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TN State Board 12th Physics Model Question Paper 5 English Medium

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
3. All questions of Part I, II, III and IV are to be attempted separately
4. Question numbers 1 to 15 in Part I are Multiple choice Questions of one mark each.  These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

PART – 1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
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

Question 2.
If λ_v , λ_x, λ_m, represent the wavelength of visible light, X-rays and microwaves, respectively, then
(a) λ_m > λ_v > λ_x
(b) λ_m > λ_x > λ_v
(c) λ_v > λ_m > λ_x
(d) λ_v > λ_x > λ_m
Answer:
(a) λ_m > λ_v > λ_x

Question 3.
The materials used in Robotics are:
(a) Aluminium and silver
(b) silver and gold
(c) Copper and gold
(d) Steel and aluminium
Answer:
(d) Steel and aluminium

Question 4.
Two wires of A and B with circular cross-section are made up of the same material with equal lengths. If R_A = 3R_B, then what is the ratio of radius of wire A to that of B?
(a) 3
(b) √3
(c) \(\frac{1}{\sqrt{3}}\)
(d) \(\frac{1}{3}\)
Answer:
(c) \(\frac{1}{\sqrt{3}}\)

Question 5.
The frequency range of 3 MHz to 30 MHz is used for
(a) Ground wave propagation
(b) Space wave propagation
(c) Sky wave propagation
(d) Satellite communication
Answer:
(c) Sky wave propagation

Question 6.
A ray of light strikes a glass plate at an angle 60°. If the reflected and refracted rays are perpendicular to each other, the refractive index of the glass is,
(a) √3
(b) \(\frac{3}{2}\)
(c) \(\sqrt{\frac{3}{2}}\)
(d) 2
Hint. Angle of refraction r = 60° ; Angle of incident i = 30°
sin i = n × sin r
n = \(\frac{\sin 30^{\circ}}{\sin 60^{\circ}}=\sqrt{3}\)
Answer:
(a) √3

Question 7.
If voltage applied on a capacitor is increased from V to 2V choose the correct conclusion
(a) Q remains the same, C is doubled
(b) Q is doubled, C doubled
(c) C remains same, Q doubled
(d) Both Q and C remain same
Answer:
(c) C remains same, Q doubled

Question 8.
The nucleus is approximately spherical in shape. Then the surface area of the nucleus having mass number A varies as
(a) A^2/3
(b) A^4/3
(c) A^1/3
(d) A^5/3
Hint. Volume of nucleus is proportional to mass number 4
\(\frac { 4 }{ 3 }\) π R³ ∝ A : R = R₀ A^1/3
So, 4 π R² = R₀ A^2/3 ⇒ 4 π R² ∝ A^2/3
Surface area is proportional to (mass number)23
Answer:
(a) A^2/3

Question 9.
The given electrical network is equivalent to

(a) AND gate
(b) OR gate
(c) NOR gate
(d) NOT gate
Answer:
(c) NOR gate

Question 10.
A wire of length / carries a current I along the Y direction and magnetic field is given by \(\vec{B}\) = \(\frac{\beta}{\sqrt{3}}\)(i +j +k)T- The magnitude of Lorentz force acting on the wire is
(a) \(\sqrt{\frac{2}{\sqrt{3}}} \beta I l\)
(b) \(\sqrt{\frac{1}{\sqrt{3}}} \beta \mathrm{Il}\)
(c) \(\sqrt{2} \beta \mathrm{Il}\)
(d) \(\sqrt{\frac{1}{2}} \beta \mathrm{Il}\)
Answer:
(a) \(\sqrt{\frac{2}{\sqrt{3}}} \beta I l\)

Question 11.
When a point charge of 6 pC is moved between two points in an electric field, the work done is 1.8 × 10^-5 J. This potential difference between the two points is :
(a) 1.08 V
(b) 1.08 μV
(c) 3 V
(d) 30 V

\(=\frac{1.8 \times 10^{-5}}{6 \times 10^{-6}}=\frac{18}{6}\)
V = 3V
Answer:
(c) 3 V

Question 12.
The wavelength λ_e of an electron and λ_p of a photon of same energy E are related by
(a) λ_p ∝ λ_e
(b) λ_p ∝ \(\sqrt{\lambda_{e}}\)
(c) λ_p ∝ \(\frac{1}{\sqrt{\lambda_{e}}}\)
(d) λ_p ∝ λ_e²
Hint:
de-Breglie wavelength of proton λ_e = \(\frac{h}{\sqrt{2 m \mathrm{E}}}\)
∴ ie λ_e ∝ \(\frac{1}{\sqrt{\mathrm{E}}}\) ⇒ λ_e² ∝ \(\frac{1}{E}\) …….. (1)
de-Breglie wave length of proton
λ_p = \(\frac{h c}{E}\)
λ_p ∝ \(\frac{1}{E}\) …….. (2)
From (1) and (2)
λ_e² ∝ λ_p i.e., λ_p ∝ λ_e²
Answer:
(d) λ_p ∝ λ_e²

Question 13.
For a myopic eye, the defect is cured by using a:
(a) convex lens
(b) concave lens
(c) cylindrical lens
(d) plane glass
Answer:
(b) concave lens

Question 14.
In a tangent galvanometer experiment, for two different values of current if the deflection are 45° and 30° respectively, then the ratio of the current is:
(a) 2 : 3
(b) 3 : 2
(c) √3 : 1
(d) 1 : √3
Hint: I₁ = tan θ₁
I₂ = tan θ₂
\(\frac{I_{1}}{I_{2}}=\frac{\tan 45^{\circ}}{\tan 30^{\circ}}=\frac{1}{\left(\frac{1}{\sqrt{3}}\right)}=\frac{\sqrt{3}}{1}\)
Answer:
(c) √3 : 1

Question 15.
If the current gain a of a transistor is 0.98, what is the value of p of the transistor?
(a) 0.49
(b) 49
(c) 4.9
(d) 5
Hint: \(\beta=\frac{\alpha}{1-\alpha}=\frac{0.98}{1-0.98}=49\)
Answer:
(b) 49

PART – II

Answer any six questions in which Q. No 24 is compulsory. [6 × 2 = 12]

Question 16.
What is meant by Fraunhofer lines?
Answer:
When the spectrum obtained from the Sun is examined, it consists of large number of dark lines (line absorption spectrum). These dark lines in the solar spectrum are known as Fraunhofer lines.

Question 17.
Why steel is preferred in making Robots?
Answer:
Steel is several time stronger. In any case, because of the inherent strength of metal, robot bodies are made using sheet, bar, rod, channel, and other shapes.

Question 18.
State Lenz’s law.
Answer:
Lenz’s law states that the direction of the induced current is such that it always opposes the cause responsible for its production.

Question 19.
Why do clouds appear white?
Answer:
Clouds have large particles like dust and water droplets which scatter light of all colours almost equally. Hence clouds generally appear white.

Question 20.
Calculate the radius of ¹⁹⁷₇₉Au nucleus.
Answer:
According to the equation R = R₀A^1/3
R = 1.2 × 10^-15 × (197)^1/3 = 6.97 × 10^-15 m (or) R = 6.97 F

Question 21.
What is the need for a feedback circuit in a transistor oscillator?
Answer:
The circuit used to feedback a portion of the output to the input is called the feedback network. If the portion of the output fed to the input is in phase with the input, then the magnitude of the input signal increases. It is necessary for sustained oscillations.

Question 22.
Show graphically the variation of electric held £ (y-axis) due to a charged infinite plane sheet with distance r (x-axis) from the plate.
Answer:
It independent of the distance. It straight line parallel to x-axis.

Question 23.
Give any two applications of internet.
(i) Search engine: The search engine is basically a web-based service tool used to search for information on World Wide Web.

(ii) Communication: It helps millions of people to connect with the use of social networking: emails, instant messaging services and social networking tools.

(iii) E-Commerce: Buying and selling of goods and services, transfer of funds are done over an electronic network.

Question 24.
Calculate the magnetic field inside a solenoid when the number of turns is halved and the length of the solenoid and the area remain the same.
Answer:
Magnetic field inside a solenoid is B = \(\frac{\mu_{0} \mathrm{NI}}{\mathrm{L}}\)
If the number of turns is havled B’ = \(\mathrm{B}^{\prime}=\frac{\mu_{0}\left(\frac{\mathrm{N}}{2}\right) \mathrm{I}}{\mathrm{L}} \Rightarrow \mathrm{B}^{\prime}=\frac{1}{2}\left(\frac{\mu_{0} \mathrm{NI}}{\mathrm{L}}\right)\)

PART – III

Answer any six questions in which Q.No. 33 is compulsory. [6 × 3 = 18]

Question 25.
Two cells each of 5V are connected in series across a 8 12 resistor and three parallel resistors of 4 12, 6 12 and 12 12. Draw a circuit diagram for the above arrangement calculate
(i) the current drawn from the cell
(ii) current through each resistor.
Answer:
V₁ = 5 V; V₂ = 5 v
R₁ = 8 Ω; R₂ = 4Ω; R₃ = 6Ω; R₄ = 12Ω
Three resistors R₂, R₃ and R₄ are connected parallel combination

\(\frac{1}{R_{p}}=\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}\)
\(=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=\frac{3}{12}+\frac{2}{12}+\frac{1}{12}=\frac{6}{12}\)
R_p = 2Ω
Resistors R₁ and R_p are connected in series combination
R_s = R₁ + R_p = 8 + 2 = 10
R₂ = 10Ω
Total voltage connected series to the circuit
V = V₁ + V₂
= 5 + 5 =10
V = 10V

(i) Current through the circuit, I = \(\frac{V}{R_{s}}=\frac{10}{10}\)
I = IA
Potential drop across the parallel combination,
V’ = I R_p = 1 × 2 .
V’ = 2 V

(ii) Current in 4Ω resistor, I = \(\frac{V^{\prime}}{R_{2}}=\frac{2}{4}\) = 0.5 A
Current in 6Ω resistor, I = \(\frac{V^{\prime}}{R_{3}}=\frac{2}{6}\) = 0.33 A.
Current in 12Ω resistor, I = \(\frac{V^{\prime}}{R_{4}}=\frac{2}{12}\) = 0.17 A

Question 26.
Explain 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.

1. 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.

2. 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.

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

Question 27.
Discuss the alpha – decay process with example.
Answer:
When unstable nuclei decay by emitting an α-particle (⁴₂He nucleus), it loses two protons and two neutrons. As a result, its atomic number Z decreases by 2, the mass number decreases by 4. We write the alpha decay process symbolically in the following way

Here X is called the parent nucleus and Y is called the daughter nucleus.
Example: Decay of Uranium ²³⁸₉₂U to thorium ²³⁴₉₀Th with the emission of ⁴₂He nucleus (α-particle)
²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He
As already mentioned, the total mass of the daughter nucleus and 2 He nucleus is always less than that of the parent nucleus. The difference in mass ( Δm = m_x – m_y m_α) is released as energy called disintegration energy Q and is given by
Q = (m_x – m_y – m_α) c²
Note that for spontaneous decay (natural radioactivity) Q > 0. In alpha decay process, the disintegration energy is certainly positive (Q > 0). In fact, the disintegration energy Q is also the net kinetic energy gained in the decay process or if the parent nucleus is at rest, Q is the total kinetic energy of daughter nucleus and the ⁴₂He nucleus. Suppose Q < 0, then the decay process cannot occur spontaneously and energy must be supplied to induce the decay.

Question 28.
Obtain the expression for energy stored in the parallel plate capacitor.
Answer:
Energy stored in the capacitor: Capacitor not only stores the charge but also it stores energy. When a battery is connected to the capacitor, electrons of total charge -Q are transferred from one plate to the other plate. To transfer the charge, work is done by the battery. This work done is stored as electrostatic potential energy in the capacitor.
To transfer an infinitesimal charge dQ for a potential difference V, the work done is given by
dW = VdQ ….(1)
Where V = \(\frac{Q}{C}\)
The total work done to charge a capacitor is
W = \(\int_{0}^{0} \frac{\mathrm{Q}}{\mathrm{C}} d \mathrm{Q}=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}\) ….. (2)
This work done is stored as electrostatic potential energy (U_E) in the capacitor.
U_E = \(\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}=\frac{1}{2} \mathrm{CV}^{2}\) (∴Q = CV) …… (3)
where Q = CV is used. This stored energy is thus directly proportional to the capacitance of the capacitor and the square of the voltage between the plates of the capacitor. But where is this energy stored in the capacitor? To understand this question, the equation (3) is rewritten as follows using the results
C = \(\frac{\varepsilon_{0} A}{d}\) and V = Ed
U_E = \(\frac{1}{2}\left(\frac{\varepsilon_{0} \mathrm{A}}{d}\right)(\mathrm{E} d)^{2}=\frac{1}{2} \varepsilon_{0}(\mathrm{A} d) \mathrm{E}^{2}\) …… (4)
where Ad = volume of the space between the capacitor plates. The energy stored per unit Volume of space is defined as energy density u_E = \(\frac{\mathrm{U}}{\text { Volume }}\) From equation (4) we get
U_E = \(\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2}\) ……. (5)
From equation (5), we infer that the energy is stored in the electric field existing between the plates of the capacitor. Once the capacitor is allowed to discharge, the energy is retrieved.

Question 29.
Explain any three recent advancements in medical technology.
Answer:
1. Virtual reality:
Medical virtual reality is effectively used to stop the brain from processing pain and cure soreness in the hospitalized patients. Virtual reality has enhanced surgeries by the use of 3D models by surgeons to plan operations. It helps in the treatment of Autism, Memory loss, and Mental illness.

2. Precision medicine:
Precision medicine is an emerging approach for disease treatment and prevention that takes into account individual variability in genes, environment, and lifestyle for each person. In this medical model it is possible to customise healthcare, with medical decisions, treatments, practices, or products which are tailored to the individual patient.

3. Health wearables:
A health wearable is a device used for tracking a wearer’s vital signs or health and fitness related data, location, etc. Medical wearables with artificial intelligence and big data provide an added value to healthcare with a focus on diagnosis, treatment, patient monitoring and prevention.
Note: Big Data: Extremely large data sets that may be analysed computationally to reveal patterns, trends, and associations, especially relating to human behaviour and interactions.

Question 30.
Two light sources with amplitudes 5 units and 3 units respectively interfere with each other. Calculate the ratio of maximum and minimum intensities.
Answer:
Amplitudes, a₁ = 5, a₂ = 3
Resultant amplitude, A = \(\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2} \cos \phi}\)
Resultant amplitude is maximum when,
Φ = 0, cos 0 = 1, A_max = \(\sqrt{a_{1}^{2}+a_{2}^{2}+2 a_{1} a_{2}}\)
A_max = \(\sqrt{\left(a_{1}+a_{2}\right)^{2}}=\sqrt{(5+3)^{2}}=\sqrt{(8)^{2}}\) = 8 units
Resultant amplitude is minimum when, Φ = π, cos π = -1, A_max = \(\sqrt{a_{1}^{2}+a_{2}^{2}-2 a_{1} a_{2}}\)
A_min = \(\sqrt{\left(a_{1}-a_{2}\right)^{2}}=\sqrt{(5-3)^{2}}=\sqrt{(2)^{2}}\) = 2 units
I ∝ A²
\(\frac{I_{\max }}{I_{\min }}=\frac{\left(A_{\max }\right)^{2}}{\left(A_{\min }\right)^{2}}\)
Substituting, \(\frac{I_{\max }}{I_{\min }}=\frac{(8)^{2}}{(2)^{2}}=\frac{64}{4}=16\) (or) I_max : I_min = 16 : 1

Question 31.
An electron moves in a circular orbit with a uniform speed v it produces a magnetic field B at the centre of the circle. Prove that the radius of the circle is proportional to \(\sqrt{\frac{v}{\mathbf{B}}}\)
Answer:
The magnetic field produce by moving electron in circular path.

Question 32.
Give the construction and working of photo emissive cell.
Answer:

Photo emissive cell:
Its working depends on the electron emission from a metal cathode due to irradiation of light or other radiations.
Construction:

• It consists of an evacuated glass or quartz bulb in which two metallic electrodes – that is, a cathode and an anode are fixed.
• The cathode C is semi-cylindrical in shape and is coated with a photo sensitive material. The anode A is a thin rod or wire kept along the axis of the semi-cylindrical cathode.
• A potential difference is applied between the anode and the cathode through a galvanometer G.

Working:

• When cathode is illuminated, electrons are emitted from it. These electrons are attracted by anode and hence a current is produced which is measured by the galvanometer.
• For a given cathode, the magnitude of the current depends on
  • the intensity to incident radiation and
  • the potential difference between anode and cathode.

Question 33.
In the circuit shown in the figure the input voltage V₁ = +5 V, V_BE = + 0.8 V and V_CE = + 0.12 V_CE. Find the values of I_B, I_C and β.
Answer:

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34.
(a) Obtain the expression for electric field due to an uniformly charged spherical shell at a distance its centre.
Answer:
Electric field due to a uniformly charged spherical shell: Consider a uniformly charged spherical shell of radius R and total charge Q. The electric field at points outside and inside the sphere is found using Gauss law.

Case (a) At a point outside the shell (r > R): Let us choose a point P outside the shell at a distance r from the center as shown in figure (a). The charge is uniformly distributed on the surface of the sphere (spherical symmetry). Hence the electric field must point radially outward if Q > 0 and point radially inward if Q < 0. So we choose a spherical Gaussian surface of radius r and the total charge enclosed by this Gaussian surface is Q. Applying Gauss law,
\(\oint \overrightarrow{\mathrm{E}} \cdot d \overrightarrow{\mathrm{A}}=\frac{\mathrm{Q}}{\varepsilon_{0}}\) …(1)

The electric field \(\vec{E}\) and \(d \vec{A}\) point in the same direction (outward normal) at all the points on the Gaussian surface. The magnitude of \(\vec{E}\) is also the same at all points due to the spherical symmetry of the charge distribution.


Substituting this value in equation (2).
E.4πr² = \(\frac{Q}{\varepsilon_{0}}\)
E.4πr² = \(\frac{Q}{\varepsilon_{0}}\) (or) E = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r^{2}}\)
In vector form \(\overrightarrow{\mathrm{E}}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{Q}}{r^{2}} \hat{r}\)

The electric field is radially outward if Q > 0 and radially inward if Q < 0. From equation (3), we infer that the electric field at a point outside the shell will be same as if the entire charge Q is concentrated at the center of the spherical shell. (A similar result is observed in gravitation, for gravitational force due to a spherical shell with mass M)

Case (b): At a point on the surface of the spherical shell (r = R): The electrical field at points on the spherical shell (r = R) is given by
\(\overrightarrow{\mathrm{E}}=\frac{Q}{4 \pi \varepsilon_{0} R^{2}} \hat{r}\) …… (4)

Case (c) At a point inside the spherical shell (r < R): Consider a point P inside the shell at a distance r from the center. A Gaussian sphere of radius r is constructed as shown in the figure (b). Applying Gauss law

E.4πr² = \(\frac{Q}{\varepsilon_{0}}\) ….. (5)
Since Gaussian surface encloses no charge, So Q = 0. The equation (5) becomes
E = 0 (r < R) …….. (6)
The electric field due to the uniformly charged spherical shell is zero at all points inside the shell.

[OR]

Question 34.
(b) Write any five properties of electromagnetic waves.
Answer
Properties of electromagnetic waves:

1. Electromagnetic waves are produced by any accelerated charge.
2. Electromagnetic waves do not require any medium for propagation. So electromagnetic wave is a non-mechanical wave.
3. Electromagnetic waves are transverse in nature. This means that the oscillating electric field vector, oscillating magnetic field vector and propagation vector (gives direction of propagation) are mutually perpendicular to each other.
4. Electromagnetic waves travel with speed which is equal to the speed of light in vacuum, or freer space, c = \(\frac{1}{\sqrt{\varepsilon_{0} \mu_{0}}}\) = 3 × 10⁸ ms^-1
5. The speed of electromagnetic wave is less than speed in free space or vacuum, that is, v < c. In a medium of refractive index,

Question 35.
(a) What is modulation? Explain the types of modulation with necessary diagrams.
Answer:
Modulation: For long distance transmission, the low frequency baseband signal (input signal) is superimposed onto a high frequency radio signal by a process called modulation.
There are 3 types of modulation based on which parameter is modified. They are
(i) Amplitude modulation,
(ii) Frequency modulation, and
(iii) Phase modulation.

(i) Amplitude Modulation (AM):
If the amplitude of the carrier signal is modified according to the instantaneous amplitude of the baseband signal, then it is called amplitude modulation. Here the frequency and the phase of the carrier signal remain constant. Amplitude modulation is used in radio and TV broadcasting.

The signal shown in figure (a) is the message signal or baseband signal that carries information, figure (b) shows the high frequency carrier signal and figure (c) gives the amplitude modulated signal. We can see clearly that the carrier wave is modified in proportion to the amplitude of the baseband signal.
Amplitude Modulation

(ii) Frequency Modulation (FM):
The frequency of the camer signal is modified according to the instantaneous amplitude of the baseband signal in frequency modulation. Here the amplitude and the phase of the carrier signal remain constant. Increase in the amplitude of the baseband signal increases the frequency of the carrier signal and vice versa. This leads to compressions and rarefactions in the frequency spectrum of the modulated wave. Louder signal leads to compressions and relatively weaker signals to rarefactions. When the amplitude of the baseband signal is zero in Figure (a), the frequency of the modulated signal is the same as the carrier signal. The frequency of the modulated wave increases when the amplitude of the baseband signal increases in the positive direction
(A, C). The increase in amplitude in the negative half cycle (B,D) reduces the frequency of the modulated wave (figure (c)).

(iii) Phase Modulation (PM):
The instantaneous amplitude of the baseband signal modifies the phase of the camer signal keeping the amplitude and frequency constant is called phase modulation. This modulation is used to generate frequency modulated signals. It is similar to frequency modulation except that the phase of the carrier is varied instead of varying frequency. The carrier phase changes according to increase or decrease in the amplitude of the baseband signal. When the modulating signal goes positive, the amount of phase
lead increases with the amplitude of the modulating signal. Due to this, the carrier signal is compressed or its frequency is increased.

On the other hand, the negative half cycle of the baseband signal produces a phase lag in the carrier signal. This appears to have stretched the frequency of the carrier wave. Hence similar to frequency modulated wave, phase modulated wave also comprises of
compressions and rarefactions. When the signal voltage is zero (A, C and E) the carrier frequency is unchanged.

[OR]

Question 35.
(b) Find the expression for the mutual inductance between a pair of coils and show that (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 q is sent through coil i₁ 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.
N₂Φ₂₁ ∝ i₁
N₂Φ₂₁ = M₂₁i₁ or M₂₁ = \(\frac{N_{2} \Phi_{21}}{i_{1}}\)
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 = 1 A, 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 i₁ changes with time, an emf ξ₂ is induced in coil 2. From Faraday’s law of electromagnetic induction, this mutually induced emf ξ₂₁ is given by

The negative sign in the above equation shows that the mutually induced emf always opposes the change in current with respect to time. If \(\frac{d i_{1}}{d t}\) = 1 As^-1, then M₂₁ = -ξ₂.
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 I is l As^-1.
Similarly, if an electric current i₂ through coil 2 changes with time, then emf ξ₁ is induced in coil 1. Therefore, N

where M₁₂ is the mutual inductance of the coil I 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 36.
(a) Derive the expression for the radius of the orbit of the electron and its velocity using Bohr atom model.
Answer:
Radius of the orbit of the electron and velocity of the electron:
Consider an atom which contains the nucleus at rest and an electron revolving around the nucleus in a circular orbit of radius rn as shown in figure. Nucleus is made up of protons and neutrons. Since proton is positively charged and neutron is electrically neutral, the charge of a nucleus is purely the total charge of protons.

Let Z be the atomic number of the atom, then +Ze is the charge of the nucleus. Let -e be the charge of the electron. From Coulomb’s law, the force of attraction between the nucleus and the electron is

\(\vec{F}\)_coloumb = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(+\mathrm{Ze})(\mathrm{e})}{\mathrm{r}_{\mathrm{n}}^{2}} \hat{\mathrm{r}}\)
\(=-\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{r_{n}^{2}} \hat{r}\)
This force provides necessary centripetal force
\(\vec{F}\)_centripetal = \(\frac{m v_{n}^{2}}{r_{n}} \hat{r}\)
where m be the mass of the electron that moves with a velocity DW in a circular orbit. Therefore,
|\(\vec{F}\)_coloumb| = |\(\vec{F}\)_centripetal|
\(\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{r_{n}^{2}}=\frac{m v_{n}^{2}}{r_{n}}\)
r_n = \(\frac{4 \pi \varepsilon_{0}\left(m \mathrm{v}_{n} r_{n}\right) 2}{Z m e^{2}}\) ……. (1)
From Bohr’s assumption, the angular momentum quantization condition mυ_nr_n = l_n = n_n


where n ∈ N . Since, ε₀, h, e and n are constants. Therefore, the radius of the orbit becomes
r_n = a₀\(\frac{n^{2}}{Z}\)
Where a₀ = \(\frac{\varepsilon_{0} h^{2}}{\pi m e^{2}}\) = 0.529Å. This is known as Bohr radius which is the smallest radius of the orbit in an atom. Bohr radius is also used as unit of length called Bohr. 1 Bohr = 0.53 Å. For hydrogen atom (Z = 1), the radius of n^th orbit is
r_n = a₀n²
For the first orbit (ground state),
r₁ = a₀ = 0.529 Å
For the second orbit (first excited state),
r₂ = 4a₀ = 2.116 Å
For the third orbit (second excited state),
r₃ = 9a₀ = 4.761 Å and so on.
Thus the radius of the orbit from centre increases with n, that is, r_n ∝ c n² as shown in figure. Further, Bohr’s angular momentum quantization condition leads to
mυ_nr_n = mυ_na₀n² = n\(\frac{h}{2 \pi}\)
υ_n = \(\frac{h}{2 \pi m a_{0}} \frac{Z}{n}\)
υ_n ∝ \(\frac{1}{n}\)
Note that the velocity of electron decreases as the principal quantum number increases as shown in figure. This curve is the rectangular hyperbola. This implies that the velocity of electron in ground state is maximum when compared to excited states.

[OR]

Question 36.
(b) Discuss the working and theory of cyclotron in detail.
Answer:

Cyclotron: Cyclotron is a device used to accelerate the charged particles to gain large kinetic energy. It is also called as high energy accelerator. It was invented by Lawrence and Livingston in 1934.

Principle: When a charged particle moves normal to the magnetic field, it experiences magnetic Lorentz force.

Construction:The particles are allowed to move in between two semicircular metal containers called Dees (hollow D – shaped objects). Dees are enclosed in an evacuated chamber and it is kept in a region with uniform magnetic field controlled by an electromagnet. The direction of magnetic field is normal to the plane of the Dees. The two Dees are kept separated with a gap and the source S (which ejects the particle to be accelerated) is placed at the center in the gap between the Dees. Dees are connected to high frequency alternating potential difference.

Working: Let us assume that the ion ejected from source S is positively charged. As soon as ion is ejected, it is accelerated towards a Dee (say, Dee – 1) which has negative potential at that time. Since the magnetic field is normal to the plane of the Dees, the ion undergoes circular path. After one semi-circular path in Dee-1, the ion reaches the gap between Dees. At this time, the polarities of the Dees are reversed so that the ion is now accelerated towards Dee-2 with a greater velocity. For this circular motion, the centripetal force of the charged particle q is provided by Lorentz force.
\(\frac{m v^{2}}{r}\) = qv B ⇒ r = \(\frac{m}{q \mathrm{B}} v\) ⇒ r ∝ v
From the equation, the increase in velocity increases the radius of circular path. This process continues and hence the particle undergoes spiral path of increasing radius. Once it reaches near the edge, it is taken out with the help of deflector plate and allowed to hit the target T. Very important condition in cyclotron operation is the resonance condition. It happens when the frequency f at which the positive ion circulates in the magnetic field must be equal to the constant frequency of the electrical oscillator f_osc From equation
f_osc = \(\frac{q \mathbf{B}}{2 \pi m} \Rightarrow \mathrm{T}=\frac{1}{f_{\mathrm{osc}}}\)
The time period of oscillation is T = \(\frac{2 \pi m}{q B}\)
The kinetic energy of the charged particle is K E = \(\frac{1}{2} m v^{2}=\frac{q^{2} \mathbf{B}^{2} r^{2}}{2 m}\)
Limitations of cyclotron
(a) the speed of the ion is limited
(b) electron cannot be accelerated
(c) uncharged particles cannot be accelerated

Question 37.
(a) Obtain lens maker’s formula and mention its significance.
Answer:
Lens maker’s formula and lens equation: Let us consider a thin lens made up of a medium of refractive index n₂ is placed in a medium of refractive index n₁ . Let R₁ and R₂ be the radii of curvature of two spherical surfaces (1) and (2) respectively and P be the pole. Consider a point object O on the principal axis. The ray which falls very close to P, after refraction at the surface (1) forms image at I’. Before it does so, it is again refracted by the surface (2). Therefore the final image is formed at I. The general equation for the refraction at a spherical surface is given by

For the refracting surface (1), the light goes from to n₁ to n₂

If the object is at infinity, the image is formed at the focus of the lens. Thus, for u = oo, v =/ Then the equation becomes.

If the refractive index of the lens is n₂ and it is placed in air, then n₂ = n and n₁ = 1. So the equation (4) becomes,
\(\frac{1}{f}=(n-1)\left(\frac{1}{\mathrm{R}_{1}}-\frac{1}{\mathrm{R}_{2}}\right)\)

The above equation is called the lens maker’s formula, because it tells the lens manufactures what curvature is needed to make a lens of desired focal length with a material of particular refractive index. This formula holds good also for a concave lens. By comparing the equations (3) and (4) we can write,
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) ……… (6)
This equation is known as lens equation which relates the object distance it and image distance v with the focal length / of the lens. This formula holds good for a any type of lens.

Question 37.
(b) Explain the construction and working of a full wave rectifier.
Answer:
Full wave rectifier:
The positive and negative half cycles of the AC input signal pass through the full wave rectifier circuit and hence it is called the full wave rectifier. It consists of two p-n junction diodes, a center tapped transfonner, and a load resistor (RL). The centre is usually taken as the ground or zero voltage reference point. Due to the centre tap transformer, the output voltage rectified by each diode is only one half of the total secondary voltage.

During positive half cycle:
When the positive half cycle of the ac input signal passes through the circuit, terminal M is positive, G’ is at zero potential and N is at negative potential. This forward biases diode D₁ and reverse biases diode D₂ Hence, being forward biased, diode D₁ conducts and current flows along the path MD₁ AGC . As a result, positive half cycle of the voltage appears across R_L in the direction G to C.

During negative half cycle:
When the negative half cycle of the ac input signal passes through the circuit, terminal N is positive, G is at zero potential and M is at negative potential. This forward biases diode D₂ and reverse biases diode D₁. Hence, being forward biased, diode D₂ conducts and current flows along the path ND₂ BGC . As a result, negative half cycle of the voltage appears across R_L in the same direction from G to C

Hence in a full wave rectifier both positive and negative half cycles of the input signal pass through the circuit in the same direction as shown in figure (b). Though both positive and negative half cycles of ac input are rectified, the output is still pulsating in nature.

