துகள் …………………………
அ) இ, ஈ
ஆ) உ, ஊ
இ) எ, ஏ
ஈ) அ, ஆ
Answer:
ஆ) உ, ஊ

Question 2.
ஆய்த எழுத்து பிறக்கும் இடம் ………………………………..
அ) மார்பு
ஆ) கழுத்து
இ) தலை
ஈ) மூக்கு
Answer:
இ) தலை

Question 3.
வல்லின எழுத்துகள் பிறக்கும் இடம் ………………………..
அ) தலை
ஆ) மார்பு
இ) மூக்கு
ஈ) கழுத்து
Answer:
ஆ) மார்பு

Question 4.
நாவின் நுனி அண்ணத்தின் நுனியைப் பொருத்துவதால் பிறக்கும் எழுத்துகள் …………………..
அ) க், ங்
ஆ) ச், ஞ்
இ) ட், ண்
ஈ) ப், ம்
Answer:
இ) ட், ண்

Question 5.
கீழ்இதழும் மேல்வாய்ப்பல்லும் இணைவதால் பிறக்கும் எழுத்து ……………………..
அ) ம்
ஆ) ப்
இ) ய்
ஈ) வ்
Answer:
ஈ) வ்

பொருத்துக.

1. க், ங் – அ) நாவின் இடை, அண்ண த்தின் இடை.
2. ச், ஞ் – ஆ) நாவின் நுனி, மேல்வாய்ப்பல்லின் அடி.
3. ட், ண் – இ) நாவின் முதல், அண்ண த்தின் அடி.
4. த்,ந் – ஈ) நாவின் நுனி, அண்ண த்தின் நுனி.
Answer:
1. இ
2. அ
3. ஈ
4. ஆ

சிறுவினா

Question 1.
எழுத்துகளின் பிறப்பு என்றால் என்ன?
Answer:
உயிரின் முயற்சியால் உடலின் உள்ளிருந்து எழும் காற்றானது மார்பு, தலை, கழுத்து, மூக்கு ஆகிய நான்கு இடங்களுள் ஒன்றில் பொருந்தி, இதழ், நாக்கு, பல், மேல்வாய் ஆகிய உறுப்புகளின் முயற்சியினால் வேறுவேறு ஒலிகளாகத் தோன்றுகின்றன. இதனையே எழுத்துகளின் பிறப்பு என்பர்.

Question 2.
மெய் எழுத்துகள் எவற்றை இடமாகக் கொண்டு பிறக்கின்றன?
Answer:

• மெய்யெழுத்துகளில், வல்லின மெய் எழுத்துகள் ஆறும் மார்பை இடமாகக் கொண்டு பிறக்கின்றன.
• மெல்லின மெய் எழுத்துகள் ஆறும் மூக்கை இடமாகக் கொண்டு பிறக்கின்றன.
• இடையின மெய் எழுத்துகள் ஆறும் கழுத்தை இடமாகக் கொண்டு பிறக்கின்றன.

Question 3.
ழகர, லகர, ளகர மெய்களின் முயற்சிப் பிறப்பு பற்றி எழுதுக.
Answer:

• ழகர மெய் மேல்வாயை நாக்கின் நுனி வருடுவதால் பிறக்கிறது.
• லகர மெய் மேல்வாய்ப் பல்லின் அடியை நாக்கின் ஓரங்கள் தடித்து நெருங்குவதால் பிறக்கிறது.
• ளகர மெய் மேல்வாயை நாக்கின் ஓரங்கள் தடித்துத் தடவுதலால் பிறக்கிறது.

மொழியை ஆள்வோம்

தமிழ்மொழியை வாழ்த்திப் பாடிய வேறு கவிஞர்களின் பாடல்களைக் கேட்டு மகிழ்க.

அருள்நெறி அறிவைத் தரலாகும்
அதுவே தமிழன் குரலாகும்

பொருள்பெற யாரையும் புகழாது
போற்றா தாரையும் இகழாது

கொல்லா விரதம் குறியாகக்
கொள்கை பொய்யா நெறியாக

எல்லா மனிதரும் இன்புறலே.
என்றும் இசைந்திடும் அன்பறமே

அன்பும் அறமும் ஊக்கிவிடும்
அச்சம் என்பதைப் போக்கிவிடும்

இன்பம் பொழிகிற வானொலியாம்
எங்கள் தமிழெனும் தேன்மொழியாம்
– நாமக்கல் கவிஞர்

கீழ்க்காணும் தலைப்பில் இரண்டு நிமிடம் பேசுக.

தமிழ் எழுத்துகளின் தோற்றமும் வளர்ச்சியும்

தொல்காப்பியம் என்னும் தொன்மைத் தமிழ்நூலில் கூறப்படும் எழுத்துகள் வட்டெழுத்துகளைக் குறிப்பனவே ஆகும். இவ்வெழுத்தினைக் கோலெழுத்து, கண்ணெழுத்து எனப் பண்டைத் தமிழ் நூல்கள் குறிப்பிட்டுள்ளன. நாம் இன்று எழுதும் எழுத்துகள் ஆப்பு வடிவ எழுத்துகளிலிருந்தே தோன்றின என்று அறிஞர் சிலர் கருதுகின்றனர். கோடுகள் மிகுதியாக இருக்கக் காரணம் இதுவே என்பர்.

எகர ஒகர எழுத்துகளைக் குறிக்க எழுத்துகளின் மேல் புள்ளி வைக்கும் வழக்கம் தொல்காப்பியர் காலம் முதல் இருந்து வந்துள்ளது. ஓலைச் சுவடிகளிலும் கல்வெட்டுகளிலும் புள்ளி பெறும் எழுத்துகளை எழுதும்போது அவை சிதைந்துவிடும் என்பதால் புள்ளி இடாமல் எழுதினர். மெய்யா? உயிர்மெய்யா? குறிலா? நெடிலா?

என உணர வேண்டிய நிலை இருந்தது. இதனால் படிப்பவர்கள் பெரிதும் இடருற்றனர். இந்நிலையில் தமிழ் எழுத்துகளில் மிகப்பெரும் சீர்திருத்தத்தைச் செய்தவர் வீரமாமுனிவர். அதன் பிறகு தந்தை பெரியாரின் சீர்திருத்தமும் பெற்று தமிழ் இன்று எழுதும் தமிழாக மாறியது. தமிழ்மொழி கணினிப் பயன்பாட்டிற்கு ஏற்ற மொழியாகவும் ஆகியிருக்கிறது. நன்றி!

அகரவரிசைப்படுத்துக.

எழுத்து, ஒலிவடிவம், அழகுணர்ச்சி, ஏழ்கடல், இரண்டல்ல, ஊழி, உரைநடை, ஔகாரம், ஓலைச்சுவடிகள், ஆரம்நீ, ஈசன், ஐயம்.

அகர வரிசைப்படுத்தியன:
அழகுணர்ச்சி, ஆரம் நீ, இரண்டல்ல, ஈசன், உரைநடை, ஊழி, எழுத்து, ஏழ்கடல், ஐயம், ஒலிவடிவம், ஓலைச்சுவடிகள், ஔகாரம்

அறிந்து பயன்படுத்துவோம்.

மரபுத் தொடர்கள்:
தமிழ் மொழிக்கெனச் சில சொல் மரபுகள் உள்ளன. அவை பழங்காலம் முதலே பின்பற்றப்படுகின்றன.

பறவைகளின் ஒலிமரபு

• ஆந்தை அலறும்
• காகம் கரையும்
• சேவல் கூவும்
• குயில் கூவும்
• கோழிகொக்கரிக்கும்
• புறா குனுகும்
• மயில் அகவும்
• கிளி பேசும்
• கூகை குழறும்

தொகை மரபு

• மக்கள் கூட்டம்
• ஆநிரை
• ஆட்டு மந்தை

வரினை மரபு

• சோறு உண்
• தண்ணீ ர் குடி
• பூக் கொய்
• முறுக்குத் தின்
• கூடை முடை
• இலை பறி
• சுவர் எழுப்பு
• பால் பருகு
• பானை வனை

சரியான மரபுச் சொல்லைத் தேர்ந்தெடுத்து எழுதுக.

1. கோழி ………………………. (கூவும் / கொக்கரிக்கும்)
2. பால் …………………….. (குடி / பருகு)
3. சோறு ……………………… (தின்/ உண்)
4. பூ ………………………. (கொய் / பறி)
5. ஆ ……………………. (நிரை / மந்தை )
Answer:
1. கொக்கரிக்கும்
2. பருகு
3. உண்
4. கொய்
5. நிரை

மரபுப் பிழையை நீக்கி எழுதுக.

சேவல் கொக்கரிக்கும் சத்தம் கேட்டுக் கயல் கண் விழித்தாள். பூப்பறிக்க நேரமாகி விட்டதை அறிந்து தோட்டத்திற்குச் சென்றாள். அங்கு மரத்தில் குயில் கரைந்து கொண்டிருந்தது. பூவைப் பறித்ததுடன், தோரணம் கட்ட மாவிலையையும் கொய்து கொண்டு வீடு திரும்பினாள். அம்மா தந்த பாலைக் குடித்துவிட்டுப் பள்ளிக்குப் புறப்பட்டாள்.

Answer:
சேவல் கூவும் சத்தம் கேட்டுக்கயல் கண்விழித்தாள். பூக்கொய்ய நேரமாகிவிட்டதை அறிந்து தோட்டத்திற்குச் சென்றாள். அங்கு மரத்தில் குயில் கூவிக் கொண்டிருந்தது. பூவைக் கொய்ததுடன், தோரணம் கட்ட மாவிலையையும் பறித்துக்கொண்டு வீடு திரும்பினாள். அம்மா தந்த பாலைப் பருகிவிட்டுப் பள்ளிக்குப் புறப்பட்டாள்.

7. கட்டுரை எழுதுக.

நான் விரும்பும் கவிஞர் – பாவேந்தர் பாரதிதாசன்

முன்னுரை :
எங்கும் தமிழ்! எதிலும் தமிழ்!’ என முழங்கும் காலம் இக்காலம். இந்த முழக்கத்திற்கு – மூலமாக இருந்த பெருமக்களுள் பாவேந்தர் பாரதிதாசனார் குறிப்பிடத்தக்கவர். இவரே நான் விரும்பும் கவிஞர் ஆவார்.

பிறப்பும் இளமையும் :
கனகசுப்புரத்தினம் என்னும் இயற்பெயர் கொண்ட பாரதிதாசனார் 29.04.1891-இல் கனகசபை – இலக்குமி அம்மையாருக்கு மகனாய்ப் புதுச்சேரியில் பிறந்தார். இளமையில் தமிழாசிரியராய் அமர்ந்தார். மகாகவி பாரதியாரிடம் கொண்ட அன்பால் தம் பெயரைப் பாரதிதாசன் என வைத்துக் கொண்டார்.

தமிழ்ப்பற்று :
‘தமிழுக்கும் அழுதென்று பேர் – அந்தத்
தமிழ் இன்பத் தமிழ் எங்கள் உயிருக்கு நேர்’
போன்ற கவிதை வரிகள் பாரதிதாசனின் தமிழ்ப்பற்றை வெளிப்படுத்துவன. தமிழரின் மேன்மையை இகழ்ந்தவனை என் தாய் தடுத்தாலும் விடேன்’ என முழங்கினார். தமிழகத்தின் தமிழ்த்தெருவில் தமிழ்தான் இல்லை ‘ என வருந்தினார்.

கவிச்சுவை :
இயற்கையில் ஈடுபாடு மிக்க பாவேந்தரின் கவிதைகள் கருத்தாழமும் கற்பனைச் சுவையும் கொண்டு கற்போரைக் களிப்புறச் செய்பவை. ‘நீலவான் ஆடைக்குள் உடல் மறைந்து நிலாவென்று காட்டுகின்றாய் ஒளிமுகத்தை’ எத்தனை அழகான கற்பனை!. இது இவரின் கவிச்சுவைக்குச் சான்று.

சமுதாயப் பார்வை:
‘சாதி இருக்கின்ற தென்பானும் இருக்கின்றானே’ எனச் சாதி வெறியைச் சாடினார்.

‘எல்லார்க்கும் எல்லாம் என்று இருப்பதான
இடம் நோக்கி நடக்கின்ற திந்த வையம்’
என்ற பொதுவுடைமைக் கருத்துக்குச் சொந்தக்காரர் பாவேந்தர்.

முடிவுரை :
உடல்வளமும் , உளத்திடமும், உண்மை உரைக்கும் பண்பும், நேர்மையும், மொழிப்பற்றும், இனப்பற்றும் கொண்ட பாவேந்தரின் கனவுகளை நனவாக்குவதே நமது கடமை.

மொழியோடு விளையாடு

பொருத்தமான பன்மை விகுதியைச் சேர்த்தெழுதுக.

கல், பூ, மரம், புல், வாழ்த்து , சொல், மாதம், கிழமை, ஈ, பசு, படம், பல், கடல், கை, பக்கம், பா.

ஒருசொல் ஒரே தொடரில் பல பொருள் தருமாறு எழுதுக.

எ.கா: அணி – பல அணிகளை அணிந்த வீரர்கள், அணி அணியாய்ச் சென்றனர்.

