Category: Class 12

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

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• 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
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

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

Question 1.
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 2.
The electric field in the region between two concentric charged spherical shells
(a) is zero
(b) increases with distance from centre
(c) is constant
(d) decreases with distance from centre
Answer:
(d) decreases with distance from centre

Question 3.
In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60W bulb for use in India is R, the resistance of a 60W bulb for use in USA will be …………
(a) R
(b) 2R
(c) \(\frac{\mathbf{R}}{4}\)
(d) \(\frac{\mathbf{R}}{2}\)
Answer:
(c) \(\frac{\mathbf{R}}{4}\)

Question 4.
The vertical component of Earth’s magnetic field at a place is equal to the horizontal component. What is the value of angle of dip at this place?
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Answer:
(b) 45°

Question 5.
A step-down transformer reduces the supply voltage from 220 V to 11 V and increase the current from 6 A to 100 A. Then its efficiency is
(a) 1.2
(b) 0.83
(c) 0.12
(d) 0.9
Answer:
(b) 0.83

Question 6.
Faraday’s law of electromagnetic induction is related to the …………….
(a) Law of conservation of charge
(b) Law of conservation of energy
(c) Third law of motion
(d) Law of conservation of angular momentum
Answer:
(b) Law of conservation of energy

Question 7.
The dimension of \(\frac{1}{\mu_{0} \varepsilon_{0}}\) is ……….
(a) [L T^-1]
(b) [L² T^-2]
(c) [L^-1 T]
(d) [L^-2 T²]
Answer:
(b) [L² T^-2]

Question 8.
In a Young’s double-slit experiment, the slit separation is doubled. To maintain the same fringe spacing on the screen, the screen-to-slit distance D must be changed to,…………..
(a) 2D
(b) \(\frac{\mathrm{D}}{2}\)
(c) \(\sqrt{2} \mathrm{D}\)
(d) \(\frac{\mathrm{D}}{\sqrt{2}}\)
Answer:
(a) 2D

Question 9.
The speed of light in an isotropic medium depends on,
(a) its intensity
(b) its wavelength
(c) the nature of propagation
(d) the motion of the source w.r.to medium
Answer:
(b) its wavelength

Question 10.
The mean wavelength of light from sun is taken to be 550 nm and its mean power is 3.8 x 10²⁶ W. The number of photons received by the human eye per second on the average from sunlight is of the order of ……………..
(a) 10⁴⁵
(b) 10⁴²
(c) 10⁵⁴
(d) 10⁵¹
Answer:
(a) 10⁴⁵
Hint:

Question 11.
As the intensity of incident light increases
(a) kinetic energy of emitted photoelectrons increases
(b) photoelectric current decreases
(c) photoelectric current increases
(d) kinetic energy of emitted photoelectrons decreases
Answer:
(c) photoelectric current increases
Hint: As the intensity of incident light increases, photoelectric current increases.

Question 12.
In J.J. Thomson e/m experiment, a beam of electron is replaced by that of muons (particle with same charge as that of electrons but mass 208 times that of electrons). No deflection condition is achieved only if
(a) B is increased by 208 times
(b) B is decreased by 208 times
(c) B is increased by 14.4 times
(d) B is decreased by 14.4 times
Answer:
(c) B is increased by 14.4 times
Hint: In the condition of no deflection \(\frac{e}{m}=\frac{\mathrm{E}^{2}}{2 v \mathrm{B}^{2}}\)
If m is increased by 208 times then B should be increased \(\sqrt{208}\) = 14.4 times

Question 13.
A forward biased diode is treated as
(a) An open switch with infinite resistance
(b) A closed switch with a voltage drop of 0V
(c) A closed switch in series with a battery voltage of 0.7V
(d) A closed switch in series with a small resistance and a battery.
Answer:
(d) A closed switch in series with a small resistance and a battery.

Question 14.
The variation of frequency of carrier wave with respect to the amplitude of the modulating
signal is called …………
(a) Amplitude modulation
(b) Frequency modulation
(c) Phase modulation
(d) Pulse width modulation
Answer:
(b) Frequency modulation

Question 15.
Who is the father of the modem robotics industry formed the world’s first robotic company in 1956?
(a) Joliot
(b) Cormark
(c) Engelberger
(d) Edward purcell
Answer:
(c) Engelberger

Part – II

Answer any six questions in which Q. No 17 is compulsory.

Question 16.
What is the general definition of electric dipole moment?
Answer:
The electric dipole moment vector lies along the line joining two charges and is directed from  – q to +q. The SI unit of dipole moment is coulomb meter (cm).
\(\vec{p}=q a \hat{i}-q a(-\hat{i})=2 q a \hat{i}\)

Question 17.
Calculate the electric flux through the rectangle of sides 5 cm and 10 cm kept in the region of a uniform electric field 100 NC^-1. The angle 0 is 60°. Suppose 0 becomes zero, what is the electric flux?
Answer:

Question 18.
What is meant by hysteresis?
Answer:
The phenomenon of lagging of magnetic induction behind the magnetising field is called hysteresis. Hysteresis means Tagging behind’.