The efficiency ( η) of full wave rectifier is twice that of a half wave rectifier and is found to be 81.2 %. It is because both the positive and negative half cycles of the ac input source are rectified.

Question 38.
(a) (i) Derive the expression for the de Broglie wavelength of an electron.
Answer:
An electron of mass m is accelerated through a potential difference of V volt. The kinetic energy acquired by the electron is given by
\(\frac{1}{2} m v^{2}\) = eV
Therefore, the speed v of the electron is v = \(\sqrt{\frac{2 e \mathrm{V}}{m}}\)
Hence, the de Broglie wavelength of the electron is λ = \(\frac{h}{m v}=\frac{h}{\sqrt{2 e m \mathrm{V}}}\)
Substituting the known values in the above equation, we get

For example, if an electron is accelerated through a potential difference of 100V, then its de Broglie wavelength is 1.227 Å . Since the kinetic energy of the electron, K = eV, then the de Broglie wavelength associated with electron can be also written as
λ = \(\frac{h}{\sqrt{2 m \mathrm{K}}}\)

(ii) An electron is accelerated through a potential difference of 81V. What is the de Broglie wavelength associated with it? To which part of the electromagnetic spectrum does this wavelength correspond?
de-Broglie wavelength of an electron beam accelerated through a potential difference of V volts is

X-ray is the part of electromagnetic spectrum does this wavelength corresponds. X-ray has the wavelengths ranging from about 10^-8 to 10^-12 m.

[OR]

Question 38.
(b) (i) How will you measure the internal resistance of a cell by potentiometer?
Answer:
Measurement of internal resistance of a cell by potentiometer:
To measure the internal resistance of a cell, the circuit connections are made as shown in figure. The end C of the potentiometer wire is connected to the positive terminal of the battery Bt and the negative terminal of the battery is connected to the end D through a key K₁ This forms the primary circuit.

The positive terminal of the cell ξ whose internal resistance is to be determined is also connected to the end C of the wire. The negative terminal of the cell ξ is connected to a jockey through a galvanometer and a high resistance. A resistance box R and key K₂ are connected across the cell ξ. With K₂ open, the balancing point J is obtained and the balancing length CJ = l₁ is measured. Since the cell is in open circuit, its emf is
ξ ∝ l₁ ….. (1)
A suitable resistance (say, 10 H) is included in the resistance box and key K₂ is closed.
Let r be the internal resistance of the cell. The current passing through the cell and the resistance R is given by
I = \(\frac{\xi}{R+r}\)
The potential difference across R is
V = \(\frac{\xi R}{R+r}\)
When this potential difference is balanced on the potentiometer wire, let /2 be the balancing length.
Then \(\frac{\xi R}{R+r} \propto l_{2}\) ….. (2)
From equations (1) and (2)

Substituting the values of the R, l₁ and l₂, the internal resistance of the cell is determined. The experiment can be repeated for different values of R. It is found that the internal resistance of the cell is not constant but increases with increase of external resistance connected across its terminals.

(ii) A cell supplies a current of 0.9 A through a 1Ω resistor and a current of 0.3 A through a 2 Ω resistor. Calculate the internal resistance of the cell.
Current from the cell 1,I₁, = 0.9 A and Resistor, R₁ = 1Ω
Current from the cell 2,I₂ = 0.3 A and Resistor, R₂ = 2Ω
Current in the circuit, I₁ = \(\frac{\xi}{r+R_{1}}\)
ξ = I₁r + I₁R₁ ….. (1)
Current in the circuit, I₂ = \(\frac{\xi}{r+R_{2}}\)
ξ = I₂r + I₂R₂ ….. (2)
From equation (1) and (2),

|r| = |-0.5| Ω
Magnitude of internal resistance r = 0.5 Ω

Students can Download Tamil Nadu 12th Biology Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Biology Model Question Paper 4 English Medium

General Instructions:

1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

I. Choose the correct answer. [8 × 1 = 8]

Question 1.
Inner most layer of anther wall is _________.
(a) Endothecium
(b) Endothelium
(c) Epithelium
(d) Tapetum
Answer:
(d) Tapetum

Question 2.
Test cross involves ______.
(a) Crossing between two genotypes with recessive trait
(b) Crossing between two F₁ hybrids
(c) Crossing between F₁ hybrid with double recessive genotype
(d) Crossing between two genotypes with dominant trait
Answer:
(c) Crossing between F₁ hybrid with double recessive genotype

Question 3.
Assertion (A): Gamma rays are generally used to induce mutation in wheat varieties.
Reason (R): Because they carry lower energy to non-ionize electrons from atom.
(a) A is correct. R is correct explanation of A.
(b) A is correct. R is not a correct explanation of A.
(c) A is correct. R is wrong explanation of A
(d) A and R are wrong.
Answer:
(c) A is correct. R is wrong explanation of A

Question 4.
Self-ligation is prevented by ________.
(a) DNA polymerase
(b) Helicase
(c) Alkaline phosphate
(d) DNA lipase
Answer:
(c) Alkaline phosphate

Question 5.
Match the following:


Answer:
(a) 1- C, 2-A, 3-D, 4-B

Question 6.
A Specific place in an ecosystem, where an organism lives and performs its functions is ______.
(a) habitat
(b) niche
(c) landscape
(d) biome
Answer:
(b) niche

Question 7.
Find the wrongly matched pair.
(a) Endemism – Species confined to a given region and not found anywhere else.
(b) Hotspots – Western ghats
(c) Ex-situ conservation – Zoological parks
(d) Sacred groves – Saintri hills of Rajasthan
(e) Alien species of India – Water hyacinth
Answer:
(d) Sacred groves – Saintri hills of Rajasthan

Question 8.
The plant source of Marijuana is ______.
(a) Andrographis paniculata
(b) Phyllanthus maderspatesis
(c) Cannabis sativa
(d) Papaver somniferum
Answer:
(c) Cannabis sativa

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
‘Pollination in Gymnosperms in different from Angiosperms’ – Give reasons.
Answer:
In gymnosperms, the ovules are exposed and the pollens are deposited directly on it. Hence the pollination is direct in gymnosperm. Whereas in angiosperms it is said to be indirect, as the pollens are deposited on stigma or the pistil.

Question 10.
A B C C B D E F G H I
From the above figure identify the type of mutation and explain it.
Answer:
In reverse tandem duplication, the duplicated segment is located immediately after the normal segment but the gene sequence other will be reversed.

Question 11.
Write the various steps involved in cell suspension culture.
Answer:
Step 1: Growing of cells/callus in medium (single or aggregates)
Step 2: Transfer of callus to a liquid medium
Step 3: Agitation of callus using rotary shaker
Step 4: Filtration and separation of cells.

Question 12.
Pyramid of energy is always upright. Give reasons.
Answer:
The energy pyramid represents a successive energy flow at each trophic level in an ecosystem. There is a gradual decrease in energy transfer at successive tropic levels from producers to higher levels, hence the pyramid of energy is always upright.

Question 13.
Write in brief about Atomic Garden.
Answer:
Atomic Garden: Is a form of mutation breeding where plants are exposed to radioactive sources typically cobalt-60 or caesium-137 in order to generate desirable mutation in crop plants.

Question 14.
Explain the law of dominance in monohybrid cross.
Answer:
Law of dominance states that the offsprings of an individual with contrasting (dissimilar) traits will only express the dominant trait in F₁ generation and both the characters are expressed in F₁ generation. This law also explains the proportion of 3 : 1 ratio in F₂ generation.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Give an account on cytoplasmic inheritance
Answer:
DNA is the universal genetic material. Genes located in nuclear chromosomes follow Mendelian inheritance. But certain traits are governed either by the chloroplast or mitochondrial genes. This phenomenon is known as extra nuclear inheritance. It is a kind of Non-Mendelian inheritance. Since it involves cytoplasmic organelles such as chloroplast and mitochondrion that act as inheritance vectors, it is also called Cytoplasmic inheritance.

Question 16.
How will you identify a vector?
Answer:

1. Vectors are able to replicate autonomously to produce multiple copies of them along with their DNA insert in the host cell.
2. It should be small in size and of low molecular weight, less than 10 Kb (kilo base pair) in size so that entry/transfer into host cell is easy.
3. Vector must contain an origin of replication so that it can independently replicate within the host.
4. It should contain a suitable marker such as antibiotic resistance, to permit its detection in transformed host cell.
5. Vector should have unique target sites for integration with DNA insert and should have the ability to integrate with DNA insert it carries into the genome of the host cell. Most of the commonly used cloning vectors have more than one restriction site. These are Multiple Cloning Site (MCS) or polylinker. Presence of MCS facilitates the use of restriction enzyme of choice.

Question 17.
What is Albedo effect? Write its effects.
Answer:
Gases let out to atmosphere causes climatic change. Emission of dust and aerosols from industries, automobiles, forest fire, SO₂ and DMS (dimethyl sulphur) play an important role in disturbing the temperature level of any region. Aerosols with small particles is reflecting the solar radiation entering the atmosphere. This is known as Albedo effect.

Question 18.
Ozone acts as a natural sun screen- Justify.
Answer:
Ozone depletion in the stratosphere results in more UV radiations especially UV B radiations (shortwaves). UV B radiation destroys biomolecules (skin ageing) and damages living tissues. UV – C is the most damaging type of UV radiation, but it is completely filtered by the atmosphere (ozone layer). UV – a contribute 95% of UV radiation which causes tanning burning of skin and enhancing skin cancer. Hence the uniform ozone layer is critical for the well being of life on earth.

Question 19.
Which is called as “The King of Bitters”? Mention their medicinal importance.
Answer:
Andrographis paniculata is called as King of Bitters. Andrographis is a potent hepatoprotective agent and is widely used to treat liver disorders. Concoction of Andrographis paniculata and eight other herbs (Nilavembu Kudineer) is effectively used to treat malaria and dengue.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) With a suitable diagram, explain the structure of an ovule.
Answer:
Structure of ovule(Megasporangium):

Ovule is also called megasporangium and is protected by one or two covering called integuments. Amature ovule consists of a stalk and a body. The stalk or the funiculus (also called funicle) is present at the base and it attaches the ovule to the placenta. The point of attachment of funicle to the body of the ovule is known as hilum. It represents the junction between ovule and funicle. In an inverted ovule, the funicle is adnate to the body of the ovule forming a ridge called raphe.

The body of the ovule is made up of a central mass of parenchymatous tissue called nucellus which has large reserve food materials. The nucellus is enveloped by one or two protective coverings called integuments. Integument encloses the nucellus completely except at the top where it is free and forms a pore called micropyle. The ovule with one or two integuments are said to be unitegmic or bitegmic ovules respectively.

The basal region of the body of the ovule where the nucellus, the integument and the funicle meet or merge is called as chalaza. There is a large, oval, sac-like structure in the nucellus toward the Raphe micropylar end called embryo sac or female \ gametophyte. It develops from the functional \ megaspore formed within the nucellus. In some species(unitegmic tenuinucellate) the inner layer of the integument may become specialized to perform the nutritive function for the embryo sac and is called as endothelium or integumentary tapetum (Example : Asteraceae).

[OR]

(b) List out the significance of ploidy.
Answer:

1. Many polyploids are more vigorous and more adaptable than diploids.
2. Many ornamental plants are autotetraploids and have larger flower and longer flowering duration than diploids.
3. Autopolyploids usually have increase in fresh weight due to more water content.
4. Aneuploids are useful to determine the phenotypic effects of loss or gain of different chromosomes.
5. Many angiosperms are allopolyploids and they play a role in an evolution of plants.

Question 21.
(a) Explain the basic concepts involved in plant tissue culture.
Answer:
Basic concepts of plant tissue culture are totipotency, differentiation, dedifferentiation and redifferentiation.

Totipotency: The property of live plant cells that they have the genetic potential when cultured in nutrient medium to give rise to a complete individual plant.

Differentiation: The process of biochemical and structural changes by which cells become specialized in form and function.

Redifferentiation: The further differentiation of already differentiated cell into another type of cell. For example, when the component cells of callus have the ability to form a whole plant in a nutrient medium, the phenomenon is called redifferentiation.

Dedifferentiation: The phenomenon of the reversion of mature cells to the meristematic state leading to the formation of callus is called dedifferentiation. These two phenomena of redifferentiation and dedifferentiation are the inherent capacities of living plant cells or tissue. This is described as totipotency.

[OR]

(b) Write a comparative note on Carbon Foot Print (CFP)
Answer:
Every human activity leaves a mark just like our footprint. This Carbon foot print is the total amount of green house gases produced by human activities such as agriculture, industries, deforestation, waste disposal, buring fossil fuels directly or indirectly. It can be measured for an individual, family, organisation like industries, state level or national level. It is usually estimated and expressed in equivalent tons of CO₂ per year.

The burning of fossil fuels releases CO₂ and other green house gases. In turn these emissions trap solar energy and thus increase the global temperature resulting in ice melting, submerging of low lying areas and inbalance in nature like cyclones, tsunamis and extreme weather conditions. To reduce the carbon foot print we can follow some practices like

• Eating indigenous fruits and products
• Reduce use of your electronic devices
• Reduce travelling
• Do not buy fast and preserved, processed, packed foods
• Plant a garden
• Less consumption of meat and sea food. Poultry requires little space, nutrients and less pollution comparing cattle farming
• reduce use of Laptops (when used for 8 hours, it releases nearly 2 kg. of CO₂ annually)
• Line dry your clothes.

(Example: If you buy imported fruit like kiwi, indirectly it increases CFP. How? The fruit has travelled a long distance in shipping or airliner thus emitting tons of CO₂)

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Match the following:


Answer:
(a) 1 – C, 2 – D, 3 – B, 4 – A

Question 2.
The male sex hormone testosterone is secreted from ___________.
(a) Sertoli cells
(b) Leydig cells
(c) Epididymis
(d) Prostate gland
Answer:
(b) Leydig cells

Question 3.
Select the incorrect action of hormonal contraceptive pills from the following
(a) Inhibition of spermatogenesis
(b) Inhibition of ovulation
(c) Changes in cervical mucus impairing its ability to allow passage and transport of sperm
(d) Alteration in uterine endometrium to make it unsustainable for implantation
Answer:
(a) Inhibition of spermatogenesis

Question 4.
Father of a child is colour blind and mother is carrier for colour blindness, the possibility of the child being colour blind is ________.
(a) 25 %
(b) 50%
(c) 100 %
(d) 75 %
Answer:
(b) 50%

Question 5.
Which of the following is the correct sequence of events with reference to the central dogma?
(a) Transcription, Translation, Replication
(b) Transcription, Replication, Translation
(c) Duplication, Translation, Transcription
(d) Replication, Transcription, Translation
Answer:
(d) Replication, Transcription, Translation

Question 6.
Who proposed the Germplasm theory?
(a) August Weismann
(b) Darwin
(c) Lamarck
(d) Alfred Wallace
Answer:
(a) August Weismann

Question 7.
AIDS virus has ________.
(a) Single stranded RNA
(b) Single stranded DNA
(c) Double stranded RNA
(d) Double stranded DNA
Answer:
(c) Double stranded RNA

Question 8.
Assertion (A): Streptomycin is an antibiotic
Reason (R): Antibiotics are microbial products inhibits the growth of pathogenic microbe.
(a) A is right, R is wrong
(b) R explain A
(c) A and R are wrong
(d) A and R are right. R does not explain A
Answer:
(b) R explain A

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
What is inhibin? State its functions.
Answer:
Inhibin is a hormone secreted by Sertoli cells of testes which is involved in the negative feedback control of sperm production.

Question 10.
Define criss-cross inheritance.
Answer:
Inheritance of genes from a male parent to female child and then to male grandchild or female parent to male child and then to female grandchild, eg., X-linked gene inheritance.

Question 11.
How does mutation theory of De vries differ from lamarck and Darwins view in the origin of new species.
Answer:
According to de Vries, sudden and large variations were responsible for the origin of new species, whereas Lamarck and Darwin believed in gradual accumulation of all variations as the causative factors in the origin of new species.

Question 12.
What is biological oxygen demand?
Answer:
The BOD (Biochemical oxygen demand or Biological oxygen demand). BOD refers to the amount of the oxygen that would be consumed, if all the organic matter in one litre of water were oxidized by bacteria. The sewage water is treated till the BOD is reduced. The greater the BOD of the waste water more is its polluting potential.

Question 13.
Differentiate between Eurytherms and Stenotherms.
Answer:

Question 14.
Name the secondary pollutant in photochemical smog. Mention the adverse effect.
Answer:
Peroxyacetyl nitrate (PAN) is a secondary pollutant present in photochemical smog. It is thermally unstable and decomposes into peroxy ethanol radicals and nitrogen dioxide gas causing eye irritation.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
What is amniocentesis? Why a statutory ban is imposed on this technique?
Answer:
Amniocentesis is a prenatal technique used to detect any chromosomal abnormalities in the foetus and it is being often misused to determine the sex of the foetus. Once the sex of the foetus is known, there may be a chance of female foeticide. Hence, a statutory ban on amniocentesis is imposed.

Question 16.
DNA acts as genetic material for majority of living organisms and not the RNA. Give reasons to support the statement.
Answer:

1. RNA was reactive and hence highly unstable.
2. Some RNA molecules acts as gene regulators by binding to DNA and affect gene expression.
3. Uracil of RNA is less stable than thymine of DNA.

Question 17.
Draw and lable the structure of Immunoglobulin
Answer:

Question 18.
Explain how “Rosie” is different from a normal cow.
Answer:
Rosie was the first transgenic cow. It produced human protein enriched milk, which contained the human alpha lactalbumin (2.4 gm. /litre). This milk was a nutritionally balanced food for infants than the normal milk of cows.

Question 19.
Extinction of a keystone species led to loss of biodiversity – Justify.
Answer:
A keystone species is an organism that helps define an entire ecosystem. Without the keystone species a particular ecosystem would be dramatically disturbed or even ceased. Keystone species either directly or indirectly affects every species in a particular ecosystem. If a keystone species is lost or removed no other organism would compensate its ecological niche.

eg: Jaguar is a keystone species. As a top predator, it plays a crucial role in ecosystem. Without jaguar there is an exponential increase in herbivoral population that would decimate the plants of the ecosystem. At one point even the herbivore populations also get declined due to the lack of vegetation. Thus jaguar acts a keystone species.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Define menstrual cycle. Explain its various phase.
Answer:
Menstrual cycle: The menstrual or ovarian cycle occurs approximately once in every 28/29 days during the reproductive life of the female from menarche (puberty) to menopause except during pregnancy. The cycle of events starting from one menstrual period till the next one is called the menstrual cycle during which cyclic changes occurs in the endometrium, every month. Cyclic menstruation is an indicator of normal reproductive phase.

Menstrual cycle comprises of the following phases

1. Menstrual phase
2. Follicular or proliferative phase
3. Ovulatory phase
4. Luteal or secretory phase

1. Menstrual phase: The cycle starts with the menstrual phase when menstrual flow occurs and lasts for 3-5 days. Menstrual flow is due to the breakdown of endometrial lining of the uterus, and its blood vessels due to decline in the level of progesterone and oestrogen. Menstruation occurs only if the released ovum is not fertilized. Absence of menstruation may be an indicator of pregnancy. However it could also be due to stress, hormonal disorder and anaemia.

2. Follicular or proliferative phase: The follicular phase extends from the 5th day of the cycle until the time of ovulation. During this phase, the primary follicle in the ovary grows to become a fully mature Graafian follicle and simultaneously, the endometrium regenerates through proliferation.

These changes in the ovary and the uterus are induced by the secretion of gonadotropins like FSH and LH, which increase gradually during the follicular phase. It stimulates follicular development and secretion of oestrogen by the follicle cells.

3. Ovulatory phase: Both LH and FSH attain peak level in the middle of the cycle (about the 14^th day). Maximum secretion of LH during the mid cycle called LH surge induces the rupture of the Graafian follicle and the release of the ovum (secondary oocyte) from the ovary wall into the peritoneal cavity. This process is called as ovulation.

4. Luteal or secretory phase: During luteal phase, the remaining part of the Graafian follicle is transformed into a transitory endocrine gland called corpus luteum. The corpus luteum secretes large amount of progesterone which is essential for the maintenance of the endometrium. If fertilization takes place, it paves way for the implantation of the fertilized ovum. The uterine wall secretes nutritious fluid in the uterus for the foetus. So, this phase is also called as secretory phase.

During pregnancy all events of menstrual cycle stop and there is no menstruation. In the absence of fertilization, the corpus luteum degenerates completely and leaves a scar tissue called corpus albicans. It also initiates the disintegration of the endometrium leading to menstruation, marking the next cycle. ,

[OR]

(b) Write the salient features of genetic code.
Answer:
The salient features of genetic code are as follows:

1. The genetic codon is a triplet code and 61 codons code for amino acids and 3 codons do not code for any amino acid and function as stop codon (Termination).

2. The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes. However similarities are more common than differences.

3. A non-overlapping codon means that the same letter is not used for two different codons. For instance, the nucleotide sequence GUU and GUC represents only two codons. .

4. It is comma less, which means that the message would be read directly from one end to the other i.e., no punctuation are needed between two codes. .

5. A degenerate code means that more than one triplet codon could code for a specific amino acid. For example, codons GUU, GUC, GUA and GUG code for valine.

6. Non-ambiguous code means that one codon will code for one amino acid.

7. The code is always read in a fixed direction i.e., from 5’ → 3’ direction called polarity.

8. AUG has dual functions. It acts as a initiator codon and also codes for the amino acid methionine.

9. UAA, UAG (tyrosine) and UGA (tryptophan) codons are designated as termination (stop) codons and also are known as “non-sense” codons.

Question 21.
(a) Suggest some of the ways to prevent drug and alcohol abuse.
Answer:
1. Effectively dealing with peer pressure:
The biggest reason for teens to start on drugs is due to their friends / peer groups imposing pressure on them. Hence, it is important to have a better group of friends to avoid such harmful drugs and alcohol.

2. Seeking help from parents and peers:
Help from parents and peer group should be sought immediately so that they can be guided appropriately. Help may even be sought from close and trusted friends. Getting proper advice to sort out their problems would help the young to vent their feelings of anxiety and guilt.

3. Education and counselling:
Education and counselling create positive attitude to deal with many problems and to accept disappointments in life.

4. Looking for danger signs:
Teachers and parents need to look for sign that indicate tendency to go in for addiction.

5. Seeking professional and medical assistance:
Assistance is available in the form of highly qualified psychologists, psychiatrists and de-addiction and rehabilitation programmes to help individuals to overcome their problems.

[OR]

(b) Mention the advantages and disadvantages of cloning.
Answer:

1. Offers benefits for clinical trials and medical research. It can help in the production of
   proteins and drugs in the field of medicine.
2. Aids stem cell research.
3. Animal cloning could help to save endangered species.
4. Animal and human activists see it as a threat to biodiversity saying that this alters evolution which will have an impact on populations and the ecosystem.
5. The process is tedious and very expensive.
6. It can cause animals to suffer.
7. Reports show that animal surrogates were manifesting adverse outcomes and cloned animals were affected with disease and have high mortality rate.
8. It might compromise human health through consumption of cloned animal meat.
9. Cloned animals age faster than normal animals and are less healthy than the parent organism as discovered in Dolly
10. Cloning can lead to occurrence of genetic disorders in animals.
11. More than 90% of cloning attempts fail to produce a viable offspring.

Students can Download Computer Science Chapter 15 Data Manipulation Through SQL Questions and Answers, Notes Pdf, Samacheer Kalvi 12th Computer Science 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 12th Computer Science Solutions Chapter 15 Data Manipulation Through SQL

Samacheer Kalvi 12th Computer Science Data Manipulation Through SQL Text Book Back Questions and Answers

PART – 1
I. Choose The Correct Answer

Question 1.
Which of the following is an organized collection of data?
(a) Database
(b) DBMS
(c) Information
(d) Records
Answer:
(a) Database

Question 2.
SQLite falls under which database system?
(a) Flat file database system
(b) Relational Database system
(c) Hierarchical database system
(d) Object oriented Database system
Answer:
(b) Relational Database system

Question 3.
Which of the following is a control structure used to traverse and fetch the records of the database?
(a) Pointer
(b) Key
(c) Cursor
(d) Insertion point
Answer:
(c) Cursor

Question 4.
Any changes made in the values of the record should be saved by the command
(a) Save
(b) Save As
(c) Commit
(d) Oblige
Answer:
(c) Commit

Question 5.
Which of the following executes the SQL command to perform some action?
(a) Execute( )
(b) Key( )
(c) Cursor( )
(d) run( )
Answer:
(a) Execute( )

Question 6.
Which of the following function retrieves the average of a selected column of rows in a table?
(a) Add( )
(b) SUM( )
(c) AVG( )
(d) AVERAGE( )
Answer:
(c) AVG( )

Question 7.
The function that returns the largest value of the selected column is ……………………………
(a) MAX( )
(b) LARGE( )
(c) HIGH( )
(d) MAXIMUM( )
Answer:
(a) MAX( )

Question 8.
Which of the following is called the master table?
(a) sqlite master
(b) sqlmaster
(c) main_master
(d) master_main
Answer:
(a) sqlite master

Question 9.
The most commonly used statement in SQL is
(a) cursor
(b) select
(c) execute
(d) commit
Answer:
(b) select

Question 10.
Which of the following clause avoid the duplicate?
(a) Distinct
(b) Remove
(c) Where
(d) GroupBy
Answer:
(a) Distinct

PART – II
II. Answer The Following Questions

Question 1.
Mention the users who uses the Database?
Answer:
Users of database can be human users, other programs or applications.

Question 2.
Which method is used to connect a database? Give an example?
Answer:
Step 1: create a connection using connect () method and pass the name of the database File
Connecting to a database in step 2 means passing the name of the database to be accessed. If the database already exists the connection will open the same. Otherwise, Python will open a new database file with the specified name.
# connecting to the database
connection = sqlite3.connect (“Academy.db”)

Question 3.
What is the advantage of declaring a column as “INTEGER PRIMARY KEY”?
Answer:
If a column of a table is declared to be an INTEGER PRIMARY KEY, then whenever a NULL . will be used as an input for this column, the NULL will be automatically converted into an integer which will one larger than the highest value so far used in that column. If the table is empty, the value 1 will be used.

Question 4.
Write the command to populate record in a table. Give an example?
Answer:
To populate (add record) the table “INSERT” command is passed to SQLite. “execute” method executes the SQL command to perform some action. In most cases, you will not literally insert data into a SQL table. You will rather have a lot of data inside of some Python data type e.g. a dictionary or a list, which has to be used as the input of the insert statement.

Question 5.
Which method is used to fetch all rows from the database table?
Answer:
Displaying all records using fetchall( )
The fetchall( ) method is used to fetch all rows from the database table
result = cursor.fetchall( )

PART – III
III. Answer The Following Questions

Question 1.
What is SQLite?What is it advantage?
Answer:

1. SQLite is a simple relational database system, which saves its data in regular data files or even in the internal memory of the computer.
2. It is designed to be embedded in applications, instead of using a separate database server program such as MySQLor Oracle.
3. SQLite is fast, rigorously tested, and flexible, making it easier to work. Python has a native library for SQLite.

Question 2.
Mention the difference between fetchone( ) and fetchmany( )
Displaying A record using fetchone( )
Answer:
The fetchone( ) method returns the next row of a query result set or None in case there is no row left.
Example
import sqlite3
connection= sqlite3 ,connect(“Academy.db”)
cursor = connection.cursor( )
cursor.execute(“SELECT * FROM student”)
print(“\nfetch one:”)
res = cursor. fetchone( )
print(res)
OUTPUT
fetch one:
(1, ‘Akshay’, ’B’, ’M’, 87.8, ‘2001-12-12’)
Displayingusing fetchmany( )
Displaying specified number of records is done by using fetchmany( ). This method returns the next number of rows (n) of the result set.
Example : Program to display the content of tuples using fetchmany( )
import sqlite3
connection = sqlite3.connect(” Academy, db”)
cursor = connection. cursor( )
cursor.execute(“SELECT * FROM student”)
print(“fetching first 3 records:”)
result= cursor. fetchmany(3)
print( result)
OUTPUT
fetching first 3 records:
[ (1, ‘Akshay’, ’B’, ‘M’, 87.8, ‘2001-12-12’), (2, ‘Aravind’, ‘A’, ’M’, 92.5, ‘2000-08-17’), (3, ‘BASKAR’, ‘C, TVT, 75.2, ‘1998-05-17’)]

Question 3.
What is the use of Where Clause.Give a python statement Using the where clause?
Answer:
The WHERE clause is used to extract only .those records that fulfill a specified condition. In this example we are going to display the different grades scored by male students from “student table” import sqlite3
connection = sqlite3.connect(“Academy.db”)
cursor = connection.cursor( )
cursor.execute(“SELECT DISTINCT (Grade) FROM student where gender=’M'”)
result = cursor. fetchall( )
print(*result,sep=”\n”)
OUTPUT
(‘B’,)
(‘A’,)
(‘C’,)
(‘D’,)

Question 4.
Read the following details.Based on that write a python script to display department wise records
database name:- organization.db
Table name:- Employee
Columns in the table:- Eno, EmpName, Esal, Dept
Display departmentwise records: import sqlite3
conn = sqlite3.connect(“organization.db”)
c = conn.execute(“select * from Employee group by dept”)
for row in c:
print(row)
conn.close( )
Output:
(1000, ‘Ram’, 5000, ’Mech’)
(2000, ‘Hari’, 6500, ‘Mech’)
(1127, ’Vijay’, 10,000, ’cs )
(1130,’Raja’, 11,000,’cs’)

Question 5.
Read the following details.Based on that write a python script to display records in
desending order of
Eno
database name:- organization.db
Table name:- Employee
Columns in the table:- Eno, EmpName, Esal, Dept
import sqlite3
connection = sqlite3.connect(“organizatoin.db”)
cursor = connection. cursor( )
cursor.execute(“select * from Employee order by Eno Desc”)
result = cursor. fetchall( )
print(result).
Output:
(2000, ‘Hari’, 6500, ‘Mech’)
(1130, ‘Raja’, 11,000, ‘cs’)
(1127, ‘Vijay’, 10,000, ‘cs’)
(1000, ‘Ram’, 5000, ‘Mech’)

PART – IV
Answer The Following Questions

Question 1.
Write in brief about SQLite and the steps used to use it?
Answer:
SQLite is a simple relational database system, which saves its data in regular data files or even in the internal memory of the computer. It is designed to be embedded in applications, instead of using a separate database server program such as MySQLor Oracle. SQLite is fast, rigorously tested, and flexible, making it easier to work. Python has a native library for SQLite. To use SQLite,
Step 1 import sqliteS
Step 2 create a connection using connect ( ) method and pass the name of the database File
Step 3 Set the cursor object cursor = connection.cursor( )

1. Connecting to a database in step2 means passing the name of the database to be accessed. If the database already exists the connection will open the same. Otherwise, Python will open a new database file with the specified name.
2. Cursor in step 3: is a control structure used to traverse and fetch the records of the database.
3. Cursor has a major role in working with Python. All the commands will be executed using cursor object only.