படி, திங்கள், ஆறு.
படி : படித்துக்கொண்டிருந்த மாலதி, மாடு கறந்த ஒரு படிப்பாலை எடுத்துக் கொண்டு, படியில் ஏறிச் சென்று தாயிடம் கொடுத்தாள்.

திங்கள் : சித்திரைத் திங்களில், முதல் திங்கள் அன்று, இரவில் திங்களைப் பார்ப்பது நல்லது.

ஆறு : ஆறுமுகம், காலையில் துவைப்பதற்காக ஆறு துணிகளை எடுத்துக்கொண்டு காவேரி ஆற்றுக்குச் சென்றான்.

சொற்களை ஒழுங்குபடுத்தி முறையான தொடராக்குக.

1. வட்டெழுத்து எனப்படும் தமிழ் கோடுகளால் வளைந்த அமைந்த எழுத்து.
2. உலகம் தமிழ்மொழி வாழட்டும் உள்ளவரையிலும்.
3. வென்றதை பரணி பகைவரை ஆகும் பாடும் இலக்கியம்.
4. கழுத்து பிறக்கும் இடம் உயிரெழுத்து ஆகும்.
5. ஏகலை கலையை அம்புவிடும் தமிழ் என்றது.

ஒழுங்குபடுத்திய தொடர் :
1. வளைந்த கோடுகளால் அமைந்த தமிழ் எழுத்து வட்டெழுத்து எனப்படும்.
2. உலகம் உள்ளவரையிலும் தமிழ்மொழி வாழட்டும்.
3. பகைவரை வென்றதைப் பாடும் இலக்கியம் பரணி ஆகும்.
4. உயிரெழுத்து பிறக்கும் இடம் கழுத்து ஆகும்.
5. தமிழ் அம்புவிடும் கலையை ஏகலை என்றது.

நிற்க அதற்குத் தக…

கலைச்சொல் அறிவோம்

1. ஒலிபிறப்பியல் – Articulatory phonetics
2. மெய்யொலி – Consonant
3. மூக்கொலி – Nasal consonant sound
4. கல்வெட்டு – Epigraph
5. உயிரொலி – Vowel
6. அகராதியியல் – Lexicography
7. ஒலியன் – Phoneme
8. சித்திர எழுத்து – Pictograph

இணையத்தில் காண்க

தமிழ் வரிவடிவ எழுத்துகளில் காலந்தோறும் ஏற்பட்டுள்ள மாற்றங்களை இணையத்தில் தேடித் தொகுத்து வருக.

1. ஓவிய எழுத்துகள் :
மனிதன் ஓவியங்கள் மூலமாக தங்களுடைய எண்ணங்களை மற்றவர்களுடன் பகிர்ந்து கொண்டான். தமிழகத்தில் மல்லப்பாடி, கீழ்வாலை, கோவை, மதுரை, நீலகிரி, தருமபுரி போன்ற மாவட்டங்களில் ஓவிய எழுத்தைக் கண்டு பிடித்துள்ளனர்.

2. உருவ எழுத்து :
பழந்தமிழகத்தில் உருவ எழுத்து உணர்வு எழுத்து, ஒலியெழுத்து, தன்மை எழுத்து என வரும் எழுத்துகள் ஒவ்வொன்றும் இருபாற்பட்டு எட்டு வகையாக விளங்கிய | பாங்கும் அதில் உருவ எழுத்து ஓவியன் கைவினைத் திறன் போன்றது என்பதும் பெறப்படும். இதனால் குழுக்களாக இருந்த பழங்குடி மக்கள் பலவகைப்பட்ட எழுத்துருக்களைப் பயன்படுத்திய பாங்கும் புலப்படும்.

3. தமிழி எழுத்து :
சங்ககாலக் கல்வெட்டுகளில் காணும் எழுத்துகளை தமிழி என்பர். தமிழுக்கே உரிய. வடிவத்தைத் தமிழி என்று கூறுவதுதான் சரி. இன்றுள்ள பழமையான எழுத்து வடிவம் தமிழியே என்பதும் அதுகாலந்தோறும் மாற்றம் பெற்றதையும் அறிஞர்கள் பட்டியல் இட்டுக் காட்டியுள்ளனர். இன்றுள்ள வடிவங்களுக்கு வித்தாக உள்ளது தமிழியே.

4. கோல் எழுத்து :
ஓவிய எழுத்தில் இருந்து கல்லில் பொறிக்கக் கோடுகளாக வரைந்த குறியீடுகளைக் கோல் எழுத்து என்றும் குறிப்பர்.

5. வட்டெழுத்து :
கல்லில் வெட்டப்பட்ட எழுத்துகளே வட்டெழுத்து எனப்பட்டது என்றும், ஓலைகளில் எழுதும்போது கோடுகள் ஓலைகளைக் கிழிக்கும் என்பதால் வட்டமாக எழுதியதால் இப்பெயர் பெற்றது என்றும் கூறுவர். பாண்டிய நாட்டில் வட்டெழுத்துகள் கி.பி. எட்டாம் நூற்றாண்டு வரை வழக்கில் இருந்ததைக் கல்வெட்டுகள் காட்டும்.

இமயமலை அடிவாரத்தில் உள்ள கோபலேசுவரர் ஆலயத்தில் வட்டெழுத்துக் கல்வெட்டு தமிழில் உள்ளது. இவ்வாறு தமிழ் வட்டெழுத்து இந்தியா முழுவதும் பரவிப் பல்வேறு எழுத்து வடிவங்கள் தோன்றக் காரணமாயின தன்மையும் அறியப்படும்.

6. கிரந்த எழுத்து :
தமிழகத்தில் பல்லவர் ஆட்சி ஏற்பட்டதும் வடமொழி ஆதிக்க மொழியாயிற்று. கிரந்த எழுத்துகள் வழக்கில் வந்தன. இதனைப் பல்லவ கிரந்தம் என்றே அழைப்பர். கிரந்தம் என்ற சொல்லிற்கு நூல் என்பது பொருள். இதுவே பின்னர் எழுத்தையும் குறித்தது. தமிழ்க் கல்வெட்டுகளில் வடமொழிப் பகுதியைக் குறிக்க இந்த எழுத்துகளையே பயன்படுத்தினர். தமிழ்நாட்டில் ஆனைமலை, அழகர்மலை, திருமயம், குடுமியான்மலை, தகடூர், பேரூர் போன்ற இடங்களில் உள்ள சோழர்காலக் கல்வெட்டுகளில் கிரந்த எழுத்துகளைக் காணலாம்.

7. முதல் அச்சு எழுத்து :
தமிழ்மொழியின் முதல் அச்சுப் புத்தகம் தம்பிரான் வணக்கம் என்னும் நூல் 1578 அக்டோபர் 20ஆம் நாள் கொல்லத்தில் அச்சிடப்பட்டது. அதில் கொல்லம் எழுத்து, மலபார் எழுத்து, கோவா எழுத்து என்பன போன்ற வேறு வேறு வகையான தமிழ் எழுத்துகள் அன்று வழக்கில் இருந்ததைக் குறித்துக் காணலாம். இவ்வாறு அச்சில் கூடப் பலவகைப்பட்ட எழுத்துகள் பழக்கத்தில் இருந்ததை நாம் அறிய முடியும்.

8. வீரமாமுனிவர் செய்த மாற்றம் :
‘எகரத்துக்கும், ஒகரத்துக்கும் உட்புள்ளியை நீக்கிக் கால் இட்டார். உயிர்மெய்களில் எகர, ஒகரங்கட்கு ஒற்றைக் கொம்பு, இரட்டைக் கொம்பு அமைத்தார். அவரே சதுர வடிவான எழுத்துகளை அமைத்தார் என்றும் கூறுவர்.

9. பெரியார் செய்த மாற்றம் :
பெரியார் தமிழ் எழுத்துகள் சீர்மை பெற வேண்டும் என்று கருதினார். அதன் விளைவாக உயிர்மெய்க்கு, மெய்யுடன் கால் சேர்த்து வழங்கும் முறையைக் கொண்டார்.

கூடுதல் வினாக்கள்

சரியான விடையைத் தேர்ந்தெடுத்து எழுதுக.

Question 1.
எழுத்துகள் ………………….. இடங்களில் பிறக்கின்றன.
அ) இரண்டு
ஆ) மூன்று
இ) நான்கு
ஈ) ஐந்து
Answer:
இ) நான்கு

Question 2.
உயிரெழுத்துகளின் பிறப்பிடம் …………………….
அ) மூக்கு
ஆ) தலை
இ) மார்பு
ஈ) கழுத்து
Answer:
ஈ) கழுத்து

Question 3.
மெல்லின எழுத்துகள் பிறக்கும் இடம் ………………….
அ) மார்பு
ஆ) தலை
இ) மூக்கு
ஈ) கழுத்து
Answer:
இ) மூக்கு

Question 4.
இடையின எழுத்துகள் பிறக்கும் இடம் ………………….
அ) தலை
ஆ) மூக்கு
இ) கழுத்து
ஈ) மார்பு
Answer:
இ) கழுத்து

Question 5.
எழுத்துகளின் பிறப்பினை ……………………. வகையாகப் பிரிப்பர்.
அ) இரண்டு
ஆ) மூன்று
இ) நான்கு
ஈ) ஐந்து
Answer:
அ) இரண்டு

குறுவினா.

Question 1.
எழுத்துகளின் பிறப்பினை எத்தனை வகையாகப் பிரிப்பர்? அவை யாவை?
Answer:

• எழுத்துகளின் பிறப்பினை இரண்டு வகையாகப் பிரிப்பர்.
• அவை, இடப்பிறப்பு, முயற்சிப் பிறப்பு ஆகும்.

Question 2.
உயிரெழுத்துகள் எவற்றை இடமாகக் கொண்டு பிறக்கின்றன?
Answer:
உயிர் எழுத்துகள் பன்னிரண்டும் கழுத்தை இடமாகக் கொண்டு பிறக்கின்றன.

Question 3.
ஆய்த எழுத்து எதை இடமாகக் கொண்டு பிறக்கிறது?
Answer:
ஆய்த எழுத்து தலையை இடமாகக் கொண்டு பிறக்கிறது.

Question 4.
உயிர் எழுத்துகளின் முயற்சிப் பிறப்பு குறித்து எழுதுக.
Answer:
(i) அ, ஆ, ஆகிய இரண்டும் வாய் திறத்தலாகிய முயற்சியால் பிறக்கின்றன. இ, ஈ, எ, ஏ, ஐ ஆகிய ஐந்தும் வாய் திறக்கும் முயற்சியுடன் நாக்கின் அடி ஓரமானது மேல்வாய்ப் பல்லைப் பொருந்தும் முயற்சியால் பிறக்கின்றன.

(ii) உ, ஊ, ஒ, ஓ, ஔ ஆகிய ஐந்தும் வாய் திறக்கும் முயற்சியுடன் இதழ்களைக் குவிப்பதால் பிறக்கின்றன.

Question 5.
க், ங் – மெய்களின் முயற்சிப் பிறப்பு யாது?
Answer:
க், ங் – ஆகிய இருமெய்களும் நாவின் முதற்பகுதி, அண்ணத்தின் அடிப்பகுதியைப் பொருந்துவதால் பிறக்கின்றன.

Question 6.
ச், ஞ் – மெய்களின் முயற்சிப் பிறப்பு யாது?
Answer:
ச், ஞ் – ஆகிய இருமெய்களும் நாவின் இடைப்பகுதி, நடு அண்ணத்தின் இடைப்பகுதியைப் பொருந்துவதால் பிறக்கின்றன.

Question 7.
ட், ண் – ஆகிய மெய்களின் முயற்சிப் பிறப்பு யாது?
Answer:
ட், ண்- ஆகிய இருமெய்களும் நாவின் நுனி, அண்ணத்தின் நுனியைப் பொருந்துவதால் பிறக்கின்றன.

Question 8.
த், ந் – மெய்களின் முயற்சிப் பிறப்பு யாது?
Answer:
த், ந் – ஆகிய இருமெய்களும் மேல்வாய்ப்பல்லின் அடியை நாக்கின் நுனி பொருந்துவதால் பிறக்கின்றன.

Question 9.
ப், ம் – மெய்களின் முயற்சிப் பிறப்பு யாது?
Answer:
ப், ம் – ஆகிய இரு மெய்களும் மேல் இதழும் கீழ் இதழும் பொருந்துவதால் பிறக்கின்றன.

Question 10.
ற், ன் – மெய்களின் முயற்சிப் பிறப்பு யாது?
Answer:
ற்,ன் – ஆகிய இருமெய்களும் மேல்வாயை நாக்கின் நுனி மிகவும் பொருந்துவதால் பிறக்கின்றன.

Question 11.
ய், ர், வ் – மெய்களின் முயற்சிப் பிறப்பு யாது?
Answer:
ய் – இது நாக்கின் அடிப்பகுதி, மேல் வாய் அடிப்பகுதியைப் பொருந்துவதால் பிறக்கிறது.
ர் – இது மேல்வாயை நாக்கின் நுனி வருடுவதால் பிறக்கிறது.
வ் – இது மேல்வாய்ப்பல்லைக் கீழ் உதடு பொருந்துவதால் பிறக்கிறது.