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

Question 20.
What is Snell’s window?
Answer:
When light entering the water from outside is seen from inside the water, the view is restricted to a particular angle equal to the critical angle i_c. The restricted illuminated circular area is called Snell’s window.

Question 21.
Define work function of a metal. Give its unit.
Answer:
The minimum energy needed for an electron to escape from the metal surface is called work function of that metal. It’s unit is electron volt (eV).

Question 22.
What is meant by radioactivity?
Answer:
The phenomenon of spontaneous emission of highly penetrating radiations such as α, β and γ rays by an element is called radioactivity.

Question 23.
What is the angular momentum of an electron in the third orbit of an atom?
Answer:
Here n = 3; h = 6.6 x 10^-34 Js
Angular momentum.

Question 24.
Explain the current flow in a NPN transistor
Answer:

• The conventional flow of current is based on the direction of the motion of holes
• In NPN transistor, current enters from the base into the emitter.

Part – III

Answer any six questions in which Q.No. 27 is compulsory. [6×3 = 18]

Question 25.
Define ‘capacitance’. Give its unit.
Answer:
The capacitance C of a capacitor is defined as the ratio of the magnitude of charge on either of the conductor plates to the potential difference existing between the conductors.
\(C=\frac{Q}{V} \text { or } Q \propto V\)
The SI unit of capacitance is coulomb per volt or farad (F).

Question 26.
Distinguish between drift velocity and mobility
Answer:

SNo.	Drift Velocity	Mobility
1.	The drift velocity is the average velocity acquired by the electrons inside the conductor when it is subjected to an electric field.	Mobility of an electron is defined as the magnitude of the drift velocity per unit electric field.
2.	\(\overrightarrow{\mathrm{V}}_{d}=\vec{a} \tau\)	\(\mu=\frac{e \tau}{m} \text { or } \mu=\frac{\left|\overrightarrow{\mathrm{V}}_ {d}\right|}{\overrightarrow{\mathrm{E}}}\)
3.	It’s unit is ms-1.	It’s unit is m2 v_1 s_1

Question 27.
A coil of a tangent galvanometer of diametre 0.24 m has 100 turns. If the horizontal component of Earth’s magnetic field is 25 x 10^-6 T then, calculate the current which gives a deflection of 60°.
Answer:
The diameter of the coil is 0.24 m. Therefore, radius of the coil is 0.12 m.
Number of turns is 100 turns. Earth’s magnetic field is 25 x 10^-6 T

Question 28.
Mention the ways of producing induced emf.
Answer:
Induced emf can be produced by changing magnetic flux in any of the following ways:

• By changing the magnetic field B
• By changing the area A of the coil and
• By changing the relative orientation 0 of the coil with magnetic field

Question 29.
A 400 mH coil of negligible resistance is connected to an AC circuit in which an effective current of 6 mA is flowing. Find out the voltage across the coil if the frequency is 1000 Hz.
Answer:
L = 400 x 10^-3 H; I_eff = 6 x 10^-3A; f= 1000 Hz
Inductive reactance, X_L = Lω = L x 2πf = 2 x 3.14 x 1000 x 0.4 = 2512Ω
Voltage across L, V – IX_L = 6 x 10^-3 x 2512 =15.072 V (RMS)

Question 30.
What are the sign conventions followed for lenses?
Answer:
The sign conventions for thin lenses differ only in the signs followed for focal lengths.
(a) The sign of focal length is not decided on the direction of measurement of the focal length from the pole of the lens as they have two focal lengths, one to the left and another to the right (primary and secondary focal lengths on either side of the lens).

(b) The focal length of the thin lens is taken as positive for a converging lens and negative for a diverging lens.

Question 31.
Write down the draw backs of Bohr atom’model.
Answer:
Limitations of Bohr atom model
The following are the drawbacks of Bohr atom model

• Bohr atom model is valid only for hydrogen atom or hydrogen like-atoms but not for
  complex atoms.
• When the spectral lines are closely examined, individual lines of hydrogen spectrum is
  accompanied by a number of faint lines. These are often called fine structure. This is not explained by Bohr atom model.
• Bohr atom model fails to explain the intensity variations in the spectral lines.
• The distribution of electrons in atoms is not completely explained by Bohr atom model.

Question 32.
What do you mean by leakage current in a diode?
Answer:
The leakage current in a diode is the current that the diode will leak when a reverse voltage is applied to it. Under the reverse bias, a very small current in μA, flows across the junction. This is due to the flow of the minority charge carriers called the leakage current or reverse saturation current.

Question 33.
Distinguish between wireline and wireless communication.
Answer:

Wireline communication	Wireless communication
It is a point-to-point communication.	It is a broadcast mode communication.
It uses mediums like wires, cable and optical fibres.	It uses free space as a communication medium.
These systems cannot be used for long distance transmission as they are connected.	These systems can be used for long distance transmission.
Ex. telephone, intercom and cable TV.	Ex. mobile, radio or TV broadcasting and satellite communication.