To create a table in the database, create an object and write the SQL command in it.
Example:- sql_comm = “SQL statement”
For executing the command use the cursor method and pass the required sql command as a parameter. Many number of commands can be stored in the sql comm and can be executed one after other. Any changes made in the values of the record should be saved by the command “Commit” before closing the “Table connection”.

Question 2.
Write the Python script to display all the records of the following table using fetchmany( )
Answer:

import sqlite3
connectoin = sqlite3.connect(“company.db”) cursor = connection.cursor() cursor.execute(“Select * from product”) print(“Displaying Records”) result = cursor.fetchmany(5) print(*result, Sep = “\n”)
Output:
Displaying records (1003, ‘Scanner’, 10500)
(1004, ’Speaker’, 3000)
(1005, ‘Printer’, 8000)
(1008,’Monitor’, 15000)
(1010, ‘Mouse’, 700)

Question 3.
What is the use of HAVING clause. Give an example python script?
Answer:
Having clause is used to filter data based on the group functions. This is similar to WHERE condition but can be used only with group functions. Group functions cannot be used in WHERE Clause but can be used in HAVING clause.
Example
import sqlite3
connection= sqlite3.connect(“Academy.db”)
cursor = connection. cursor( )
cursor.execute(“SELECT GENDER,COUNT(GENDER) FROM Student GROUP BY GENDER HAVING COUNT(GENDER)>3 “)
result = cursor. fetchall( )
co= [i[0] for i in cursor, description]
print(co)
print( result)
OUTPUT
[‘gender’, ‘COUNT(GENDER)’]
[(‘M’, 5)]

Question 4.
Write a Python script to create a table called ITEM with following specification?
Add one record to the table.
Name of the database:- ABC
Name of the table:- Item
Column name and speculation:-

import sqlite 3
connection = sqlite3.connect(“ABC.db”)
cursor = connectoin.cursor( )
sql_command -“””CREATE TABLE Item
(
Icode INTEGER PRIMARY KEY,
ItemName VARCHAR(25), Rate INTEGER);”””
cursor.execute(sql_command)
sql_command=””” INSERT INTO ITEM(lcode, ItemName, Rate) VALUES(1008, ‘Monitor’, 15000);”””
cursor.execute(sql_command)
connection.commit( )
connection.close( )
print(“Table created”)
Output:
Table created

Question 5.
Consider the following table Supplier and item .Write a python script for (i) to (ii)
Answer:

1. Display Name, City and Itemname of suppliers who do not reside in Delhi.
2. Increment the SuppQty of Akila by 40

Answer:
1. import sqlite3
connection = sqlite3.connect(“ABC.db”)
cursor = connection,cursor( )
cursor.execute(“SELECT Supplier.Name, Supplier.city, Item.ItemName FROM Supplier, Item
where Supplier.Icode = Item.Icode AND Supplier.city NOT IN ‘Delhi'”)
co=[i[0] for i in cursor.description]
print(co)
result = cursor. fetchall( )
for r in result:
print(r).
Output:
[‘Name’, ‘City’, ‘ItemName’]
[‘Anu’, ‘Bangalore’, ‘Mouse’]
[‘Shahid’, ‘Bangalore’, ‘Monitor’]
[‘Akila’, ‘Hydrabad’, ‘Printer’]
[‘Girish’, ‘Hydrabad’, ‘Scanner’]
[‘Shylaja’, ‘Chennai’, ‘Monitor’]
[‘Lavanya’, ‘Mumbai’, ‘Printer’]

2. Increment the suppQty of Akila by 40 import sqlite3
connection = sqlite3.connect(“ABC.db”)
cursor = connection.cursor( )
cursor.execute(“UPDATE Supplier SET SuppQty = SuppQty + 40 where Name = ‘Akila'”)
cursor.commit( )
result = cursor. fetchall( )
print(result)
connection.close( )
Output:
(S004, ‘Akila’, ‘Hydrabad’, 1005, 235)

Practice Programs

Question 1.
Create an interactive program to accept the details from user and store it in a csv file using Python for the following table?
Answer:
Database name;- DB1
Table name : Customer

import sqlite3 import io import csv
d = open(‘c:/pyprg/sql.csv’, ‘w’) c = csv.writer(d)
connection = sqlite3.connect(“dbl.db”)
cursor = connection.cursor( )
cursor.execute(“create table customer(cust_Id, cust_Name, Address, Phone_no, City)”)
print(“Enter 3 customer details:”)
print(“Enter 3 customer Id :”)
cid = [int(input( )) for i in range (3)]
print(“Enter customer names :”)
cname = [input( ) for i in range (3)]
print(“Enter their Address:”)
add = [input( ) for i in range (3)]
int(“Enter their phone numbers:”)
ph = [int(input( )) for i in range (3)]
print(“Enter their cities:”)
city = [input( ) for i in range (3)]
n = len(cname)
for i in range (n):
cursor.execute(“insert into customer values (?,?,?,?,?)”, (cid[i], cname[i], add[i], ph[i], city[i]))
cursor.execute(“Select * from customer “)
co = [i[0] for i in cursor, description]
c.writerow(co)
data = cursor. fetchall( )
for item in data:
c.writerow(item)
– d.close( )
with open(‘c:/pyprg/sql.csv’, “r”, newline = None) as fd:
for line in fd:
line = line.replace(“\n”,” “)
print(line)
cursor. close( )
connection. close( )
OUTPUT:
Enter 3 customer details:
Enter 3 customer Id:
C008
C010
C012
Enter customer names:
Sandeep
Anurag Basu
Hrithik
Enter their Address:
14/1 Pritam Pura
15/A, Park Road
7/2 Vasant Nagar
Enter their Phone Numbers:
41206819
61281921
26121949
Enter their cities:
Delhi
Kolkata
Delhi
Displaying Data:
(‘cust_Id’, ‘cust_Name’, ‘Address’, ‘Phone_no’, ‘city’)
(C008, ‘Sandeep’, ’14/1 Pritampura’, ‘41206819’, ’Delhi’)
(C010, ‘Anurag Basu’, ’15A, Park Road’, ’61281921’, ’Kolkata’)
(C012, ’Hrithik’, ’7/2 Vasant Nagar’, ’26121949’, ’Delhi’)

Question 2.
Consider the following table GAMES. Write a python program to display the records for question (i) to (iv) and give outputs for SQL queries (v) to (viii)?
Table: GAMES

1. To display the name of all Games with their Geodes in descending order of their schedule date.
2. To display details of those games which are having Prize Money more than 7000.
3. To display the name and gamename of the Players in the ascending order of Gamename.
4. To display sum of PrizeMoney for each of the Numberof participation groupings (as shown in column Number 4)
5. Display all the records based on GameName

1. To display the name of all Games with their Geodes in descending order of their schedule date.
import sqlite3
conn = sqlite3.connect(“Games.db”)
cursor = conn.cursor( )
cursor.execute(“Select GameName, Geode from Games order by ScheduleDate Desc “)
result = cursor,fetchall( )
print(*result, sep = “\n”)
conn.close( )
Output:
(’Table Tennis’, 108)
(’Table Tennis’, 103)
(’Carom Board’, 101)
(’Carom Board’, 105)
(’Badminton1, 102)

2. To display details of those games which are having Prize Money more than 7000.
import sqlite3
conn = sqlit3.connect(“Games.db”)
cursor = conn.cursor( )
cursor.execute(” Select * from Games where prize money > 7000″)
result = cursor.fetchall( )
print(*result, sep = “\n”)
conn.close( )
Output:
(102, ‘Vidhya’, ‘Badminton’, 2, 12000, ’12-12-2013′)
(103, ‘Guru’, ’Table Tennis’, 4, 8000, ’02-14-2014’)
(105, ‘Keerthana’, ‘Carom Board’, 2, 9000, ’01-01-2014′)
(108, ’Krishna’, ’Table Tennis’, 4, 25000, ’03-19-2014′)

3. To display the name and gamename of the Players in the ascending order of Gamename.
import sqlite3
conn = sqlite3.connect(“Games.db”)
cursor = conn.cursor( )
cursor.execute(” Select Name, GameName from games order by GameName “)
result = cursor.fetchall( )
print(*result, sep = “\n”)
conn.close( )
Output:
(‘Vidhya’, ‘Badminton’)
(‘Padmaja’, ‘Carom Board’)
(‘Keerthana’, ‘Carom Board’)
(‘Guru’, ‘Table Tennis’)
(‘Krishna’, ‘Table Tennis’)

4. To display sum of PrizeMoney for each of the Numberof participation groupings (as shown in column Number 4)
import sqlite3
conn = sqlite3.connect(“Games.db”)
cursor = conn.cursor( )
cursor.execute(“Select Sum(Number * Prizemoney) from games”)
result = cursor, fetchall( )
print(result)
conn.close( )
Output:
[(184000)]

5. Display all the records based on GameName
import sqlite3
conn = sqlite3.connect(“Games.db”)
cursor = conn.cursor( )
cursor.execute(“Select * from games group by gamename”)
result = cursor. fetchall( )
print(*result, sep = “\n”)
conn.close( )
Output:
(‘Carom Board’, 101, ‘Padmaja’, 2, 5000, ’01-23-2014′)
(’Carom Board’, 105, ‘Keerthana’, 2, 9000, ’01-01-2014′)
(‘Badminton’, 102, ‘Vidhya’, 2, 12000, ’12-12-2013′)
(‘Table Tennis’, 103, ‘Guru’, 4, 8000, ’02-14-2014′)
(’Table Tennis’, 108, ‘Krishna’, 4, 25000, ’03-19-2014′)

Samacheer kalvi 12th Computer Science Data Manipulation Through SQL Additional Questions and Answers

PART – 1
I. Choose The Correct Answer

Question 1.
The ………………………….. is a software application for the interaction between users and the databases.
Answer:
Database Management System

Question 2.
Which has a native library for sqlite?
(a) C
(b) C++
(c) Java
(d) Python
Answer:
(d) Python

Question 3.
Which method is used to create a connection with database file?
(a) connect( )
(b) pass( )
(c) link( )
(d) create( )
Answer:
(a) connect( )

Question 4.
All the commands will be executed using ……………………………. object only.
Answer:
cursor

Question 5.
How many commands can be stored in the sql_comm?
(a) 1
(b) 2
(c) 3
(d) Many
Answer:
(d) Many

Question 6.
What is the extension for database files?
(a) dot
(b) database
(c) .db
(d) .dot
Answer:
(c) .db

Question 7.
The SQL commands have to be defined with ……………………………. quotes.
(a) single
(b) double
(c) triple
(d) No quotes
Answer:
(c) triple

Question 8.
A column which is labelled like ………………………………. is automatically auto incremented in sqlite3.
Answer:
Integer primary key

Question 9.
What will be the value assigned to the empty table if it is given Integer Primary Key?
(a) 0
(b) 1
(c) 2
(d) -1
Answer:
(b) 1

Question 10.
Which command is used to populate the records in the table?
(a) populate
(b) create
(c) pop
(d) Insert
Answer:
(d) Insert

Question 11.
Which command in SQL is used to retrieve or fetch data from a table in the database?
(a) Select
(b) Fetch
(c) retrieve
(d) create
Answer:
(a) Select

Question 12.
Which method is used to fetch all rows from the database table?
(a) Fetch( )
(b) fetchall( )
(c) printall( )
(d) retrieveall( )
Answer:
(b) fetchall( )

Question 13.
Which method returns the next row of a query result set?
(a) Fetch ne( )
(b) fetch all( )
(c) fetch next( )
(d) fetch last( )
Answer:
(a) Fetch ne( )

Question 14.
What will be the result of fetchone( ) method if no row is left?
(a) 1
(b) 2
(c) 3
(d) none
Answer:
(d) none

Question 15.
Which one of the following methods displays the specified number of records?
(a) fetchone( )
(b) fetchmany( )
(c) fetchall( )
(d) fetchsome( )
Answer:
(b) fetchmany( )

Question 16.
Which one of the following is new line character?
(a) \n
(b) \r
(c) \t
(d) \nl
Answer:
(a) \n

Question 17.
Which one of the following is used to print all elements separated by space?
(a) ,
(b) .
(c) :
(d) ;
Answer:
(a) ,

Question 18.
SQL provides various clauses that can be used in the …………………………….. statements.
Answer:
SELECT

Question 19.
The clauses in SQL can be called through
(a) C
(b) C++
(c) Python script
(d) DOS
Answer:
(c) Python script

Question 20.
In SQL, the …………………………. clause is used to extract only those records that fulfill a specified condition.
(a) why
(b) what
(c) where
(d) how
Answer:
(c) where

Question 21.
Which clause returns are record for each group?
(a) Select
(b) group in
(c) group with
(d) group by
Answer:
(d) group by

Question 22.
Identify the statement which is wrong?
(a) Group by clause is used with aggregate functions
(b) group by clause groups records into summary rows
(c) group by clause is used to filter data
Answer:
(c) group by clause is used to filter data

Question 23.
Pick the odd one out.
(i) count, max, min, OR, SUM
(ii) AND, OR, MAX, NOT
(iii) COUNT, NULL, AVG, SUM
Answer:
(i) OR, (ii) MAX, (iii) NULL

Question 24.
How many types of sorting are there?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 25.
Which command is used to display the records in asscending or descending order.
(a) Group by
(b) order by
(c) group with
(d) order with
Answer:
(b) order by

Question 26.
Identify the Incorrect pair
(a) Group by – 1. Aggregate functions
(b) order by – 2. Sortind data
(c) Having – 3. filter data
(d) where – 4.Max, min
(a) 2
(b) (1)
(c) (4)
(d) (3)
Answer:
(c) d

Question 27.
Find the Incorrect statement?
(a) The WHERE clause can be combined with AND, OR, NOT
(b) Having clause is used to filter data based on the group function
(c) WHERE cannot be used with NOT
Answer:
(c) WHERE cannot be used with NOT

Question 28.
Which operators are used to filter records based on more than one condition?
(a) AND
(b) NOT
(c) OR
(d) a & c
Answer:
(d) a & c

Question 29.
How many values are returned from the aggregate functions?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1

Question 30.
Find the correct answer.
(i) count functions returns the number of rows in a table satisfying the criteria
(ii) count returns 0 if there were no matching rows
(iii) Null values are counted
(a) (i), (ii) – True
(b) (ii), (iii) – true
(c) (i), (ii), (iii) – True
(d) (i),.(ii), (iii) – False
Answer:
(a) (i), (ii) – True

Question 31.
Find the correct statement.
(a) A record can be deleted using sql command
(b) A record can be deleted with python
(c) both are true
Answer:
(c) both are true

Question 32.
Which command is used to accpet data during run time in python?
(a) Insert( )
(b) input( )
(c) create( )
(d) update( )
Answer:
(b) input( )

Question 33.
The two kinds Of placeholders in sqlite3 module are ………………………….. and ………………………….
Answer:
qmarkstyle, namedstyle

Question 34.
…………………………… is used to display the field names of the table
Answer:
cursor description

Question 35.
In cursor description o_index refers to the column name
Answer:
O

Question 36.
A) Query result can be stored in csv file.
R) To display the query output in a tabular form
(a) R is the reason for A
(b) R is wrong
(c) A is wrong
(d) both are not related
Answer:
(a) R is the reason for A

Question 37.
The path of a python file can be represented as ……………………….. and ………………………………
(a) /, //
(b) \, \\
(c) /, \\
(d) \, //
Answer:
(c) /, \\

Question 38.
Which table holds the key information about the database tables?
(a) page
(b) select
(c) primary
(d) Master
Answer:
(d) Master

Question 39.
………………………. functions are used to do operations from the values of the column and a single value is returned.
Answer:
Aggregate

Question 40.
Which function returns the smallest value of the selected columns?
(a) MIN( )
(b) MINIMUM( )
(c) SMALL( )
(d) LEAST( )
Answer:
(a) MIN( )

PART – II
II. Answer The Following Questions

Question 1.
Write note on “Select” command?
Answer:
“Select” is the most commonly used statement in SQL. The SELECT Statement in SQL is used to retrieve or fetch data from a table in a database. The syntax for using this statement is “Select*from table_name” and all the table data can be fetched in an object in the form of list of lists.

Question 2.
Mention the frequently used clauses in SQL?
Answer:

1. DISTINCT
2. WHERE
3. GROUP BY
4. ORDER BY
5. HAVING

Question 3.
What is the use of distinct clause in SQL?
Answer:
The distinct clause is helpful when there is need of avoiding the duplicate values present in any specific columns/table. When we use distinct keyword only the unique values are fetched.

Question 4.
List some aggregate functions in SQL?
Answer:

1. COUNTO
2. SUM( )
3. MIN( )
4. AVG( )
5. MAX( )

PART – III
III. Answer The Following Questions

Question 1.
What is meant by cursor? How is it created?
Answer:
A cursor in SQL and databases is a control structure to traverse over the records in a database. So it’s used for the fetching of the results.
Cursor is used for performing all SQL commands.
The cursor object is created by calling the cursor( ) method of connection. The cursor is used to traverse the records from the result set.

Question 2.
Write a short note on fetchall( ), fetchone( ) and fetchmany( ) commands?
Answer:
cursor. fetchall( ) -fetchall ( )method is to fetch all rows from the database table
cursor. fetchone( ) – The fetchone ( ) method returns the next row of a query result set or None in case there is no row left.
cursor.fetchmany( ) method that returns the next number of rows (n) of the result set.

Question 3.
How will you use order by clause in SQL. Explain with sample program?
Answer:
The ORDER BY Clause can be used along with the SELECT statement to sort the data of specific fields in an ordered way. It is used to sort the result-set in ascending or descending order. In this example name and Rollno of the students are displayed in alphabetical order of names.
Example
import sqlite3
connection= sqlite3.connect(“Academy.db”)
cursor = connection.cursor( )
cursor.execute(“SELECT Rollno,sname FROM student Order BY sname”)
result = cursor.fetchall( )
print(*result,sep=” \n”)
OUTPUT
(1, ‘Akshay’)
(2, ‘Aravind’)
(3, ‘BASKAR’)
(6, ‘PRIYA’)
(4, ‘SAJINI’)
(7, ‘TARUN’)
(5, ‘VARUN’)

Question 4.
Write a program to count the number of male and female students from the student table
Example
Answer:
import sqlite3
connection= sqlite3.connect(“Academy.db”)
cursor = connection.cursor( )
cursor.execute(“SELECT gender,count(gender) FROM student Group BY gender”)
result = cursor. fetchall( )
print(*result,sep=”\n”)
OUTPUT
(‘F’, 2)
(‘M’, 5)

Question 5.
Write a program to count the number of records in a table?
Example 1 :In this example we are going to count the number of records( rows)
import sqlite3
connection = sqlite3.connect(“Academy.db”)
cursor = connection.cursor( )
cursor.execute(“SELECT COUNT(*) FROM student”)
result = cursor. fetchall( )
print( result)
Output:
[(7,) ]

Question 6.
Write a program to display list of tables created in a database?
Example
Answer:
import sqlite3
con= sqlite3 .connect(‘Academy.db’)
cursor = con.cursor( )
cursor.execute(“SELECT name FROM sqlitemaster WHERE type-table’;”)
print(cursor. fetchall( ))
OUTPUT
[(‘Student’,), (‘Appointment’,), (‘Person’,)]

PART – IV
IV. Answer The Following Questions

Question 1.
Explain OR, AND and NOT operator in SQL?
Answer:
The WHERE clause can be combined with AND, OR, and NOT operators. The AND and OR operators are used to filter records based on more than one condition. In this example you are going to display the details of students who have scored other than ‘A’, or ‘B’ from the “student table”
Example for WHERE WITH NOT Operator
Example
import sqlite3
connection= sqlite3.connect(“Academy.db”)
cursor = connection.cursor( )
cursor.execute(“SELECT *FROM student where grade< >’A’ and Grade< >’B'”)
result = cursor. fetchall( )
print(*result,sep=”\n”)
OUTPUT
(3, ‘BASKAR’, ’C, ’M’, 75.2, T998-05-17′)
(7, ‘TARUN’, ‘D’, ‘M’, 62.3, ‘1999-02-01’)
Example for WHERE WITH AND Operator
In this example we are going to display the name, Rollno and Average of students who have scored an average between 80 to 90% (both limits are inclusive)
Example
import sqlite3
connection = sqlite3.connect(“Academy.db”)
cursor = connection.cursor( )
cursor.execute(“SELECT Rollno, Same, Average FROM student WHERE (Average>=80 AND Average<=90)”)
result = cursor.fetchall( )
print(*result,sep=”\n”)
OUTPUT
(1, ‘Akshay’, 87.8)
(5, ‘VARUN’, 80.6)
Example for WHERE WITH OR Operator
In this example we are going to display the name and Rollno of students who have not scored an average between 60 to 70%
Example import sqlite3
connection = sqlite3.connect(“Academy.db”)
cursor = connection.cursor( )
cursor.execute(“SELECT Rollno,sname FROM student WHERE (Average<60 OR Average> 70)”)
result = cursor. fetchall( )
print(*result,sep=”\n”)
OUTPUT
(1, ‘Akshay’)
(2, ‘Aravind’)
(3, ‘BASKAR’)
(4, ‘SAJINI’)
(5, ‘VARUN’)
(6, ‘PRIYA’)

Question 2.
Write python program to accept 5 students names, their ages and ids during run time and display all the records from the table?
Answer:
In this example we are going to accept data using Python input() command during runtime and then going to write in the Table called “Person”
Example
# code for executing query using input data
import sqlite3
#creates a database in RAM
con =sqlite3.connect(“Academy,db”)
cur =con.cursor( )
cur.execute(“DROP Table person”)
cur.execute(“create table person (name, age, id)”)
print(“Enter 5 students names:”)
who =[input( ) for i in range(5)]
print(“Enter their ages respectively:”)
age =[int(input()) for i in range(5)]
print(“Enter their ids respectively:”)
p_d =[int(input( ))for i in range(5)]
n =len(who)
for i in range(n):
#This is the q-mark style:
cur.execute(“insert into person values(?,?,?)”, (who[i], age[i], p_id[i]))
#And this is the named style:
cur.execute(“select *from person”)
#Fetches all entries from table
print(“Displaying All the Records From Person Table”)
print (*cur.fetchall(), sep=’\n’)
OUTPUT
Enter 5 students names:
RAM
KEERTHANA
KRISHNA
HARISH
GIRISH
Enter their ages respectively:
28
12
21
18
16
Enter their ids respectively:
1
2
3
4
5
Displaying All the Records From Person Table
(‘RAM’, 28, 1)
(‘KEERTHANA’, 12, 2)
(‘KRISHNA’, 21, 3)
(‘HARISH’, 18,4)
(‘GIRISH’, 16, 5)

Question 3.
Considers the following tables appointments and students and write the python script to display the following output?
Answer:
Given:
Student → Table Rollno, Sname, grade, gender, average, birth-date
Appointment → rollno, duty, age
Example
import sqlite3
connection = sqlite3.connect(“Academy.db”)
cursor = connection.cursor( )
cursor.execute(“””DROP TABLE Appointment;”””)
sql_command = “””
CREATE TABLE Appointment(rollnointprimarkey,Dutyvarchar(10),age int)”””
cursor.execute(sqlcommand)
sql command – ‘””INSERT INTO Appointment (Rollno,Duty ,age)
VALUES (“1”, “Prefect”, “17”);”””
cursor, execute(sql_command)
sql_command =”””INSERT INTO Appointment (Rollno, Duty, age)
VALUES (“2”, “Secretary”, “16”);”””
cursor.execute(sql_command)
# never forget this, if you want the changes to be saved:
connection.commit( )
cursor.execute(“SELECT student.rollno,student.sname,Appointment.
Duty,Appointment.Age FROM student,Appointment where student.
rollno= Appointment.rollno”)
#print (cursor.description) to display the field names of the table
co= [i[0] for i in cursor.description]
print(co)
# Field informations can be readfrom cursor.description.
result = cursor. fetchall( )
for r in result:
print(r)
OUTPUT
[‘Rollno’, ‘Sname’, ‘Duty’, ‘age’]
(1, ‘Akshay’, ‘Prefect’, 17)
(2,’Aravind’,’Secretary’, 16)

Students can Download Tamil Nadu 12th Maths Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Maths Model Question Paper 3 English Medium

Instructions:

1.  The question paper comprises of four parts.
2.  You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. questions of Part I, II. III and IV are to be attempted separately
4. Question numbers 1 to 20 in Part I are objective type questions of one -mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Parr III are three-marks questions, These are to be answered in about three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-marks questions. These are to be answered) in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 90

Part – I

I. Choose the correct answer. Answer all the questions.  [20 × 1 = 20]

Question 1.
If (QAB)^-1 = \(\left[\begin{array}{cc}
12 & -17 \\
-19 & 27
\end{array}\right]\) and A^-1 = \(\left[\begin{array}{cc}
1 & -1 \\
-2 & 3
\end{array}\right]\) then B^-1 = ……………

Answer:
(a) \(\left[\begin{array}{cc}
2 & -5 \\
-3 & 8
\end{array}\right]\)

Question 2.
If z is a complex number such that z ∈ C\R and \(z+\frac{1}{z}\) ∈ R, then |z| is ……………
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
(b) 1

Question 3.
Let A = \(\left[\begin{array}{ccc}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]\) and 4B = \(\left[\begin{array}{ccc}
3 & 1 & -1 \\
1 & 3 & x \\
-1 & 1 & 3
\end{array}\right]\). If B is the inverse of A, then the value of x is ……………
(a) 2
(b) 4
(c) 3
(d) 1
Answer:
(d) 1

Question 4.
If sin^-1x + sin^-1y = \(\frac{2 \pi}{3}\) then cos^-1x + cos^-1y is equal to ……………
(a) \(\frac{2 \pi}{3}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{\pi}{6}\)
(d) π
Answer:
(b) \(\frac{\pi}{3}\)

Question 5.
The eccentricity of the ellipse (x – 3)² + (y – 4)² = \(\frac{y^{2}}{9}\) is ……………
(a) √3/2
(b) 1/3
(c) 3√2
(d) 1/√3
Answer:
(b) 1/3

Question 6.
The directrix of the parabola y² = 4x is ……………
(a) y = -1
(b) x = -1
(c) y = 1
(d) x = 1
Answer:
(b) x = -1

Question 7.
The angle between the line \(\vec{r}=(\hat{i}+2 \hat{j}-3 \hat{k})+t(2 \hat{i}+\hat{j}-2 \hat{k})\) and the plane \(\vec{r} \cdot(\hat{i}+\hat{j})+4=0\) is ……………
(a) 0°
(b) 30°
(c) 45°
(d) 90°
Answer:
(c) 45°

Question 8.
The d.c.s. of a vector whose direction ratios are 2, 3, -6 are

Answer:
(a) \(\left(\frac{2}{7}, \frac{3}{7}, \frac{6}{7}\right)\)

Question 9.
A balloon rises straight up at 10 m/s. An observer is 40 m away from the spot where the balloon left the ground. Find the rate of change of the balloon’s angle of elevation in radian per second when the balloon is 30 metres above the ground.
(a) 1/25 radians/sec
(b) 4/25 radians/sec
(c) 1/5 radians/sec
(d) 1/3 radians/sec
Answer:
(b) 4/25 radians/sec

Question 10.
The asymptote to the curve y²(2 +x) = x² (6 – x) is
(a) x = 2
(b) x = -2
(c) x = 6
(d) x = -6
Answer:
(b) x = -2

Question 11.
The change in the surface area S = 6x² of a cube when the edge length varies from x₀ to x₀ + dx is
(a) 12x₀ + dx
(b) 12x₀dx
(c) 6x₀dx
(d) 6x₀ + dx
Answer:
(b) 12x₀dx

Question 12.
The differential of y if y = sin 2x is
(a) 2 cos 2x
(b) 2 cos 2x.dx
(c) -2 cos lx.dx
(d) cos2x.dx
Answer:
(b) 2 cos 2x.dx

Question 13.
The value of \(\int_{-4}^{4}\left[\tan ^{-1}\left(\frac{x^{2}}{x^{4}+1}\right)+\tan ^{-1}\left(\frac{x^{4}+1}{x^{2}}\right)\right] d x\) is
(a) π
(b) 2π
(c) 3π
(d) 4π
Answer:
(d) 4π

Question 14.
\(\begin{aligned}
&\int^{b} f(x) d x=\\
&a
\end{aligned}\) ………

Answer:
(b) \(-\int_{b}^{a} f(x) d x\)

Question 15.
The integrating factor of the differential equation \(\frac{d y}{d x}+y=\frac{1+y}{x}\) is
(a) \(\frac{x}{e^{\lambda}}\)
(b) \(\frac{e^{x}}{x}\)
(c) λe^x
(d) e^x

Question 16.
The differential equation representing the family of curves y = A cos (x + B), where A and B are parameters, is

Answer:
(b)\(\frac{d^{2} y}{d x^{2}}+y=0\)

Question 17.
A rod of length 21 is broken into two pieces at random. The probability density function of
the shorter of the two pieces is \(f(x)=\left\{\begin{array}{ll}
\frac{1}{l}, & 0< x, \mathrm{x}=0, & l \leq x<2 l
\end{array}\right.\)
The mean and variance of the shorter of the two pieces are respectively ,

Answer:

Question 18.
A computer salesperson knows from his past experience that he sells computers to one in every twenty customers who enter the showroom. What is the probability that he will sell a computer to exactly two of the next three customer?

Answer:
(a) \(\frac{57}{20^{3}}\)

Question 19.
The operation * defined by a * b = \(\frac{a b}{7}\) is not a binary operation on
(a) Q⁺
(b) Z
(c) R
(d) C

Question 20.
If X is a discrete random variable then P(X >a) =
(a) P(X < a)
(b) 1 – P(X < a)
(c) 1 – P(X < a)
(D) = 0

Part – II

II. Answer any seven questions. Question No. 30 is compulsory.[7 x 2 = 14]

Question 21.
State the reason for \(\cos ^{-1}\left[\cos \left(-\frac{\pi}{6}\right)\right] \neq-\frac{\pi}{6}\)
Answer:
We know cos(-θ) = cos θ

Question 22.
Find the equations of the tangent and normal to the circle x² + y² = 25 at P(-3, 4).
Answer:
Equation of tangent to the circle at P (x₁, y₁) is xx₁ +yy₁ = a².
That is, x(-3) + y(4) = 25
-3x + 4y = 25
Equation of normal is xy₁ – yx₁ = 0
That is, 4x + 3y = 0.

Question 23.
Determine whether the three vectors \(2 \hat{i}+3 \hat{j}+\hat{k}, \hat{i}-2 \hat{j}+2 \hat{k}\) and \(3 \hat{i}+\hat{j}+3 \hat{k}\) are coplanar.
Answer:

Question 24.
Find the intervals of monotonicities and hence find the local extremum for the following
functions f(x) = \(\frac{x}{x-5}\)
Answer:
f(x) = \(\frac{x}{x-5}\)


(when x ≠ 5)
f(x) is strictly decreasing on (-∞, 5) and (5, ∞)
And there is no local extremum

Question 25.
Find the linear approximation for f(x) = [altex]\sqrt{1+x}[/latex] , x ≥ -1, at x₀ = 3. Use the linear approximation to estimate f(3.2).
Answer:
We know from that L(x) = f(x₀) + f'(x₀)(x – x₀). We have x₀ = 3, ∆x = 0.2 and hence

Actually, if we use a calculator to calculate we get \(\sqrt{4.2}=2.04939\)

Question 26.
Evaluate \(\int_{0}^{2 \pi} \sin ^{4} x \cos ^{3} d x\)
Answer:
f(x) = sin⁴x cos³x
f(2π – x) = sin⁴(2π – x) cos³ (2π- x)
= sin⁴x cos³x = f(x)
\(2\int_{0}^{\pi} \sin ^{4} x \cos ^{3} d x\)
Again f(x) = sin⁴x cos³x
f(π – x) = sin⁴(π – x) cos³ (π – x)
f(π – x) = – sin⁴(π -x) cos³(π -x) = -f(x) = 0.