Question 12.
சார்பு எழுத்துகளின் பிறப்பு குறித்து எழுதுக.
Answer:
ஆய்த எழுத்து வாயைத் திறந்து ஒலிக்கும் முயற்சியால் பிறக்கிறது. பிற சார்பெழுத்துகள் யாவும் தத்தம் முதலெழுத்துகள் தோன்றும் இடங்களிலேயே அவை பிறப்பதற்கு உரிய முயற்சிகளைக் கொண்டு தாமும் பிறக்கின்றன.

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Unit Exercise 3

10th Maths Unit Exercise 3 Question 1.
Solve \(\frac{1}{3}\) (x + y – 5) = y – z = 2x – 11 = 9 – (x + 2 z).
Solution:
Given

3z = 3 ⇒ z = 1
(3) becomes, 3x + 2 = 20 ⇒ 3x = 20 – 2 = 18
x = \(\frac{18}{3}\) = 6
(1) becomes, 6 – 2y + 3(1) = 5 ⇒ 9 – 2y = 5
⇒ 9 – 5 = 2y ⇒ 2y = 4
∴ y = \(\frac{4}{2}\) = 2
∴ Solution set is {6, 2, 1}

Unit Exercise 3 Question 2.
One hundred and fifty students are admitted to a school. They are distributed over three sections A, B and C. If 6 students are shifted from section A to section C, the sections will have equal number of students. If 4 times of students of section C exceeds the number of students of section A by the number of students in section B, find the number of students in the three sections.
Solution:
Let the students in section A, B, C be a, b, c, respectively.

Samacheer Kalvi 10th Maths Exercise 3.1 Question 3.
In a three-digit number, when the tens and the hundreds digit are interchanged the new number is 54 more than three times the original number. If 198 is added to the number, the digits are reversed. The tens digit exceeds the hundreds digit by twice as that of the tens digit exceeds the unit digit. Find the original number.
Solution:
Let the three digits numbers be 100a +10b + c.
100b + 10a + c = 3(100a + 10b + c) + 54 ………. (1)
100a + 106 + c + 198 = 100c + 106 + a ……… (2)
(b – a) = 2(b – c) ……… (3)
(1) ⇒ 100b + 10a + c = 300a + 30b + 3c + 54
⇒ 290a – 70b + 2c = -54
(2) ⇒ 99a – 99c = -198 ⇒ a – c = -2
⇒ a = c – 2
(3) ⇒ a + b – 2c = 0 ⇒ a + b = 2c
⇒ b = 2c – c + 2
⇒ b = c + 2
Substituting a, b in (1)
290(c – 2) – 70 (c + 2) + 2c = -54
290c – 580 – 70c – 140 + 2c = -54
222c = 666 ⇒ c = 3
a = 1, 6 = 5
∴ The number is 153.

10th Maths Unit Exercise 2 Question 4.
Find the least common multiple of
xy (k² + 1) + k(x² + y²) and
xy(k² – 1) + k (x² – y²)
Answer:
xy (k² + 1) + k(x² + y²) = k²xy + xy + kx² + ky²
= (k²xy + kx²) + (ky² + xy)
= kx(ky + x) + y (ky + x)
= (ky + x) (kx + y)
xy (k² – 1) + k(x² – y²) = k²xy – xy + kx² – ky²
= (k²xy + kx²) – xy – ky²
= kx(ky + x) -y (ky + x)
= (ky + x) (kx – y)
L.C.M. = (ky + x) (kx + y) (kx – y)
= (ky + x)(k²x² – y²)
The least common multiple is
(ky + x) (k²x² – y²)

Unit Exercise Question 5.
Find the GCD of the following by division algorithm 2x⁴ + 13.x³ + 21 x² + 23x + 7, x³ + 3x² + 3x + 1, x² + 2x + 1.
Solution:
2x⁴ + 13x³ + 27x² + 23x + 7,
x³ + 3x² + 3x + 1, x² + 2x + 1.
By division algorithm, first divide

∴ (x + 1)² is G.C.D of x³ + 3x² + 3x + 1 and x² + 2x + 1.
Next let us divide
2x⁴ + 13x³ + 27x² + 23x + 7 by x² + 2x + 1

∴ G.C.D of 2x⁴ + 13x³ + 21 x² + 23x + 7, x³ + 3x² + 3x + 1, x² + 2x + 1 is (x + 1)².

10th Maths Example Sums Question 6.
Reduce the given Rational expressions to its lowest form

Solution:

Samacheer Kalvi 10th Maths Solution Question 7.

Solution:

Samacheer Kalvi 10th Maths Answers Question 8.
Arul, Ravi and Ram working together can clean a store in 6 hours. Working alone, Ravi takes twice as long to clean the store as Arul does. Ram needs three times as long as Arul does. How long would it take each if they are working alone?
Solution:
Let Aral’s speed of working be x
Let Ravi’s speed of working be y
Let Ram’s speed of working be z
given that they are working together. ,
Let V be the quantum of work, x + y + z = \(\frac{w}{6}\) …………. (1)
Also given that Ravi takes twice the time as Aral for finishing the work.

Also Ram takes 3 times the time as Aral for finishing the work.
∴ \(\frac{w}{z}\) = 3 × \(\frac{w}{x}\)
∴ x = 3z ∴ z = \(\frac{x}{3}\)
Substitute (2) and (3) in (1),

Exercise 1.6 Class 10 Maths Samacheer Question 9.
Find the square root of 289x⁴ – 612x³ + 970x² – 684x + 361
Solution:

10th Maths Exercise 1.6 Answers Question 10.
Solve \(\sqrt{y+1}+\sqrt{2 y-5}\) = 3.
Solution:
Squaring both sides

9y² – 78y + 169 = 4 (y + 1)(2y – 5)
9y² – 78y + 169 = 4 (2y² + 2y – 5y – 5)
9y² – 78y + 169 = 8y² + 8y – 20y – 20
9y² – 78y + 169 – 8y² + 12y + 20 = 0
y² – 66y + 189 = 0
y² – 63y – 3y + 189 = 0
y(y – 63) – 3(y – 63) = 0
(y – 63)(y – 3) = 0
y = 63, 3

10th Maths Book Example Sums Question 11.
A boat takes 1.6 hours longer to go 36 kms up a river than down the river. If the speed of the water current is 4 km per hr, what is the speed of the boat in still water?
Solution:
Let the speed of boat in still water be ‘v’

⇒ 36(v + 4) – 36(v – 4) = \(\frac{8}{5}\) (v – 4) (v + 4)
⇒ 36v + 144 – 36v + 144 = \(\frac{8}{5}\) (v² – 4v + 4v – 16)
⇒ 288 = \(\frac{8}{5}\) v² – \(\frac{128}{5}\) ⇒ 8v² – 128 = 1440
⇒ 8v² = 1568 ⇒ v² = 196 v = ±14
∴ Speed of the boat = 14 km/hr. (∵ speed cannot be -ve)

10th Algebra Solutions Question 12.
Is it possible to design a rectangular park of perimeter 320 m and area 4800 m²? If so find its length and breadth.
Solution:
Let the length and breadth of the rectangle be lm and bm
Given 2(1 + b)
⇒ l + b = 160 ………. (1)
Also l b = 4800

∴ Length and breadth of the rectangular park is 120m and 40 m

10th Maths Samacheer Question 13.
At t minutes past 2 pm, the time needed to 3 pm is 3 minutes less than \(\frac{t^{2}}{4}\) Find t.
Solution:
60 – t = \(\frac{t^{2}}{4}\) – 3
⇒ t² – 12 = 240 – 4t
⇒ t² + 4t – 252 = 0
⇒ t² + 18t – 14t – 252 = 0
⇒ t(t + 18) – 14(t + 18) = 0
⇒ (t + 18) (t – 14) = 0
∴ t = 14 or t = -18 is not possible.

Samacheer Kalvi 10th Solutions Question 14.
The number of seats in a row is equal to the total number of rows in a hall. The total number of seats in the hall will increase by 375 if the number of rows is doubled and the number of seats in each row is reduced by 5. Find the number of rows in the hall at the beginning.
Solution:
Let the no of seats in each row be x

⇒ 2x² – 10x = x² + 375
⇒ x² – 10x – 375 = 0
⇒ x² – 25x + 15x – 375 = 0
⇒ x (x – 25) + 15 (x – 25) = 0
⇒ (x – 25) (x + 15) = 0
⇒ x = 25, x = -15, x > 0
∴ 25 rows are in the hall.

3rd Standard Samacheer Maths Book Solutions Question 15.
If a and b are the roots of the polynomial f(x) = x² – 2x + 3, find the polynomial whose roots are
(i) α + 2, β + 2
(ii) \(\frac{\alpha-1}{\alpha+1}, \frac{\beta-1}{\beta+1}\)
Solution:

(i) α + 2, β + 2 are the roots (given)
Sum of the roots = α + 2 + β + 2
= α + β + 4
= 2 + 4 = 6
Product of the roots = (α + 2) (β + 2)
= αβ + 2α + 2β + 4
= αβ + 2(α + β) + 4
= 3 + 2 × 2 + 4
= 3 + 4 + 4 = 11
∴ The required equation = x² – 6x + 11 = 0.

⇒ 3x² – 2x + 1 = 0

Class 3 Samacheer Kalvi Solutions Question 16.
If -4 is a root of the equation
x² + px – 4 = 0 and if the equation
x² + px + q = 0 has equal roots, find the values of p and q.
Answer:
Let p(x) = x² + px – 4
– 4 is the root of the equation
P(-4) = 0
16 – 4p – 4 = 0
-4p + 12 = 0
-4p = -12
p = \(\frac { 12 }{ 4 } \) = 3
The equation x² + px + q = 0 has equal roots
x² + 3 x + q = 0
Here a = 1, b = 3, c = q
since the roots are real and equal
b² – 4 ac = 0
3² – 4(1)(q) = 0
9 – 4q = 0
9 = 4q
q = \(\frac { 9 }{ 4 } \)
The value of p = 3 and q = \(\frac { 9 }{ 4 } \)

10thmaths Question 17.
Two farmers Senthil and Ravi cultivates three varieties of grains namely rice, wheat and ragi. If the sale (in ₹) of three varieties of grains by both the farmers in the month of April is given by the matrix.

May month sale (in ₹) is exactly twice as that of the April month sale for each variety.
(i) What is the average sales of the months April and May.
(ii) If the sales continues to increase in the same way in the successive months, what will be sales in the month of August?
Solution:

Samacheer Kalvi 10th Maths Book Answers Question 18.

Solution:

10th Maths Question 19.

Solution:

10th Algebra Chapter 3 Question 20.

Solution:

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.

You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.4

10th Maths Exercise 3.4 Samacheer Kalvi Question 1.
Reduce each of the following rational expressions to its lowest form.


Solution:

Ex 3.4 Class 10 Samacheer Question 2.
Find the excluded values, if any of the following expressions.

Solution:
(i) \(\frac{y}{y^{2}-25}=\frac{y}{(y+5)(y-5)}\) is undefined when (y + 5) (y – 5) = 0 that is y = -5, 5
∴ The excluded values are -5, 5

(ii) \(\frac{t}{t^{2}-5 t+6}\) is undefined when t² – 5t + 6 = 0 i.e.
(t – 3) (t – 2) = 0 ⇒ t = 3, 2
∴ The excluded values are 3, 2

(iii) \(\frac{x^{2}+6 x+8}{x^{2}+x-2}\) is undefined when x² + x – 2 = 0 i.e.
(x + 2) (x + 1) = 0
∴ The excluded values are 2, 1

(iv) \(\frac{x^{3}-27}{x^{3}+x^{2}-6 x}\) is undefined when x³ + x² – 6x = 0, i.e
x(x² + x – 6) = 0
x(x + 3) (x – 2) = 0
∴ The excluded values are -3, 2

Guys who are planning to learn and understand the topics of 10th Social Science Civics can grab this Tamilnadu State board solutions for Chapter 2 Central Government Questions and Answers from this page for free of cost. Make sure you use them as reference material at the time of preparation & score good grades in the final exams.

Students who feel tough to learn concepts can take help from this Samacheer Kalvi 10th Social Science Book Solutions Guide Pdf, all the Questions and Answers can easily refer in the exams. Go to the below sections and get 10th Social Science Civics Chapter 2 Central Government Tamilnadu State Board Solutions PDF.

Tamilnadu Samacheer Kalvi 10th Social Science Civics Solutions Chapter 2 Central Government

Do you feel scoring more marks in the 10th Social Science Civics Grammar sections and passage sections are so difficult? Then, you have the simplest way to understand the question from each concept & answer it in the examination. This can be only possible by reading the passages and topics involved in the 10th Social Science Civics Board solutions for Chapter 2 Central Government Questions and Answers. All the Solutions are covered as per the latest syllabus guidelines. Check out the links available here and download 10th Social Science Civics Chapter 2 textbook solutions for Tamilnadu State Board.