Part – IV

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

Question 34.
(a) Derive an expression for the torque experienced by a dipole due to a uniform electric field.
Answer:
Torque experienced by an electric dipole in the uniform electric field: Consider an electric dipole of dipole moment \(\vec{p}\) placed in a uniform electric field \(\vec{E}\) whose field lines are equally spaced and point in the same direction. The charge +q will experience a force q\(\vec{E}\)  in the direction of the field and charge -q will experience a force -q \(\vec{E}\) in a direction opposite to the field. Since the external field \(\vec{E}\) is uniform, the total force acting on the dipole is zero. These two forces acting at different points will constitute a couple and the dipole experience a torque. This torque tends to rotate the dipole. (Note that electric field lines of a uniform field are equally spaced and point in the same direction). The total torque on the dipole about the point O.

Using right-hand corkscrew rule, it is found that total torque is perpendicular to the plane of the paper and is directed into it.
The magnitude of the total torque


where θ is the angle made by \(\vec{p} \text { with } \overrightarrow{\mathrm{E}}\) .
Since p = 2aq, the torque is written in terms of the vector product as \(\vec{\tau}=\vec{p} \times \vec{E}\)
The magnitude of this torque is τ = pE sin θ and is maximum
when θ =90°.
This torque tends to rotate the dipole and align it with the electric field [/latex]\overrightarrow{\mathrm{E}}[/latex] . Once \(\overrightarrow{\mathrm{p}}\) is aligned with \(\overrightarrow{\mathrm{E}}\), the total torque on the dipole becomes zero.

[OR]

(b) Explain the equivalent resistance of a parallel resistor network.
Answer:
Resistors in parallel: Resistors are in parallel when they are connected across the same potential difference as shown in fig. (a).
In this case, the total current I that leaves the battery in split into three separate paths. Let I₁, I₂ and I₃ be the current through the resistors R_{1 ,}R₂ and R₃ respectively. Due to the conservation of charge, total current in the circuit I is equal to sum of the currents through each of the three resistors.
I = I_{1 +} I₂ + I₃ ………………… (1)

Since the voltage across each resistor is the same, applying Ohm’s law to each resistor, we have

Substituting these values in equation (1) we get.

Here R_p is the equivalent resistance of the parallel combination of the resistors. Thus, when a number of resistors are connected in parallel, the sum of the reciprocal of the values of resistance of the individual resistor is equal to the reciprocal of the effective resistance of the combination as shown in the fig. (b).
Note: The value of equivalent resistance in parallel connection will be lesser than each individual resistance.

Question 35.
(a) Obtain a relation for the magnetic induction at a point along the axis of a circular coil ‘ carrying current.
Answer:
Magnetic field produced along the axis of the current carrying circular coil: Consider a current carrying circular loop of radius R and let I be the current flowing through the wire in the direction. The magnetic field at a point P on the axis of the circular coil at a distance z from its center of the coil O. It is computed by taking two diametrically opposite line elements of the coil each of length \(d \bar{l} \) at C and D Let r be the vector joining the current element (I \(d \bar{l}\)) at C to the point P.
PC = PD = r = \(r=\sqrt{\mathrm{R}^{2}+\mathrm{Z}^{2}}\)
PC = PD = r = \(r=\sqrt{\mathrm{R}^{2}+\mathrm{Z}^{2}}\)
angle ∠CPO = ∠DPO = θ
According to Biot-Savart’s law, the magnetic field at P due to the current element \(d \bar{l} \) is


The magnitude of magnetic field due to current element I \(d \bar{l} \) at C and D are equal because of equal distance from the coil. The magnetic field dB due to each current element I dl is resolved into two components; dB sin θ along y-direction and dB cos θ along the z-direction. Horizontal components of each current element cancels out while the vertical components (dB cos θ \(\hat{k}\) ) alone contribute to total magnetic field at the point P.
If we integrate \(d \bar{l} \) around the loop, \(d \bar{B} \) sweeps out a cone, then the net magnetic field \(\overrightarrow{\mathrm{B}}\) at point P is

Using Pythagorous theorem r² = R² + Z² and integrating line element from 0 to 2πR, we get

Note that the magnetic field \(\bar{B} \) points along the direction from the point O to P. Suppose if the current flows in clockwise direction, then magnetic field points in the direction from the point P to O.

[OR]

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

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

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

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

When field magnet rotates through 90°, magnetic field becomes parallel to PQRS. The induced emfs across PQ and RS would become maximum. Since they are connected in series, emfs are added up and the direction of total induced emf is given by Fleming’s right hand rule.

Care has to be taken while applying this rule; the thumb indicates the direction of the motion of the conductor with respect to field. For clockwise rotating poles, the conductor appears to be rotating anti-clockwise. Flence, thumb should point to the left. The direction of the induced emf is at right angles to the plane of the paper. For PQ, it is downwards and for RS upwards. Therefore, the current flows along PQRS. The point A in the graph represents this maximum emf.