Question 27.
Solve the differential equation \(\frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}\)
Answer:

Question 28.
Suppose the amount of milk sold daily at a milk booth is distributed with a minimum of 200 litres and a maximum of 600 litres with probability density function
\(f(x)=\left\{\begin{array}{lr}
k & 200 \leq x \leq 600 \\
0 & \text { otherwise }
\end{array}\right.\)
Find the value of k.
Answer:

Question 29.
Let * be defined on R by (a*b) = a + b + ab – 7. Is * binary on R ? If so, find \(3*\left(\frac{-7}{15}\right)\)
Answer:
a*b = a + b + ab – 7
Now when a, b ∈ R, then ab ∈ R also a + b ∈ R.
So, a + b + ab ∈ R.
We know – 7 ∈ R.
So, a + b + ab – 7 ∈ R.
(i.e.) a * b ∈ R

Question 30.
Examine for the rational roots of x⁸ – 3x + 1 = 0.
Answer:
x⁸ – 3x + 1 = 0……..(1)
Here a_n = – 1, a₀ = 1
If p/q is a rational root of (1)
Then q is a factor a_n, p is a factor of aQ
The possible values of p and q are ±1. Among the possible values 1,-1, [(p,q) = 1]
None of them satisfy the equation (1)
.’. The above equation has no rational roots.

Part – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 x 3 = 21]

Question 31.
Solve the following system of linear equations by matrix inversion method:
2x + 5y = -2, x + 2y = -3
Answer:

Question 32.
If (x₁ + iy₁)(x₂ + iy₂)(x₃ + iy₃)…………(x_n +iy_n) = a + ib , show that
(i) (x₁² + y₁²)(x₂² + y₂²)(x₃² + y₃²)…….(x_n² + y_n²) = a² + b²
(ii)

Answer:

Question 33.
Find the sum of squares of roots of the equation 2x⁴ – 5x³ + 6x² – 3 = 0.
Answer:
The given equation is 2x⁴ – 8x³ + 6x² – 3 = 0.

Let the roots be α β γ δ
α + β + γ + δ = -b = 4
α β + βγ + γδ + aδ + aγ + βδ ) = c = 3
α βγ + βγδ + γαδ = -d = 0
αβγδ? = -3/2
To Find α² + β² + γ² + δ ² = ( α + β + γ + δ)² – 2 (αβ + βγ + γδ + αδ + αγ +βδ)
= (4)² – 2(3) = 16 – 6 = 10

Question 34.
Find the non-parametric form of vector equation, and Cartesian equation of the plane
passing through the point (0, 1, -5) and parallel to the straight lines .
\(\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+s(2 \hat{i}+3 \hat{j}+6 \hat{k})\) and \(\hat{r}=(\hat{i}-3 \hat{j}+5 \hat{k})+t(\hat{i}+\hat{j}-\hat{k})\)
Answer:
We observe that the required plane is parallel to the vectors \(\vec{b}=2 \hat{i}+3 \hat{j}+6 \hat{k}, \vec{c}=\hat{i}+\hat{j}-\hat{k}\) and passing through the point (0, 1, —5) with position vector \(\vec{a}\). We observe that \(\vec{b}\) is not parallel to \(\vec{c}\)
Then the vector equation of the plane in non-parametric form is given by

If \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) is the position vector of an arbitrary point on the plane, then from the above equation, we get the Cartesian equation of the plane as -9x + 8y – z = 13 or 9x – 8y + z +13 = 0.

Question 35.
Find ‘C’ of Lagrange’s mean value theorem for the function f(x) = x³ – 5x² – 3x in [1,3]
Answer:
f(x) = x³ – 5x² – 3x
a = 1, b = 3
f(a) = f(1)= 1 – 5 – 3 = -7
f(b) = f (3) = 21 – 5(9) – 3(3)
= 27 – 45 – 9 = -27

f(x) = x³ – 5x²– 3x
f(x) = 3² – 10x – 3
f'(c) = 3c² – 10c – 3 ………(2)
From (1) and (2),
3c² – 10c – 3
3c² – 10c – 3 + 10 = 0
3c² – 10c + 7 =0
3c² – 3c – 7c + 7 = 0

So, Lagrange’s mean value theorem is true with c = 7/3

Question 36.
Using differentials, find the approximate value of each of the following upto 3 places of decimal. (255)^1/4
Answer:

Question 37.
Find the area of the region bounded by the line 6x + 5y = 30, x – axis and the lines x = – 1 and x = 3.
Answer:
The region is sketched in Fig. It lies above the x – axis.
Hence, the required area is given by

Question 38.
A six sided die is marked ‘2’ on one face, ‘3’ on two of its faces, and ‘4’ on remaining three faces. The die is thrown twice. If X denotes the total score in two throws, find the values of the random variable and number of points in its inverse images.
Answer:
Six sided die marked ‘2’ on one face, ‘3’ on two faces and ‘4’ on three faces.
When it is thrown twice, we get 36 sample points.
‘X’ denotes sum of the face numbers and the possible values of ‘X’ are 4, 5, 6, 7 and 8
For X = 4, the sample point is (2, 2)
For X = 5, the sample points are (2, 3), (3, 2)
For X = 6, the sample points are (3, 3), (2, 4), (4, 2)
For X = 7, the sample points are (3, 4), (4, 3)
For X = 8, the sample point is (4, 4)

Question 39.
Check whether the statement p —>• (q —► p) is a tautology or a contradiction without using the truth table.
Answer:

Question 40.
Solve: \(\left(1-x^{2}\right) \frac{d y}{d x}+2 x y=x \sqrt{1-x^{2}}\)
Answer:

Part – IV

IV. Answer all the questions. [7 x 5 = 35]

Question 41.
(a) Solve the following system of homogeneous equation.
3x + 2y + 7z – 0, 4x – 3y – 2z – 0, 5x + 9y + 23z = 0,
Answer:
The matrix form of the above equations is

The above matrix is in echelon form. Here ρ(A, B) = ρ( A) < number of unknowns.
⇒ The system is consistent with infinite number of solutions. To find the solutions.
Writing the equivalent equations.
We get 3x + 2y + 7z = 0
-17y – 34z = 0
Taking z = t in (2) we get -17y – 34t = 0

Taking z = t; y = -2t in (1) we get
3x + 2 (-2t) + 7t= 0
3x – 4f + 7f = 0 ⇒ 3x = -3f ⇒ x = -t
So the solution is x = -t; y = -2t; and z=t, t∈R [OR]

(b) Two cards are drawn with replacement from a well shuffled deck of 52 cards. Find the mean and variance for the number of aces.
Answer:
n (S) = 52
Number of aces = n (A) = 4

Let X denotes number of aces when two cards are drawn.

a) Show thatis purely imaginary.

Question 42.
(a) Show that \(\left(\frac{19+9 i}{5-3 i}\right)^{15}-\left(\frac{8+i}{1+2 i}\right)^{15}\) is purely imaginary.
Answer:

[OR]

(b) (i) Define an operation * on Q as follows: a*b = \(\left(\frac{a+b}{2}\right) \) a,b ∈ Q . Examine the closure, commutative, and associative properties satisfied by * on Q
(ii) Define an operation * on Q as follows: a*b = \(\left(\frac{a+b}{2}\right) \) a,b ∈ Q
existence of identity and the existence of inverse for the operation * on Q
Answer:
(i) 1. Closure property:
Let a, b ∈ Q.
a*b = \(\left(\frac{a+b}{2}\right) \) ∈ Q (∵ a, b, 2 ∈ Q)
So, closure property is satisfied.

2. Commutative property:
Let a,b ∈ Q.
a*b = \(\left(\frac{a+b}{2}\right) \) …..(1)
b*a = \(\left(\frac{a+b}{2}\right) \) ……(2)
(1) = (2) ⇒ Now a * b = b * a
⇒ Commutative property is satisfied.

3. Associative property:
Let a, b,c ∈ Q.
To prove associative property we have to prove that a * (b * c) = (a * b) * c

So, associative property is not satisfied.

(ii) a* b = \(\left(\frac{a+b}{2}\right) \)
Let e ∈ Q be the identity element. Then a * e = a
To find e : a * e = a
(i.e.) \(\left(\frac{a+b}{2}\right) \) = a ⇒a + e = 2a ⇒ e = 2a- a
(i.e.) the identity element e = a which is not possible.
So, the identity element does not exist and so inverse does not exist.

Question 43.
(a) Solve (v – 3) (x – 6) (x – 1) (x + 2) + 54 = 0.
Answer:
(x – 3) (x – 1) (x – 6) (x + 2) + 54 = 0
(x²– 4x + 3) (x² – 4x – 12) + 54 = 0
Put x² – 4x = y
(y + 3)(y – 12) + 54 = 0
y²– 9y – 36 + 54 = 0
y² – 9y + 18 = 0
(y – 3) (y – 6) = 0
(y – 3) = 0
x² – 4x – 3 = 0


[OR]

(b) Find the local extrema of the function f(x) = 4x⁶ – 6x⁴
Answer:
Differentiating with respect to x, we get
f (x) = 24x⁵ – 24x³
= 24x³ (x² – 1)
= 24x³ (x + 1) (x – 1)
f’ (x) = 0 ⇒ x = -1, 0, 1. Hence the critical
numbers are x = -1, 0, 1
Now, f'”(x) = 120x⁴ – 72x² = 24x² (5x² – 3)
⇒ f” (-1)= 48, f” (0) = 0, f”(1) = 48.
As f” (-1) and f”(1) are positive by the second derivative test, the function f(x) has local minimum. But at x = 0, f ” (0) = 0. That is the second derivative test does not give any information about local extrema at x = 0 . Therefore, we need to go back to the first derivative test. The intervals of monotonicity is tabulated in the table.

Question 44.
(a) Prove that \(\tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2}\)
Answer:

(b) If v(x, y) = \(\log \left(\frac{x^{2}+y^{2}}{x+y}\right)\), prove that \(x \frac{\partial v}{\partial x}+y \frac{\partial v}{\partial y}=1\)
Answer:

‘ f ‘ is a homogeneous function of degree 1. By Euler’s theorem, we have

Question 45.
(a) Find the equations of the two tangents that can be drawn from (5, 2) to the ellipse 2x² + 7y² = 14 .
Answer:
2x² + 7y² = 14

The equation of tangent to the above ellipse will be of the form

Here the tangents are drawn from the point (5, 2)

Squaring on both sides we get
(2 – 5m)² = 7m² + 2
25m² + 4 – 20m – 7m² – 2 = 0
18m² – 20m + 2 = 0
(= by 2) => 9m² – 10m + 1 = 0
(9m – 1) (m – 1) = 0
m = 1 (or) m = 1/9
When m = 1, the equation of tangent is
y = x + 3 or x-y + 3 = 0
When m = 1/9 the equation of tangent is 9

[OR]

(b) Find the coordinates of the foot of the perpendicular and length of the perpendicular from the point (4, 3, 2) to the plane x + 2y + 3z = 2.
Answer:

Direction of the normal plane (1, 2, 3) JP(4, 3,2)
d.c.s of the PQ is\(\frac{x_{1}-4}{1}=\frac{y_{1}-3}{2}=\frac{z_{1}-2}{3}=k\)
x₁k + 4, y₁ = 2k + 3, z₁ = 3k + 2
This passes through the plane x + 2y + 3z = 2
k + 4 + 2(2 k + 3) + 3(3 k + 2) = 2
k+ 4 + 4k + 6 + 9k + 6 = 2
14k = 2 – 16
14k = -14
k = -1
∴ The coordinate of the foot of the perpendicular is (3, 1, -1)
∴ Length of the perpendicular to the plane is

Question 46.
(a) Solve (x + y)²\(\frac{d y}{d x}\) = a² dx
Answer:
Put x+y = z. Differentiating with respect to x we get

(b) A police jeep, approaching an orthogonal intersection from the northern direction, is chasing a speeding car that has turned and moving straight east. When the jeep is 0.6 km north of the intersection and the car is 0.8 km to the east. The police determine with a radar that the distance between them and the car is increasing at 20 km/hr. If the jeep is moving at 60 km/hr at the instant of measurement, what is the speed of the car?
Answer:

Question 47.
(a) Let w(x, y’ z) = \(\frac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}(x, y, z)\) ≠ (0, 0, 0). Show that \(\frac{\partial^{2} w}{\partial x^{2}}+\frac{\partial^{2} w}{\partial y^{2}}+\frac{\partial^{2} w}{\partial z^{2}}=0\)
Answer:

(b) Evaluate \(\int_{0}^{\frac{\pi}{2}} \frac{e^{-\tan x}}{\cos ^{6} \pi x} d x\)
Answer:

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TN State Board 12th Commerce Model Question Paper 1 English Medium

Instructions:

1. The question paper comprises of four parts.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II. III and IV are to be attempted separately.
4. Question numbers 1 to 20 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 21 to 30 in Part II are two-mark questions. These are to be answered in about 50 words.
6. Question numbers 31 to 40 in Part III are three-mark questions. These are to be answered in about 150 words.
7. Question numbers 41 to 47 in Part IV are five-mark questions. These are to be answered in about 250 words. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 90

Part – I

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
With a wider span, there will be ………… hierarchical levels.
(a) more
(b) less
(c) multiple
(d) additional
Answer:
(b) less

Question 2.
Spot market is a market when the delivery of the financial instrument and payment of cash occurs.
(a) Immediately
(b) In the future
(c) Uncertain
(d) After one mouth
Answer:
(a) Immediately

Question 3.
How many times a security can be sold in a secondary market?
(a) only one time
(b) two time
(c) three times
(d) multiple times
Answer:
(d) multiple times

Question 4.
Money market provides ……………
(a) medium term funds
(b) short term funds
(c) long term funds
(d) shares
Answer:
(b) short term funds

Question 5.
Jobbers transact in a stock exchange.
(a) For their Clients
(b) For their Own Transactions
(c) For other Brokers
(d) For other Members
Answer:
(b) For their Own Transactions

Question 6.
Registering and controlling the functioning of collective investment schemes as ………..
(a) Mutual funds
(b) Listing
(c) Rematerialisation
(d) Dematerialisation
Answer:
(d) Dematerialisation

Question 7.
Human resource management determines the …… relationship.
(a) internal, external
(b) employer, employee
(c) owner, servant
(d) principle, agent
Answer:
(b) employer, employee

Question 8.
Advertisement is a ………. source of recruitment.
(a) internal
(b) external
(c) agent
(d) outsourcing
Answer:
(b) external

Question 9.
Selection is usually considered as a,process.
(a) positive
(b) negative
(c) natural
(d) none of these
Answer:
(b) negative

Question 10.
Improves skill levels of employees to ensure better job performance.
(a) Training
(b) Selection
(c) Recruitment
(d) Performance appraisal
Answer:
(d) Performance appraisal

Question 11.
Which one of the market deals in the purchase and sale of shares and debentures?
(a) stock exchange market
(b) Manufactured goods market
(c) local market
(d) family market
Answer:
(a) stock exchange market

Question 12.
In the following variables which one is not the variable of marketing mix?
(a) place variable
(b) product variable
(c) program variable
(d) price variable
Answer:
(c) program variable

Question 13.
Social marketing deals with
(a) Society
(b) Social clan
(c) Social change
(d) Social evil
Answer:
(c) Social change

Question 14.
The main objective of all business enterprises is …………….
(a) Providing service
(b) Providing better standard of life
(c) Providing necessities to the society
(d) Earn profit
Answer:
(d) Earn profit

Question 15.
As the consumer is having the rights, they are also having
(a) measures
(b) Promotion
(c) responsibilities
(d) duties
Answer:
(c) responsibilities

Question 16.
Which of the following is not a consumer right summed up by John.F.Kennedy.
(a) Right to safety
(b) Right to choose
(c) Right to consume
(d) Right to be informed
Answer:
(a) Right to safety

Question 17.
The chairman of the District Forum is
(a) District Judge
(b) High court Judge
(c) Supreme court Judge
(d) None of the above
Answer:
(a) District Judge

Question 18
ownership makes bold management decisions due to their strong foundation in the intimation level.
(a) Private
(b) Public
(c) Corporate
(d) MNC’s
Answer:
(a) Private

Question 19.
The property in the goods means the
(a) Possession of goods
(b) Custody of goods
(c) Ownership of goods
(d) Both (a) and (b)
Answer:
(c) Ownership of goods

Question 20.
Number of parties is a bill of exchange are
(a) 2
(b) 6
(c) 3
(d) 4
Answer:
(c) 3

Part – II

Answer any seven in which question No. 30 is compulsory. [7 x 2 = 14]

Question 21.
What are the objectives of MBO?
Answer:
Management by objectives is intended primarily:

• To measure and judge performance.
• To relate individual performance to organisational goals.
• To clarify both the job to be done and the expectations of accomplishment.
• To foster the increasing competence and growth of the subordinates.

Question 22.
Write a note on OTCEI.
Answer:
The OTCEI was set up by a premier financial institution to allow the trading of securities across the electronic counters throughout the country.

Question 23.
Define Stock Exchange.
Answer:
According to Husband and Dockerary, “Stock exchanges are privately organized markets which are used to facilitate trading in securities.”

Question 24.
Give the meaning of Human Resource.
Answer:
In an organisation, the human resource are the employees who are inevitable for the survival t and success of the enterprise.

Question 25.
What is service marketing?
Answer:
Service marketing denotes the processing of selling service goods like telecommunication, banking, insurance, car rentals, healthcare, tourism, professional services, repairs etc.

Question 26.
Write short notes on: “Right to be informed.”
Answer:
Consumers should be given all the relevant facts about the product. This implies that the manufacturer and the dealer are expected to disclose all the material facts relevant and relating to the product.

Question 27.
What is GST?
Answer:
GST is the indirect tax levied on goods and services across the country. It is a comprehensive, multi-stage, destination-based tax that is levied on every value addition.

Question 28.
What is a contract of sale of goods?
Answer:
Contract of sale of goods is a contract whereby the seller transfers or agrees to transfer the property (ownership) of the goods to the buyer for a price.

Question 29.
What is the other name of business entrepreneur?
Answer:
The term entrepreneur means the person who takes steps for commencing the business. So he is otherwise called as organiser or proprietor.

Question 30.
Define Director.
Answer:
The Companies Act 2013 section 2 (34) defines a director appointed to the board of a Company is “A Person who is appointed or elected member of the Board of Directors of a company and has the responsibility of determining and implementing policies along with others in the board”.

Part – III

Answer any seven in which question No. 31 is compulsory. [7 x 3 = 21]

Question 31.
What are the process involved in MBO?
Answer:

• Defining Organisational Objectives
• Goals of Each Section
• Fixing Key Result Areas
• Setting subordinate objectives or targets
• Matching Resources with objective
• Periodical Review meetings
• Appraisal of Activities
• Reappraisal of objectives

Question 32.
Differentiate spot market from future market.
Answer:

Question 33.
What are the documents required for a Demat account?
Answer:
For opening a demat account, we submit proof of identity and address along with a passport size photo and the account opening form.

• Documents for Identity: Pan card, Voters ID, Passport, Driver’s License, IT Returns, Electricity and Telephone Bills are the Identity documents.
• Documents for Address: Ration card, Passport, Voter’s ID card, Driving License, Telephone Bills, Electricity bills are the address documents.

Question 34.
Mention any three Role of Marketer.
Answer:
(1) Instigator: As an instigator, marketer keenly watches the developments taking place in the market and identifies marketing opportunities emerging in the ever changing market.

(2) Integrator: Marketer plays a role of integrator in the sense that he collects feedback or vital inputs from channel members and consumers.

(3) Implemented Marketer plays a role of implementer when he/she actually converts marketing opportunities into marketable product.

Question 35.
Discuss the objectives E-Marketing.
Answer:
The following are the objectives of E-Marketing:

• Expansion of market share
• Reduction of distribution and promotional expenses
• Achieving higher brand awareness
• Strengthening database

Question 36.
What do you understand by “Right to redressal”?
Answer:
The complaints and protests are not just to be heard: but the aggrieved party is to be granted compensation within a reasonable time period . There should be prompt settlement of complaints and claims lodged by the aggrieved customers.

Question 37.
Explain the concept of Privatisation.
Answer:
Privatisation means permitting the private sector to set up industries which were previously reserved for the public, sector. Under this policy many Public Sector Units (PSUs) were sold to private sector. The main reason for privatisation was that PSUs were running in losses due to mismanagement and political interference.

Question 38.
What are the characteristics of a bill of exchange?
Answer:
Characteristics of a Bill of Exchange:

• A bill of exchange is a document in writing.
• The document must contain an order to pay.
• The order must be unconditional.
• The instrument must be signed by the person who draws it.
• The name of the person on whom the bill is drawn must be specified in the bill itself.

Question 39.
Distinguish between Entrepreneur and Manager.

Question 40.
When are alternative directors appointed?
Answer:
Alternate director is appointed by the Board of Directors, as a substitute to a director who may be absent from India, for a period which is not less than three months.

Part – IV

Answer all the following questions. [7 x 5 = 35]

Question 41
(a) Explain the various functions of management.
Answer:

• Planning:- Planning is the primary function of management. Planning is a constructive reviewing of future needs so that present actions can be adjusted in view of the established goal.
• Organising:- Organising is the process of establishing harmonious relationship among the members of an organisation and the creation of network of relationship among them.
• Staffing:- Staffing function comprises the activities of selection and placement of competent personnel.
• Directing:- Directing denotes motivating, leading, guiding and communicating with subordinates on an ongoing basis in order to accomplish pre-set goals.
• Controlling:- Controlling is performed to evaluate the performance of employees and deciding increments and promotion decisions.
• Co-ordination:- Co-ordination is the synchronization of the actions of all individuals, working in the enterprise in different capacities.
• Motivating:- The goals are achieved with the help of motivation. Motivation includes increasing the speed of performance of a work and developing a willingness on the part of workers.

Subsidiary Functions:

• Innovation:- Innovation includes developing new material, new products, new techniques in production, new package, new design of a product and cost reduction.
• Representation:- A manager has to act as representative of a company. It is the duty of every manager to have good relation with others.
• Decision-making:- Every employee of an organisation has to take a number of decisions every day. Decision-making helps in the smooth functioning of an organisation.
• Communication:- Communication is the transmission of human thoughts, views or opinions from one person to another person. Communication helps the regulation of job and co-ordinates the activities.

[OR]

(b) Compare the concept of social marketing with service marketing.
Answer:
Social marketing: Social marketing is a new marketing tool.
It is the systematic application of marketing philosophy to achieve social good.
The primary aim of social marketing is ‘social good’ such as anti-tobacco, anti-drug, anti-pollution, anti-dowry, road safety, protection of girl child.

• Service marketing: A service is any activity or benefit that one party can offer to another which is essentially intangible.
• Service marketing is a specialized branch of marketing.
• Service marketing denotes the process of selling service goods like telecommunication, banking, insurance, car rentals, healthcare, tourism and repairs.

Question 42
(a) Explain the various disadvantages of MBO.
Answer:

• MBO fails to explain the philosophy; most of the executives do not know how MBO works? what is MBO? and why is MBO necessary? and how participants can benefit by MBO.
• MBO is a time consuming process. Much time is needed by senior people for framing the MBO. Next, it leads to heavy expenditure and also requires heavy paper work.
• MBO emphasises only on short-term objectives and does not consider the long-term objectives.
• The status of subordinates is necessary for proper objectives setting. But, this is not possible in the process of MBO.
• MBO is rigid one. Objectives should be changed according to the changed circumstances, external or internal. If it is not done, the planned results cannot be obtained.

[OR]
(b) Explain the duties of consumers.
Answer:
Apart from rights, there are certain duties imposed on the consumer. The following are the duties of consumers.

• Buying Quality Products at Reasonable Price: It is the duty of a consumer to purchase a product after gaining a thorough knowledge of its price, quality and other terms and conditions.
• Ensure the Weights and Measurement before Purchase: The consumer should ensure that he/she is getting the product of exact weight and. measure.
• Reading the Label Carefully: It is the duty of the consumer to read the label of the product thoroughly.
• Beware of False and Attractive Advertisements: It is the prime duty of the consumer about the genuineness of the advertisement, before purchasing the product.
• Ensuring the Receipt of Cash Bill: It is a legitimate duty of consumers to get the cash receipt and warranty card supplied along with the bill.

Question 43.
(a) Explain the Instruments of Money Market.
Answer:
(i) Treasury Bills: Treasury bills are very popular and enjoy a higher degree of liquidity since they are issued by the Government. A Treasury bill is nothing but a promissory note issued for a specified period stated therein. The Government promises to pay the specified amount mentioned therein to the bearer of the instrument on the due date. The period does not exceed a period of one year.

(ii) Certificate of Deposits: Certificate of Deposits are short-term deposit instruments issued by banks and financial institutions to raise large sums of money. The Certificate of Deposit is transferable from one party to another. Due to their negotiable feature, they are also known as negotiable certificate of deposit.

(iii) Commercial Bills: The Commercial Bill is an instrument drawn by a seller of goods on a buyer of goods. It possesses the advantages like self-liquidating in nature, recourse to two parties, knowing exact date of transactions, transparency of transactions etc.

[OR]

(b) Explain the overall performance of State Commission.
The State Commission is to be appointed by the State Government. The person who is a judge or retired judge of high court is the president of the state commission.

Performances:

• The compensation of the value should not exceed Rs.20 lakhs and below Rs.1 crore.
• The state commission has the power to call for the records and pass orders in the district forum.
• To furnish the information which is required for the purpose of the Act to any officer.

Question 44.
(a) Explain the powers of SEBI.
Answer:
The various powers of a SEBI are explained below:

• Powers Relating to Stock Exchanges and Intermediaries: SEBI has wide powers to get the information from the stock exchange and intermediaries regarding their business transactions for inspection.
• Power to Impose Monetary Penalties: SEBI has power to impose monetary penalties on capital market intermediaries for violations.
• Power to Initiate Actions in Functions Assigned
• Power to Regulate Insider Trading: SEBI has power to regulate insider trading or can regulate the functions of merchant banker.
• Powers Under Securities Contracts Act: For the regulation of stock exchange, the Ministry of Finance issued a notification, for delegating several of its powers under the securities contract Act.

[OR]

(b) Explain the impact of LPG on Indian Economy.
Impact of Liberalization:

• Liberalization has opened up new business opportunities abroad and increased foreign direct investment.
• New market for various goods came into existence and resulted not only in urban but also in rural development.
• It became very easy to obtain loans from banks for business expansion.

Impact of Privatization:

• Privatization has a positive impact on the financial growth by decreasing the deficits and debts.
• Increase in the efficiency of government undertakings.
• Provide better goods and services to the consumers.

Impact of Globalization:

• Multinational corporations can manufacture, buy and sell goods worldwide.
• Globalization has led to a boom in consumer products market.
• The advent of foreign companies and growth in economy has led to job creation.

Question 45
(a) Describe the significance of External source of recruitment.
Answer:
External sources of recruitment include sources that lie outside the organisation. This provides a wider collection of potential employees with the necessary skillset, especially for managerial and technical positions. Existing employees can recommend suitable candidates, which may lead to a higher level of teamwork and synchronisation among employees. Hiring new employees can lead to the introduction of new blood and thus the introduction of a new set of skills and ideas. External sources of recruitment offer jobs to unskilled, semi-skilled and skilled workers.

[OR]

(b) Discuss in detail the features of a cheque.
Answer:
A cheque is a negotiable instrument drawn on a particular banker.
Features:
(i) Instrument in Writings
A cheque or a bill or a promissory note must be an instrument in writing. Though the law does not prohibit a cheque being written in pencil, bankers never accept it because of risks involved. Alternation is quite easy but detection is impossible in such cases.

(ii) Unconditional Orders
The instrument must contain an order to pay money. It is not necessary that the word ‘order’ or its equivalent must be used to make the document a cheque. It does not cease to be a cheque just because the world ‘please’ is used before the word pay. Further the order must be unconditional.

(iii) Drawn on a Specified Banker Only
The cheque is always drawn on a specified banker. A cheque vitally differs from a bill in this respect as latter can be drawn on any person including a banker. The customer of a banker can draw the cheque only on the particular branch of the bank where he has an account.

(iv) A Certain Sum of Money Only
The order must be for payment of only money. If the banker is asked to deliver securities, the document cannot be called a cheque. Further, the sum of money must be certain.

(v) Payee to be Certain
The cheque must be made payable to a certain person or to the order of a certain person . or to the bearer of the instrument. The word, person includes corporate bodies, local authorities, associations, holders of the office of an institution etc.

(vi) Signed by the Drawer
The cheque is to be signed by the drawer. Further, it should tally with specimen signature furnished to the bank at the time of opening the account

Question 46.
(a) What are the differences between on the job training and off the job training?
Answer:

[OR]

(b) What are the various kinds of Debentures?
Answer:
Debenture is a document issued by the company for acknowledging the loan from the public. Debentures are classified into different categories on the basis of:

• Convertibility of the Instrument
• Security of the Instrument;
• Redemption ability; and
• Registration of Instrument.

On the basis of convertibility:

• Non-Convertible Debentures: These instruments cannot be converted into equity shares.
• Partly Convertible Debentures: A part of these instruments are converted into equity shares.
• Fully Convertible Debentures: These are fully convertible into equity shares.
• Optionally Convertible Debentures: The investor can have the option to either convert the debentures at a price decided by the issuer or agreed upon at the time of issue.

On the basis of Security:

• Secured Debentures: These instruments are secured by a charge on the fixed assets of the issuer company.
• Unsecured Debentures: These instruments are unsecured against the assets.

On the basis of Redeemability:

• Redeemable Debentures: It refers to the debentures which will be redeemed in the future.
• Irredeemable Debentures: It is a debenture, in which no specific time is specified by the companies to pay back the money.

On the basis of Registration:

• Registered Debentures: These are issued in the name of a particular person, who is registered by the company.
• Bearer Debentures: These are issued to the bearer and are negotiable instruments, and are transferred by mere delivery.

Question 41.
(a) Why the marketing is important to the society and individual firm? Explain.
Answer:
Importance of Marketing:
To the Society:

• Marketing is a connecting link between the consumer and the producer.
• Marketing helps in increasing the living standard of people.
• Marketing helps to increase the nation’s income.
• Marketing process increases employment opportunities.
• Marketing helps to maintain economic stability and rapid development in various countries.

To the Individual Firms:

• Marketing generates revenue to firms.
• Marketing gives information to the top management for taking overall decisions on production.
• Marketing and innovation are the two basic functions of all businesses.

[OR]

(b) State the qualification of directors.
Qualifications of Director: As regards to the qualification of directors, there is no direct provision in the Companies Act, 2013. In general, a director shall possess appropriate skills, experience and knowledge in the fields of finance, law, management, sales, marketing, research and other disciplines related to the business.