Central Government Textual Exercise

I. Choose the correct answer.

Samacheer Kalvi Guru 10th Social Question 1.
The Constitutional Head of the Union is …………..
(a) The President
(b) The Chief Justice
(c) The Prime Minister
(d) Council of Ministers
Answer:
(a) The President

Social Solutions For Class 10 Samacheer Kalvi Question 2.
Who is the real executive in a Parliamentary type of-government?
(a) Army
(b) The Prime Minister
(c) The President
(d) Judiciary
Answer:
(b) The Prime Minister

Kalvi Guru 10th Social Question 3.
Who among the following decides whether a Bill is a Money Bill or not?
(a) President
(b) Attorney General
(c) Parliamentary Affairs Minister
(d) Speaker of Lok Sabha
Answer:
(d) Speaker of Lok Sabha

Class 10 Social Science Civics Chapter 2 Question 4.
The Council of Ministers is collectively responsible to the:
(a) The President
(b) Lok Sabha
(c) The Prime Minister
(d) Rajya Sabha
Answer:
(b) Lok Sabha

Samacheerkalvi.Guru Social Science Question 5.
The Joint sittings of Indian Parliament for transacting legislative business are presided over by?
(a) Senior most member of Parliament
(b) Speaker of the Lok Sabha
(c) The President of India
(d) The Chairman of the Rajya Sabha
Answer:
(b) Speaker of the Lok Sabha

Samacheer Kalvi.Guru 10th Social Question 6.
What is minimum age laid down for a candidate to seek election to the Lok Sabha?
(a) 18 years
(b) 21 years
(c) 25 years
(d) 30 years
Answer:
(c) 25 years

Question 7.
The authority to alter the boundaries of state in India rest with?
(a) The President
(b) The Prime Minister
(c) State Government
(d) Parliament
Answer:
(d) Parliament

Question 8.
Under which Article the President is vested with the power to proclaim Financial Emergency
(a) Article 352
(b) Article 360
(c) Article 356
(d) Article 365
Answer:
(b) Article 360

Question 9.
The Chief Justice and other Judges of the Supreme court are appointed by …………….
(a) The President
(b) The Attorney General
(c) The Governor
(d) The Prime Minister
Answer:
(a) The President

Question 10.
Dispute between States of India comes to the Supreme Court under:
(a) Appellate Jurisdiction
(b) Original Jurisdiction
(c) Advisory Jurisdiction
(d) None of these
Answer:
(b) Original Jurisdiction

Question 11.
If you are elected as the President of India, which of the following decision can you take on your own?
(a) Nominate the leaders of your choice to the council of minister
(b) Ask for reconsideration of a bill passed by both the Houses
(c) Select the person you like as Prime Minister
(d) Dismiss a Prime Minister who has a majority in the Lok Sabha
Answer:
(b) Ask for reconsideration of a bill passed by both the Houses

II. Fill in the Blanks.

1. ………….. Bill cannot be introduced in the Parliament without President’s approval.
2. ………….. is the leader of the nation and chief spokesperson of the country.
3. ………….. is the Ex-officio Chairperson of the Rajya Sabha.
4. The President generally nominates two members belonging to the …………. community to the Lok Sabha.
5. ……………. has the right to speak and to take part in the proceedings of both Houses of the Parliament.
6. The Chief Justice and other judges of the Supreme Court hold the office up to the age of ………….. years.
7. ………….. is the Guardian of the Constitution.
8. At present, the Supreme Court consists of ………….. judges including the Chief Justice.
Answers:
1. Money
2. Prime Minister
3. Vice-President
4. Anglo-Indian
5. Attorney General of India
6. 65
7. Supreme Court
8. 28

III. Choose the Correct Statement.

Question 1.
(i) Total members of the Rajya Sabha is 250.
(ii) The 12 nominated members shall be chosen by the President from amongst persons experience in the field of literature, science, art, or social service.
(iii) The Members of the Rajya Sabha should not be less than 30 years of age.
(iv) The members of the Rajya Sabha are directly elected by the peoples.
(a) ii and iv are correct
(b) iii and iv are correct
(c) i and iv are correct
(d) i, ii and iii are correct
Answer:
(d) i, ii and iii are correct

Question 2.
(i) The Chief Justice and other judges of the Supreme Court hold the office up to the age of 62 years.
(ii) Judiciary is the third organ of the government.
(iii) The cases involving fundamental rights come under the Appellate jurisdiction of the Supreme Court.
(iv) The law declared by Supreme Court is binding on all courts within the territory of India.
(a) ii and iv are correct
(b) iii and iv are correct
(c) i and iv are correct
(d) i, and ii are correct
Answer:
(a) ii and iv are correct

Question 3.
Assertion (A): The Rajya Sabha is a permanent house and it cannot be dissolved.
Reason (R): One third of the members of Rajya Sabha retire every two years, and new members are elected to fill the seats thus vacated.
(a) (A) is false but R is true
(b) (A) is false but (R) is true
(c) Both (A) and (R) are true and (R) is the correct reason for( A)
(d) Both (A) and (R) are true and (R) is not the correct reason for (A)
Answer:
(c) Both (A) and (R) are true and (R) is the correct reason for (A)

IV. Match the Following.

Answer:
1. (c)
2. (d)
3. (a)
4. (e)
5. (b)

V. Answer the brief questions.

Question 1.
How is President of India elected?
Answer:
The President is elected by an electoral college with the system of proportional representation by means of single transferable vote.

Question 2.
What are the different categories of Ministers at the Union level?
Answer:
The different categories of Ministers at the Union level are –

1. Cabinet Ministers
2. Ministers of States and
3. Deputy Ministers

Question 3.
What is the qualification of Judges of the Supreme Court?
Answer:

1. He must be a citizen of India.
2. He should have worked as a judge of a High court for atleast 5 years.
3. He should have worked as an advocate of High Court for atleast 10 years.
4. He is in the opinion of the President a distinguished Jurist.

Question 4.
Write a short note on Speaker of the Lok Sabha?
Answer:
The office of the Speaker occupies an important position in our Parliamentary democracy. He presides over a joint sitting of the two Houses of Parliament. He has the power to decide whether a Bill is Money Bill or an ordinary one. The Speaker continues to be in the office even if the Houses dissolved, till a new speaker is elected by the new’ Lok Sabha.

Question 5.
What is Money Bill?
Answer:

1. Money Bill in general covers issue of . receipt and spending of money, borrowing by the Government, tax laws and expenditure of the Government and prevention of Black money etc.
2. Money Bill can be introduced in the Parliament (the Lok Sabha) Lower House only.

Question 6.
List out any two special powers of the Attorney General of India?
Answer:
(i) The Attorney General of India has the right to speak and to take part in the proceedings of both Houses of the Parliament or their joint sitting and any committee of the Parliament of which he may be named as a member, but without a right to vote.

(ii) In the performance of his official duties, Attorney General of India has the right of audience ‘ in all Courts in the territory of India.

VI. Answer in detail.

Question 1.
Describe the powers and functions of the President of India.
Answer:
According to Article 53 of the Constitution the executive power of the Union shall be vested in the President which shall be exercised by him directly or through subordinates to him in accordance with Constitution.

Powers of the President: Executive power.

1. Article 77 specifies that every executive action of the union shall be taken in the name of the President.
2. He appoints Prime Minister and the other council of minister on the advice of the Prime Minister, distributing Port folios to them .Wide variety of appointments made by him include appointment of Governors of States, the Chief Justice and the other Judges of the Supreme Court and High courts, the Attorney General, the comptroller and Auditor General, the Chief Election Commissioner, the Chairman and other members of the Union Public Service Commission, Ambassadors and High commissioners to other countries.
3. He appoints a commission to investigate into the conditions of SCs, STs and other backward classes.
4. The President appoints the chiefs of defence forces – Army, Navy and Air Force.

Legislative power:

1. After the General election (Election to the Lok Sabha) and also at the beginning of the first session each year the President inaugurates the session by addressing.
2. The President has to summon the Parliament.
3. All bills passed in the Parliament become “Laws of Acts” only after getting the assent of the President.
4. Money bills cannot be introduced in the Lok Sabha without his approval.
5. He has the power to dissolve the Lok Sabha before the expiry term of the House.
6. He nominates 12 persons who are eminent in various fields to Rajya Sabha.
7. He also nominates 2 persons from Anglo-Indian community to the Lok Sabha.

Financial Power:

1. Money bill can be introduced in the Parliament only with his approval.
2. Annual Budget is presented in the Parliament in the name of the President.
3. Contingency Fund is at his disposal.
4. He can make advances out of it to meet any unforeseen expenditure.
5. He constitutes a finance commission after every five years for the distribution of revenue between Centre and the State.

Judicial Power:

1. President has the power to grant pardon, reprieves, respites or remmission of punishment.
2. He is not answerable to any court of law.
3. But subjected to Impeachment.
4. Military power: President is the Supreme Commander of the Defence Forces.
5. He can declare war against a country or make peace.

Diplomatic powers:

1. The President appoints Indian diplomats to other countries and receives foreign diplomats.
2. All treaties and agreements with foreign states are entered into, in the name of the President.
3. Emergency powers: President has empowered by the Parliament to proclaim emergency arose out of war or armed rebellion, Constitution failure in the state and a threat to financial stability.
4. Being the First Citizen of India, he is the Constitutional Head and Nominal head of the country.

Question 2.
Explain the Jurisdiction and powers of the Supreme Court of India.
Answer:
The Supreme Court is the guardian of the constitution. He performs the following functions.
(i) Original Jurisdiction: The cases which are brought directly in the first instance to the Supreme Court come under original jurisdiction. These may be – (a) dispute between the government of India and one or more States, (b) dispute between two or more States and (c) the cases involving fundamental rights come under the jurisdiction of the Supreme Court.

(ii) Appellate Jurisdiction: The Supreme Court is the final appellate Court in the country. It hears appeals against the decisions of High Court in “civil, criminal and constitutional” cases with a certificate from the High Court that it is fit to appeal in the Supreme Court.

(iii) Advisory Jurisdiction The Constitution confers on the President the power to refer to the Supreme Court any question of law or fact which in his opinion is of public importance

(iv) The Law declared by the Supreme Court is binding on all courts within the territory of India.
(v) The Supreme Court also enjoys the power of judicial review.

Question 3.
What are the duties and functions of Prime Minister of India?
Answer:
The leader of the majority party in the Lok Sabha is appointed as the Prime Minister by the President.
The duties and functions of Prime Minister:

1. The Prime Minister decides the rank of his ministers and distributes various departments.
2. He decides the dates and the agenda of the meeting of the cabinet he presides.
3. The Prime Minister is the Head of the Cabinet and the other ministers are his colleagues.
4. The Prime Minister supervises the work of various ministers.
5. He consults informally two or three of his senior colleagues when he does not convene a Cabinet meeting.
6. Prime Minister should inform the affairs of the Union and proposals of legislation, all decisions of the council of ministers connecting to the government.
7. He act as the link between die President and the Council of ministers.
8. He is the leader of the nation and chief spokes person of the country.
9. As the leader of the nation the Prime Minister represents our nation at all International conferences such as SAARC, Common Wealth and Summit of Non-aligned nations.

Question 4.
Critically examine the Powers and Functions of the Parliament.
Answer:

1. The Parliament is the highest law making body in our country. It makes laws for the entire country.
2. It provides a platform where public grievances are discussed and their solutions are made
3. The Parliament is vested with powers to impeach the President and to remove judges of the Supreme Court and High Courts, Chief Election Commissioner and Comptroller and Auditor-General of India in accordance with the procedure laid down in the Constitution.
4. The Parliament exercises control over the executive by asking questions and supplementary questions, moving motions of adjournment, discussing and passing resolutions, discussing and pushing censure motion or vote of no-confidence.
5. It has the power to change the boundaries of the States.

Question 5.
List out the functions of the Lok Sabha and the Rajya Sabha.
Answer:
Functions of the Lok Sabha:

1. Any bill can be introduced and passed in the Lok Sabha (including Money bill)
2. It has the power in passing any bill for Constitutional Amendment.
3. Lok Sabha member have the power to elect the President and the Vice President.
4. It can participate in case of impeachment of the President and the judges of the Supreme Court.
5. Motion of no confidence can only be introduced in Lok Sabha. If it is passed then the Prime Minister and the other council of ministers need to resign from their post.

Functions of Rajya Sabha:

1. Any bill (except the money bill) needs to be approved by Rajya Sabha to get passed.
2. If the bill is not approved for more than six months then President calls for a joint session of both the houses to resolve the dead lock.
3. It has the same power like Lok Sabha for passing any bills for Constitutional Amendment.
4. Together with the members of Lok Sabha and all the State Legislative Assemblies they elect the President and Vice President.
5. It has the power in the impeachment procedure of the President and judges of the Supreme Court and High Court.
6. Rajya Sabha can also create or abolish an All India service, and has the power to make a state list subject into National importance with two third majority to support it.