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

On completion of 360°, the induced emf becomes zero and is represented by the point D. From the graph, it is clear that emf induced in PQRS is alternating in nature.

Therefore, when field magnet completes one rotation, induced emf in PQRS finishes one cycle. For this construction, the frequency of the induced emf depends on the speed at which the field magnet rotates.

Question 36.
(a) What is emission spectra? Give their types.
Answer:
Emission spectra: When the spectrum of self luminous source is taken, we get emission spectrum. Each source has its own characteristic emission spectrum. The emission spectrum _ can be divided into three types:

(i) Continuous emission spectra (or continuous spectra): If the light from incandescent lamp filament bulb) is allowed to pass through prism (simplest spectroscope), it splits into seven colours. Thus, it consists of wavelengths containing all the visible colours ranging from violet to red. Examples: spectrum obtained from carbon arc, incandescent solids, liquids gives continuous spectra.

(ii) Line emission spectrum (or line spectrum): Suppose light from hot gas is allowed to pass through prism, line spectrum is observed. Line spectra are also known as discontinuous spectra. The line spectra are sharp lines of definite wavelengths or frequencies. Such spectra arise due to excited atoms of elements. These lines are the characteristics of the element which means it is different for different elements. Examples: spectra of atomic hydrogen, helium, etc.

(iii) Band emission spectrum (or band spectrum): Band spectrum consists of several number of very closely spaced spectral lines which overlapped together forming specific bands which are separated by dark spaces, known as band spectra. This spectrum has a sharp edge at one end and fades out at the other end. Such spectra arise when the molecules are excited. Band spectrum is the characteristic of the molecule hence, the structure of the molecules can be studied using their band spectra. Examples, spectra of hydrogen gas, ammonia gas in the discharge tube etc.

[Or]

(b) Describe the Fizeau’s method to determine speed of light.
Answer:
Fizeau’s method to determine speed of light:
Apparatus: The light from the source S was first allowed to fall on a partially silvered glass plate G kept at an angle of 45° to the incident light from the source. The light then was allowed to pass through a rotating toothed-wheel with N teeth and N cuts of equal widths whose speed of rotation could be varied through an external mechanism.

The light passing through one cut in the wheel will, get reflected by a mirror M kept at a long distance d, about 8 km from the toothed wheel. If the toothed wheel was not rotating, the reflected light from the mirror would again pass through the same cut and reach the eyes of the observer through the partially silvered glass plate.

Working: The angular speed c rotation of the toothed wheel was increased from zero to a value co until light passing through one cut would completely be blocked by the adjacent tooth. This is ensured by the disappearance of light while looking through the partially silvered glass plate.

Expression for speed of light: The speed of light in air v is equal to the ratio of the distance the light travelled from the toothed wheel to the mirror and back 2d to the time taken t.
\(v=\frac{2 d}{t}\) ………………… (1)

The distance d is a known value from the arrangement. The time taken t for the light to travel the distance to and fro is calculated from the angular speed ω of the toothed wheel. The angular speed ω of the toothed wheel when the light disappeared for the first time is,
\(\omega=\frac{\theta}{t}\) ………………….. (2)

Here, θ is the angle between the tooth and the slot which is rotated by the toothed wheel within that time t

Rewriting the above equation for t
\(t=\frac{\pi}{\mathrm{N} \omega}\) ……………….. (3)
Substituting t from equation (3) in equation (1)
\(v=\frac{2 d}{\pi / \mathrm{N} \omega}\)
\(v=\frac{2 d \mathrm{N} \omega}{\pi}\) …………….(4)
Fizeau had some difficulty to visually estimate the minimum intensity blocked by the adjacent tooth, and his value for speed of light was very value.

Question 37.
(a) List out the laws of photoelectric effect.
Answer:
Laws of photoelectric effect:

• For a given frequency of incident light, the number of photoelectrons emitted is directly proportional to the intensity of the incident light. The saturation current is also directly proportional to the intensity of incident light.
• Maximum kinetic energy of the photo electrons is independent of intensity of the incident light.
• Maximum kinetic energy of the photo electrons from a given metal is directly proportional to the frequency of incident light.
• For a given surface, the emission of photoelectrons takes place only if the frequency of incident light is greater than a certain minimum frequency called the threshold frequency.
• There is no time lag between incidence of light and ejection of photoelectrons.

[OR]

(b) Describe the working of nuclear reactor with a block diagram.
Answer:
Nuclear reactor:

• Nuclear reactor is a system in which the nuclear fission takes place in a self-sustained controlled manner and the energy produced is used either for research purpose or for power generation.
• The main parts of a nuclear reactor are fuel, moderator and control rods. In addition to this, there is a cooling system which is connected with power generation set up.