The following are the qualifications:

• A director must be a person of sound mind.
• A director must hold share qualification if the article of association provides such.
• A director must be an individual.
• A director should be a solvent person.
• A director should not be convicted by the Court for any offence, etc.

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PART – I
I. Choose The Best Answer

Question 1.
Which of the following is not a scripting language?
(a) JavaScript
(b) PHP
(c) Perl
(d) HTML
Answer:
(d) HTML

Question 2.
Importing C++ program in a Python program is called …………………………
(a) wrapping
(b) Downloading
(c) Interconnecting
(d) Parsing
Answer:
(a) wrapping

Question 3.
The expansion of API is ……………………….
(a) Application Programming Interpreter
(b) Application Programming Interface
(c) Application Performing Interface
(d) Application Programming Interlink
Answer:
(b) Application Programming Interface

Question 4.
A framework for interfacing Python and C++ is ………………………….
(a) Ctypes
(b) SWIG
(c) Cython
(d) Boost
Answer:
(d) Boost

Question 5.
Which of the following is a software design technique to split your code into separate parts?
(a) Object oriented Programming
(b) Modular programming
(c) Low Level Programming
(d) Procedure oriented Programming
Answer:
(b) Modular programming

Question 6.
The module which allows you to interface with the Windows operating system is ………………………….
(a) OS module
(b) sys module
(c) csv module
(d) getopt module
Answer:
(a) OS module

Question 7.
getopt( ) will return an empty array if there is no error in splitting strings to …………………………..
(a) argv variable
(b) opt variable
(c) args variable
(d) ifile variable
Answer:
(c) args variable

Question 8.
Identify the function call statement in the following snippet.
if_name ==’_main_’:
main(sys.argv[l:])
(a) main(sys.argvfl:])
(b) _name_
(c) _main_
(d) argv
Answer:
(b) _name_

Question 9.
Which of the following can be used for processing text, numbers, images, and scientific data?
(a) HTML
(b) C
(c) C++
(d) PYTHON
Answer:
(d) PYTHON

Question 10.
What does name contains?
(a) C++ filename
(b) main( ) name
(c) python filename
(d) os module name
Answer:
(c) python filename

PART – II
II. Answer The Following Questions

Question 1.
What is the theoretical difference between Scripting language and other programming language?
Answer:
The theoretical difference between the two is that scripting languages do not require the 228 compilation step and are rather interpreted. For example, normally, a C++ program needs to be compiled before running whereas, a scripting language like JavaScript or Python need not be compiled. A scripting language requires an interpreter while a programming language requires a compiler.

Question 2.
Differentiate compiler and interpreter?
Answer:
Compiler:

1. It converts the whole program at a time
2. It is faster
3. Error detection is difficult. Eg. C++

Interpreter:

1. line by line execution of the source code.
2. It is slow
3. It is easy Eg. Python

Question 3.
Write the expansion of

1. SWIG
2. MinGW

Answer:
SWIG (Simplified Wrapper Interface Generator. Both C and C++)
MinGW (Minimalist GNU for Windows)

Question 4.
What is the use of modules?
Answer:
We use modules to break down large programs into small manageable and organized files. Furthermore, modules provide reusability of code. We can define our most used functions in a module and import it, instead of copying their definitions into different programs.

Question 5.
What is the use of cd command. Give an example?
Answer:
The syntax to change from c:\> to the folder where Python is located is
cd <absolute path>
where “cd” command refers to change directory and absolute path refers to the complete path where Python is installed.

In this Example to go to the folder where Python is located we should type the following command “cd C:\Program Files\OpenOffiice 4\Program”:

PART – III
III. Answer The Following Questions

Question 1.
Differentiate PYTHON and C++?
Answer:
PYTHON:

1. Python is typically an “interpreted” language
2. Python is a dynamic-typed language
3. Data type is not required while declaring variable
4. It can act both as scripting and general purpose language

C++:

1. C++ is typically a “compiled” language
2. C++ is compiled statically typed language
3. Data type is required while declaring variable
4. It is a general purpose language

Question 2.
What are the applications of scripting language?
Answer:
Applications of Scripting Languages

1. To automate certain tasks in a program
2. Extracting information from a data set
3. Less code intensive as compared to traditional programming language
4. can bring new functions to applications and glue complex systems together

Question 3.
What is MinGW? What is its use?
Answer:
MinGW refers to a set of runtime header files, used in compiling and linking the code of C, C++ and FORTRAN to be run on Windows Operating System.

MinGw-W64 (versionofMinGW) is the best compiler for C++ on Windows. To compile and execute the C++ program, you need ‘g++’ for Windows. MinGW allows to compile and execute C++ program dynamically through Python program using g++.

Python program that contains the C++ coding can be executed only through minGW-w64 project run terminal. The run terminal open the command-line window through which Python program should be executed.

Question 4.
Identify the module, operator, definition name for the following
welcome.display( )
Answer:
welcome – module name
– dot operator
display( ) – function name

Question 5.
What is sys.argv? What does it contain?
Answer:
sys.argv is the list of command-line arguments passed to the Python program, argv contains all the items that come along via the command-line input, it’s basically an array holding the command-line arguments of the program.
To use sys.argv, you will first have to import sys. The first argument, sys.argv[0], is always the name of the program as it was invoked, and sys.argv[l] is the first argument you pass to the program (here it is the C++ file).
For example:
main(sys.args[1]) Accepts the program file (Python program) and the input file (C++ file) as a list(array). argv[0] contains the Python program which is need not to be passed because by default _main_ contains source code reference and argv[l] contains the name of the C++ file which is to be processed.

PART – IV
IV. Answer The Following Questions

Question 1.
Write any 5 features of Python?
Answer:

1. Python uses Automatic Garbage Collection
2. Python is a dynamically typed language.
3. Python runs through an interpreter.
4. Python code tends to be 5 to 10 times shorter than that written in C++.
5. In Python, there is no need to declare types explicitly
6. In Python, a function may accept an argument of any type, and return multiple values without any kind of declaration beforehand.

Question 2.
Explain each word of the following command?
Answer:
The syntax to execute the Python program is
Python <filename.py> -i <C++filename without cpp extension>
Where,

Question 3.
What is the purpose of sys,os,getopt module in Python. Explain?
Answer:
1. Python’s sys module
This module provides access to some variables used by the interpreter and to functions that interact strongly with the interpreter.

sys.argv:
sys.argv is the list of command-line arguments passed to the Python program, argv contains all the items that come along via the command-line input, it’s basically an array holding the command-line arguments of the program.

To use sys.argv, you will first have to import sys. The first argument, sys.argv[0], is always the name of the program as it was invoked, and sys.argv) 1] is the first argument you pass to the program (here it is the C++ file). For example

main(sys.argv[ 1 ]) Accepts the program file (Python program) and the input file (C++ file) as a list(array). argv[0] contains the Python program which is need not to be passed because by default main contains source code reference and argv[l] contains the name of the C++ file which is to be processed.

2. Python’s OS Module
The OS module in Python provides a way of using operating system dependent functionality. The functions that the OS module allows you to interface with the Windows operating system where Python is running on.
os.system( ): Execute the C++ compiling command (a string contains Unix, C command which also supports C++ command) in the shell (Here it is Command Window). For Example to compile C++ program g++ compiler should be invoked. To do so the following command is used.
os.system (g++’ + <varciiable_namel> ‘-<mode>’+ <variable_name2>
where,

os.system(‘g++ ‘+ cpp_file + ‘-o ‘+ exe_file) g++ compiler compiles the file_cpp_file and -o (output) send to exe file
Note
‘+’ in os.system( ) indicates that all strings are concatenated as a single string and send that as a List.

3. Python getopt module
The getopt module of Python helps you to parse (split) command-line options and arguments. This module provides two functions to enable command-line argument parsing, getopt.getopt method
This method parses command-line options and parameter list. Following is the syntax for this method
<opts>,<args>=getopt.getopt(argv, options, {long_options])
Here is the detail of the parameters –
argv – This is the argument list of values to be parsed (splited). In our program the complete command will be passed as a list.
options – This is string of option letters that the Python program recognize as, for input or for output, with options (like ‘i’ or ‘o’) that followed by a colon (:). Here colon is used to denote the mode.
long options – This parameter is passed with a list of strings. Argument of Long options should be followed by an equal sign (‘=’). In our program the C++ fde name will be passed as string and ‘i’ also will be passed along with to indicate it as the input file.
getopt( ) method returns value consisting of two elements. Each of these values are stored separately in two different list (arrays) opts and args. Opts contains list of splitted strings like mode, path and args contains any string if at all not splitted because of wrong path or mode, args will be an empty array if there is no error in splitting strings by getopt( ).
For example The Python code which is going to execute the C++ file p4 in command line will have the getopt( ) method like the following one. opts, args = getopt.getopt (argv, “i:”,[‘ifile=]) where opto contains [(‘-i’, ‘c:\ \pyprg\\p4’)]
-i: – option nothing but mode should be followed by:
‘c: \ \pyprg\\p4’ value nothing but the absolute path of C++ file.
In our examples since the entire command line commands are parsed and no leftover argument, the second argument args will be empty [ ]. If args is displayed using print( ) command it displays the output as [ ].
>>>print(args)
[ ]

Question 4.
Write the syntax for getopt( ) and explain its arguments and return values?
Answer:
Python getopt module
The getopt module of Python helps you to parse (split) command-line options and arguments. This module provides two functions to enable command-line argument parsing, getopt.getopt method
This method parses command-line options and parameter list. Following is the syntax for this method –
<opts>,<args>=getopt.getopt(argx>, options, {.long_options])
Here is the detail of the parameters –

getopt( ) method returns value consisting of two elements. Each of these values are storec separately in two different list (arrays) opts and args .Opts contains list of splitted strings like mode, path and args contains any string if at all not splitted because of wrong path or mode, args will be an empty array if there is no error in splitting strings by getopt( ).
For example The Python code which is going to execute the C++ file p4 in command line will have the getopt( ) method like the following one. opts, args = getopt.getopt (argv, “i:”,[‘ifile=’])

In our examples since the entire command line commands are parsed and no leftover argument, the second argument args will be empty [ ]. If args is displayed using print( ) command it display the output as [ ].
>>>print(args)
[ ]
Some more command for wrapping C++ code
if_name_==’_main_’
main(sys.argv[1:])
_name_(A Special variable) in Python
Since there is no main() function in Python, when the command to run a Python program is given to the interpreter, the code that is at level 0 indentation is to be executed. However, before doing that, interpreter will define a few special variables. _name_ is one such special variable which by default stores the name of the file. If the source file is executed as the main program, the interpreter sets the _name_ variable to have a value as “_main_” name is a built-in variable which evaluates to the name of the current module. Thus it can be used to check whether the current script is being run on its own.
For example consider the following
if_name_ ==’_ main_’
main(sys.argv[1:])
if the command line Python program itself is going to execute first, then _main_ contains the name of that Python program and the Python special variable _name_ also contain the Python program name. If the condition is true it calls the main which is passed with C++ file as argument.

Question 5.
Write a Python program to execute the following C++ coding?
Answer:
#include <iostream>
using namespace std;
int main( )
{ cout«“WELCOME”;
return (0);
}
The above C++ program is saved in a file welcome.cpp
Python program
Type in notepad and save as welcome.cpp
#include<iostream>
using namespace std;
int main( )
{
cout<<“WELCOME”;
retum(0);
}
Open Notepad and type python program and save as welcome.py
import sys,os,getopt
def main(argv):
cppfile =”
exefile = ”
opts, args = getopt.getopt(argv, “i:”, [ifile = ‘])
for o,a in opts:
if o in (“_i”,” ifile “):
cpp_file = a+ ‘.cpp’
exe_file = a+ ‘.exe’
run(cpp_file, exe_file)
def run(cpp_file, exe_file):
print(“compiling” +cpp_file)
os.system(‘g++’ +cpp_file + ‘_o’ + exe_file)
print(“Running” + exe_file)
print(“………………………….”)
print
os.system(exe_file)
print
if — name — == ‘–main –‘;
main(sys.argv[1:])
Output:
——————-
WELCOME
——————-

Practice Programs

Question 1.
Write a C++ program to create a class called Student with the following details?
Answer:
Protected member
Rno integer
Public members
void Readno(int); to accept roll number and assign to Rno
void Writeno( ); To display Rno.
The class Test is derived Publically from the Student class contains the following details
Protected member
Mark1 float
Mark2 float
Public members
void Readmark(float, float); To accept mark1 and mark2
void Writemark( ); To display the marks
Create a class called Sports with the following detail
Protected members
score integer
Public members
void Readscore(int); To accept the score
void Writescore( ); To display the score
The class Result is derived Publically from Test and Sports class contains the following – details
Private member
Total float
Public member
void display( ) assign the sum of mark1, mark2, score in total
invokeWriteno( ), Writemark( ) and Writescore( ). Display the total also.
Save the C++ program in a file called hybrid. Write a python program to execute the hybrid.cpp
Answer:
In Notepad, type the C++ program
#include<iostream>
using namespace std;
class student
{
protected:
int mo;
public:
void readno(int rollno)
{
mo = rollno;
}
void writeno( )
{
cout<< “\n Roll no:” <<rno;
}};
class test: public student
{
protected:
float mark1,mark2;
public:
void readmark(float m1, float m2)
{
mark1 = m1;
mark2 = m2;
}
void writemark( )
{
cout<< “\n mark1 ” << mark1;
cout<< “\n mark2 ” <<mark2;
}};
class sports
{
protected:
int score;
public:
void readscore(int s)
{
score = s;
}
void writescore( )
{
cout<< “SCORE : ” <<score;
}};
class result: public test, public sports
{
private: float total; public:
void display( )
{
total = mark1 + mark2;
cout<< “TOTAL MARKS: ” <<total;
}};
int main( )
{
result r;
r.readno(5);
r.readmark(100,100);
r.readscore(200);
r.writeno( );
r.writemark( );
r.display( );
r.writescore( );
}
save this file as hybrid.cpp
Now type the python program in New Notepad file.
# python hybrid.py -i hybrid.cpp
import sys,os,getopt def main(argv): cpp Jile =” exe_file =”
opts, args = getopt.getopt(argv, “i:”,[‘ifile-])
for o, a in opts:
if o in(“-i”, “–file”):
cpp_file =”a+ ‘.cpp’
exe_file = a+ ‘.exe’
run(cpp_file, exefile)
def run(cpp_file, exe_file):
print(“Compiling” + cpp_file)
os.system(‘g++ ‘+ cpp_file + ‘-o ‘+ exe_file)
print(“Running” + exefile)
print(“———————“)
print
os.system(excfile)
print
if name ==’ main
main(sys.argv[1:])
Output:
Rollno : 5
Mark1 : 100
Mark2 : 100
TOTAL MARKS : 200
SCORE : 200

Question 2.
Write a C++ program to print boundary elements of a matrix and name the file as Border.cpp. Write a python program to execute the Border.cpp?
Answer:
Select File → New in Notepad and type the C++ program.
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
const int MAX =100;
void printBoundary(int a[][max], int m, int n)
{
for (int i=0; i < m; i++)
{
for(int j=0; j < n; j++)
{
if(i= =0 || j= =0 || i= =n-l ||j==n-l)
cout <<a[i][j]
<< ” else
cout <<“”
cout<<”
}
cout <<” \n”;
}}
int main( )
{
int a[4][MAX] = { {1,2,3,4}, {5,6,7,8}, {1,2,3,4}, {5,6,7,8}};
print Boundary(a,4,4);
return 0;
}
save it as Border.cpp
open a New notepad file and type the python program to execute border.cpp
#python border.py -i border.cpp
import sys,os,getopt
def main(argv):
cpp_file =”
exe_file =”
opts, args = getopt.getopt(argv, “i:”,[‘ifile=’])
for o, a in opts;
if o in(“-i”, “-ifile”):
cpp_file =a+ ‘.cpp’
exe_file = a+ ‘.exe’
run(cpp_file, exe_file)
def run(cpp_file, exe_file):
print(“Compiling” + cpp_file)
os.system(‘g++ ‘+ cpp_file + ‘-o ‘+ exe_file)
print(“Running” + exe_file)
print(“———————“)
print
os.system(exe_file)
print
if_name_==’_main_’:
main(sys.argv[1:])
Output:

Samacheer kalvi 12th Computer Science Importing C++ Programs in Python Additional Questions and Answers

I. Choose The Correct Answer

Question 1.
Which one of the following language act as both scripting and general purpose language?
(a) python
(b) c
(c) C++
(d) html
Answer:
(a) python

Question 2.
Find the correct statement.
(a) C++ is a dynamic typed language
(b) python is a dynamic typed language
Answer:
(b) python is a dynamic typed language

Question 3.
Identify the script language from the following.
(a) Html
(b) Java
(c) Ruby
(d) C++
Answer:
(c) Ruby

Question 4.
Pick the odd one out Perl, Ruby, ASP, Tel, Java
Answer:
Java

Question 5.
Read the statement given below and choose the correct option.
(I) All scripting languages are progamming languages.
(II) A scripting language requires an interpreter
(III) Programming language requires a compiler
(a) (I) – False, (II), (III) – True
(b) (I), (II) – False
(c) All are true
(d) All are false
Answer:
(c) All are true

Question 6.
How many values can be returned by a function in python?
(a) 1
(b) 2
(c) 3
(d) many
Answer:
(d) many

Question 7.
The process by which python periodically frees and reclaims block of memory that no longer are in use is called ……………………………..
Answer:
Garbage Collection

Question 8.
………………………. is a python like language for writing c extensions.
Answer:
cython

Question 9.
Which of the following language codes are linked by minGW on windows OS?
(a) C
(b) C++
(c) FORTRAN
(d) all of these
Answer:
(d) all of these

Question 10.
……………………. is the best compiler for C++ on windows.
(a) MinGW-w64
(b) MinGW-w63
(c) MinGW-w62
(d) MinGW-w60
Answer:
(a) MinGW-w64

Question 11.
MinGW allows to compile and execute C++ program dynamically through python program using ……………………….
(a) C++
(b) g++
(c) a++
(d) x++
Answer:
(b) g++

Question 12.
g++ is a program that calls GNU C compiler called …………………………….
(a) GCC
(b) CCC
(c) GGC
(d) GGG
Answer:
(a) GCC

Question 13.
The command to change directory is …………………………….
(a) cc
(b) cd
(c) dc
(d) changed
Answer:
(b) cd

Question 14.
………………………. keyword is used to execute the python program from command line
Answer:
Python

Question 15.
The input mode in python command is given by ………………………….
(a) -i
(b) o
(c) -p
Answer:
(a) -i

Question 16.
Which command is used to clear the screen?
(a) clear
(b) clean
(c) cls
(d) clrscr
Answer:
(c) cls

Question 17.
……………………… refer to a file containing python statements and definitions.
Answer:
Modules

Question 18.
pythons’ …………………………… provides access to same variables and functions that interact with the interpreter.
Answer:
sys module

Question 19.
Which is the list of command-line arguments passed to the python program.
(a) sys.ar
(b) sys.argv
(c) sys.sys
(d) sys.opt
Answer:
(b) sys.argv

Question 20.
Match the following related to OS module.
1. g++ – (i) Name of C++file
2. variable namel – (ii) General compiler
3. mode – (iii) input output mode
4. variable_name2 – (iv) Name of exe file
(a) 1-i, 2-ii, 3-iii, 4-iv
(b) 1-ii, 2-i, 3-iii, 4-iv
(c) 1-iv, 2-iii, 3-ii, 4-i
(d) 1-iv, 2-i, 3-iii, 4-ii
Answer:
(b) 1-ii, 2-i, 3-iii, 4-iv

Question 21.
Which in os.system( ) indicates that all strings are concatenated?
(a) +
(b) –
(c) #
(d) *
Answer:
(a) +

Question 22.
Which module helps to parse(split) command line options and arguments?
(a) getopt
(b) argopt
(c) get
(d) putopt
Answer:
(a) getopt

Question 23.
How many functions are there in getopt module to enable command line argument parsing?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 24.
getopt mode is given by ………………………….
(a) ;
(b) =
(c) #
(d) :
Answer:
(d) :

Question 25.
Argument of long options in the getopt method is followed by ………………………….
(a) ‘
(b) =
(c) #
(d) :
Answer:
(b) =

Question 26.
When the command to run a python program is given to interpreter, code at …………………………. indentation is executed.
(a) level 0
(b) level 1
(c) level 2
(d) level 3
Answer:
(a) level 0

Question 27.
Which one is a special variable which by default stores the name of the file.
(a) main
(b) none
(c) _name_
(d) get
Answer:
(c) _name_

Question 28.
The line number starts from …………………………… script.
Answer:
python

Question 29.
The …………………………… definition invoke the g++ compiler and creates the exe file?
Answer:

Question 30.
Which command of os module executes the exe file?
(a) run
(b) system( )
(c) main
(d) name
Answer:
(a) run

Question 31.
Which opeator is used to access the functions of a imported value?
(a) +
(b) *
(c) .
(d) /
Answer:
(c) .

Question 32.
………………………….. refers to a set of runtime header files used in compiling and linking the code of c, C++, Fortran to run on window os
(a) MaxGW
(b) CountGW
(c) MinGW
(d) AvgGW
Answer:
(c) MinGW

PART – II
II. Answer The Following Questions

Question 1.
Write note on scripting language?
Answer:
A scripting language is a programming language designed for integrating and communicating with other programming languages. Some of the most widely used scripting languages are JavaScript, VBScript, PHP, Perl, Python, Ruby, ASP and Tel.

Question 2.
Write note on Modular programming?
Answer:
Modular programming is a software design technique to split your code into separate parts. These parts are called modules.The focus for this separation should have modules with no or just few dependencies upon other modules.

PART – III
III. Answer The Following Questions

Question 1.
Name some commonly used Interfaces for importing C++ files on python?
Answer:
Importing C++ Files in Pythona:
Importing C++ program in a Python program is called wrapping up of C++ in Python. Wrapping or creating Python interfaces for C++ programs are done in many ways. The commonly used interfaces are

1. Python-C-API (API-Application Programming Interface for interfacing with C programs)
2. Ctypes (for interfacing with c programs)
3. SWIG (Simplified Wrapper Interface Generator.Both C and C++)
4. Cython (Cython is both a Python-like language for writing C-extensions)
5. Boost. Python (a framework for interfacing Python and C++)
6. MinGW (Minimalist GNU for Windows)

Question 2.
How will you import modules in python?
Answer:
We can import the definitions inside a module to another module. We use the import keyword to do this. To import the module factorial we type the following in the Python prompt.
>>> import factorial
Using the module name we can access the functions defined inside the module. The dot(.)
operator is used to access the functions. The syntax for accessing the functions from the module is
<module name>.<function name>
For example:
>>> factorial.fact(5)
120

Question 3.
Give the syntax for getopt module?
Answer:
The syntax for this method
<opts> ,<args>=getopt.getopt(argv, options, [long_options])
Here is the detail of the parameters –

Question 4.
Write the steps for executing the C++ program to check whether a given number is palindrome or not?
Answer:

PART – IV
IV. Answer The Following Questions

Question 1.
Explain Pythons OS Module?
Answer:
Python’s OS Module:
The OS module in Python provides a way of using operating system dependent functionality. The functions that the OS module allows you to interface with the Windows operating system where Python is running on.
os.system( ): Execute the C++ compiling command (a string contains Unix, C command which also supports C++ command) in the shell (Here it is Command Window). For Example to – compile C++ program g++ compiler should be invoked. To do so the following command is used.
os.system (g++’ + <varaiable_namel> ‘-<mode>’+ <variable_name2>
where,

For example the command to compile and execute C++ program is given below
os.system(’g++ ’+ cpp_file + ‘-o ‘+ exe_file) g++ compiler compiles the file cpp_file and -0(output) send to exe_file

Question 2.
Write a C++ program to check whether the given number is palindrome or not. Write a program to execute it?
Answer:
Example:- Write a C++ program to enter any number and check whether the number is palindrome or not using while loop.
/*. To check whether the number is palindrome or not using while loop.*/
//Now select File-> New in Notepad and type the C++ program
#include <iostream>
using namespace std; intmain( )
{
int n, num, digit, rev = 0;
cout << “Enter a positive number:”;
cin>>num;
n = num;
while(num)
{ digit=num% 10;
rev= (rev* 10) +digit;
num = num/10;
cout << “The reverse of the number is:”<<rev <<end1;
if (n ==rev)
cout<< “The number is a palindrome”;
else
cout<< “The number is a palindrome”;
return 0;
}
//save this file as pali_cpp.cpp
#Now select File → New in Notepad and type the Python program
#Save the File as pali.py.Program that complies and executes a .cpp file
import sys, os, getopt
def main(argv);
cpp_file=”
exe_file=”
opts.args = getopt.getopt(argv, “i:”,[‘ifile-])
for o, a in opts:
if o in (“-i”, “—file”):
cpp_file =a+ ‘.cpp’
exe_file =a+ ‘.exe’
run(cpp_file, exe_file)
def run(cpp_file, exe_file):
print(“Compiling” + eppfile)
os.system(‘g++’ + cpp_file +’ -o’ + exe file)
print(“Running” + exe_file)
print(“——————“)
print
os.system(exefile)
print
if_name_==’_main_’: #program starts executing from here
main(sys.argv[1:])
Output of the above program
C:\Program Files\OpenOffice 4\program>Python c:\pyprg\pali.py -i c:\pyprg\pali_cpp
Compiling c:\pyprg\pali_cpp.cpp
Running c:\pyprg\pali_cpp.exe
———————————–
Enter a positive number: 56765
The reverse of the number is: 56765
The number is a palindrome
C:\Program Files\OpenOffice 4\program>Python c:\pyprg\pali.py -i c:\pyprg\pali_cpp Compiling c:\pyprg\pali_cpp.cpp Running c:\pyprg\pali_cpp.exe
Enter a positive number: 56756
The reverse of the number is: 65765
The number is not a palindrome
————————

Question 3.
Write a C++ program to check whether the given number is palindrome or not. Write a program to execute it?
Write a C++ program to enter any number and check whether the number is palindrome or not using while loop?
Answer:
//Now select File-> New in Notepad and type the C++ program
#include <iostream>
using namespace std;
intmain( )
{
int a[3][3], i, j;
for(i=o; j<3;i++)
{
for(j=0;j<3;j++)
{count<<“enter the value for array [”<<i+1<<“]<<”<<“[“<<j+1<<“]”;
cin>>a[i][j];
}
}
system(“cls”); .
cout<<“\n \nOriginal Array\n”;
for(i=0; i<3; i++)
{
for(j=0; j<3; j++)
cout<<a[i] [j]
cout<<end1;
}
cout<<end1;
}
return 0;
}
// Save this file as trans_cpp.cpp
// Now select File → New in Notepad and type the Python program
#Save the File as transpose.py.Program that compiles and executes a .cpp file
#Python tanspose.py -i trans_cpp
import sys, os, getopt
def main(argv):
cpp_file =”
exefile =”
opts, args = getopt.getopt(argv, “i:”,[‘ifile=’])
for o, a in opts:
if o in (“-i”,–ifile”):
cpp_file =a+ ‘.cpp’
exe_file = a+ ‘.exe’
run(cpp_file, exe_file)
def run(cpp_file, exe file):
print(“Compiling” + cppfile)
os.system(‘g++’ + cpp_file + ‘-o ‘+ exe_file)
print(“Running” + exefile)
print(“———————-“)
print
os.system(exe_file)
print
if_name_==’_main_’:
main(sys.argv[1:])
Output of the above program
Original Array
1 2 3
4 5 6
7 8 9
The Transpose of Matrix
1 4 7
2 5 8
3 6 9

Question 4.
Write a C++ program to find cube of a number. Write a python program to execute it?
Answer:
/* Write a C++program using a user definedfunction to function cube of a number. */
//Now select File → New in Notepad and type the C++program
#include <iostream>
using namespace std;
//Function declaration
int cube(int num); int main( )
{
int num; int c;
cout<<“Enter any number: “<<end1;
cin>>num;
c = cube(num);
cout<<“Cube of’ <<num<<” is “<<c;
return 0;
}
//Function to find cube of any number
int cube(int num)
{
return (num * num * num);
}
//Save this file as cube_file.cpp
#Now select File → New in Notepad and type the Python program
#Save the File as fun.py
#Program that compiles and executes a .cpp file
#Python fun.py -i c:\pyprg\cuheJile.cpp
import sys, os, getopt
def main(argv):
cpp_Jile =”
exe_file =”
opts, args = getopt.getopt(argv, “i:”,[‘ifile-])
for o, a in opts:
if o in(“-i”, —ifile”):
cpp_file =a+ ‘.cpp’
exe_file = a+ ‘.exe’
run(cpp_file, exe_file)
def run(cpp_file, exe file):
print(“Compiling” + cppfile)
os.system(‘g++’+ cpp file + ‘-o ‘+ exe file)
print(“Running” + exe_file)
print(“——————–“)
print
os. sy stem(exe_file)
print
if_name_==’_main_’:
main(sys.argv[1:])
Output of the above program
Compiling c:\pyprg\cube_file.cpp
Running c:\pyprg\cube_file.exe
———————-
Enter any number:
5
Cube of 5 is 125

Question 5.
Write a C++ program to implement multilevel inheritance. Write a python program to execute it?
Answer:
// C++program to implement Multilevel Inheritance
//Now select File → New in Notepad and type the C++program
#include <iostream>
using namespace std;
// base class
class Vehicle
{
public:
Vehicle( )
{
cout<< “This is a Vehicle” <<end1;
}
};
class threeWheeler: public Vehicle
{public:
threeWheeler( )
{
cout<<“Objects with 3 wheels are vehicles”<<end1;
}
};
// sub class derivedfrom two base classes
class Auto: public threeWheeler
{
public:
Auto( )
{
cout<<“Auto has 3 Wheels”<<end1;
}
// main function
int main( )
{
//creating object of sub class will invoke the constructor of base classes
Auto obj;
return 0;
}
// Save this file as inheri_cpp.cpp
//Now select File —* New in Notepad and type the Python program
#Save the File as classpy.py
#Python classpy.py -i inheri_cpp command to execute C++ program
import sys, os, getopt
def main (argv):
cpp_file =”
exe file= 11
opts, args = getopt.getopt (argv, 11i: 11 ,[‘ifile=’])
for o, a in opts:
if o in (“-i”, —ifile”):
cpp_file =a+ ‘.cpp’
exe_file = a+ ‘.exe’
run (cpp file, exe_file)
def run(cpp_file, exe_file):
print (”Compiling 11 + cpp file)
os.system (‘g++’ + cpp_file +’ -o ‘+ exe_file)
print (“Running”+ exe file)
print(“———————“)
print
os.system (exefile) print
if_name_==’_main_’:
main (sys.argv[1:])
Output of the above program
Compiling c:\pyprg\class_file.cpp
Running c:\pyprg\class_file.exe
This is a Vehicle
Objects with 3 wheels are vehicles
Auto has 3 Wheels

Students can Download Tamil Nadu 12th Maths Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Maths Model Question Paper 2 English Medium

Instructions:

1.  The question paper comprises of four parts.
2.  You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. questions of Part I, II. III and IV are to be attempted separately
4. Question numbers 1 to 20 in Pan I are objective type questions of one -mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 21 to 30 in Part II are two-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 31 to 40 in Parr III are three-marks questions, These are to be answered in about three to five short sentences.
7. Question numbers 41 to 47 in Part IV are five-marks questions. These are to be answered) in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 90

Part-I

I. Choose the correct answer. Answer all the questions.  [20 × 1 = 20]

Question 1.
If A = \(\left[\begin{array}{ll}
7 & 3 \\
4 & 2
\end{array}\right]\) then 9I₂ – A =
(a) A^-1
(b) \(\frac{A^{-1}}{2}\)
(c) 3A^-1
(d) 2A^-1
Answer:
(d) 2A^-1

Question 2.
If (1+ i) (1+ 2i) (1+ 3i)…(1+ ni) = x + iy, then 2-5T0…(1 + n² ) is ..
(a) 1
(b) i
(c) x² + y²
(d) 1 + n²
Answer:
(c) x² + y²

Question 3.
If p + iq = (2 – 3i) (4 + 2i) then q is
(a) 14
(b) -14
(c) -8
(d) 8
Answer:
(c) -8

Question 4.
A zero of x³ + 64 is
(a) 0
(b) 4
(c) 4i
(d) -4
Answer:
(d) -4

Question 5.
sin^-1(2COS²A – 1) + cos^-1(1 – 2 sin²x) = .