Central Government Additional Questions

I. Choose the correct answer.

Question 1.
The Council of Ministers is headed by the ……………….
(a) Prime Minister
(b) Speaker
(c) President
(d) Vice – President
Answer:
(a) Prime Minister

Question 2.
Our country has …………….. system of government.
(a) Aristocratic
(b) Monarchial
(c) Parliamentary
(d) Oligarchy
Answer:
(c) Parliamentary

Question 3.
Which are the two states of India where the President’s rule was imposed for maximum number of times?
(a) Kerala and Maharashtra
(b) Kerala and Punjab
(c) Punjab and Haryana
(d) Goa and Tamil Nadu
Answer:
(b) Kerala and Punjab

Question 4.
The President of India resides at …………….. in New Delhi.
(a) Raj Bhavan
(b) Sastri Bhavan
(c) Rashtrapathi Bhavan
(d) Vignyan Bhavan
Answer:
(c) Rashtrapathi Bhavan

Question 5.
The members of the Rajya Sabha are elected for a term of ……………
(a) Six years
(b) Five Years
(c) Four Years
(d) Seven years
Answer:
(a) Six years

Question 6.
The …………….. occupies the second highest office in the country.
(a) Vice-President
(b) Prime Minister
(c) Chief Minister
(d) Cabinet Minister
Answer:
(a) Vice-President

Question 7.
Oath of the President is administered by ……………
(a) Vice President
(b) Lok Sabha Speaker
(c) Chief Justice of India
Answer:
(c) Chief Justice of India

Question 8.
The leader of the majority party in the …………….. is appointed by the President as the Prime Minister.
(a) Committee
(b) Rajya Sabha
(c) Lok Sabha
(d) Cabinet
Answer:
(c) Lok Sabha

Question 9.
Judges of Supreme Court is appointed by ……………
(a) Lok Sabha Speaker
(b) Vice President
(c) President
Answer:
(c) President

Question 10.
The first Prime Minister of India was ……………..
(a) Pandit Jawaharlal Nehru
(b) Thiru. Lai Bahadur Sastri
(c) Mrs Indira Gandhi
(d) Thiru Narendra Modi
Answer:
(a) Pandit Jawaharlal Nehru

II. Fill in the blanks.

1. The Lok Sabha can be dissolved by the ………….. before the expiry of its term on the advice of the Prime Minister.
2. The Rajya Sabha does not have any power to amend or reject the …………… Bill.
3. The salaries and allowances of the Prime Minister and the Ministers are determined by the ………………
4. The Supreme Court is the final ……………. court in the country.
5. The ………….. of India is the highest law officer in the country.
6. A joint meeting of Lok Sabha and Rajya Sabha is presided by ……………
7. The annual financial statement is caused to be laid before both Houses of Parliament by ……………
8. The Lower House of the Indian Parliament is called ……………….
9. The …………. is a permanent House and it can not be dissolved.
10. Presently the Lok Sabha consists of ………… members.
11. The Lok Sabha is presided over by the …………..
12. …………. is the third organ of the Government.
13. The ………….. is the final appellate court in the country.
14. There are ………….. different levels of court in our country.
15. President of India is elected by elected members of …………., ………….. and ………….
Answer:
1. President
2. Money
3. Parliament
4. Appellate
5. Attorney General
6. Speaker of Lok Sabha
7. President
8. The Lok Sabha
9. Rajya Sabha
10. 545 Members
11. Speaker
12. Judiciary
13. Supreme Court
14. Three
15. Lok Sabha, Rajya Sabha and State Assemblies

III. Choose the correct statement.

Question 1.
(i) The power of the judiciary to declare a law as unconstitutional is known as judicial review.
(ii) Our Constitution provides for a democratic form of Government.
(iii) All Bills passed by the Parliament become “Law of Acts” only after getting the assent of the President.
(iv) The leader of the majority party in the Lok Sabha is appointed by the Chief Justice of the Supreme Court as the Prime Minister
(a) i, ii and iv are correct
(b) ii and iv are correct
(c) i, ii and iii are correct
(d) i, iii and iv are correct
Answer:
(c) i, ii and iii are correct

Question 2.
(i) The Lok Sabha is the popular House of the Indian Parliament and contains elected representatives of the people.
(ii) At present, the Lok Sabha consists of 547 Members.
(iii) The Lok sabha is a permanent House and it cannot be dissolved.
(iv) The Lok Sabha can only introduce Money Bill
(a) i and ii are correct
(b) i, ii and iv are correct
(c) ii, iii and iv are correct
(d) i, and iv are correct
Answer:
(d) i, and iv are correct

IV. Match the following.

A.

Answer:
1. (c)
2. (d)
3. (e)
4. (b)
5. (a)

B.

Answer:
1. (b)
2. (d)
3. (e)
4. (a)
5. (c)

C.

Answer:
1. (c)
2. (e)
3. (a)
4. (b)
5. (d)

V. True or False.

1. The Constitution of Independent India adopted the principle of Universal Adult Franchise.
2. The opposition parties play no role in the functioning of a Government.
3. Not a single seat is reserved in the Parliament for SCs and STs.
4. The total membership of the Lok Sabha is 500.
5. When the Parliament is in session, it begins with a question hour.
6. The Supreme Court is presided over by the Chief Justice of India.
7. Poor people can approach only the lower courts.
8. The judiciary acts as per the wishes of the Government.
9. The public interest litigation plays an important role in facilitating justice to all.
10. The civil cases usually begin with the lodging of a First Information Report (FIR) with
police who investigate the crime after which a case is filed in the court.
Answers:
1. True
2. False
3. False
4. False
5. True
6. True
7. False
8. False
9. True
10. False

VI. Answer in brief.

Question 1.
What does the Electoral College consist of?
Answer:
The Electoral College consists of the elected members of both the houses of the Parliament, the elected members of the states and elected members of National Capital Territory of Delhi and Puducherry.

Question 2.
What do you know about the term of the Rajya Sabha?
Answer:
The Rajya Sabha is a permanent House. It means that it cannot be dissolved. The members of the Rajya Sabha are elected for a term of six years. One third of its members retire every two years and new members are elecied to till the seats. The Vice – President of India is the Ex- officio Chairperson of the Rajya Sabha.

Question 3.
Our Parliament is a “Bicameral Legislature”. How?
Answer:
If any legislature has two Houses the Upper House and the Lower House it is said to be Bicameral Legislature. Since our Parliament has two Houses namely Lok Sabha (House of the elected people) and the Rajya Sabha (House of the States) it is termed as “Bicameral Legislature”

Question 4.
What is the qualification of Attorney General of India?
(OR)
Who can be appointed as the Attorney General of India?
Answer:
One who wants to become the Attorney General of India must

1. be a citizen of India
2. have been a judge of some High Court for five years or an advocate of some High Court for 10 years or eminent jurist, in the opinion of the President.

Question 5.
How the members of Lok Sabha are elected?
Answer:

1. The members of the Lok Sabha are directly elected by the people of the Constituencies on the basis of population.
2. Universal Adult Franchise is followed while electing the members of the Lok Sabha.
3. All Indian citizens above 18 years of age who are registered as voters will vote for their representatives.

Question 6.
What is Parliament?
Answer:
The Legislature is known as the Parliament. It consists of two houses, namely Rajya Sabha and the Lok Sabha.

Question 7.
What does the Collegium consist of?
Answer:

1. Collegium consists of four senior-most judges of the Supreme Court and 3 members of concerned High court with Chief Justice as Head.
2. The appointment and transfer of the judges in Supreme Court and high courts is taken by Collegium.

Question 8.
How is the Vice-President of India elected?
Answer:
The Vice-President occupies the second highest office in the country. He is not elected directly by the people but by the method of indirect election. He is elected by the members of an electoral college consisting of the member of both Houses of Parliament.

Question 9.
What are the three sessions of the Parliament?
Answer:
Parliament sessions should be held at least twice in a year.
The three sessions of the Parliament

1. Budget session – From February to May
2. Monsoon Session – From July to September
3. Winter session – From November to December.

Question 10.
Name the different Parliament Sessions.
Answer:

1. Budget Session from February to May
2. Monsoon Session from July to September
3. Winter Session from November to December

Question 11.
What do you mean by integrated Judiciary?
Answer:
An integrated Judiciary means a single judicial hierarchy for the whole country.

Question 12.
What do you mean by the executive?
Answer:
An executive is a group of people who work together to implement the laws made by Parliament.

Question 13.
Who is the leader of the ruling party in the Lok Sabha?
Answer:
The Prime Minister is the leader of the ruling party in the Lok Sabha.

Question 14.
Name the ministries which are housed in the North Block.
Answer:
The ministries in the housed in the North Block are The Ministry of Finance and the Ministry of Home Affairs.

Question 15.
Who nominates the 12 members of the Rajya Sabha?
Answer:
The President of India nominates the 12 members of the Rajya Sabha.

VII. Answer in detail.

Question 1.
Describe briefly the different categories of the ministers at the Union level.
Answer:
The ministers at the Union level are classified under three ranks:
(i) Cabinet Ministers:- The Cabinet is an informal body of senior ministers who form the nucleus of administration. All the important decisions of the Government are taken by the Cabinet, such as defence, finance, external affairs and home. The Cabinet recommends to the President to promulgate an ordinance. It is instrumental in moving amendments to the Constitution. These Finance Bills have their origin in the Cabinet and they are introduced in Lok Sabha with the President’s recommendations.

(ii) Ministers of State:- These ministers belong to the second category of ministers in the council. They are also in charge of ministries or departments but they do not participate in the meetings of the Cabinet unless invited to do so.

(iii) Deputy Ministers:- They are the lowest-ranked ministers in the Cabinet. They assist either the ministers of Cabinet or State in the performance of the duties entrusted to them.

Question 2.
How the Vice President elected? What are the qualification needed to elect as Vice President?
Answer:

1. He should be a citizen of India.
2. He must have completed the age of thirty five years.
3. He must not hold any office of profit under the Union, State or Local government.
4. He should have the other qualifications required to become a member of the Rajya Sabha. Election and term of-the Vice-President is five years. His office may terminate earlier than the fixed term either by resignation, death or by removal.
5. He is eligible for re-election.
6. The Constitution does not provide a mechanism of succession to the office of the Vice President.
7. The Deputy Chairman of the Rajya Sabha can perform the duties of the Chairman of the Rajya Sabha (the vice president).

We think the data given here clarify all your queries of Chapter 2 and make you feel confident to attempt all questions in the examination. So, practice more & more from Tamilnadu State Board solutions for 10th Social Science History Chapter 2 Central Government Questions and Answers & score well. Need any information regarding this then ask us through comments & we’ll give the best possible answers very soon.

We think the data given here clarify all your queries of Chapter 2 and make you feel confident to attempt all questions in the examination. So, practice more & more from Tamilnadu State Board solutions for 10th Social Science Civics Chapter 2 Central Government Questions and Answers & score well. Need any information regarding this then ask us through comments & we’ll give the best possible answers very soon.

Guys who are planning to learn and understand the topics of 10th Social Science Geography can grab this Tamilnadu State board solutions for Chapter 7 Human Geography of Tamil Nadu Questions and Answers from this page for free of cost. Make sure you use them as reference material at the time of preparation & score good grades in the final exams.

Students who feel tough to learn concepts can take help from this Samacheer Kalvi 10th Social Science Book Solutions Guide Pdf, all the Questions and Answers can easily refer in the exams. Go to the below sections and get 10th Social Science Geography Chapter 7 Human Geography of Tamil Nadu Tamilnadu State Board Solutions PDF.

Tamilnadu Samacheer Kalvi 10th Social Science Geography Solutions Chapter 7 Human Geography of Tamil Nadu

Do you feel scoring more marks in the 10th Social Science Geography Grammar sections and passage sections are so difficult? Then, you have the simplest way to understand the question from each concept & answer it in the examination. This can be only possible by reading the passages and topics involved in the 10th Social Science Geography Board solutions for Chapter 7 Human Geography of Tamil Nadu Questions and Answers. All the Solutions are covered as per the latest syllabus guidelines. Check out the links available here and download 10th Social Science Geography Chapter 7 textbook solutions for Tamilnadu State Board.

Human Geography of Tamil Nadu Textual Exercise

I. Choose the correct answer.

Human Geography Of Tamil Nadu Question 1.
The delta which is known as Granary of South India is ……………….
(a) Cauvery delta
(b) Mahanadi delta
(c) Godavari delta
(d) Krishna delta
Answer:
(a) Cauvery delta

Samacheer Kalvi 10th Geography Book Question 2.
Second staple food of the people of Tamil Nadu is:
(a) Pulses
(b) Millets
(c) Oilseeds
(d) Rice
Answer:
(b) Millets

Question 3.
Literacy rate of Tamil Nadu as per 2011 census is ……………
(a) 80.32%
(b) 62.33%
(c) 73.45%
(d) 80.33%
Answer:
(d) 80.33%

Question 4.
A major hydro-electric power project of Tamil Nadu is:
(a) Mettur
(b) Papanasam
(c) Sathanur
(d) Tungabhadra
Answer:
(a) Mettur

Question 5.
Number of major and minor ports in Tamil Nadu are
(a) 3 and 15
(b) 4 and 15
(c) 3 and 16
(d) 4 and 16
Answer:
(a) 3 and 15

II. Fill in the blanks.

1. Agriculture of Tamil Nadu constitutes …………….. % of its economy.
2. Sathanur dam is constructed across the river …………..
3. Tamil Nadu ranks …………….. in India with a share of over 20% in total road projects under operation in the Public – Private Partnership (PPP).
4. ………….. is the third largest airport in India after Mumbai and Delhi.
5. The difference between the value of exports and imports is called ………….
Answers:
1. 21
2. Thenpennai
3. Second
4. Chennai International Airport
5. Balance of trade

III. Match the following.

Answer:
1. (b)
2. (d)
3. (a)
4. (c)

IV. Questions 1-2 are assertion and reasoning type

Question 1.
Assertion (A): Coimbatore, Tiruppur and Erode region is called as The Textile Valley of Tamil Nadu.
Reasoning (R): They contribute a major share to the state’s economy through textiles.
(a) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true but, (R) does not explain (A)
(c) (A) is true but (R) is false
(d) (A) is false but (R) is true
Answer:
(a) Both (A) and (R) are true and (R) explains (A)

Question 2.
Assertion (A): The Nilgiris is the least populated district of Tamil Nadu
Reasoning (R): It is located in the western most part of Tamil Nadu.
(а) Both (A) and (R) are true and (R) explains (A)
(b) Both (A) and (R) are true but, (R) does not explain (A)
(c) (A) is true but (R) is false
(d) (A) is false but (R) is false
Answer:
(b) Both (A) and (R) are true but, (R) does not explain (A)

V. Answer the following in briefly.

Question 1.
Explain the cropping seasons of Tamil Nadu.
Answer:

Question 2.
Why is Coimbatore called the Manchester of Tamil Nadu?
Answer:
Coimbatore has the ideal conditions for cotton cultivation – Humid weather in the early stages and hot weather during the harvest period. Cotton cultivation and the textile industries are the reasons to call Coimbatore as the “Manchester of Tamil Nadu”.