Fuel:

• The fuel is fissionable material, usually uranium or plutonium. Naturally occurring uranium contains only \(0.7 \% \text { of }_{92}^{235} \mathrm{U} \text { and } 99.3 \% \text { are only }^{238} \mathrm{g}_{2} \mathrm{U}\) . So the \(_{ 92 }^{ 235 }{ U }\) must be enriched such that it contains at least 2 to 4% of \(_{ 92 }^{ 235 }{ U }\) .
• In addition to this, a neutron source is required to initiate the chain reaction for the first time. A mixture of beryllium with plutonium or polonium is used as the neutron source. During fission of \(_{ 92 }^{ 235 }{ U }\), only fast neutrons are emitted but the probability of initiating fission by it in another nucleus is very low. Therefore, slow neutrons are preferred for sustained nuclear reactions.

Moderators:

• The moderator is a material used to convert fast neutrons into slow neutrons. Usually the moderators are chosen in such a way that it must be very light nucleus having mass comparable to that of neutrons. Hence, these light nuclei undergo collision with fast neutrons and the speed of the neutron is reduced
• Most of the reactors use water, heavy water (D₂0) and graphite as moderators. The blocks of uranium stacked together with blocks of graphite (the moderator) to form a large pile.

 

Control rods:

• The control rods are used to adjust the reaction rate. During each fission, on an average 2.5 neutrons are emitted and in order to have the controlled chain reactions, only one neutron is allowed to cause another fission and the remaining neutrons are absorbed by the control rods.
• Usually cadmium or boron acts as control rod material and these rods are inserted into the uranium blocks. Depending on the insertion depth of control rod into the uranium, the average number of neutrons produced per fission is set to be equal to one or greater than one.
• If the average number of neutrons produced per fission is equal to one, then reactor is said to be in critical state. In fact, all the nuclear reactors are maintained in critical state by suitable adjustment of control rods. If it is greater than one, then reactor is said to be in super-critical and it may explode sooner or may cause massive destruction.

Shielding:

• For a protection against harmful radiations, the nuclear reactor is surrounded by a concrete wall of thickness of about 2 to 2.5 m.

Cooling system:

• The cooling system removes the heat generated in the reactor core. Ordinary water, heavy water and liquid sodium are used as coolant since they have very high specific heat capacity and have large boiling point under high pressure.
• This coolant passes through the fuel block and carries away the heat to the steam generator through heat exchanger. The steam runs the turbines which produces electricity in power reactors.

Question 38.
(a) Explain the working principle of a solar cell. Mention its applications.
Answer:
Solar cell:
A solar cell, also known as photovoltaic cell, converts light energy directly into electricity or electric potential difference by photovoltaic effect. It is basically a p-n junction which generates emf when solar radiation falls on the p-n junction. A solar cell is of two types: p-type and n-type.

Both types use a combination of p-type and n-type Silicon which together forms the p-n junction of the solar cell. The difference is that p-type solar cells use p-type Silicon as the base with an ultra-thin layer of n-type Silicon as shown in Figure, while n-type solar cell uses the opposite combination. The other side of the p-Silicon is coated with metal which forms the back electrical contact. On top of the n-type Silicon, metal grid is deposited which acts as the front electrical contact. The top of the solar cell is coated with anti-reflection coating and toughened glass.

In a solar cell, electron-hole pairs are generated due to the absorption of light near the junction. Then the charge carriers are separated due to the electric field of the depletion region. Electrons move towards n-type Silicon and holes move towards p-type Silicon layer.

The electrons reaching the n-side are collected by the front contact and holes reaching p-side are collected by the back electrical contact. Thus a potential difference is developed across solar cell. When an external load is connected to the solar cell, photocurrent flows through the load.

Many solar cells are connected together either in series or in parallel combination to form solar panel or module. Many solar panels are connected with each other to form solar arrays. For  high power applications, solar panels and solar arrays are used.

Applications:

• Solar cells are widely used in calculators, watches, toys, portable power supplies, etc.
• Solar cells are used in satellites and space applications
• Solar panels are used to generate electricity.

(b) Elaborate on the basic elements of communication system with the necessary block diagram.
Answer:
Elements of an electronic communication system
Information (Baseband or input signal): Information can be in the form of a sound signal like speech, music, pictures, or computer data which is given as input to the input transducer.

Input transducer
A transducer is a device that converts variations in a physical quantity (pressure, temperature, sound) into an equivalent electrical signal or vice versa. In communication system, the transducer converts the information which is in the form of sound, music, pictures or computer data into corresponding electrical signals. The electrical equivalent of the original information is called the baseband signal. The best example for the transducer is the microphone that converts sound energy into electrical energy.

Transmitter:
It feeds the electrical signal from the transducer to the communication channel. It consists of circuits such as amplifier, oscillator, modulator, and power amplifier. The transmitter is located at the broadcasting station.

Amplifier:
The transducer output is very weak and is amplified by the amplifier. Oscillator: It generates high-frequency carrier wave (a sinusoidal wave) for long distance transmission into space. As the energy of a wave is proportional to its frequency, the carrier wave has very high energy.