(a) π/2
(b) π /3
(c) π/4
(d) π/6
Answer:
(a) π/2

Question 6.
If cot^-1 x = \(\frac{2 \pi}{5}\) for some x ∈ R, the value of tan^-1 x is
(a) \(-\frac{\pi}{10}\)
(b) \(\frac{\pi}{5}\)
(c) \(\frac{\pi}{10}\)
(d) \(-\frac{\pi}{5}\)
Answer:
(c) \(\frac{\pi}{10}\)

Question 7.
An ellipse has OB as semi minor axes, F and F’ its foci and the angle FBF’ is a right angle. Then the eccentricity of the ellipse is

Answer:
(a) \(\frac{1}{\sqrt{2}}\)

Question 8.
The focus of the parabola x² = 20 y is
(a) (0, 0)
(b) (5, 0)
(c) (0, 5)
(d) (-5, 0)
Answer:
(c) (0, 5)

Question 9.
If the planes \(\vec{r} \cdot(2 \hat{i}-\lambda \hat{j}+\hat{k})=3\) and \((4 \hat{i}+\hat{j}-\mu \hat{k})=5\) are parallel, then the value of λ and μ are

Answer:
(c) \(-\frac{1}{2}\) , -2

Question 10.
The unit normal vectors to the plane 2x – y + 2z = 5 are

Question 11.
The slope of the line normal to the curve f(x) – 2cos 4x at x = \(\frac{\pi}{2}\) is ……………….
(a) -4√3
(b) -4
(c) √3/12
(d) 4√3
Answer:
(c) √3/12

Question 12.
The curve y² = x² ( 1 – x) has
(a) only one loop between x = -1 and x = 0
(b) only one loop between x = 0 and x = 1
(c) two loops between x = -1 and x = 1
(d) no loop
Answer:
(b) only one loop between x = 0 and x = 1

Question 13.
If w (x, y) = x^y, x > 0, then \(\frac{\partial w}{\partial x}\) is equal to
(a) x^y logx
(b) y logx
(c) yx^y-1
(d) xlogy
Answer:
(c) yx^y-1

Question 14.
If (x, y, z) = xy + yz + zx, then f_x – f_z is equal to
(a) z – x
(b) y – z
(c) x – z
(d) y – x
Answer:
(a) z – x

Question 15.
If \(\frac{\Gamma(n+2)}{\Gamma(n)}\) = 90 then n is
(a) 10
(b) 5
(c) 8
(d) 9
Answer:
(d) 9

Question 16.
If n is even then \(\int_{0}^{\pi / 2} \sin ^{n} x d x\) is ………………

Answer:
(b)

Question 17.
The slope at any point of a curve y =/(x) is given by \(\frac{d y}{d x}=3 x^{2}\) and it passes through (-1,1).
Then the equation of the curve is
(a) y = x³ + 2
(b) y = 3x² + 4
(c) y = 3x³ + 4
(d) y = x³ + 5
Answer:
(a) y = x³ + 2

Question 18.
The solution of the differential equation \(\frac{d y}{d x}+\frac{1}{\sqrt{1-x^{2}}}=0\) is
(a) y + sin^-1x = c
(b) x + sin^-1y = 0
(c) y² + 2sin^-1x = c
(d) x² + 2sin^-1y = 0

Question 19.
Suppose that x takes on one of the values 0, 1, and 2. If for some constant k,
P(X = i) = k P(x = i – 1) for i = 1, 2 P(x = 0) = 1/7. Then the value of k is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 20.
Determine the truth value of each of the following statements:
(a) 4 + 2 = 5 and 6 + 3 = 9
(b) 3 + 2 = 5 and 6 + 1 = 7
(c) 4 + 5 = 9 and 1 + 2 = 4
(d) 3 + 2 = 5 and4 + 7 = 11

Answer:
(1) (a) F
(b) T
(c) F
(d) T

Part – II

II. Answer any seven questions. Question No. 30 is compulsory. [7 × 2 = 14]

Question 21.
If A = \(\left[\begin{array}{ll}
4 & 3 \\
2 & 5
\end{array}\right]\) , find x and y such that A² + xA + yI₂ = O₂. Hence , find A^-1
Answer:

So, we get 22 + 4x + y = 0, 31+ 5x + y – 0, 27 + 3x = 0 and 18 + 2x = 0.
Hence x = -9 and y = 14. Then, we get A² – 9A + 14I₂ = O₂
Post-multiplying this equation by A^-1, we get A – 9I + 14A^-1 – I₂ = O₂. Hence, we get

Question 22.
Find the principal argument arg z, when z = \(\frac{-2}{1+i \sqrt{3}}\)
Answer:

This implies that one of the values of arg z is \(\frac{2 \pi}{3}\)
Since \(\frac{2 \pi}{3}\) lies between -π and π , the principal argument arg z is \(\frac{2 \pi}{3}\)

Question 23.
Verify whether the line \(\frac{x-3}{-4}=\frac{y-4}{-7}=\frac{z+3}{12}\) lies in the plane 5x – y + z = 8 .
Answer:
Here, (x₁, y₁, z₁) = (3,4, -3) and direction ratios of the given straight line are (a, b, c) – (-4, -7,12).
Direction ratios of the normal to the given plane are (A, B, C) = (5, -1,1).
We observe that, the given point (x₁, y₁, z₁) = (3,4, -3) satisfies the given plane 5x – y + z = 8
Next, aA+ bB + cC = (-4)(5) + (-7)(-l) + (12)( 1) = -1 ≠ 0. So, the normal to the plane is not perpendicular to the line. Hence, the given line does not lie in the plane.

Question 24.
Evaluate \(\begin{aligned}
&\lim\\
&x \rightarrow 0^{+} \quad x^{2} \log e^{x}
\end{aligned}\)
Answer:

Question 25.
Find a linear approximation for the function at the indicated points.
g(x) = \(\sqrt{x^{2}+9}\), x₀ = -4

The required linear approximation L(x) = g (x₀) + g’ (x₀) (x – x₀)

Question 26.
Evaluate: \(\int_{0}^{2 \pi} x^{2} \sin n x d x\) , Where n is a positive integer.
Answer:
Taking u = x² and v = sin nx, and applying the Bernoulli’s formula, we get

Question 27.
Form the differential equation from the equation y² = 4a (x – a)
Answer:
y²= 4 a(x-a) …(1)
Differentiating, 2yy’ = 4a …(2)
Eliminating a between (1) and (2) we get

(yy ‘)² – 2xyy’ + y² = 0

Question 28.
Compute P(X = k) for the binomial distribution, B(n,p) where n = 6, p = 1/3, k = 3
Answer:
Given n = 6, p = 1/3,k = 3
∴ q = 1 – p = \(1-\frac{1}{3}=\frac{2}{3}\)
P(X = x) = nC_xp^xq^n-x, x = 0, 1,2,…………n
P (X = k) = P (X = 3)

Question 29.
Let A = {a + √5b : a, b∈Z} . Check whether the usual multiplication is a binary operation on A.
Answer:
Let A = a + √5 b and B = c + √5 d, where a, b,c,d∈ R.
Now A * B = (a + √5 b) (c + √5 d)
= ac + √5 ad + √5 bc + √5 b √5 d
= (ac + 5 bd) + 45 (ad + bc) ∈ A .
Where a, b,c,d ∈ Z.
So * is a binary operation.

Question 30.
Find the value of \(\sin ^{-1}\left(\sin \frac{5 \pi}{9} \cos \frac{\pi}{9}+\cos \frac{5 \pi}{9} s \quad \frac{\pi}{9}\right)\)
Answer:

Part – III

III. Answer any seven questions. Question No. 40 is compulsory. [7 x 3 = 21]

Question 31.
Solve the following system of linear equations, using matrix inversion method:
5x + 2y – 3, 3x + 2y – 5.
Answer:
The matrix form of the system is AX = B, where A :\(\left[\begin{array}{ll}
5 & 2 \\
3 & 2
\end{array}\right]\) , X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\), B = \(\left[\begin{array}{l}
3 \\
5
\end{array}\right]\)

We find |A| = \(\left|\begin{array}{ll}
5 & 2 \\
3 & 2
\end{array}\right|\) = 10 – 6 = 4 ≠ 0
So A^-1 exists A^-1 = \(\frac{1}{4}\left[\begin{array}{rr}
2 & -2 \\
-3 & 5
\end{array}\right]\)
Then, applying the formula X = A 1B, we get

So the solution is (x = -1 ,y = 4).

Question 32.
If z₁ = 2 – i and z₂ = -4 + 3i, find the inverse of z₁ z₂ and \(\frac{z_{1}}{z_{2}}\)
Answer:
z₁ = 2 – i, z₂ = -4 + 3i

Question 33.
If k is real, discuss the nature of the roots of the polynomial equation 2x² + kx + k = 0, in terms of k.
Answer:
The given quadratic equation is 2x² + kx + k = 0
a = 2,b = k,c = k
∆ = b² – 4ac = k² – 4(2) k = k² – 8k
(i) If the roots are equal
k² – 8k = 0 ⇒ k(k -8) = 0
k = 0, k = 8
(it) If the roots are real
k² – 8k > 0
k (k – 8) > 0
k ∈ (-∞, 0) ∪ (8, ∞)
(iii) ‘If this roots are imaginary
k² – 8k < 0 ⇒ f ∈ (0, 8)

Question 34.
Find the number of solution of the equation tan^-1 (x – 1) + tan^-1 x + tan^-1 (x + 1) = tan^-1(3x).
Answer:
tan^-1(x – 1) + tan^-1x + tan^-1(x + 1) = tan^-1(3x)
tan^-1(x – 1) + tan^-1(x + 1) = tan^-1 3x – tan^-1 x

So, the equation has 2 solutions.

Question 35.
Find the non-parametric form of vector equation, and Cartesian equations of the plane \(\vec{r}=(6 \hat{i}-\hat{j}+\hat{k})+s(-\hat{i}+2 \hat{j}+\hat{k})+t(-5 \hat{i}-4 \hat{j}-5 \hat{k})\)
Answer:

Non-parametric form of vector equation

Question 36.
Find two positive numbers whose sum is 12 and their product is maximum.
Answer:
Let the two numbers be x, 12 – x.
Their product p = x (12 – x) – 12x – x²
To find the maximum product.
p’ (x) = 12 – 2x
p”(x) = -2
p’ (x) = 0 ⇒ 12 – 2x = 0 2x = 12 => x = 6
at x = 6, p”(x) = -2 = -ve
⇒ p is maximum at x = 6
when x = 6, 12-x = 12 – 6 = 6
So the two numbers are 6, 6

Question 37.
Find the differential equation that will represent family of all circles having centres on the x-axis and the radius is unity.
Answer:
Equation of a circle with centre on x-axis and radius 1 unit is
(x-a)² + y² = 1 …(1)
Differentiating with respect to x,
2 (x-a) + 2yy’ = 0
⇒ 2 (x – a) = – 2yy’
(or) x-a = -yy’ …(2)
Substituting (2) in (1), we get,
(-yy’)² + y² = 1
(i.e. y² (y’)² + y² = 1 ⇒ y² [1 + (y’)²] = 1

Question 38.
The probability that a certain kind of component will survive a electrical test is \(\frac{3}{4}\). Find
the probability that exactly 3 of the 5 components tested survive.
Answer:
Given n = 5
Probability that a component survive in a test =p = \(\frac{3}{4}\)
∴ q = 1 – p = \(1-\frac{3}{4}=\frac{1}{4}\)
Let ‘X’ be the random variable denotes the number of components survived in a test.
Probability of ‘x’ successes in V trials is
P (X = x) = nC_xp^xq^n-x, x = 0, 1,2,………n
Probability that exactly 3 components survive

Question 39.
Fill in the following table so that the binary operation * on A = {a,b,c} is commutative.

Answer:
Given that the binary operation * is Commutative.
To find a * b :
a * b = b * a (∵ * is a Commutative)
Here b * a = c. So a * b = c
To find a * c :
a * c = c * a (∵ * is a Commutative)
c * a = a. (Given)
So a * c = a
To find c * b :
c * b = b * c
Here b * c = a. So c * b = a

Question 40.
Find the area of the region enclosed by y² = x and y = x – 2.
Answer:
The points of intersection of the parabola y² = x and the line y = x – 2 are (1, – 1) and (4, 2)
To compute the region [shown in the figure] by integrating with respect to x, we would have to split the region into two parts, because the equation of the lower boundary changes at x = 1. However if we integrate with respect to y no splitting is necessary.

Part – IV

IV. Answer all the questions. [ 7 x 5 = 35]

Question 41.
(a) Use product \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right]\left[\begin{array}{rrr}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right]\)
to solve the system of equations x – y + 2z = 1, 2y – 3z = 1, 3x – 2y + 4z = 2.
Answer:

The system of equations can be written in the form AX = C, where

Thus x = 0, y = 5 and z = 3

(b) Find, by integration, the volume of the container which is in the shape of a right circular conical frustum as shown in the Figure.
Answer:

Question 42.
(a) Simplify (-√3 + 3i)³¹
Answer:
Let – √3 + 3i = r (cos θ + z sin θ). Then, we get

(b) Solve the following equation x⁴ – 10x³ + 26x² – 10x + 1 = 0.
Answer:
This equation is Type I even degree reciprocal equation. Hence it can be rewritten as


[(y² – 2) – 10y + 26] = 0 ⇒ (y² – 10y + 24) = 0 ⇒ (y – 6) {y – 4) = 0 ⇒ y = 6 (or) 4

Hence the roots are 3 ± 2 √2, 2 ± √3

Question 43.
(a) Solve

Answer:

(b) Suppose a person deposits 10,000 Indian rupees in a bank account at the rate of 5% per annum compounded continuously. How much money will be in his bank account 18 months later?
Answer:
Let P be the principal
Rate of interest 5 %

Given, when t = 0, P = 10000
⇒ c = 10000

Question 44.
(a) Find the equation of the circle through the points (1, 0), (-1, 0), and (0, 1).
Answer:
Let the required circle be
x² + y² + 2gx + 2fy + c = 0 ……(A)
The circle passes through (1, 0), (-1, 0) and (0, 1)
(1,0) ⇒ 1 +0 + 2g(1) + 2f(0) + c = 0 -(A)
2g + c = -1
(-1, 0) ⇒ 1 + 0 + 2g (-1) + 2/(0) + c = 0 …(1)
-2g + c = -1
(0,1)⇒ 0+ 1 +2g(0) + 2f(l) + c = 0 …(2)
2 + c = -1
Now solving (1), (2) and (3) …(3)
2g + c = -1 …(1)
-2g + c = -1 (2)
(1) + (2) ⇒ 2c = -2 ⇒ c = -1
Substituting c = -1 in (1) we get
2g – 1 = -1 2g = -1 + 1= 0 ⇒ g = 0
Substituting c = -1 in (3) we get
2f – 1 = -1 ⇒ 2f = -1 + 1 = 0 ⇒ f = 0
So we get g = 0,f= 0 and c = -1
So the required circle will be
x² + y² + 2(0)x + 2(0)y – 1 = 0
(i.e) x² + y² – 1 = 0 ⇒ x² + y² = 1

[OR]

(b) Find the area of the loop of the curve 3ay² = x(x – a)².
Answer:
Put y = 0; we get x = 0, a
It meets the x – axis at x = 0 and x = a
∴ Here a loop is formed between the points (0, 0) and (a, 0) about x-axis. Since the curve is symmetrical about x-axis, the area of the loop is twice the area of the portion above the x – axis.

Question 45.
Show that the lines \(\frac{x-3}{3}=\frac{y-3}{-1}, z-1=0\) and \(\frac{x-6}{2}=\frac{z-1}{3}, y-2=0\) interest. Also find the point of intersection.
Answer:

(x₁, y₁, z₁) = (3, 3,1) and (x₂, y₂, z₂) = (6,2,1)
(b₁ b₂, b₃) = (3, -1, 0) and (d₁ d₂, d₃) = (2, 0, 3)
Condition for intersection of two lines

Given two lines are intersecting lines.
Any point on the first line

(3λ + 3, -λ + 3,1)
Any point on the Second line

(2µ + 6, 2, 3µ + 1)
∴ 3µ + 1 = 1
3µ = 0
µ = 0

-λ + 3 = 2
-λ = -1
λ = 1
∴ The required point of intersection is (6, 2, 1)

(b) Prove that \(p \rightarrow(\neg q \vee r) \equiv \neg p \vee(\neg q \vee r)\) using truth table.
Answer:

The entries in the column corresponding to \(p \rightarrow(\neg q \vee r) \text { and } \neg p \vee(\neg q \vee r)\) are identical.
Hence they are equivalent.

Question 46.
(a) A water tank has the shape of an inverted circular cone with base radius 2 metres and height 4 metres. If water is being pumped into the tank at a rate of 2m3/min, find the rate at which the water level is rising when the water is 3m deep.
Answer:
We first sketch the cone and label it as in diagram. Let V, r and h be respectively the volume of the water, the radius of the cone and the height at time t, where t is measured in minutes.
We are given that \(\frac{d \mathrm{V}}{d t}\) = 2m³/min and we are asked to find
\(\frac{d h}{d t}\) where h is 3m
The quantities V and h are related by the equation \(\mathrm{V}=\frac{1}{3} \pi r^{2} h\). But it is very useful to express V as function of h alone.
In order to eliminate r we use similar triangles in diagram to write \(\frac{r}{h}=\frac{2}{4} \Rightarrow r=\frac{h}{2}\) and the expression for V becomes \(\mathrm{V}=\frac{1}{3} \pi\left(\frac{h}{2}\right)^{2} h=\frac{\pi}{12} h^{3}\)
Now we can differentiate each side with respect to t and we have

(b) Let U (x, y) = e^x sin y, where x = st², y = s²t s, t e R. Find \(\frac{\partial \boldsymbol{U}}{\partial \boldsymbol{s}}, \frac{\partial \boldsymbol{U}}{\partial t}\) and evaluate them at s = t = 1.
Answer:

Question 47.
(a) The probability density function of X is given by f(x) = \(\begin{array}{c}
f(x)=\left\{\begin{array}{cc}
k e^{-\frac{x}{3}} & \text { for } x>0 \\
0 & \text { for } x \leq 0
\end{array}\right.
\end{array}\)
Find (i) the value of k
(ii) the distribution function
(iii) P(X < 3)
(iv) P(5 ≤ X) (v) P(X ≤ 4) .
Answer:

[OR]

(b) Solve [y(1 – xtanx) + x²cosx]dx – xdy = 0.

Students can Download Tamil Nadu 12th Biology Model Question Paper 1 English Medium Pdf, Tamil Nadu 12th Biology Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Biology Model Question Paper 1 English Medium

General Instructions:

1. The question paper comprises of four parts. Questions for Botany and Zoology are asked separately.
2. You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
3. All questions of Part I, II, III and IV are to be attempted separately.
4. Question numbers 1 to 8 in Part I are Multiple Choice Questions of one mark each. These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer.
5. Question numbers 9 to 14 in Part II are two-marks questions. These are to be answered in about one or two sentences.
6. Question numbers 15 to 19 in Part III are three-marks questions. These are to be answered in about three to five short sentences.
7. Question numbers 20 and 21 in Part IV are five-marks questions. These are to be answered in detail. Draw diagrams wherever necessary.

Time: 2.30 Hours
Maximum Marks: 70

Bio-Botany [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Identify plant species which is popularly called as “Terror of Bengal”.
(a) Eichornia Crassipes
(b) Vallisneria spiralis
(c) Pistia stratiotes
(d) Zostera marina
Answer:
(a) Eichornia Crassipes

Question 2.
‘Gametes are never hybrid’ is concluded by __________.
(a) Law of dominance
(b) Law of segregation
(c) Law of lethality
(d) Law of independent assortment
Answer:
(b) Law of segregation

Question 3.
Match list I with list II

(a) A-i, B-iii, C-ii, D-iv
(b) A-ii, B-iii, C-iv, D-i
(c) A-ii, B-iii, C-i, D-iv
(d) A-iii, B-ii, C-i, D-iv
Answer:
(c) A-ii, B-iii, C-i, D-iv

Question 4.
EcoRI cleaves the DNA at _______.
(a) AGGGTT
(b) GTATATC
(c) GAATTC
(d) TATAGC
Answer:
(c) GAATTC

Question 5.
________ is the climax community of hydrosere.
(a) Reed Swamp stage
(b) Marsh meadow stage
(c) Shrub stage
(d) Forest stage
Answer:
(d) Forest stage

Question 6.
People’s movement for the protection of environment in Sirsi of Karnataka is ________.
(a) Chipko movement
(b) Appiko movement
(c) Amirtha Devi Bishwas movement
(d) None of the above
Answer:
(b) Appiko movement

Question 7.
Which of the following is incorrectly paired?
(a) Wheat – Himgiri
(b) Rice – Ratna
(c) Milch breed – sahiwal
(d) Pusa komal – Brassica
Answer:
(d) Pusa komal – Brassica

Question 8.
Tectona grandis is coming under the family _______.
(a) Lamiaceae
(b) Fabaceae
(c) Dipterocaipaceae
(d) Ebenaceae
Answer:
(a) Lamiaceae

Part – II

Answer any four of the following questions. [4 x 2 = 8]

Question 9.
Draw and label the structure of a mature embryo sac of angiosperm.
Answer:

Question 10.
What are plasmogenes?
Answer:
Plasmogenes are independent, self-replicating, extra-chromosomal units located in cytoplasmic organelles, chloroplast and mitochondrion.

Question 11.
Define the terms
(a) Bioventing
(b) Bioaugmentation
Answer:
(a) Bioventing is the process that increases the oxygen or air flow to accelerate the degradation of environmental pollutants.
(b) Bioaugmentation is the addition of selected microbes to speed up degradation process.

Question 12.
Differentiate between Eurythermal animals and Stenothermal animals.
Answer:
Eurythermal: Organisms which can tolerate a wide range of temperature fluctuations. Example: Zostera.
Stenothermal: Organisms which can tolerate only small range of temperature variations. Example: Mango.

Question 13.
What are Blue carbon ecosystems?
Answer:
Sea grasses and mangroves of Estuarine and coastal ecosystems are the most efficient in carbon sequestration. Hence, these ecosystems are called as “ Blue carbon ecosystems”.

Question 14.
What is silvopasture system? How it helps economy?
Answer:
The production of woody plants combined with pasture is referred to silvopasture system. The trees and shrubs may be used primarily to produce fodder for livestock or they may be grown for timber, fuel wood and fruit or to improve the soil.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Given an account an king of spices and its uses.
Answer:
King of Spices:
Pepper is one of the most important Indian spices referred to as the “King of Spices” and also termed as “Black Gold of India”. Kerala, Karnataka and Tamil Nadu are the top producers in India. The characteristic pungency of the pepper is due to the presence of alkaloid Pipeline. There are two types of pepper available in the market namely black and white pepper.

Uses:
It is used for flavouring in the preparation of sauces, soups, curry powder and pickles. It is used in medicine as an aromatic stimulant for enhancing salivary and gastric secretions and also as a stomachic. Pepper also enhances the bio-absorption of medicines.

Question 16.
Distinguish between primary production and secondary production.
Answer:

Question 17.
What is co-evolution? Explain with example.
Answer:
The interaction between organisms, when continues for generations, involves reciprocal changes in genetic and morphological characters of both organisms. This type of evolution is called Co-evolution. It is a kind of co-adaptation and mutual change among interactive species.

Examples:

• Corolla length and proboscis length of butterflies and moths (Habenaria and Moth).
• Bird’s beak shape and flower shape and size.

Question 18.
What do you mean by Germplasm conservation? Describe it.
Answer:
Germplasm conservation refers to the conservation of living genetic resources like pollen, seeds or tissue of plant material maintained for the purpose of selective plant breeding, preservation in live condition and used for many research works.

Germplasm conservation resources is a part of collection of seeds and pollen that are stored in seed or pollen banks, so as to maintain their viability and fertility for any later use such as hybridization and crop improvement. Germplasm conservation may also involve a gene bank, DNA bank of elite breeding lines of plant resources for the maintenance of biological diversity and also for food security.

Question 19.
Point out the reasons for Mendels’ success in his breeding experiment.
Answer:

1. He applied mathematics and statistical methods to biology and laws of probability to his breeding experiments.
2. He followed scientific methods and kept accurate and detailed records that include quantitative data of the outcome of his crosses.
3. His experiments were carefully planned and he used large samples.
4. The pairs of contrasting characters which were controlled by factor (genes) were present on separate chromosomes.
5. The parents selected by Mendel were pure breed lines and the purity was tested by self crossing the progeny for many generations.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Explain how Nicotiana exhibit self-compatibility in detail.
Answer:

Self-sterility means that the pollen from a plant is unable to germinate on its own stigma and will not be able to bring about fertilization in the ovules of the same plant. East (1925) observed multiple alleles in Nicotiana which are responsible for self-incompatibility or self-sterility. The gene for self-incompatibility can be designated as S, which has allelic series S₁, S₂, S₃, S₄ and S₅

The cross-fertilizing tobacco plants were not always homozygous as S₁S₁ or S₂S₂, but all plants were heterozygous as S₁S₂, S₃S₄ and S₅S₆. When crosses were made between different S₁S₂ plants, the pollen tube did not develop normally. But effective pollen tube development was observed when crossing was made with other than S₁S₂ for example S₃S₄.

When crosses were made between seed parents with S₁S₂ and pollen parents with S₂S₃, two kinds of pollen tubes were distinguished. Pollen grains carrying S₃ were not effective, but the pollen grains carrying S₃ were capable of fertilization. Thus, from the cross S₁, S₂ X S₃, S₄, all the pollens were effective and four kinds of progency resulted: S₁S₃, S₁S₄, S₂S₃ and S₂S₄.

[OR]

(b) Point out the applications of plant tissue culture.
Answer:
Plant tissue culture techniques have several applications such as:

• Improved hybrids production through somatic hybridization.
• Somatic embryoids can be encapsulated into synthetic seeds (synseeds). These encapsulated seeds or synthetic seeds help in conservation of plant biodiversity.
• Production of disease resistant plants through meristem and shoot tip culture.
• Production of stress resistant plants like herbicide tolerant, heat tolerant plants.
• Micro-propagation technique to obtain large numbers of plantlets of both crop and tree species useful in forestry within a short span of time and all through the year.
• Production of secondary metabolites from cell culture utilized in pharmaceutical, cosmetic and food industries.

Question 21.
(a) Describe the various stages of decomposition process.
Answer:
(1) Fragmentation: The breaking down of detritus into smaller particles by detritivores like bacteria, fungi and earth worm is known as fragmentation. These detritivores secrete certain substances to enhance the fragmentation process and increase the surface area of detritus particles.

(2) Catabolism: The decomposers produce some extracellular enzymes in their surroundings to break down complex organic and inorganic compounds in to simpler ones. This is called catabolism.

(3) Leaching or Eluviation: The movement of decomposed, water soluble organic and inorganic compounds from the surface to the lower layer of soil or the carrying away of the same by water is called leaching or eluviation.

(4) Humification: It is a process by which simplified detritus is changed into dark coloured amorphous substance called humus. It is highly resistant to microbial action, therefore decomposition is very slow. It is the reservoir of nutrients.

(5) Mineralisation: Some microbes are involved in the release of inorganic nutrients from the humus of the soil, such process is called mineralisation.

[OR]

(b) Explain the steps involved in hybridization.
Answer:
Steps involved in hybridization are as follows:

• Selection of Parents: Male and female plants of the desired characters are selected. It should be tested for their homozygosity.
• Emasculation: It is a process of removal of anthers to prevent self pollination before anthesis (period of opening of a flower).
• Bagging: The stigma of the flower is protected against any undesirable pollen grains, by covering it with a bag.
• Crossing: Transfer of pollen grains from selected male flower to the stigma of the female emasculated flower.
• Harvesting seeds and raising plants: The pollination leads to fertilization and finally seed formation takes place. The seeds are grown into new generation which are called hybrid.

Bio-Zoology [Maximum Marks: 35]

Part – I

Choose the correct answer. [8 × 1 = 8]

Question 1.
Assertion (A): In bee society, all the members are diploid, except drones.
Reason (R): Drones are produced by parthenogenesis.
(a) A and R are true, R is the correct explanation for A
(b) A and R are true, R is not the correct explanation for A
(c) A is true, R is false
(d) Both A and R are false
Answer:
(a) A and R are true, R is the correct explanation for A

Question 2.
Colotrum is rich in ______.
(a) IgE
(b) IgA
(c) IgD
(d) IgM
Answer:
(b) IgA

Question 3.
Match column I with column II

(a) A-iv, B-ii, C-i, D-iii
(b) A-iv, B-i, C-ii, D-iii
(c) A-iv, B-i, C-ii, D-iii
(d) A-i, B-iv, C-iii, D-ii
Answer:
(b) A-iv, B-i, C-ii, D-iii

Question 4.
Which of the following is the correct sequence of event with reference to central dogma?
(a) Transcription, Translation, Replication
(b) Transcription, Replication, Translation
(c) Duplication, Transcription, Translation
(d) Replication, Transcription, Translation
Answer:
(d) Replication, Transcription, Translation

Question 5.
Spread of cancerous cells to distant sites is termed as ______.
(a) Metastasis
(b) Oncogenes
(c) Proto-oncogenes
(d) Malignant neoplasm
Answer:
(a) Metastasis

Question 6.
How many amino acids are arranged in the two chains of Insulin?
(a) Chain A has 12 and Chain B has 13 amino acids
(b) Chain A has 21 and Chain B has 30 amino acids
(c) Chain A has 20 and chain B has 30 amino acids
(d) Chain A has 12 and chain B has 20 amino acids
Answer:
(b) Chain A has 21 and Chain B has 30 amino acids

Question 7.
Who introduced the term biodiversity?
(a) Edward Wilson
(b) Walter Rosen
(c) Norman Myers
(a) Alice Norman
Answer:
(b) Walter Rosen

Question 8.
What is the name of the action plan for sustainable development framed in Rio conference in 1992?
(a) Action 21
(b) Agenda 21
(c) Declaration 21
(d) Protocol 21
Answer:
(b) Agenda 21

Part – II

Answer any four of the following questions. [4 × 2 = 8]

Question 9.
How is polyspermy avoided in humans?
Answer:
Once fertilization is accomplished, cortical granules from the cytoplasm of the ovum form a barrier called the fertilization membrane around the ovum preventing further penetration of other sperms. Thus polyspermy is prevented.