Question 3.
Name the important multipurpose projects of Tamil Nadu.
Answer:
Mettur Dam, Amaravathi Dam, Papanasam Dam, Bhavani Sagar Dam.

Question 4.
What is MRTS?
Answer:

1. Mass Rapid Transport System (MRTS) a well established suburban railway network. Currently developing a Metro system, with its first underground operation since May 2017.
2. It is mainly started to manage the crowd during peak hours. The elevated metro system. Connects the heart of the city from North to South.

Question 5.
List out the air ports and sea ports of Tamil Nadu.
Answer:
Airports:

1. Chennai International Airport
2. Coimbatore International Airport
3. Madurai International Airport
4. Tiruchirapalli International Airport

Domestic Airports:

1. Tuticorin and Salem

Sea Port: Major Sea Ports are:

1. Chennai
2. Ennore and
3. Tuticorin
   Intermediate port at Nagapattinam and 15 minor ports.

Question 6.
Have you heard about any stampede in your district? write about that incident briefly.
Answer:
During Athivaradhar festival minor stampede that brokeout on Thursday (18^th July 2019) the crowd flocked to Kancheepuram as it was an auspicious day after the lunar eclipse. Four people died.

On 21^st April 2019 atleast seven people died in stampede in Tamil Nadu during a temple festival at Muthuaiyampalayam Karuppusamy temple in Thuraiyur a village near Trichy (Sunday).

VI. Distinguish between the following.

Question 1.
Marine Fishing and Inland fishing
Answer:

Question 2.
Food Crops and Non- Food Crops
Answer:

Question 3.
Surface Water and Ground Water
Answer:

VII. Give reasons for the following

Question 1.
Farmers switch over from inorganic to organic farming.
Answer:
By continuous usage of chemical fertilizers, pesticides and insecticides the soil became unfit for cultivation after some years. It causes a threat to agriculture. So to bring back the soil to fertility and to make it fit for cultivation farmers switch over from inorganic to organic farming. Organic wastes, biological pest control crop residues and animal manure only used under organic farming.

Question 2.
Cities are densely populated than the villages.
Answer:
Agriculture, job opportunities and industrial development are the main causes of population density in the cities then the villages.

Question 3.
Karur is Called the Textile Capital of Tamil Nadu.
Answer:
Karur is an industry town and is very famous for cottage and hand-loom textile industries. Moreover state’s most of the textile goods exports is from Karur district. That is why Karur is called the textile capital of Tamil Nadu.

Question 4.
Mostly stampede occurs in Temples.
Answer:
Most of the religious festivals are located in areas like banks of rivers hilly terrains or mountain tops. These areas lack proper pathways, posing a geographical risk to the pilgrims. It often disrupts the orderly movement of crowds resulting in irrational and dangerous movement for self-protection, leading to injuries and fatalities. Because of the lack of physical infrastructure, the mob behaviour increases the sense of threat, which results in a stampede.

VIII. Answer the following in a paragraph.

Question 1.
Write about the plantation farming of Tamil Nadu.
Answer:
Plantation Crops: Tea, coffee, cashew, rubber, pepper and cinchona.

The hill slopes with laterite soil and acidic-nature is ideal for the plantation crops.

Tea: Tea plantations are found in the hills of the Nilgins and Coimbatore. Notable region for tea estates Nilgiris.

Coffee: Coffee plants are grown in the hills of Western Ghats as well as Eastern Ghats. It is also found in the hilly slopes of Dindigui, Madurai, Theni and Salem districts.
Notable places: Yercaud, Kolli hills and Kodaikanal,

Rubber: Rubber plantations are significant in Kanyakumari.

Cashew: It is extensively cultivated in Cuddalore district.

Pepper: Confined to the warm and wet slopes of Eastern and Western Ghats of Tamil Nadu.

Cinchona: It is planted at heights varying from 1060 to 1280 metres in Anaimalai hills.

Cardamom: Cardamom estates are located at few places in the hills of Madurai region at an elevation of 915 to 1525 metres.

Position: Tamil Nadu stands second in area and production of tea and coffee next to Assam and Karnataka respectively.

Question 2.
Give an account on water resources of Tamil Nadu.
Answer:

1. Tamil Nadu constitutes 4% of India’s land area and is inhabited by 6% of India’s population, but has-only 2.5% of India’s water resources.
2. More than 95% of the surface water and 80% of the ground water have already been put into use.
3. Major uses of water include human/animal consumption irrigation and industrial use.
4. The state is heavily dependent on monsoon rains.
5. The annual average rainfall is around 930 mm (47% during the northeast monsoon, 35% during the south west monsoon, 14% in summer and 4% in winter.
   

Question 3.
Bring out the mineral distribution in Tamil Nadu.
Answer:
(i) Tamil Nadu is the leading holder of country’s resources of vermiculite, magnetite, dunite, rutile, garnet, molybdenum and ilmenite.

(ii) The state accounts for the country’s 55.3% of lignite, 75% of vermiculite, 69% of dunite, 59% of garnet, 52% of molybdenum and 30% of titanium mineral resources.
Important minerals are found in the state are as follows:

1. Neyveli has large lignite resources.
2. Coal is available in Ramanathapuram
3. Oil gas are found in the Cauvery basin
4. Iron deposits are found in Kanjamalai region in Salem district and Kalrayan Malai region of Tiruvannamalai district.
5. Magnesite ores are available near Salem.
6. Bauxite is found in Servarayan Hills, Kotagiri, Udagamandalam, Palani and Kollimalai areas.
7. Gypsum is obtained in Tiruchirappalli, Tirunelveli, Thoothukudi and Virudhunagar districts.
8. Ilmenite and rutile are found in the sands of Kanyakumari beach.
9. Limestone is available in Coimbatore, Cuddalore, Dindigul, Kancheepuram, Karur, Madurai, Nagapattinam, Namakkal, Perambalur, Ramanathapuram, Salem and Tiruvallur districts.
10. Magnesite is obtained in Coimbatore, Dharmapuri, Karur, Namakkal, the Nilgiris, Salem.
11. Feldspar, quartz, copper and lead are found in some parts of the state.

Question 4.
State the densely populated regions of Tamil Nadu and account for its high density.
Answer:

1. The number of persons living per square km is referred as population density.
2. A place is said to have high density of population if number of persons living per sq km is above 800.
3. As per 2011 census the density of population in Tamil Nadu is 555 per sq.km

Densely populated regions of Tamil Nadu:
Chennai is the densest district – 26,903 persons per sq.km followed by Kanyakumari (1106), Tiruvallur (1049), Kancheepuram (927), Madurai (823), Coimbatore (748), Cuddalore (1702), Thanjavur (691), Nagapattinam (668-), Salem (663), Vellore (646), Tiruchirappalli (602).

Question 5.
Explain the different modes of transport available in Tamil Nadu.
Answer:
Roadways:
The State has a total road length of 167,000 km, In which 60,628km are maintained by state Highways Department. It ranks second in India with a share of over 20% in total road projects under operation in the Public-Private Partnership (PPP) model.

Railways:

• Tamil Nadu has a well-developed rail network as part of Southern Railway, headquartered at Chennai.
  The present Southern Railway network extends over a large area of India’s southern peninsula, covering Tamil Nadu, Kerala, Puducherry, minor portions of Karnataka and Andhra Pradesh.
• Tamil Nadu has a total railway track length of 6,693 km with 690 railway stations in the state.
• The system connects it with most of the major cities in India.
• Main rail junctions in the state include Chennai, Coimbatore, Erode, Madurai, Salem, Tiruchirappalli and Tirunelveli.
• Chennai has a well-established suburban railway network, a Mass Rapid Transport System (MRTS) and is currently developing a Metro system, with its first underground stretch in operation since May 2017.

Airways:

• Tamil Nadu has four major international airports. Chennai International Airport is currently the third largest airport in India after Mumbai and Delhi.

Water ways:

• Tamil Nadu has 3 major ports. They are Chennai, Ennore and Tuticorin. It has an intermediate port at Nagapattinam and 15 minor ports. Ennore intermediate port was recently converted as a major port and handles the major coal and ore traffics in Tamil Nadu.

Question 6.
Write about Road safety rules.
Answer:
In recent years the number of road accidents in Tamil Nadu has been increasing. It is reported that 15% of the road accidents of the country takes place in Tamil Nadu.

Basic road safety rules:

1. Aware of the road signals
2. Stop, look and cross
3. Listen and ensure whether a vehicle is approaching
4. Don’t rush on roads
5. Crossroads in pedestrian crossings
6. Don’t stretch hands while driving vehicles.
7. Never crossroad at bends and stay safe in a moving vehicle.
8. Avoid speeding, Drunk and driving, use helmets and seat belts and follow traffic rules.

“Know – Risk! No – risk!”

IX. Map Study

Question 1.
Mark the areas of major crops, minerals, dams, airport and seaports.
Answer:

Human Geography of Tamil Nadu Additional Questions

I. Choose the correct answer.

Question 1.
Tea, Coffee, rubber, pepper and cashew are …………. crops.
(a) Food
(b) Fiber
(c) Plantation
Answer:
(c) Plantation

Question 2.
The word Agriculture is derived from the ……………….. words “ager and Cultura”.
(a) Latin
(b) Greek
(c) Tamil
(d) Spanish
Answer:
(a) Latin

Question 3.
Rice, Cotton, Sugercane are grown with …………….
(a) Drying farming
(b) Irrigation farming
(c) Cattle reasing
Answer:
(b) Irrigation farming

Question 4.
In Tamil Nadu Rice Research Institute is situated at:
(a) Aduthurai
(b) Madurai
(c) Thiruvallur
(d) Chennai
Answer:
(a) Aduthurai

Question 5.
Millets are ……………… crops.
(a) Wet crops
(b) Dry crops
(c) Plantation
Answer:
(b) Dry crops

Question 6.
To promote ……………….. a central scheme named ‘National Project on Organic Farming’was launched.
(a) mixed farming
(b) organic farming
(c) aquaculture
(d) plantation farming
Answer:
(b) organic farming

Question 7.
Which is known as the Textile capital of Tamil Nadu?
(a) Salem
(b) Karur
(c) Tirunelveli
Answer:
(b) Karur

Question 8.
……………….. is also known as ‘poor man’s cow’.
(a) Goat
(b) Sheep
(c) Buffalo
(d) Horse
Answer:
(a) Goat

Question 9.
The most populated district in Tamil Nadu is …………….
(a) Madurai
(b) Tiruvallur
(c) Chennai
Answer:
(c) Chennai

Question 10.
districts lead in the Inland fish production.
(a) Chennai
(b) Cuddalore
(c) Sivagangai
(d) Vellore
Answer:
(d) Vellore

II. Fill in the blanks:

1. …………. irrigation is most predominant irrigation system in TamilNadu.
2. ……….. Dam is one of the biggest earthen dams in the country.
3. Amaravathi reservoir is notable for the …………….
4. Papanasam dam is also known as ………….. dam.
5. Parappalar project is located near …………….
6. Tamil Nadu occupies ……………… position in the country in silk production.
7. The Vellore district is the top exporter of finished …………. in the country.
8. Tamil Nadu accounts for about ………….. of India’s …………. exports.
9. Tamil Nadu ranks among the Indian states in population density.
10. ………… in considered the ‘fire works capital’ of India.
Answers:
1. Well
2. Bhavani
3. Mugger crocodiles
4. Karaiyar
5. Ottanchatram
6. Fourth
7. leather goods
8. 17%, Software
9. 12th
10. Sivakasi

III. Match the following.

a.

Answer:
1. (c)
2. (a)
3.(d)
4.(b)

b.

Answer:
1. (b)
2. (a)
3. (d)
4. (e)
5. (c)

IV. Answer briefly:

Question 1.
What is Agriculture?
Answer:
Agriculture is a practice of farming that includes cultivation of crops, rearing of animals, birds, forestry, fisheries and other related activities.

Question 2.
What is the Multipurpose River Valley Projects?
Answer:
Multipurpose river valley projects are basically designed for the development of irrigation for agriculture and hydropower generation they are used for many other purposes as well.