Modulator:
It superimposes the baseband signal onto the carrier signal and generates the modulated signal.

Power amplifier:
It increases the power level of the electrical signal in order to cover a large distance.

Transmitting antenna:
It radiates the radio signal into space in all directions. It travels in the form of electromagnetic waves with the velocity of light (3 ^x 10⁸ ms^-1).

Communication channel:
Communication channel is used to carry the electrical signal from transmitter to receiver with less noise or distortion. The communication medium is basically of two types: wireline communication and wireless communication.

Noise:
It is the undesirable electrical signal that interfaces with the transmitted signal. Noise attenuates or reduces the quality of the transmitted signal. It may be man-made (automobiles, welding machines, electric motors etc .) or natural (lightening, radiation from sun and stars and environmental effects). Noise cannot be completely eliminated. However, it can be reduced using various techniques.

Receiver:
The signals that are transmitted through the communication medium are received with the help of a receiving antenna and are fed into the receiver. The receiver consists of electronic circuits like demodulator, amplifier, detector etc. The demodulator extracts the baseband signal from the carrier signal. Then the baseband signal is detected and amplified using amplifiers. Finally, it is fed to the output transducer.

Repeaters:
Repeaters are used to increase the range or distance through which the signals are sent. It is a combination of transmitter and receiver. The signals are received, amplified, and retransmitted with a carrier signal of different frequency to the destination. The best example is the communication satellite in space.

Output transducer:
It converts the electrical signal back to its original form such as sound, music, pictures or data. Examples of output transducers are loudspeakers, picture tubes, computer monitor, etc.

Attenuation:
The loss of strength of a signal while propagating through a medium is known as attenuation.

Range:
It is the maximum distance between the source and the destination up to which the signal is received with sufficient strength.

Students can Download Tamil Nadu 12th Maths Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Maths Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

TN State Board 12th Maths Model Question Paper 4 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 A^T A^-1 is symmetric, then A² = _______
(a) A^-1
(b) (A^T)²
(c) A^T
(d) (A^-1)²
Answer:
(b) (AT)2

Question 2.
If p + iq = \(\frac{a+i b}{a-i b}\), then p² + q² = ________.
(a) 0
(b) 2
(c) 1
(d) -1
Answer:
(c) 1

Question 3.
If ω ≠ 1 is a cubic root of unity and \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -\omega^{2}-1 & \omega^{2} \\
1 & \omega^{2} & \omega^{7}
\end{array}\right|\) = 3k, then k is equal to _______.
(a) 1
(b) -l
(c) \(\sqrt{3} i\)
(d) \(-\sqrt{3} i\)
Answer:
(d) \(-\sqrt{3} i\)

Question 4.
The value of sin^-1 (cos x), 0 ≤ x ≤ π is _______.
(a) π – x
(b) x – \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{2} – x\)
(d) π – x
Answer:
(c) \(\frac{\pi}{2} – x\)

Question 5.
The radius of the circle 3x² + by² + 4bx – 6by + b² = 0 is ________.
(a) 1
(b) 3
(c) \(\sqrt{10}\)
(d) \(\sqrt{11}\)
Answer:
(c) \(\sqrt{10}\)

Question 6.
The equation of the directrix of the parabola y² = -8x is ______.
(a) y + 2 = 0
(b) x – 2 = 0
(c) y – 2 = 0
(d) x + 2 = 0
Answer:
(b) x – 2 = 0

Question 7.
If \(\vec{a}\) and \(\vec{b}\) are parallel vector, then \([\vec{a}, \vec{c}, \vec{b}]\) is equal to _____
(a) 2
(b) -1
(c) 1
(d) 0
Answer:
(d) 0

Question 8.
The length of the perpendicular from the origin to the plane \(\vec{r} \cdot(3 \vec{i}+4 \vec{j}+12 \vec{k})=26\) is _______.
(a) 26
(b) \(\frac{26}{169}\)
(c) 2
(d) \(\frac{1}{2}\)
Answer:
(c) 2

Question 9.
The curve y = ax⁴ + bx² with ab > 0
(a) has no horizontal tangent
(b) is concave up
(c) is concave down
(d) has no points of inflection
Answer:
(d) has no points of inflection

Question 10.
The asymptote to the curve y² (1 + x) = x² (1 – x) is _______.
(a) x = 1
(b) y = 1
(c) y = -1
(d) x = -1
Answer:
(d) x = -1

Question 11.
If f(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 12.
If f(x, y) = e^xy , then \(\frac{\partial^{2} f}{\partial x \partial y}\) is equal to ________.
(a) xye^xy
(b) (1 + xy) e^xy
(c) (1 + y) e^xy
(d) (1 + x) e^xy
Answer:
(b) (1 + xy) e^xy

Question 13.
The value of \(\int_{0}^{\frac{\pi}{6}} \cos ^{3} 3 x d x\) is _______.
(a) \(\frac{2}{3}\)
(b) \(\frac{2}{9}\)
( c) \(\frac{1}{9}\)
(d) \(\frac{1}{3}\)
Answer:
(b) \(\frac{2}{9}\)

Question 14.
If f(x) is even then \(\int_{-a}^{a} f(x) d x \) _______.