Question 10.
What are holandric genes?
Answer:
The genes present in the differential region of Y chromosome are called Y- linked or holandric genes. The Y linked genes have no corresponding allele in X chromosome.

Question 11.
What are connecting links? Give example.
Answer:
The organisms which possess the characters of two different groups (transitional stage) are called connecting links. Example Peripatus (connecting link between Annelida and Arthropoda) Archaeopteryx (connecting link between Reptiles and Aves).

Question 12.
Saccharomyces cerevisiae is called as brewer’s yeast. Justify.
Answer:
Saccharomyces cerevisiae commonly called brewer’s yeast is used for fermenting malted cereals and fruit juices to produce various alcoholic beverages. Wine and beer are produced without distillation, whereas whisky, brandy and rum are obtained by fermentation and distillation.

Question 13.
Give the diagnostic characters of a Biome.
Answer:

• Location, Geographical position (Latitude and Longitude)
• Climate and physiochemical environment
• Predominant plant and animal life
• Boundaries between biomes are not always sharply defined. Transition or transient zones are seen.

Question 14.
What would Earth be like without the greenhouse effect?
Answer:
Greenhouse effect is vital for the sustenance of life. Greenhouse gases like CO₂, water vapour etc absorb some of the reflected sun’s radiation and radiate back it to the Earth surface, thus maintaining the Earth’s warm condition. Without this effect, life on Earth would be difficult or rather impossible for existence or become hostile to most living organisms.

Part – III

Answer any three questions in which question number 19 is compulsory. [3 × 3 = 9]

Question 15.
Write a short note on phases of life cycle.
Answer:

• Juvenile phase – Period of growth between birth of an individual and reproductive maturity.
• Reproductive phase – Period of growth when an organism attain reproductive maturity and produces new offsprings.
• Senescent plane – Period of growth when the structure and functioning of body starts degenerating.

Question 16.
What is MTP? Add a note on it.
Answer:
Medical Termination of Pregnancy (MTP): Medical method of abortion is a voluntary or intentional termination of pregnancy in a non-surgical or non-invasive way. Early medical termination is extremely safe upto 12 weeks (the first trimester) of pregnancy and generally has no impact on a women’s fertility. Abortion during the second trimester is more risky as the foetus becomes intimately associated with the maternal tissue.

Question 17.
Genetic code is ‘universal’. Give reason.
Answer:
The genetic code is universal. It means that all known living systems use nucleic acids and the same three base codons (triplet codon) direct the synthesis of protein from amino acids. For example, the mRNA (UUU) codon codes for phenylalanine in all cells of all organisms. Some exceptions are reported in prokaryotic, mitochondrial and chloroplast genomes. However similarities are more common than differences.

Question 18.
Autoimmunity is a misdirected immune response. Justify.
Answer:
Autoimmune diseases : Autoimmunity is due to an abnormal immune response in which the immune system fails to properly distinguish between self and non-self and attacks its own body. Our body produces antibodies (auto antibodies) and cytotoxic T cells that destroy our own tissues. If a disease-state results, it is referred to as auto-immune disease. Thus, autoimmunity is a misdirected immune response.

Question 19.
PCR is a useful tool for early diagnosis of an Infectious disease. Elaborate.
Answer:
The specificity and sensitivity of PCR is useful for the diagnosis of inherited disorders (genetic diseases), viral diseases, bacterial diseases, etc., The diagnosis and treatment of a particular disease often requires identifying a particular pathogen. Traditional methods of identification involve culturing these organisms from clinical specimens and performing metabolic and other tests to identify them.

The concept behind PCR based diagnosis of infectious diseases is simple – if the pathogen is present in a clinical specimen its DNA will be present. Its DNA has unique sequences that can be detected by PCR, often using the clinical specimen (for example, blood, stool, spinal fluid, or sputum) in the PCR mixture.

Part – IV

Answer all the questions. [2 × 5 = 10]

Question 20.
(a) Give an detailed account on various natural methods of contraception.
Answer:
Natural method is used to prevent meeting of sperm with ovum, i.e., Rhythm method (safe period), coitus interruptus, continuous abstinence and lactational amenorrhoea.

1. Periodic abstinence/rhythm method: Ovulation occurs at about the 14th day of the menstrual cycle. Ovum survives for about two days and sperm remains alive for about 72 hours in the female reproductive tract. Coitus is to be avoided during this time.

2. Continuous abstinence is the simplest and most reliable way to avoid pregnancy is not to have coitus for a defined period that facilitates conception.

3. Coitus interruptus is the oldest family planning method. The male partner withdraws his penis before ejaculation, thereby preventing deposition of semen into the vagina.

4. Lactational amenorrhoea : Menstrual cycles resume as early as 6 to 8 weeks, from parturition. However, the reappearance of normal ovarian cycles may be delayed for six months during breast-feeding. This delay in ovarian cycles is called lactational amenorrhoea. It serves as a natural, but an unreliable form of birth control. Suckling by the baby during breast-feeding stimulates the pituitary to secrete increased prolactin hormone in order to increase milk production.

This high prolactin concentration in the mother’s blood may prevent menstrual cycle by suppressing the release of GnRH (Gonadotropin Releasing Hormone) from hypothalamus and gonadotropin secretion from the pituitary.

[OR]

(b) Explain the three major categories in which fossilization occur.
Answer:
(1) Actual remains is the most common method of fossilization. When marine animals die, their hard parts such as bones and shells, etc. are covered with sediments and are protected from further deterioration. They get preserved as such as they are preserved in vast ocean; the salinity in them prevents decay.

The sediments become hardened to form definite layers or strata. For example, Woolly Mammoth that lived 22 thousand years ago were preserved in the frozen coast of Siberia as such. Several human beings and animals living in the ancient city of Pompeii were preserved intact by volcanic ash which gushed out from Mount Vesuvius.

(2) Petrifaction – When animals die the original portion of their body may be replaced . molecule for molecule by minerals and the original substance being lost through disintegration. This method of fossilization is called petrifaction. The principle minerals involved in this type fossilization are iron pyrites, silica, calcium carbonate and bicarbonates of calcium and magnesium.

(3) Natural moulds and casts – Even after disintegration, the body of an animal might leave indelible impression on the soft mud which later becomes hardened into stones. Such impressions are called moulds. The cavities of the moulds may get filled up by hard minerals and get fossilized, which are called casts. Hardened faecal matter termed as coprolites occur as tiny pellets. Analysis of the coprolites enables us to understand the nature of diet, the prehistoric animals thrived.

Question 21.
(a) Explain in detail about stem cell therapy.
Answer:
Stem cells are undifferentiated cells found in most of the multi cellular animals. These cells maintain their undifferentiated state even after undergoing numerous mitotic divisions.

Stem cell research has the potential to revolutionize the future of medicine with the ability to regenerate damaged and diseased organs. Stem cells are capable of self renewal and exhibit ‘cellular potency’. Stem cells can differentiate into all types of cells that are derived from any of the three germ layers ectoderm, endoderm and mesoderm.

In mammals there are two main types of stem cells – embryonic stem cells (ES cells) and adult stem cells. ES cells are pluripotent and can produce the three primary germ layers ectoderm, mesoderm and endoderm. Embryonic stem cells are multipotent stem cells that can differentiate into a number of types of cells. ES cells are isolated from the epiblast tissue of the inner cell mass of a blastocyst. When stimulated ES can develop into more than 200 cells types of the adult body. ES cells are immortal i.e. they can proliferate in a sterile culture medium and maintain their undifferentiated state.

Adult stem cells are found in various tissues of children as well as adults. An adult stem cell or somatic stem cell can divide and create another cell similar to it. Most of the adult stem cells are multipotent and can act as a repair system of the body, replenishing adult tissues.The red bone marrow is a rich source of adult stem cells.

The most important and potential application of human stem cells is the generation of cells and tissues that could be used for cell based therapies. Human stem cells could be used to test new drugs.

[OR]

(b) Explain in detail about various types of extinctions.
Answer:
There are three types of Extinctions
(1) Natural extinction: It is a slow process of replacement of existing species with better adapted species due to changes in environmental conditions, evolutionary changes, predators and diseases. A small population can get extinct sooner than the large population due to inbreeding depression (less adaptivity and variation)

(2) Mass extinction: The Earth has experienced quite a few mass extinctions due to environmental catastrophes. A mass extinction occurred about 225 million years ago during the Permian, where 90% of shallow water marine invertebrates disappeared.

(3) Anthropogenic extinctions: These are abetted by human activities like hunting, habitat destruction, over exploitation, urbanization and industrialization. Some examples of extinctions are Dodo of Mauritius and Steller’s sea cow of Russia. Amphibians seem to be at higher risk of extinction because of habitat destruction. The most serious aspect of the loss of biodiversity is the extinction of species. The unique information contained in its genetic material (DNA) and the niche it possesses are lost forever.

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 5 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Chemistry Model Question Paper 5 English Medium

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
3. All questions of Part I, II, III and IV are to be attempted separately
4. Question numbers 1 to 15 in Part I are Multiple choice Questions of one mark each.  These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Maximum Marks: 70

Part – I

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
Which one of the following ore is best concentrated by froath – floatation method?
(a) Magnetite
(b) Haematite
(c) Galena
(d) Cassiterite
Answer:
(c) Galena

Question 2.
Which compound is used as flux in metallurgy?
(a) Boric acid
(b) Borax
(c) Diborane
(d) BF₃
Answer:
(b) Borax

Question 3.
The shape of XeOF₄ is
(a) T Shaped
(b) Pyramidal
(c) Square planar
(d) Square pyramidal
Answer:
(d) Square pyramidal

Question 4.
How many moles of acidified KMnO₄ required to oxidise one mole of oxalic acid?
(a) 5
(b) 0.6
(c) 1.5
(s) 0.4
Answer:
(b) 0.6

Question 5.
The type of isomerism exhibited by [Pt(NH₃)₂ Cl₂] ?
(a) coordination isomerism
(b) linkage isomerism
(c) optical isomerism
(d) geometrical isomerism
Answer:
(d) geometrical isomerism

Question 6.
The fraction of the total volume occupied by the atoms in a fcc is

Answer:
(a) \(\frac{\pi \sqrt{2}}{6}\)

Question 7.
The half life period of a radioactive element is 140 days. After 280 days 1g of element will be
reduced to which amount of the following?
(a) 1/4
(b) 1/16
(c) 1/8
(d) 1/2
Answer:
(a) 1/4
Solution:

Question 8.
Which is not a Lewis base?
(a) BF₃
(b) PF₃
(c) CO
(d) F
Answer:
(a) BF₃

Question 9.
During electrolysis of molten copper chloride, the time required to produce 0.2 mole of chlorine gas using a current of 2A is ………..
(a) 32.66 min
(b) 321.66 min
(c) 378 min
(d) 260 min
Solution:
m = ZIt (mass of 1 mole of Cl2 gas = 71) m
t = \(\frac{\mathrm{m}}{\mathrm{Zl}}\) (mass of 0.2 mole of Cl2 gas = 14.2 g)

Question 10.
Smoke is a colloidal solution of …………
(a) Solid in gas
(b) Gas in gas
(c) Liquid in gas
(d) Gas in liquid
Answer:
(c) Liquid in gas

Question 11.
Iso propyl benzene on oxidation in presence of air and dilute acid gives
(a) C₆H₅COOH
(b) C₆H₅COCH₃
(c) C₆H₅COC₆H₅
(d) C₆H₅OH
Answer:
(d) C₆H₅OH

Question 12.
But-2 ene on ozonolyis followed by subsequent cleavage with Zn and water gives …………….
(a) ethanal
(b) Propanal
(c) Propanone
(d) Methanal
Answer:
(a) ethanal

Question 13.
3(i)
This reaction is known as
(a) Friedal- craft’s reaction
(b) HVZ reaction
(c) Schotten – Baumann reaction
(d) Cannizaro reaction
Answer:
(c) Schotten – Baumann reaction

Question 14.
The pyrimidine bases present in DNA are
(a) Cytosine and Adenine
(b) Cytosine and Guanine
(c) Cytosine and Thiamine
(d) Cytosine and Uracil
Answer:
(c) Cytosine and Thiamine

Question 15.
Nylon is an example of
(a) Polyamide
(b) Polythene
(c) Polyester
(d) Polysaccharide
Answer:
(a) Polyamide

Part – II

Answer any six questions. Question No. 24 is compulsory. [6 × 2 = 12]

Question 16.
Write a test to identify borate radical?
Answer:
When boric acid or borate salt is heated with ethyl alcohol in presence of concentrated H₂SO₄, an ester triethyl borate is formed. The Vapour of this ester bums with a green edged flame and this reaction is used to identify the presence of borate.

Question 17.
How is pure phosphine prepared from phosphorous acid ?
Answer:
Phosphine is prepared in pure form by heating phosphorous acid.

Question 18.
What are ionisation isomers? Explain with an example.
Answer:
1. Ionisation isomerism arises when an ionisable counter ion (simple ion) itself can act as a ligand.

2. The exchange of such counter ions with one or more ligands in the coordination entity will result in ionisation isomers. These isomers will give different ions in solution.

3. For example, consider the coordination compound [Pt (en)₂ Cl₂]Br₂. In this compound, both Br^– and Cl^– have the ability to act as a ligand and the exchange of these two ions result in a different isomer [Pt(en)₂Br₂]Cl₂. In solution, the first compound gives Br^– ions while the later gives Cl ions and hence these compounds are called ionization isomers.

Question 19.
What is pseudo first order reaction? Give one example.
Answer:
A second order reaction can be altered to a first order reaction by taking one of the reactant in large excess, such reaction is called pseudo first order reaction.
Let us consider the acid hydrolysis of an ester,

If the reaction is carried out with the large excess of water, there is no significant change in the concentration of water during hydrolysis, i.e., concentration of water remains almost a constant.
Now we can define k [H₂O] = k’
.’. The above rate equation becomes, Rate = k’ [CH₃COOCH₃]
Thus it follows first order kinetics.

Question 20.
State Faraday’s second law of electrolysis.
Answer:
When the same quantity of charge is passed through the solutions of different electrolytes, the , amount of substances liberated at the respective electrodes are directly proportional to their electrochemical equivalents.

Question 21.
How will you convert glycerol into acrolein?
Answer:

Question 22.
Give any four differences between DNA and RNA.
DNA

1. It is mainly present in nucleus, mitochondria and chloroplast
2. It contains deoxyribose sugar
3. It’s life time is high
4. It is stable and not hydrolysed easily by alkalis
5. It can replicate itself

RNA

1. It is mainly present in cytoplasm, nucleolus and ribosomes
2. It contains ribose sugar
3. It is Short lived
4. It is unstable and hydrolyzed easily by alkalis
5. It cannot replicate itself. It is formed from DNA.

Question 23.
Write short notes on Antioxidants.
Answer:

• Antioxidants are substances which retard the oxidative deteriotations of food. Food containing fats and oils is easily oxidised and turn rancid.
• To prevent the oxidation of fats and oils, chemical BHT (butyl hydroxy toluene), BHA (butylated hydroxy anisole) are added as antioxidants.
• These materials readily undergo oxidation by reacting with free radicals generated by the oxidation of oils there by stop the chain reaction of oxidation of food.
• Sulphur dioxide, sulphites are also used as antioxidant and also act as antimicrobial agents and
  enzyme inhibitors.

Question 24.
50ml of 0.05M HNO₃ is added to 50ml of 0.025M KOH. Calculate the pH of the resultant solution.
Answer:
Number of moles of HNO₃ = 0.05 × 50 × 10^-3 = 2.5 × 10^-3
Number of moles of KOH = 0.025 × 50 × 10^-3 = 1.25 × 10^-3
Number of moles of HNC₃ after mixing = 2.5 × 10^-3 – 1.5 × 10 ^-3 = 1.25 × 10^-3

After mixing, total volume = 100 ml = 100 × 10^-3 L

pH = – log [H⁺]
pH = -log(l.25 × 10^-2) = 2 – 0.0969
= 1.9031

Part – III

Answer any six questions. Question No. 33 is compulsory. [6 × 3 = 18]

Question 25.
Explain the electro metallurgy of aluminium.
Answer:
Electrochemical extraction of aluminium -Hall-Herold process: In this method, electrolysis is carried out in an iron tank lined with carbon, which acts as a cathode. The carbon blocks immersed in the electrolyte acts as a anode. A 20% solution of alumina, obtained from the bauxite ore is mixed with molten cyrolite and is taken in the electrolysis chamber. About 10% calcium chloride is also added to the solution. Here calcium chloride helps to lower the melting point of the mixture. The fused mixture is maintained at a temperature of above 1270 K. The chemical reactions involved in this process are as follows:

Ionisation of alumina : Al₂O₃ → 2Al³⁺ + 3O^2-
Reaction at cathode : 2Al³⁺ (melt) + 6e^– → 2Al _(l)
Reaction at anode : 6O^2- (melt) → 3O₂ + 12e^–
Since carbon acts as anode the following reaction also takes place on it.
C (s) + O^2- (melt) → CO + 2e^– ; C_(s) + 2O^2- (melt) → CO₂ + 4e^–

Due to the above two reactions, anodes are slowly consumed during the electrolysis. The pure aluminium is formed at the cathode and settles at the bottom. The net electrolysis reaction can be written as follows:

4Al³⁺ (melt) + 6O^2- (melt) + 3C_(s) → 4Al_(l) + 3CO_2(g)

Question 26.
Give the uses of helium.
Answer:

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

Question 27.
Explain chromyl chloride Test.
Answer:
(i) When potassium dichromate is heated with any chloride salt in the presence of Conc.H₂SO₄, orange red vapours of chromyl chloride (CrO₂Cl₂) is evolved. This reaction is used to confirm the presence of chloride ion in inorganic qualitative analysis.

(ii) The chromyl chloride vapours are dissolved in sodium hydroxide solution and then acidified with acetic acid and treated with lead acetate. A yellow precipitate of lead chromate is obtained. Cr02Cl2 + 4NaOH -► Na2Cr04

Question 28.
A face centred cubic solid of an element (atomic mass 60 g mol ) has a cube edge of 4A. Calculate its density.
Answer:
For FCC unit cell n 4
Edge length(a) = 4A = 4 × 10^-8cm
Mass (M) = 60 g mol^-1
Dcnsity(ρ) = ?

Question 29.
Describe the construction of Daniel cell and write its cell reaction.
Answer:
The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells.

Oxidation half cell: The metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker.
Reduction half cell: A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker.
Joining the half cell : The zinc and copper strips are externally connected using a wire through a switch (k) and a load (example: volt meter). The electrolytic solution present in the cathodic and anodic compartment are connected using an inverted U tube containing a agar-agar gel mixed with an inert electrolyte such as KCl, Na₂SO₄ etc., The ions of inert electrolyte do not react with other ions present in the half cells and they are not either oxidised (or) reduced at the electrodes. The solution in the salt bridge cannot get poured out, but through which the ions can move into (or) out of the half cells.

When the switch (k) closes the circuit, the electrons flows from zinc strip to copper strip. This is due to the following redox reactions which are taking place at the respective electrodes.

Anodic oxidation:- The electrode at which the oxidation occur is called the anode. In Daniel cell, the oxidation take place at zinc electrode, i.e., zinc is oxidised to Zr²⁺ ions and the electrons. The Zn²⁺ ions enters the solution and the electrons enter the zinc metal, then flow through the external wire and then enter the copper strip. Electrons are liberated at zinc electrode and hence it is negative (- ve).
Zn(s) → Zn²⁺+(aq) + 2e^–
(loss of electron-oxidation)

Cathodic reduction:
As discussed earlier, the electrons flow through the circuit from zinc to copper, where the Cu ions in the solution accept the electrons, get reduced to copper and the same get deposited on the electrode. Here, the electrons are consumed and hence it is positive (+ve).

cu²⁺ 2e^– cu_(s)( gain of electron – reduction)

Salt bridge: -The electrolytes present in two half cells are connected using a salt bridge. We have learnt that the anodic oxidation of zinc electrodes results in the increase in concentration of Zn²⁺ in solution, i.e., the solution contains more number of Zn²⁺ ions as compared to SO^2-₄ and hence the solution in the anodic compartment would become positively charged. Similarly, the solution in the cathodic compartment would become negatively charged as the Cu²⁺ ions are reduced to copper i.e., the cathodic solution contain more number of SO^2-₄ ions compared to Cu²⁺ .

Completion of circuit:- Electrons flow from the negatively charged zinc anode into the positively charged copper electrode through the external wire, at the same time, anions move towards anode and cations move towards the cathode compartment. This completes the circuit.

Consumption of Electrodes :- As the Daniel cell operates, the mass of zinc electrode gradually decreases while the mass of the copper electrode increases and hence the cell will function until the entire metallic zinc electrode is converted into Zn²⁺ or the entire Cu²⁺ ions are converted into mettalic copper.
Daniel cell is represents as

Question 30.
Write short notes on (i) Negative catalyst (ii) Phase transfer catalyst
Answer:
(i) Negative catalyst:
(i) In certain reactions, presence of certain substances decreases the rate of the reaction. Such
substances are called negative catalyst and the process is called negative catalysis.

(ii) In oxidation of chloroform, ethanol decreases the rate of the reaction and ethanol act as negative catalyst.

(iii) Decomposition of H202 rate is decreased by glycerol and it act as negative catalyst

(ii) Phase transfer catalyst:

(i) Consider the reactant of a reaction is present in one solvent and the other reactant is present in an another solvent. The reaction between them is very slow, if the solvents are immisible.

(ii) As the solvents form separate phases, the reactants have to migrate across the boundary to react. But migration of reactants across the boundary is not easy. For such situation, a third solvent is added which is miscible with both. So, the phase boundary is eliminated the reactants freely mix and react fast.

(iii) But for large scale preparation of any product, use of a third solvent is not convenient as it . may be expensive. For such problems, phase transfer catalysis provides a simple solution, which avoids the use of solvents.

(iv) It directs the use of a phase transfer catalyst (a phase transfer reagent) to facilitate transport of a reactant in one solvent to other solvent where the second reactant is present. As the reactants are now brought together, they rapidly react and form the product.

(v) Example : Substitution of Cl^– and CN^– in the following reaction.

R Cl = 1- chloro octane
R CN = 1- cyano octane
(vi) By direct heating of two phase mixture of organic 1- chloro octane with aqueous sodium cyanide for several days, 1- cyano octane is not obtained. However, if a small amount of quartemary ammonium salt like tetra alkyl ammonium cation which has hydrophobic and hydrophilic ends, transports CN- from the aqueous phase to the organic phase using its hydrophilic end and facilitates the reactions with 1- chloro octane as shown below

(vii) So phase transfer catalyst, speeds up the reaction by transporting one reactant from one phase to another.

Question 31.
Explain the mechanism of Aldol condensation of acetaldehyde.
Answer:
In presence of dilute base NaOH, or KOH, two molecules of an aldehyde or ketone having α – hydrogen add together to give β- hydroxyl aldehyde (aldol) or β – hydroxyl ketone (ketol). The reaction is called aldol condensation reaction.

Mechanism
The mechanism of aldol condensation Of acetaldehyde takes place in three steps.
Step 1: The carbanion is formed as the a – hydrogen atom is removed as a proton by the base.

Step 2: The carbanion attacks the carbonyl carbon of another unionized aldehyde to form an alkoxide ion.

Step 3: The alkoxide ion formed is protonated by water to form aldol.

The aldol rapidly undergoes dehydration on heating with acid to form a, p unsaturated aldehyde.

Question 32.
Explain the preparation of Nylon – 6,6 and Buna- S.
Answer:
Nylon 6,6 can be prepared by mixing equimolar adipic acid and hexamethylene diamine. With the elimination of water to form amide bonds.

Buna – S is prepared by the polymerisation of buta -1,3 – diene and styrene in the ratio of 3:1 in the presence of sodium.

Question 33.
Identify A to C in the following sequence.

Part – IV

Answer all the questions. [5 × 5 = 25]

Question 34.
(a) (i) Explain how gold ore is leached by cyanide process
(ii) Explain the classification of lnosilicates

[OR]

(b) (i) What are interhalogen compounds? Give examples
(ii) Explain the preparation of KMn04.
Answer:
(a) (i) 4Au (s) + 8CN^– (aq) + 2H₂O(aq) + O₂(g)→ 4Au(CN)₂]^–(aq) + 4OH^–
2[Au(CN)₂]^–(aq) + Zn (s) → 2Au (s) + [Zn(CN)₄]^-2(aq)
In the first reaction Au changes into Au+, i.e. its oxidation takes place. In the second reaction: Au⁺ → Au° (i.e. ) reduction takes place.

(ii) Ino silicones: Silicates which contain V number of silicate units liked by sharing two or more oxygen atoms are called inosilicates. They are further classified as chain silicates and double chain silicates.

Chain silicates (or pyroxenes): These silicates contain [(SiO₃)_n]^2n- ions formed by linking ‘ri number of tetrahedral [SiO₄]^4- units linearly. Each silicate unit shares two of its oxygen atoms with other units.

Example: Spodumene – LiAl(SiO₃)₂ Double chain silicates (or amphiboles): These silicates contains [Si₄O₁₁] _n^6n-. ions. In these silicates there are two different types of tetrahedra : (i) Those sharing 3 vertices (ii) those sharing only 2 vertices.

Examples:
Asbestos: These are fibrous and noncombustible silicates. Therefore they are used for thermal insulation material, brake linings, construction material and filters. Asbestos being carcinogenic silicates, their applications are restricted.

[OR]

(b) (i) Each halogen combines with other halogens to form a series of compounds called interhalogen compounds, e.g., ClF₃, IF₅, IF₇
(ii) Potassium permanganate is prepared from pyrolusite (MnO₂) ore. The preparation involves the following steps.

Conversion of MnO₂ to potassium manganate:
Powdered ore is fused with KOH in the presence of air or oxidising agents like KNO₃ A green coloured potassium manganate is formed.

Oxidation of potassium manganate to potassium permanganate:
Potassium manganate can be oxidised in two ways, either by chemical oxidation or electrolytic oxidation.

Chemical oxidation: In this method potassium manganate is treated with ozone (O₃) or chlorine to get potassium permanganate.

Electrolytic oxidation: In this method aqueous solution of potassium manganate is electrolyzed in the presence of little alkali.
K₂MnO₄ ⇌ 2K⁺ + MnO₄^2-
H₂O ⇌ H⁺ + OH^–

Manganate ions are converted into permanganate ions at anode.

The purple coloured solution is concentrated by evaporation and forms crystals of potassium permanganate on cooling.

Question 35.
(a) (i) Explain [Fe(CN)₆]^3- is paramagnetic, using Crystal Field theory
(ii) What is schottky detect?

[OR]

(A) (0 Derive Henderson – Hassel balch equation
(ii) What is kohlraush’s law?
Answer:
(a) (i)

(ii) Schottky defect arises due to the missing of equal number of cations and anions from the crystal lattice. This effect does not change the stoichiometry of the crystal.Ionic solids in which the cation and anion are of almost of similar size show schottky defect. Example: NaCl.

Presence of large number of schottky defects in a crystal, lowers its density.
For example, the theoretical density of vanadium monoxide (VO) calculated using the edge length of the unit cell is 6.5 g cm^-3 , but the actual experimental density is 5.6 g cm^-3 .It indicates that there is approximately 14% Schottky defect in VO crystal. Presence of Schottky defect in the crystal provides a simple way by which atoms or ions can move within the crystal lattice.

[OR]

(b) (i) Henderson – Hasselbalch equation:
1. The concentration of hydronium ion in acidic buffer solution depends on the ratio of concentration of the weak acid to the concentration of its conjugate base present in the solution, i.e.,

2. The weak acid is dissociated only to a small extent. Moreover due to common ion effect, the dissociation is further suppressed and hence the equilibrium concentration of the acid is nearly equal to the initial concentration of the unionised acid. Similarly the concentration of the conjugate base is nearly equal to the initial concentration of the added salt.

3. [Acid] and [Salt] represent the initial concentration of the acid and salt, respectively used to prepare the buffer solution.

4. Taking logarithm on both sides of the equation

5. Reverse the sign on both sides

Equation (6) & (7) are called Henderson – Hasselbalch equations.

(ii) Kohlrauselrs law: It is defined as, at infinite dilution the limiting molar conductivity of an electrolyte is equal to the sum of the limiting molar conductivities of its constituent ions.

Question 36.
(a) (i) Explain Intermediate compound formation theory.
(ii) Write short notes on ultra filtration.

[OR]

(b) How the following conversions are effected?
(i) Phenol → Salicylaldehvde
(ii) Phenol → Phenolphthalein
(iii) Glycol → 1.4dioxane
Answer:
(a) (i) The intermediate compound formation theory:
A catalyst acts by providing a new path with low energy of activation.. In homogeneous catalysed reactions a catalyst may combine with one or more reactant to form an intermediate which reacts with other reactant or decompose to give products and the
catalyst is regenerated.
Consider the reactions:
A+B → AB (1)
A+C →AC (intermediate) (2)
C is the catalyst
AC+B → AB+C (3)

Activation energies for the reactions (2) and (3) are lowered compared to that of (1). Hence the formation and decomposition of the intermediate accelerate the rate of the reaction.

(ii) Ultrafiltration:

• The pores of ordinary filter papers permit the passage of colloidal solutions. In ultrafilteration, the membranes are made by using collodion, cellophane or visiking.
• When a colloidal solution is filtered using such a filter, colloidal particles are separated on the filter and the impurities are removed as washings.
• This process is quickened by application of pressure. The separation of sol particles from electrolyte by filtration through an ultrafilter is called ultrafiltration.
• Collodian is 4% solution of nitrocellulose in a mixture of alcohol and water.

[OR]

(b) (i) Phenol → Salicylaldehyde:

(ii) Phenol → Phenolphthalcin:
On heating phenol with phthalic anhydride in presence of con.H2S04, phenolphthalein
is obtained.

(iii) Glycol →1,4 dioxane:
When distilled with Conc.112S04, glycol forms 1, 4-dioxane

Question 37.
(a) Write short notes on
(i) Mustard oil reactions
(ii) Carbylamine reaction
(iii) Gabriel pathalimide synthesis

[OR]

(b) Explain the structure of Fructose.
Answer:
(a) (i) Mustard oil reactions:
When primary amines are treated with carbon disulphide (CS₂), N – alkyldithio carbonic acid is formed which on subsequent treatment with HgCl₂, give an alkyl isothiocyanate.

(ii) Carbylamine reaction:
Aliphatic (or) aromatic primary amines react with chloroform and alcoholic KOH to give isocyanides (carbylamines), which has an unpleasant smell. This reaction is known as carbylamines test. This test is used to identify the primary amines.