Question 3.
List out the major food crops and commercial crops of the state.
Answer:

1. The principal food crops – paddy, millets and pulses.
2. The commercial crops – sugarcane, cotton, oilseeds, spices, tea, and coffee and cashew.

Question 4.
Define Trade.
Answer:
Trade-in an exchange of goods and commodities either within the country or between countries.

Question 5.
Write a note on Tamil Nadu Rice Research Institute.
Answer:

1. Tamil Nadu Rice Research Institute was established in April 1985 under Tamil Nadu Agricultural University.
2. It is situated at AduthUrai (Thanjavur).
3. It’s function is to perform lead function for rice and rice based cropping system research with the help of existing agriculture colleges and research centers.

Question 6.
Name any four districts with low density of population.
Answer:
Nilgiris, Perambalur, Dharmapuri and Sivagangai.

Question 7.
What are minor ports?
Answer:
Minor ports are anchorage ports where cargo is transhipped from the vessel to the store.
Ex. Cuddalore, Nagapattinam, Kolachal and Rameshwaram.

Question 8.
What are the uses of Chinchona?
Answer:

• Chinchona is a forest product.
• From chinchona quinine a drug is extracted.
• Malaria is treated with this medicine.

Question 9.
Write the expansion of these abbreviated terms.

1. TNAU
2. TANTEA
3. TNPL
4. TANCEM
5. Gl
6. CLRI
7. SEZ
8. TTDC

Answer:

1. TNAU: Tamil Nadu Agriculture University
2. TANTEA: Tamil Nadu Tea Plantation Corporation Limited
3. TNPL: Tamil Nadu Newsprint and Papers Limited
4. TANCEM: Tamil Nadu Cements Corporation Limited
5. GI: Geographical Indication
6. CLRI: Central Leather Research Institute –
7. SEZ: Special Economic Zone
8. TTDC: Tamil Nadu Tourism-Development Corporation

Question 10.
Name the factors influencing agriculture.
Answer:
The factors influencing agriculture can be classified as Physical, Social and economic factors.

• Physical factors includes soil, temperature, rainfall, humidity, climate and slope of land.
• The social factors includes traditional knowledge belief and myths of farmers, farm size and holdings and fanners acceptance towards innovation.
• Economic factors are market loan assistance, government subsidy and incentives.

V. Distinguish:

Question 1.
Wet farming and Dry farming
Answer:

Question 2.
Commercial crops and Plantation crops
Answer:

Question 3.
Imports and Exports
Answer:

VI. Answer the following in detail.

Question 1.
Tourism is considered as an industry – justify. Give an account on Tourism in Tamil Nadu.
Answer:

1. Tourism is considered as an industry because of its enormous potential in creating employment for a large number of people.
2. Approximately 28 lakh foreign and 11 crore domestic tourists visit our state (Tamil Nadu) annually.
3. Reasons: The presence of ancient temples and monuments, hill stations pilgrim centres, a variety of natural land scapes, long coast line along with rich culture and heritage make Tamil Nadu the best destination for tourists.
4. Promoter: Tourism in Tamil Nadu is promoted by Tamil Nadu Tourism Development Corporation (TTDC).
5. The state currently ranks the highest among Indian states with about 25 crore arrivals (in 2013).

Question 2.
Give an account for the distribution of Textile industry in Tamil Nadu.
Answer:
Textile industry is one of the traditionally well-developed industries in Tamil Nadu. Tamil Nadu has a major share in the Indian Textile industry in terms of production and export of yams, fabrics, knitwear and garments. Tamil Nadu contributes 30% of India’s share in export of cotton, yam and fabrics.

Tropical climate availability of raw materials, demand for cotton in market, power supply from numerous power projects and abundant cheap labour are favourable factors for widespread distribution of textile industries in Tamil Nadu. The textile mills are concentrated in Coimbatore, Tirupur, Salem, Palladam, Kamr, Dindigul, Vimdhunagar, Tirunelveli, Thoothukudi, Madurai and Erode. Tamil Nadu has about 3,50,000 power looms manufacturing cotton fabrics.

Maximum units are concentrated in and around Coimbatore region for this region it is known as the “Manchester of South India”. Timpur and Erode contributes much for the state economy, therefore they are referred to as ‘Textile Valley of Tamil Nadu’.

Timpur alone contributes 70% of export of knitwear of Tamil Nadu. Erode specializes in garments and bedspreads. The city of Kamr is known as ‘Textile capital of Tamil Nadu Silk Textiles’. Tamil Nadu occupies fourth place in silk textile production in our country Kancheepuram silk is unique in its quality and is known for its traditional value all over the world. Arani, Rasipuram and Thimbuvanam are other silk centres of Tamil Nadu.

Sericulture Training Institute in Hosur training farmers to adopt agriculture along with farm work to accelerate rural industrialization Mettur, Madurai and Ramanathapuram are specialized areas for manufacturing synthetic clothes.

We think the data given here clarify all your queries of Chapter 7 and make you feel confident to attempt all questions in the examination. So, practice more & more from Tamilnadu State Board solutions for 10th Social Science Geography Chapter 7 Human Geography of Tamil Nadu Questions and Answers & score well. Need any information regarding this then ask us through comments & we’ll give the best possible answers very soon.

Students can Download Tamil Chapter 1.5 அணி இலக்கணம் Questions and Answers, Summary, Notes Pdf, Samacheer Kalvi 7th Tamil Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 7th Tamil Solutions Term 3 Chapter 1.5 அணி இலக்கணம்

மதிப்பீடு

குறுவினா

Question 1.
உவமை, உவமேயம், உவம உருபு விளக்குக.
Answer:
ஒரு சொல்லை மற்றொரு சொல்லோடு ஒப்பிட்டுக் கூறுவது உவமை அல்லது உவமானம் ஆகும். உவமையால் விளக்கப்படும் பொருள் உவமேயம் ஆகும். உவமை உவமேயம் இரண்டுக்கும் இடையில் வரும் உருபு உவம உருபு ஆகும்.

Question 2.
உவமை அணிக்கும் எடுத்துக்காட்டு உவமை அணிக்கும் உள்ள வேறுபாடு யாது?
Answer:

கற்பவை கற்றபின்

பின்வரும் தொடர்களில் உள்ள உவமை, உவமேயம், உவம உருபு ஆகியவற்றைக் கண்டறிந்து எழுதுக.

கூடுதல் வினாக்கள்

நிரப்புக.

1. அணி என்னும் சொல்லுக்கு அழகு என்பது பொருள்.
2. ஒரு செய்யுளைச் சொல்லாலும், பொருளாலும் அழகு பெறச் செய்தல் அணி எனப்படும்.
3. போல, புரைய , அன்ன, இன்ன, அற்று, இற்று, மான, கடுப்ப, ஒப்ப, உறழ போன்றவை உவம உருபுகள் ஆகும்.
4. ஒரு பாடலில் உவமையும் உவமேயமும் வந்து உவம உருபு வெளிப்படையாக வந்தால் அது உவம அணி எனப்படும்.
5. ஒரு பாடலில் உவமை ஒரு தொடராகவும் உவமேயம் ஒரு தொடராகவும் வந்து உவம உருபு மறைந்து வந்தால் அது எடுத்துக்காட்டு உவமை அணி எனப்படும்.
6. உலகில் இல்லாத ஒன்றை உவமையாகக் கூறுவது இல்பொருள் உவமை அணி ஆகும்.

விடையளி :

Question 1.
வமையணி விளக்குக.
Answer:
அணி விளக்கம்: ஒரு பாடலில் உவமையும், உவமேயமும் வந்து உவம உருபு வெளிப்படையாக வருவது உவமையணி ஆகும்.
அணி அமைந்த பாடல்:
அகழ்வாரைத் தாங்கும் நிலம்போலத் தம்மை
இகழ்வார்ப் பொறுத்தல் தலை.
பாடல் பொருள் : பூமி தன்னைத் தோண்டுபவரைப் பொறுத்துக் கொள்வதுபோல நாம் நம்மை இகழ்ந்து பேசுபவரைப் பொறுத்து கொள்ள வேண்டும்.
அணி பொருத்தம் : பூமி தன்னைத் தோண்டுபவரைப் பொறுத்துக் கொள்ளுதல் என்பது உவமை. நாம் நம்மை இகழ்ந்து பேசுபவரைப் பொறுத்துக் கொள்ள வேண்டும் என்பது உவமேயம். போல’ என்பது உவம உருபு.

Question 2.
எடுத்துக்காட்டு உவமையணி விளக்குக.
Answer:
அணி இலக்கணம் : உவமை ஒரு தொடராகவும் உவமேயம் ஒரு தொடராகவும் வந்து உவம உருபு மறைந்து வருவது எடுத்துக்காட்டு உவமை அணி ஆகும்.
எ.கா. :
தொட்டனைத்து ஊறும் மணற்கேணி மாந்தர்க்குக்
கற்றனைத்து ஊறும் அறிவு.

பாடலின் பொருள் : மணற்கேணியில் தோண்டிய அளவிற்கு நீர் ஊறும். மனிதர்கள் கற்கும் அளவிற்கு ஏற்ப அறிவு பெருகும் என்பதாகும்.

அணி பொருத்தம் :
தொட்டனைத்து ஊறும் மணற்கேணி என்பது உவமை.
மாந்தர்க்குக் கற்றனைத்து ஊறும் அறிவு என்பது உவமேயம்.
இடையில் அதுபோல ன்னும் உவம உருபு மறைந்து வந்துள்ளது.

Question 3.
இல்பொருள் உவமையணி விளக்குக.
Answer:
அணி இலக்கணம் : உலகில் இல்லாத ஒன்றை உவமையாகக் கூறுவது இல்பொருள் உவமை அணி ஆகும்.

எ.கா. :
1. மாலை வெயிலில் மழைத்தூறல் பொன்மழை பொழிந்தது போல் தோன்றியது.
2. காளை கொம்பு முளைத்த குதிரை போலப் பாய்ந்து வந்தது.

அள்ளி பொருத்தம் : உலகில் பொன் மழையாகப் பொழிவது இல்லை, கொம்பு முளைத்த குதிரையும் இல்லை. இவ்வாறு உலகில் இல்லாத ஒன்றை உவமையாகக் கூறப்பட்டதால் இல்பொருள் உவமையணிக்குச் சான்றாயின.

மொழியை ஆள்வோம்

கேட்க.

புகழ்பெற்ற கவிஞர் ஒருவரது உரையின் ஒலிப்பதிவைக் கேட்டு கிழ் . மாணவர்கள் தாங்களாகவே புகழ்பெற்ற கவிஞர் ஒருவரது உரையின் ஒலிப்பதிவை கேட்டு மகிழ வேண்டும்.

கீழ்க்காணும் தலைப்புகளுள் ஒன்று பற்றி இரண்டு நிமிடம் பேசுக

Question 1.
நான் விரும்பும் கவிஞர்.
Answer:
நான் விரும்பும் கவிஞர் – மகாகவி பாரதியார் :
நான் விரும்பும் கவிஞரான பாரதியாரைப் பற்றிப் பேச வந்துள்ளேன். இந்தியத் தாயின் மடியில் ஆனந்தமாய்த் தவழ வேண்டிய நாம் அடிமைகளாய்ச் சுருண்டு கிடந்ததைப் பொறுக்காமல், நம்மை மீட்க நம்மிடையே சுதந்திர உணர்வையும் எழுச்சியையும் வீரம் மிகுந்த தன்னுடைய பாடல்களால் ஏற்படுத்தியவர் பாரதியார். இவர் சாதி, மதம், இனம், மொழி, மத வேறுபாடுகளின்றி அனைவரையும் தம் பாடல்களினால் ஒன்றிணைத்தவர்.

மகாகவி பாரதியார் தூத்துக்குடி மாவட்டம் எட்டையபுரத்தில் 1882 ஆம் ஆண்டு டிசம்பர் கு 11 ஆம் தேதி சின்னசாமி ஐயர் – இலக்குமியம்மாள் தம்பதியருக்கு மகனாகப் பிறந்தார். இவரது இயற்பெயர் சுப்பிரமணியன் ஆகும்.

தன் இளம் வயதிலேயே தமிழில் கவிதைகள் பாடி அனைவரையும் கவர்ந்தவர். இவரது 16ஆவது வயதிலேயே எட்டையபுரம் அரசவையில் கவிதை பாடி பட்டத்தைப் பெற்றார். பாரதி என்றால் கலைமகள் என்பது பொருளாகும். தமிழ், வடமொழி, ஆங்கிலம் என்ற பல மொழிகளில் தன் புலமையினை மேம்படுத்திக் கொண்டவர்.

பாரதியாரின் உள்ளத்தில் விடுதலை வேட்கை எப்போதும் கொழுந்து விட்டு எரிந்து கொண்டே இருக்கும். அதன் வெளிப்பாடே அவருடைய பாடல்கள். அவர் தம் பேச்சாலும் எழுத்தாலும் விடுதலை வேட்கையைத் தூண்டியவர். அதுமட்டுமா? வ.உ.சி. சுப்பிரமணிய சிவா முதலான விடுதலைப் போராட்ட வீரர்களுடன் சேர்ந்து விடுதலை பற்றியே கலந்துரையாடுவார். விடுதலைப் போராட்டத்தில் முழு மூச்சாக ஈடுபட்டார். அவர் பாடிய பாடல்களும் கட்டுரைகளும் விடுதலையைப் பற்றி மட்டுமே எதிரொலித்தன.