Answer:
(b) \(2 \int_{0}^{a} f(x) d x\)

Question 15.
The order and degree of the differential equation \(\sqrt{\sin x}\)(dx + dy) = \(\sqrt{\sin x}\) (dx- dy) is ________.
(a) 1, 2
(b) 2, 2
(c) 1, 1
(d) 2, 1
Answer:
(c) 1, 1

Question 16.
The solution of the differential equation \(\frac{d y}{d x}\) = 2xy is _______.
(a) y = c e^x2
(b) y = 2x² + c
(c) = ce^-x2 + c
(d) y = x² + c
Answer:
(a) y = c e^x2

Question 17.
If P{X = 0} = 1 – P{X = 1}. If E[X] = 3Var(X), then P{X = 0} ________.
(a) \(\frac{2}{3}\)
(b) \(\frac{2}{5}\)
(c) \(\frac{1}{5}\)
(d) \(\frac{1}{3}\)
Answer:
(d) \(\frac{1}{3}\)

Question 18.
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed n times. Then the possible values of X are _________.
(a) i + 2n, i = 0, 1, 2 … n
(b) 2i – n, i = 0, 1, 2 … n
(c) n – i, i = 0, 1, 2 … n
(d) 2i + 2n, i = 0, 1, 2 … n
Answer:
(b) 2i – n, i = 0, 1, 2 … n

Question 19.
In the set Q define a Θ b= a + b + ab. For what value of y, 3 Θ (y Θ 5) = 7 ?

Answer:
(b) y = \(\frac{-2}{3}\)

Question 20.
If X is a continuous random variable then P(X > a) =
(a) P (X < a)
(b) 1 – P (X > a)
(c) P (X > a)
(d) 1 – P (x ≥ a)
Answer:
(c) P (X > a)

Part – II

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

Question 21.
Reduce the matrix \(\left[\begin{array}{ccc}
3 & -1 & 2 \\
-6 & 2 & 4 \\
-3 & 1 & 2
\end{array}\right]\) to a row-echelon form.
Answer:

Question 22.
Find the least positive integer n such that \(\left(\frac{1+i}{1-i}\right)^{n}=1\)
Answer:

Question 23.
Find the value of \(\sin ^{-1}\left(\sin \left(\frac{5 \pi}{4}\right)\right)\)
Answer:

Question 24.
Identify the type of conic section for the equation 3x² + 3y² – 4x + 3y + 10 = 0
Answer:
Comparing this equation with the general equation of the conic
Ax² + Bxy + cy² + Dx + Ey +F = 0
We get A = C also B = 0
So the given conic is a circle.

Question 25.
If U(x, y, z) = log(x³ + y³ + z³), find \(\frac{\partial \mathrm{U}}{\partial x}+\frac{\partial \mathrm{U}}{\partial y}\) and \(\frac{\partial U}{\partial z}\)
Answer:

Question 26.
Find, by integration, the volume of the solid generated by revolving about the x-axis, the region enclosed by y = 2x², y = 0 and x = 1.
Answer:

Question 27.
Solve the differential equation: \(\frac{d y}{d x}-x \sqrt{25-x^{2}}=0\)
Answer:

Question 28.
Three fair coins are tossed simultaneously. Find the probability mass function for number of heads occurred.
Answer:
When three coins are tossed, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
‘X’ is the random variable denotes the number of heads.
∴ ‘X’ can take the values of 0, 1, 2 and 3
Hence, the probabilities
P(X = 0) = P (No heads) = \(\frac{1}{8}\);
P(X = 1) = P (1 head) = \(\frac{3}{8}\);
P(X = 2) = P (2 heads) = \(\frac{3}{8}\);
P(X = 3) = P (3 heads)= \(\frac{1}{8}\);
∴ The probability mass function is
\(f(x)=\left\{\begin{array}{lll}
1 / 8 & \text { for } & x=0,3 \\
3 / 8 & \text { for } & x=1,2
\end{array}\right.\)

Question 29.
Construct the truth table for the following statements. \(\neg p \wedge \neg q\)
Answer:
Truth table for \(\neg p \wedge \neg q\)

Question 30.
Write the Maclaurin series expansion of the function: ex
Answer:
f (x) = e^x; f (0) = e⁰ = 1
f’ (x) = e^x; f’ (0) = 1
f”(x) = e^x; f”(0) = 1

Part – III

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

Question 31.
If A = \(\left[\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right]\)show that A^-1 = \(\frac{1}{2}\) (A² – 3I).

Question 32.
Find the values of the real numbers x and y, if the complex numbers.
(3 – i)x – (2 – i) y + 2i + 5 and 2x + (-1 + 2i) y + 3 + 2i are equal.