(iii) Gabriel pathalimide synthesis:
Gabriel synthesis is used for the preparation of Aliphatic primary amines. Phthalimide on treatment with ethanolic KOH forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives primary amine.

[OR]

Structure of Fructose:

• Elemental analysis and molecular weight determination of fructose show that it has the molecular formula C₆H₁₂O₆.
• Fructose on reduction with HI and red phosphorus gives a mixture of n – hexane (major product) and 2 – iodohexane (minor product). This reaction indicates that the six carbon atoms in fructose are in a straight chain.
  
• Fructose reacts with NH₂OH and HCN. It shows the presence of carbonyl groups in the molecules of fructose.
• Fructose reacts with acetic anhydride in the presence of pyridine to form penta acetate. This reaction indicates the presence of five hydroxyl groups in a fructose molecule.
• Fructose is not oxidized by bromine water. This rules out the possibility of presence of an aldehyde (-CHO) group.
• Partial reduction of fructose with sodium amalgam and water produces mixtures of sorbitol and mannitol which are epimers at second carbon. New asymmetric carbon is formed at C-2. This confirms the presence of keto group.
  

On oxidation with nitric acid, it gives glycolic acid and tartaric acids which contain smaller number of carbon atoms than in fructose.

This shows that a keto group is present in C-2. It also shows the presence of 10 alcoholic groups at C- 1 and C- 6. From the above reaction the structure of fructose is

Question 38.
(a) (i) A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?
(ii) K_sp of Ag₂CrO₄ is 1.1 x 10^-12. What is the solubility of Ag₂CrO₄ in 0.1M K₂CrO₄?
[OR]
(b) Compound A of molecular formula C₇H₆O reduces Tollen’s reagent when A reacts with 50% NaOH gives compound B of molecular formula C₇H₈O and C of molecular formula C₇H₅O₇Na. Compound C on treatment with dil.HCl gives compound D of molecular formula C₇H₆O₂. When D is heated with sodalime gives compound E.
Identify A,B,C,D & E. Write the corresponding equations.
Answer:
(a) (i) 1. For the first order reaction k =\(\frac{2.303}{t} \log \frac{a}{(a-x)}\)
Assume, a = 100 %, x = 40%, t = 50 minutes
Therefore, a – x = 100 – 40 = 60.
k = (2.303/ 50) log (100/ 60)
k = 0.010216 min^-1
Hence the value of the rate constant is 0.010216 min^-1
2. t = ?, when x = 80%
Therefore, a – x = 100 – 80 = 20
From above, k = 0.010216 min^-1
t = (2.303/0.010216) log (100/20)
t = 157.58 min
The time at which the reaction will be 80% complete is 157.58 min.

(ii)

[OR]

(b)

Students can Download Tamil Nadu 12th Physics Model Question Paper 3 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

TN State Board 12th Physics Model Question Paper 3 English Medium

Instructions:

1. The question paper comprises of four parts
2. You are to attempt all the parts. An internal choice of questions is provided wherever: applicable
3. All questions of Part I, II, III and IV are to be attempted separately
4. Question numbers 1 to 15 in Part I are Multiple choice Questions of one mark each.  These are to be answered by choosing the most suitable answer from the given four alternatives and writing the option code and the corresponding answer
5. Question numbers 16 to 24 in Part II are two-mark questions. These are lo be answered in about one or two sentences.
6. Question numbers 25 to 33 in Part III are three-mark questions. These are lo be answered in about three to five short sentences.
7. Question numbers 34 to 38 in Part IV are five-mark questions. These are lo be answered in detail. Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

PART -1

Answer all the questions. Choose the correct answer. [15 × 1 = 15]

Question 1.
Two metallic spheres of radii 1 cm and 3 cm are given charges of -1 × 10^-2 C and 5 × 10^-2 C respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is
(a) 3 × 10^-2 C
(b) 4 × 10^-2 C
(c) 1 × 10^-2 C
(d) 2 × 10^-2 C
Answer:
(a) 3 × 10^-2 C

Question 2.
What is the value of resistance of the following resistor?

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

Question 3.
A flow of 10⁷ electrons per second in a conduction wire constitutes a
(a) 1.6 × 10^-26A
(b) 1.6 × 10¹²A
(c) 1.6 × 10^-12A
(d) 1.6 × 10²⁶A
Hint:

Answer:
(c) 1.6 x 10^-12A

Question 4.
The force experienced by a particle having mass m and charge q accelerated through a potential difference V when it is kept under perpendicular magnetic field B is ……………….. .
(a) \(\sqrt{\frac{2 q^{3} \mathrm{BV}}{m}}\)
(b) \(\sqrt{\frac{q^{3} \mathrm{B}^{2} V}{2 m}}\)
(c) \(\sqrt{\frac{2 q^{3} \mathbf{B}^{2} \mathbf{V}}{m}}\)
(d) \(\sqrt{\frac{2 q^{3} B V}{m^{3}}}\)
Answer:
(c) \(\sqrt{\frac{2 q^{3} \mathbf{B}^{2} \mathbf{V}}{m}}\)

Question 5.
A magnetic needle is kept in a non-uniform magnetic field. It experiences
(a) a force and a torque
(b) a force but not a torque
(c) a torque but not a force
(d) neither a force nor a torque
Answer:
(a) a force and a torque

Question 6.
The instantaneous values of alternating current and voltage in a circuit are i = \(\frac{1}{\sqrt{2}}\) sin( 100πt) A and v = \(\frac{1}{\sqrt{2}} \sin \left(100 \pi t+\frac{\pi}{3}\right) \mathrm{V}\) The average power in watts consumed in the circuit is
(a) \(\frac{1}{4}\)
(b) \(\frac{\sqrt{3}}{4}\)
(c) \(\frac{1}{2}\)
(d) \(\frac{1}{8}\)
Answer:
(d) \(\frac{1}{8}\)

Question 7.
Which of the following is an electromagnetic wave’?
(a) α – rays
(b) β – rays
(c) γ – rays
(d) all of them
Answer:
(c) γ – rays

Question 8.
When light is incident on a soap film of thickness 5 × 10^-5 cm, the wavelength of light reflected maximum in the visible region is 5320 Å. Refractive index of the film will be.
(a) 1.22
(b) 1.33
(c) 1.51
(d) 1.83
Hint. The condition for constructive interference, (for reflection)

Answer:
(b) 1.33

Question 9.
A plane glass is placed over a various coloured letters (violet, green, yellow, red) The letter which appears to be raised more is,
(a) red
(b) yellow
(c) green
(d) violet
Hint. Letters appear to be raised depending upon the refractive index of the material. Since violet has a higher refractive index than red (the index increases with frequency), red will be the lowermost.
Answer:
(d) violet

Question 10.
In photoelectric emission, a radiation whose frequency is 4 times threshold frequency of a certain metal is incident on the metal. Then the maximum possible velocity of the emitted electron will be
(a) \(\sqrt{\frac{h v_{0}}{m}}\)
(b) \(\sqrt{\frac{6 h u_{0}}{m}}\)
(c) \(2 \sqrt{\frac{h v_{0}}{m}}\)
(d) \(\sqrt{\frac{h v_{0}}{2 m}}\)

Hint: From Einstein’s photoelectric equation
K_max = hυ – hυ₀ [υ = 4υ₀]
\(\frac { 1 }{ 2 }\) mV²_max = 4hυ₀ – hυ₀
V²_max = \(\frac{6 h v_{0}}{m}\)
V_max = \(\sqrt{\frac{6 h v_{0}}{m}}\)
Answer:
(b) \(\sqrt{\frac{6 h u_{0}}{m}}\)

Question 11.
The number of photo-electrons emitted for light of a frequency u (higher than the threshold frequency υ₀) is proportional to
(a) Threshold frequency (υ₀)
(b) Intensity of light
(c) Frequency of light (υ)
(d) υ – υ₀
Hint: Photoelectric current of Intensity of incident light
Answer:
(b) Intensity of light

Question 12.
Atomic number of H-like atom with ionization potential 122.4 V for n = 1 is
(a) 1
(b) 2
(c) 3
(d) 4
Hint: The ionisation energy of a hydrogen atom is, IE = \(\frac{13.6 z^{2}}{n^{2}}\)

Answer:
(c) 3

Question 13.
The given electrical network is equivalent to ………….. .
(а) AND gate
(b) OR gate
(c) NOR gate
(d) NOT gate

Answer:
(c) NOR gate

Question 14.
The frequency range of 3 MHz to 30 MHz is used for
(a) Ground wave propagation
(b) Space wave propagation
(c) Sky wave propagation
(d) Satellite communication
Answer:
(c) Sky wave propagation

Question 15.
Which one of the following is the natural nanomaterial?
(a) Peacock feather
(b) peacock beak
(c) Grain of sand
(d) Skin of the wale
Answer:
(a) Peacock feather

PART – II

Answer any six questions in which Q. No 21 is compulsory. [6 × 2 = 12]

Question 16.
When two objects are rubbed with each other, approximately a charge of 50 nC can be produced in each object. Calculate the number of electrons that must be transferred to produce this charge.
Answer:
Charge produced in each object q = 50 nC
q = 50 × 10^-9C
Charge of electron (e) = 1.6 × 10^-19C
Number of electron transferred, n = \(\frac{q}{e}=\frac{50 \times 10^{-9}}{1.6 \times 10^{-19}}\) = ,n =31.25 × 10^-9 × 10¹⁹
n = 31.25 × 1o¹⁰ electrons

Question 17.
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 18.
Define magnetic dipole moment.
Answer:
The magnetic dipole moment is defined as the product of its pole strength and magnetic length.
\(\overrightarrow{\mathrm{P}}\) = q_m\(\overrightarrow{\mathrm{d}}\)

Question 19.
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_RMSI_RMS cos Φ . So that it is also known as ‘WattfuT current.

Question 20.
What are electromagnetic waves?
Answer:
Electromagnetic waves are non-mechanical waves which move with speed equals to the speed of light (in vacuum).

Question 21.
Two light sources have intensity of light as 10. What is the resultant intensity at a point where the two light waves have a phase difference of π/3?
Answer:
Let the intensities be I₀.
The resultant intensity is, I = 4 I₀ cos² (φ/2)
Resultant intensity when, Φ = π / 3, is I = 4I₀ cos² (π / 6)
I = 4I₀(√3/2)² = 3I₀

Question 22.
State de Broglie hypothesis.
Answer:
De Broglie hypothesis, all matter particles like electrons, protons, neutrons in motion are associated with waves.

Question 23.
Calculate the energy equivalent of 1 atomic mass unit.
Answer:
We take, m = 1 amu = 1.66 × 10²⁷ kg
c = 3 × 10⁸ ms^-1
Then, E = mc² = 1.66 × 10²⁷ × (3 × 1o⁸)²⁷ J

E ≈ 981 MeV
∴ 1 amu = 931 MeV

Question 24.
hat do you mean by doping?
Answer:
The process of adding impurities to the intrinsic semiconductor is called doping.

PART III

Answer any six questions in which Q.No. 32 is compulsory. [6 × 3 = 18]

Question 25.
What are the differences between Coulomb force and gravitational force?
Answer:

• The gravitational force between two masses is always attractive but Coulomb force between two charges can be attractive or repulsive, depending on the nature of charges.
• The value of the gravitational constant G = 6.626 × 10¹¹ N m² kg^-2. The value of the constant k in Coulomb law is k = 9 × 10⁹ N m² C^-2.
• The gravitational force between two masses is independent of the medium. The electrostatic force between the two charges depends on nature of the medium in which the two charges are kept at rest.
• The gravitational force between two point masses is the same whether two masses are at rest or in motion. If the charges are in motion, yet another force (Lorentz force) comes into play in addition to coulomb force.

Question 26.
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.
Answer:
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, R_T = ?
Temperature of boiling point of water, T = 100 °C
R_T = R₀ ( 1 + α T) = 10[1 + (0.004 × 100)]
R_T = 10(1 + 0.4) = 10 × 1.4
R_T = 14Ω
As the temperature increases the resistance of the wire also increases.

Question 27.
What is meant by magnetic induction?
Answer:
The magnetic induction (total magnetic held) inside the specimen \(\overrightarrow{\mathrm{B}}\) is equal to the sum of the magnetic field \(\overrightarrow{\mathrm{B}}\)₀ produced in vacuum due to the magnetising field and the magnetic field \(\overrightarrow{\mathrm{B}}\)_m due to the induced magnetisation of the substance.

Question 28.
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 29.
Derive the relation between/and R for a spherical mirror.
Answer:
Let C be the centre of curvature of the mirror. Consider a light ray parallel to the principal axis is incident on the mirror at M and passes through the principal focus F after reflection. The geometry of reflection of the incident ray is shown in figure. The line CM is the normal to the mirror at M. Let i be the angle of incidence and the same will be the angle of reflection.
If MP is the perpendicular from M on the principal axis, then from the geometry, The angles

∠MCP = i and ∠MFP = 2i From right angle triangles ∆MCP and ∆MFP,
tan i = \(\frac{P M}{P C}\) and tan 2i = \(\frac{P M}{P F}\)
As the angles are small, tan i ≈ i, i = \(\frac{P M}{P C}\) and 2i = \(\frac{P M}{P F}\)
Simplifying further, 2 \(\frac{\mathrm{PM}}{\mathrm{PC}}=\frac{\mathrm{PM}}{\mathrm{PF}}\) ; 2PF = PC
PF is focal length f and PC is the radius of curvature R.
2f = R (or) f = \(\frac{R}{2}\)
f = \(\frac{R}{2}\) is the relation between/and R.

Question 30.
What is a photo cell? Mention the different types of photocells.
Answer:
photocells: Photo electric cell or photo cell is a device which converts light energy into electrical energy. It works on the principle of photo electric effect.
Types:

• Photo emissive cell
• Photo voltaic cell
• Photo conductive cell

Question 31.
Show that nuclear density is almost constant for nuclei with Z > 10.
Answer:

The expression shows that the nuclear density is independent of the mass number A. In other words, all the nuclei (Z >10) have the same density and it is an important characteristics of the nuclei.

Question 32.
Calculate the range of the variable capacitor that is to be used in a tuned-collector oscillator which has a fixed inductance of 150 pH. The frequency band is from 500 kHz to 1500 kHz.
Answer:
Resonant frequency f₀ = \(\frac{1}{2 \pi \sqrt{\mathrm{LC}}}\)
On simplifying we get C = \(\frac{1}{4 \pi^{2} f_{0}^{2} L}\)
When frequency is equal to 500 kHz

When frequency is equal to 1500 kHz

There fore, the capacitor range is 75 – 676 pF

Question 33.
Distinguish between wireline and wireless communication.
Answer:

PART – IV

Answer all the questions. [5 × 5 = 25]

Question 34.
(a) Explain in detail the construction and working of a Van de Graaff generator.
Answer:
Principle: Electrostatic induction and action at points.
Construction: A large hollow spherical conductor is fixed on the insulating stand. A pulley B is mounted at the center of the hollow sphere and another pulley C is fixed at the bottom. A belt made up of insulating materials like silk or rubber runs over both pulleys. The pulley C is driven continuously by the electric motor. Two comb shaped metallic conductors E and D are ‘ fixed near the pulleys.
The comb D is maintained at a positive potential of 10⁴ V by a power supply. The upper comb E is connected to the inner side of the hollow metal sphere.

Working: Due to the high electric field near comb D, air between the belt and comb D gets ionized. The positive charges are pushed towards the belt and negative charges are attracted towards the comb D. The positive charges stick to the belt and move up. When the positive charges reach the comb E, a large amount of negative and positive charges are induced on either side of comb E due to electrostatic induction. As a result, the positive charges are pushed away from the comb E and they reach the outer surface of the sphere. Since the sphere is a conductor, the positive charges are distributed uniformly on the outer surface of the hollow sphere. At the same time, the negative charges nullify the positive charges in the belt due to corona discharge before it passes over the pulley.

When the belt descends, it has almost no net charge. At the bottom, it again gains a large positive charge. The belt goes up and delivers the positive charges to the outer surface of the sphere. This process continues until the outer surface produces the potential difference of the order of 107 V which is the limiting value. We cannot store charges beyond this limit since the extra charge starts leaking to the surroundings due to ionization of air. The leakage of charges can be reduced by enclosing the machine in a gas filled steel chamber at very high pressure. Uses: The high voltage produced in this Van de Graaff generator is used to accelerate positive ions (protons and deuterons) for nuclear disintegrations and other applications.

[OR]

Question 34.
(b) Explain the equivalent resistance of a series resistor network.
Answer:
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 = IR₁ 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 is the equivalent resistance,
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.

Question 35.
(a) Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.
Answer:
Magnetic field due to long straight conductor carrying current: Consider a long straight wire NM with current I flowing from N to M. Let P be the point at a distance a from point O. Consider an element of length dl of the wire at a distance l from point O and \(\vec{r}\) be the vector joining the element dl with the point P. Let θ be the angle between dl and \(\vec{r}\) Then, the magnetic field at P due to the element is
\(d \overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{Id} \vec{l}}{4 \pi r^{2}} \sin \theta\) (unit vector perpendicular to \(d \vec{l}\) and \(\vec{r}\) )-(1)
The direction of the field is perpendicular to the plane of the paper and going into it. This can be determined by taking the cross product between two vectors \(d \vec{l}\) and \(\vec{r}\) (let it be n̂). The net magnetic field can be determined by integrating equation with proper limits.
\(\vec{B}\) = ∫d\(\vec{B}\)
From the figure, in a right angle triangle PAO,

tan (π – θ) = \(\frac{a}{l}\)
l = \(-\frac{a}{\tan \theta}\) (since tan (π – θ) = – tan θ) ⇒ \(\frac{1}{\tan \theta}\) = cot θ
l = -a cot θ and r = a cosecθ

Differentiating,
dl = a cosec²θ dθ

This is the magnetic field at a point P due to the current in small elemental length. Note that we have expressed the magnetic field OP in terms of angular coordinate i.e. θ. Therefore, the net magnetic field at the point P which can be obtained by integrating \(d \vec{B}\) by varying the angle from θ = φ₁ to θ = φ₂ is

For a an infinitely long straight wire, l = 0 and 2 = , the magnetic field is
\(\overrightarrow{\mathrm{B}}=\frac{\mu_{0} I}{2 \pi a} \hat{n}\) …… (3)
Note that here n̂ represents the unit vector from the point O to P.

[OR]

Question 35.
(b) 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 q is sent through coil i₁ 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.
N₂Φ₂₁ ∝ i₁
N₂Φ₂₁ = M₂₁i₁ or M₂₁ = \(\frac{N_{2} \Phi_{21}}{i_{1}}\)
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 = 1 A, 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 i₁ changes with time, an emf ξ₂ is induced in coil 2. From Faraday’s law of electromagnetic induction, this mutually induced emf ξ₂₁ is given by

The negative sign in the above equation shows that the mutually induced emf always opposes the change in current with respect to time. If \(\frac{d i_{1}}{d t}\) = 1 As^-1, then M₂₁ = -ξ₂.
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 I is l As^-1.
Similarly, if an electric current i₂ through coil 2 changes with time, then emf ξ₁ is induced in coil 1. Therefore, N

where M₁₂ is the mutual inductance of the coil I 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 36.
(a) Derive the equation for refraction at single spherical surface. Equation for refraction at single spherical surface:
Answer:
Let us consider two transparent media having refractive indices n₁ and n₂ are separated by a spherical surface. Let C be the centre of curvature of the spherical surface. Let a point object O be in the medium n₁. The line OC cuts the spherical surface at the pole P of the surface. As the rays considered are paraxial rays, the perpendicular dropped for the point of incidence to the principal axis is very close to the pole or passes through the pole itself.

Light from O falls on the refracting surface at N. The normal drawn at the point of incidence passes through the centre of curvature C. As n₂ > n₁, light in the denser medium deviates towards the normal and meets the principal axis at I where the image is formed. Snell’s law in product form for the refraction at the point N could be written as,
n₁ sin i = n₂ sin r …… (1)
As the angles are small, sin of the angle could be approximated to the angle itself.
n₁i = n₂r ……… (2)
Let the angles
∠NOP = α ∠NCP = β ∠NIP = γ
tan α = \(\frac{P N}{P O}\) tan β = \(\frac{P N}{P C}\) tan γ = \(\frac{P N}{P I}\)
As these angles are small, tan of the angle could be approximated to the angle itself.
α = \(\frac{P N}{P O}\) ; β = \(\frac{P N}{P C}\) ; γ = \(\frac{P N}{P I}\) …… (3)
For the triangle, ∆ONC,
i = α + β ….. (4)
For the triangle, ∆INC,
β = r + γ (or) r = β – γ …… (5)
Substituting for i and r from equations (4) and (5) in the equation (2).
n₁ (α + β) = n₂ (β – γ)
Rearranging,
n₁α + n₂γ = (n₂ – n₁)β
Substituting for α, β and γ from equation (3)
\(n_{1}\left(\frac{\mathrm{PN}}{\mathrm{PO}}\right)+n_{2}\left(\frac{\mathrm{PN}}{\mathrm{PI}}\right)=\left(n_{2}-n_{1}\right)\left(\frac{\mathrm{PN}}{\mathrm{PC}}\right)\)
Further simplifying by cancelling method
\(\frac{n_{1}}{\mathrm{PO}}+\frac{n_{2}}{\mathrm{PI}}=\frac{n_{2}-n_{1}}{\mathrm{PC}}\) ….. (6)
Following sign conventions, PO = -υ , PI = +ν and PC = +R in equation (6),
\(\frac{n_{1}}{-u}+\frac{n_{2}}{v}=\frac{\left(n_{2}-n_{1}\right)}{\mathrm{R}}\)
After rearranging, finally we get,
\(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{\left(n_{2}-n_{1}\right)}{\mathrm{R}}\) ……… (7)
Equation (7) gives the relation among the object distance υ, image distance ν , refractive indices of the two media (nl and n2) and the radius of curvature R of the spherical surface. It holds for any spherical surface.
If the first medium is air then, n₁ = 1 and the second medium is taken just as n2 = n, then the equation is reduced to,
\(\frac{n}{v}-\frac{1}{u}=\frac{(n-1)}{\mathrm{R}}\) ……… (8)

[OR]

Question 36.
(b) What do you mean by electron emission? Explain briefly various methods of electron emission.
Answer:
Electron emission:

• Free electrons possess some kinetic energy and this energy is different for different electrons. The kinetic energy of the free electrons is not sufficient to overcome the surface barrier.
• Whenever an additional energy is given to the free electrons, they will have sufficient energy to cross the surface barrier. And they escape from the metallic surface.
• The liberation of electrons from any surface of a substance is called electron emission. There are mainly four types of electron emission which are given below.

(i) Thermionic emission:
When a metal is heated to a high temperature, the free electrons on the surface of the metal get sufficient energy in the form of thermal energy so that they are emitted from the metallic surface. This type of emission is known as thermionic emission.

The intensity of the thermionic emission (the number of electrons emitted) depends on the metal used and its temperature.

Examples: cathode ray tubes, electron microscopes, X-ray tubes etc.

(ii) Field emission:
Electric field emission occurs when a very strong electric field is applied across the metal. This strong field pulls the free electrons and helps them to overcome the surface barrier of the metal.

Field emission
Examples: Field emission scanning electron microscopes, Field-emission display etc.

(iii) Photo electric emission:
When an electromagnetic radiation of suitable frequency is incident on the surface of the metal, the energy is transferred from the radiation to the.free electrons. Hence, the free electrons get sufficient energy to cross the surface barrier and the photo electric emission takes place. The number of electrons emitted depends on the intensity of the incident radiation.

Examples: Photo diodes, photo electric cells etc.

(iv) Secondary emission:
When a beam of fast moving electrons strikes the surface of the metal, the kinetic energy of the striking electrons is transferred to the free electrons on the metal surface. Thus the free electrons get sufficient kinetic energy so that the secondary emission of electron occurs.

Examples: Image intensifires, photo multiplier tubes etc.

Question 37.
(a) Discuss the spectral series of hydrogen atom.
Answer:
The spectral lines of hydrogen are grouped in separate series. In each series, the distance of separation between the consecutive wavelengths decreases from higher wavelength to the lower wavelength, and also wavelength in each series approach a limiting value known as the series limit. These series are named as Lyman series. Balmer series, Paschen series, Brackett series, Pfund series, etc. The wavelengths of these spectral lines perfectly agree with the equation derived from Bohr atom model.
\(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{n^{2}}-\frac{1}{m^{2}}\right)=\bar{v}\) …… (1)
where ν̂ is known as wave number which is inverse of wavelength, R is known as Rydberg constant whose value is 1.09737 × 10⁷ m^-1 and m and n are positive integers such that m > n. The various spectral series are discussed below:

(a) Lyman series:
Put n = 1 and m = 2,3,4 in equation (I). The wave number or wavelength of spectral lines of Lyman series which lies in ultra-violet region is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{1^{2}}-\frac{1}{m^{2}}\right)\)

(b) Balmer series:
Put n = 2 and m = 3,4,5 in equation (I). The wave number or wavelength of spectral lines of Balmer series which lies in visible region is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{m^{2}}\right)\)

(c) Paschen series:
Put n = 3 and m = 4,5,6 in equation (I). The wave number or wavelength of spectral lines of Paschen series which lies in infra-red region (near IR) is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{3^{2}}-\frac{1}{m^{2}}\right)\)

(d) Brackett series:
Put n = 4 and m = 5,6,7 in equation (I). The wave number or wavelength of spectral lines of Brackett series which lies in infra-red region (middle IR) is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{m^{2}}\right)\)

(e) Pfund series:
Put n = 5 and m = 6,7,8 in equation (I). The wave number or wavelength of spectral lines of Pfund series which lies in infra-red region (far IR) is
\(\bar{v}=\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{5^{2}}-\frac{1}{m^{2}}\right)\)

[OR]

Question 37.
(b) State and prove De Morgan’s First and Second theorems.
Answer:
De Morgan’s First Theorem:
The first theorem states that the complement of the sum of two logical inputs is equal to the product of its complements.

Proof:
The Boolean equation for NOR gate is Y = \(\overline{\mathrm{A}+\mathrm{B}}\) . The Boolean equation for a bubbled AND gate is Y = \(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\) Both cases generate same outputs for same inputs. It can be verified using the following truth table.

From the above truth table, we can conclude \(\overline{\mathrm{A}+\mathrm{B}}=\overline{\mathrm{A}} \cdot \overline{\mathrm{B}}\). Thus De Morgan’s First Theorem is proved. It also says that a NOR gate is equal to a bubbled AND gate. The corresponding logic circuit diagram is shown in figure.

De Morgan’s Second Theorem:
The second theorem states that the complement of the product of two inputs is equal to the sum of its complements.

Proof
The Boolean equation for NAND gate is Y = \(\overline{\mathrm{AB}}\)
The Boolean equation for bubbled OR gate is Y = \(\overline{\mathrm{A}}+\overline{\mathrm{B}}\) . A and B are the inputs and Y is the output. The above two equations produces the same output for the same inputs. It can be verified by using the truth table.

From the above truth table we can conclude \(\overline{\mathrm{A} . \mathrm{B}}=\overline{\mathrm{A}}+\overline{\mathrm{B}}\). Thus De Morgan’s Second Theorem is proved. It also says, a NAND gate is equal to a bubbled OR gate. The corresponding logic circuit diagram is shown in figure.

Question 38.
(a) What is modulation? Explain the types of modulation with necessary diagrams.
Answer:
Modulation: For long distance transmission, the low frequency baseband signal (input signal) is superimposed onto a high frequency radio signal by a process called modulation.
There are 3 types of modulation based on which parameter is modified. They are
(i) Amplitude modulation,
(ii) Frequency modulation, and
(iii) Phase modulation.

(i) Amplitude Modulation (AM):
If the amplitude of the carrier signal is modified according to the instantaneous amplitude of the baseband signal, then it is called amplitude modulation. Here the frequency and the phase of the carrier signal remain constant. Amplitude modulation is used in radio and TV broadcasting.

The signal shown in figure (a) is the message signal or baseband signal that carries information, figure (b) shows the high frequency carrier signal and figure (c) gives the amplitude modulated signal. We can see clearly that the carrier wave is modified in proportion to the amplitude of the baseband signal.
Amplitude Modulation

(ii) Frequency Modulation (FM):
The frequency of the camer signal is modified according to the instantaneous amplitude of the baseband signal in frequency modulation. Here the amplitude and the phase of the carrier signal remain constant. Increase in the amplitude of the baseband signal increases the frequency of the carrier signal and vice versa. This leads to compressions and rarefactions in the frequency spectrum of the modulated wave. Louder signal leads to compressions and relatively weaker signals to rarefactions. When the amplitude of the baseband signal is zero in Figure (a), the frequency of the modulated signal is the same as the carrier signal. The frequency of the modulated wave increases when the amplitude of the baseband signal increases in the positive direction
(A, C). The increase in amplitude in the negative half cycle (B,D) reduces the frequency of the modulated wave (figure (c)).

(iii) Phase Modulation (PM):
The instantaneous amplitude of the baseband signal modifies the phase of the camer signal keeping the amplitude and frequency constant is called phase modulation. This modulation is used to generate frequency modulated signals. It is similar to frequency modulation except that the phase of the carrier is varied instead of varying frequency. The carrier phase changes according to increase or decrease in the amplitude of the baseband signal. When the modulating signal goes positive, the amount of phase
lead increases with the amplitude of the modulating signal. Due to this, the carrier signal is compressed or its frequency is increased.

On the other hand, the negative half cycle of the baseband signal produces a phase lag in the carrier signal. This appears to have stretched the frequency of the carrier wave. Hence similar to frequency modulated wave, phase modulated wave also comprises of
compressions and rarefactions. When the signal voltage is zero (A, C and E) the carrier frequency is unchanged.

[OR]

Question 38.
(b) Elaborate any two types of Robots with relevant examples.
Answer:
(i) Human Robot:
Certain robots are made to resemble humans in appearance and replicate the human activities like walking, lifting, and sensing, etc.

• Power conversion unit: Robots are powered by batteries, solar power, and hydraulics.
• Actuators: Converts energy into movement. The majority of the actuators produce rotational or linear motion.
• Electric motors: They are used to actuate the parts of the robots like wheels, arms, fingers, legs, sensors, camera, weapon systems etc. Different types of electric motors are used. The most often used ones are AC motor, Brushed DC motor, Brushless DC motor, Geared DC motor, etc.
• Pneumatic Air Muscles: They are devices that can contract and expand when air is pumped inside. It can replicate the function of a human muscle. They contract almost 40% when the air is sucked inside them.
• Muscle wires: They are thin strands of wire made of shape memory alloys. They can contract by 5% when electric current is passed through them.
• Piezo Motors and Ultrasonic Motors: Basically, we use it for industrial robots.
• Sensors: Generally used in task environments as it provides information of real-time knowledge.
• Robot locomotion: Provides the types of movements to a robot. The different types are (a) Legged (b) Wheeled (c) Combination of Legged and Wheeled Locomotion (d) Tracked slip/skid

(ii) Industrial Robots:
Six main types of industrial robots:

1. Cartesian
2. SCARA (Selective Compliance Assembly Robot Arm)
3. Cylindrical
4. Delta
5. Polar
6. Vertically articulated

Six-axis robots are ideal for:

• Arc Welding
• Spot Welding
• Material Handling
• Machine Tending
• Other Applications