பாரதியாருடைய அகன்று விரிந்த விசாலப் பார்வை இந்திய நதிகளை இணைத்தது. மாநிலங்களை இணைத்தது. அதற்குச் சான்று சிந்து நதியின் மிசை நிலவினிலே…. என்ற பாடல் வரிகள் தான். கங்கை நதிப்புறத்துக் கோதுமையைப் பெற்றுக் கொண்டு, காவிரி வெற்றிலையை மாறு கொள்வோம்’ என்று பாடியுள்ளார்.

இந்தியாவின் ஒன்றுபட்ட நிலைக்குத் தமிழ்ச் சமுதாயம் இணைந்து செயல்பட தம் பாடல்களைப் பார் முழுவதும் பரவச் செய்தவர் பாரதியார். இவர் விடுதலைப் போராட்டக் காலத்தில் மக்களை நோக்கி, இந்த நாடு நம் எல்லோருக்கும் சொந்தம், நாம் எல்லோரும் இந்த நாட்டுக்குச் சொந்தம் என்று பாடினார்.

பாரதியார் நம்பிக்கையின் மறு உருவம் என்று சொன்னால் அது மிகையாகாது. ஏனெனில் இந்தியா விடுதலை கிடைப்பதற்கு முன்பாகவே
“ஆடுவோமே பள்ளுப் படுவோமே
ஆனந்த சுதந்திரம் அடைந்துவிட் டோமென்று” – என்று ஆடிப் பாடினார்.

பல மொழிகளைக் கற்றறிந்த பாரதி, “யாமறிந்த மொழிகளிலே தமிழ்மொழி போல் இனிதாவ தெங்கும் காணோம்” என்று தமிழ் மொழியின் சிறப்பை எடுத்துரைத்துள்ளார்.

சாதிக் கொடுமைகள், பெண்ணடிமை, சமூக ஏற்றத்தாழ்வுகள் என்று அனைத்தையுமே இந்நாட்டிலிருந்து நீக்க வேண்டும் என்பதில் உறுதியாக இருந்தவர்.

வளமான, வலிமையான பாரதத்திற்குத் தேவையான சிறந்த வழிகள் யாவும் அவருடைய பாடல்களில் உள்ளன. அதை நாம் பின்பற்றினால் அவர் கனவில் கண்ட பாரதத்தை நாம் நிகழ்காலத்தில் உருவாக்க முடியும் என்பதைக் கூறி வாய்ப்பளித்தமைக்கு நன்றி கூறி விடை பெறுகிறேன்.

Question 2.
எனக்குப் பிடித்த பாடல்.
Answer:
அனைவருக்கும் வணக்கம் ! எனக்குப் பிடித்த நாட்டுப்புறப் பாடல் பற்றி இங்குப் பேச வந்துள்ளேன்.
நாட்டுப்புறப் பாடல்கள் ஏட்டில் எழுதப்படாத ஓர் இலக்கியம். வழி வழியாக முன்னோர்கள் பாடியதைக் கேட்டும், அதனோடு தங்கள் இரசனைக்கேற்றவாறு அவற்றில் வார்த்தைகளைச் சேர்த்தும் நாட்டுப்புறப் பாடல்கள் உருவாயின. நாட்டுப்புறப் பாடல்களில் சொல்லப்படாத கருத்துகள் கிடையாது. மிகக் கடினமான கருத்துகளைக் கூட மிகச் சுலபமாக பாடிவிடுவார்கள். ஆழமான கருத்துகளுடன் ழகர, லகர, ளகர எழுத்துகளின் உச்சரிப்பு மற்றும் றகர, ரகர உச்சரிப்புகள், நாப்பிறழ்ச்சி இல்லாமல் பாடும் பயிற்சி
ஆகியவை நாட்டுப்புறப் பாடல்களினால் கிடைக்கின்றன.

(i) “கடலையிலே ஒரு உரல்
உரளுது பெரளுது
தத்தளிக்குது தாளம் போடுது
கடலை தளர உழுது
கல கடலை விதைச்சேன்
கல கடலையும் கல கடலையாச்சு.”

(ii) “வியாழக்கிழமை ஏழைக்கிழவன்
வாழைப்பழம் வழுக்கி
கீழே விழுந்தான்.”

வெள்ளையர் ஆட்சியை உள்ளே நுழையவிட்டது நம் தவறு. அதை வெகு அழகாகக் ஒரு நாட்டுப்புறப் பாடல் கூறுகிறது :

“ஓரான் ஓரான் தோட்டத்திலே
ஒருவன் போட்டது வெள்ளரிக்காய்
காசுக்கு ரெண்டு விற்கச் சொல்லிக்
காயிதம் போட்டானாம் வெள்ளைக்காரன்
வெள்ளைக்காரன் பணம் வெள்ளிப்பணம்
வேடிக்கைக் காட்டுதாம் சின்ன பணம்.”

இவ்வாறு நாட்டுப்புற இலக்கியங்கள் பல செய்திகளை எடுத்துக் கூறுகின்றன. இவ்விலக்கியம் எளிமையான சொற்களால் அமைக்கப்பட்டுள்ளதால், அனைவராலும் இதனை படித்து உணர முடியும். எனக்கு பேச வாய்ப்பளித்தமைக்கு நன்றி கூறி விடைபெறுகிறேன்.

சொல்லக் கேட்டு எழுதுக

1. மாடுகள் கொண்டு நிலத்தை உழுதனர்.
2. நீர்வளம் மிக்க ஊர் திருநெல்வேலி.
3. நெல்லையில் தமிழ்க் கவிஞர் பலர் வாழ்ந்தனர்.
4. அகத்தியர் வாழ்ந்த மலை பொதிகை மலை.
5. இல்லாத பொருளை உவமையாக்குவது இல்பொருள் உவமை அணி.

பாடலைப் படித்து வினாக்களுக்கு விடையளிக்க

பனை மரமே பனை மரமே
ஏன் வளந்தே இத் தூரம்?
குடிக்கப் பதனியானேன்!
கொண்டு விற்க நுங்கானேன்! தூரத்து மக்களுக்குத்
தூதோலை நானானேன் !
அழுகிற பிள்ளைகட்குக்
கிலுகிலுப்பை நானானேன்!
கைதிரிக்கும் கயிறுமானேன் !
கன்று கட்டத் தும்புமானேன்! ……….(- நாட்டுப்புறப்பாடல் )

வினாக்கள் :

Question 1.
பனை மரம் தரும் உணவுப்பொருள்கள் யாவை?
Answer:
பதனி, நுங்கு ஆகியவை பனை மரம் தரும் உணவுப் பொருள்களாகும்.

Question 2.
பனை மரம் யாருக்குக் கிலுகிலுப்பைத் தரும்?
Answer:
பனை மரமானது, அழுகிற பிள்ளைக்குக் கிலுகிலுப்பையைத் தருகிறது.

Question 3.
தூதோலை’ என்னும் சொல்லைப் பிரித்து எழுதுக.
Answer:
தூதோலை = தூது + ஓலை.

Question 4.
பனைமரம் மூலம் நமக்குக் கிடைக்கும் பொருள்களைப் பட்டியலிடுக.
Answer:
பதனி, நுங்கு, தூதோலை, கிலுகிலுப்பை, கயிறு, தும்பு முதலியன பனைமரம் மூலம்
நமக்குக் கிடைக்கும் பொருள்களாகும்.

Question 5.
பாடலுக்கு ஏற்ற தலைப்பை எழுதுக.
Answer:
பனை தரும் வரம்.

பின்வரும் தலைப்பில் கட்டுரை எழுதுக

என்னைக் கவர்ந்த நூல்

என்னைக் கவர்ந்த நூல் – திருக்குறள் :
கல் தோன்றி மண் தோன்றாக் காலத்தே முன் தோன்றிய மூத்தகுடி தமிழ்குடி’ என்பது முன்னோர் வாக்கு. உலகின் மிகத் தொன்மையான தமிழ்மொழி பண்டைக் காலத்திலிருந்து தற்காலம் வரை நமக்கு பல நூல்களைத் தந்து கொண்டேயிருக்கிறது. ஆனாலும் இன்றளவும் உலக மக்களால் போற்றப்படும் உலகப் பொதுமறையாம் திருக்குறளைப் பற்றி இக்கட்டுரையில் காண்போம்.

உலகப்பொதுமறை :
அறம், பொருள், இன்பம் ஆகியவற்றை செம்மையுற நமக்குக் கூறும் திருக்குறள் குறிப்பிட்ட ஒரு நாட்டினருக்கோ , மொழியினருக்கோ மட்டும் உரித்தன்று. உலகம் முழுவதிற்கும் சொந்தமானது. உலகிலேயே அதிக மொழிகளில் மொழிபெயர்க்கப்பட்ட நூல்களில் திருக்குறள் மூன்றாம் இடத்தில் உள்ளது. இதுவரை 107 மொழிகளில் மொழிபெயர்க்கப்பட்டுள்ளது. வெளிநாட்டினருக்கும் இங்கு வந்து தமிழ் கற்று திருக்குறளைப் படித்து, பின் தங்கள் மொழிகளில் மொழிபெயர்த்துள்ளனர்.

அறங்கள் கூறும் திருக்குறள் :
திருக்குறளில் மனிதனுக்குச் சொல்லாத அறங்களே கிடையாது. சாதாரண மனிதன் முதல் மன்னன் வரை அனைவருக்கும் கூறப்பட்டுள்ள கருத்துகள் பொதுவானவை; எந்தக் காலத்திற்கும் பொருந்தக் கூடியவை. திருவள்ளுவர் சாதாரணக் குடிமகனாக வாழ்ந்தவர்தான். ஆனால் அரசன், துறவி, குடும்பத்தலைவன் என்று அனைவருக்கும் வாழ்வியல் நெறிகளைக் கூறியுள்ளார். திருக்குறளைப் படிக்க படிக்க இன்பமும் பண்பும் வளரும்.

முடிவுரை :

“வள்ளுவன் தன்னை உலகினுக்கே தந்து
வான் புகழ் கொண்ட தமிழ்நாடு’ என்று பாரதி புகழ்ந்துள்ளார். ஏழே சீர்களில் உலக நீதியைச் சொல்லும் திருக்குறளே நான் இன்றும் என்றும் விரும்பும் நூலாகும்.

மொழியோடு விளையாடு

குறுக்கெழுத்துப் புதிர்.

தமிழ்நாட்டில் உள்ள ஊர்ப் பெயர்களையும் அவற்றின் சிறப்பையும் அறிவோம்.


Answer:

தொடருக்குப் பொருத்தமான உவமையை எடுத்து எழுதுக

Question 1.
என் தாயார் என்னை ……………. காத்து வளர்த்தார்.
(கண்ணை இமை காப்பது போல/ தாயைக் கண்ட சேயைப் போல)
Answer:
கண்ணை இமை காப்பது போல

Question 2.
நானும் என் தோழியும் ……………… இணைந்து இருப்போம்.
(இஞ்சி தின்ற குரங்கு 9 போல/ நகமும் சதையும் போல)
Answer:
நகமும் சதையும் போல

Question 3.
திருவள்ளுவரின் புகழை ………….. உலகமே அறிந்துள்ளது.
(எலியும் பூனையும் போல/ உள்ளங்கை நெல்லிக்கனி போல)
Answer:
உள்ளங்கை நெல்லிக்கனி போல

Question 4.
அப்துல் கலாமின் புகழ் ………………… உலகெங்கும் பரவியது.
(குன்றி மேலிட்ட விளக்கு
போல/ குடத்துள் இட்ட விளக்கு போல)
Answer:
குன்றின் மேலிட்ட விளக்கு போல

Question 5.
சிறு வயதில் நான் பார்த்த நிகழ்ச்சிகள். ………….. என் மனத்தில் பதிந்தன.
(கிணற்றுத்தவளை போல/ பசுமரத்தாணி போல)
Answer:
பசுமரத்தாணி போல

கொடுக்கப்பட்டுள்ள ஊரின் பெயர்களில் இருந்து புதிய சொற்களை உருவாக்குக

(எ.கா.) திருநெல்வேலி – திரு, நெல், வேலி, வேல்
1. நாகப்பட்டினம் – நாகம், பட்டினம், படி, பட்டி, கப்பம், நாடி, நா
2. கன்னியாகுமரி – கன்னி, குமரி, கனி, கரி, யா, கயா
3. செங்கல்பட்டு – செங்கல், பட்டு, படு, செல், பல், கல், பகல், பகட்டு
4. உதகமண்டலம் – மண், மண்டலம், மடம், உண், கண், தடம், கமண்ட லம், கடம்
5. பட்டுக்கோட்டை – பட்டு, கோட்டை, கோடை, படை, பட்டை, கோ, கோடு

நிற்க அதற்குத் தக

என் பொறுப்புகள்

1. நகரங்களின் சிறப்புத் தன்மையை அறிந்து போற்றுவேன்.
2. ஒவ்வொரு ஊரிலும் நடைபெறும் தொழில்களின் சிறப்பை அறிந்து தொழில் செய்வோரை மதிப்பேன்.

கலைச்சொல் அறிவோம்