Question 33.
It is known that the roots of the equation x³ – 6x² – 4x + 24 = 0 are in arithmetic progression. Find its roots.

Question 34.
Prove that: \(\cos \left[\tan ^{-1}\left\{\sin \left(\cot ^{-1} x\right)\right\}\right]=\sqrt{\frac{x^{2}+1}{x^{2}+2}}\)

Question 35.
Find the equation of a circle of radius 5 whose centre lies on x-axis and which passes through the point (2, 3).

Question 36.
Using the l’ Hopital Rule prove that, \(\lim _{x \rightarrow 0^{+}}(1+x)^{\frac{1}{x}}=e\)

Question 37.
If v(x, y) = x² – xy + \(\frac{1}{4}\) y² + 7, x, y ∈ R, find the differential dv.

Question 38.
Find the area of the region bounded by 2x – y + 1 =0, y = – 1, y = 3 and y-axis..

Question 39.
Solve: \(\frac{d y}{d x}\) + 2y cot x = 3x² cosec²x

Question 40.
If the straight lines \(\frac{x-5}{5 m+2}=\frac{2-y}{5}=\frac{1-z}{-1}\) and x = \(\frac{2 y+1}{4 m}=\frac{1-z}{-3}\) are perpendicular to each other, find the value of m.

Part – IV

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

Question 41.
(a) Investigate the values of X and p the system of linear equations.
2x + 3y + 5z = 9, 7x + 3y – 5z = 8, 2x + 3y + λz = µ, have
(i) no solution (ii) a unique solution (iii) an infinite number of solutions.
[OR]
(b) If z(x, y) = x tan^-1 (xy), x = t², y = s e^t, s, t ∈ R, Find \(\frac{\partial z}{\partial t}\) and \(\frac{\partial z}{\partial t}\) at s = t = 1.

Question 42.
(a) Form the equation whose roots are the squares of the roots of the cubic equation
x³ + ax² + bx + c = 0.
[OR]
(b) Find the intervals of concavity and the points of inflection of the function.
f(θ) = sin 2θ in (0, π)

Question 43.
(a) If a = cos 2α + i sin 2α, b = cos 2β + i sin 2β and c = cos 2γ + i sin 2γ, prove that.

(b) A closed (cuboid) box with a square base is to have a volume of 2000 c.c. The material for the top and bottom of the box is to cost Rs. 3 per square cm and the material for the sides is to cost Rs. 1.50 per square cm. If the cost of the materials is to be the least, find the dimensions of the box.

Question 44.
(a) Prove that a straight line and parabola cannot intersect at more than two points.
[OR]
(b) Solve \(\left(y-e^{\sin ^{-1} x}\right) \frac{d x}{d y}+\sqrt{1-x^{2}}=0\)

Question 45.
(a) Solve \(\tan ^{-1}\left(\frac{x-1}{x-2}\right)+\tan ^{-1}\left(\frac{x+1}{x+2}\right)=\frac{\pi}{4}\)
[OR]
(b) Show that the lines \(\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}\) and \(\frac{x-4}{2}=\frac{y}{0}=\frac{z+1}{3}\) the point of intersection.

Question 46.
(a) A tank initially contains 50 liters of pure water. Starting at time t = 0 a brine containing with 2 grams of dissolved salt per litre flows into the tank at the rate of 3 liters per minute. The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. Find the amount of salt present in the tank at any time t > 0.
[OR]
(b) If X ~ B(n, p) such that 4P (X = 4) = P (x = 2) and n = 6 . Find the distribution, mean and standard deviation.

Question 47.
(a) Find the centre, foci, and eccentricity of the hyperbola 11x² – 25y² – 44x + 50y – 256 = 0
[OR]
(b) Verify (i) closure property (ii) commutative property (iii) associative property (iv) existence of identity and (v) existence of inverse for the operation +5 on Z5 using table corresponding to addition modulo 5.

Students can Download Tamil Nadu 12th Physics Model Question Paper 4 English Medium Pdf, Tamil Nadu 12th Physics Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• 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
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

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

Question 1.
Which charge configuration produces a uniform electric field?
(a) point charge
(b) infinite uniform line charge
(c) uniformly charged infinite plane
(d) uniformly charged spherical shell
Answer:
(c) uniformly charged infinite plane

Question 2.
The work done in carrying a charge Q, once round a circle of radius R with a charge Q₂ at the centre is  ………….
(a) \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}^{2}}\)
(b) Zero
(c)  \(\frac{\mathrm{Q}_{1} \mathrm{Q}_{2}}{4 \pi \varepsilon_{0} \mathrm{R}} \)
(d) infinite
Answer:
(b) Zero
Hint: The electric field is conservative. Therefore, no work is done in moving a charge around a closed path in a electric field.

Question 3.
The internal resistance of a 2.1 V cell which gives a current of 0.2 A through a resistance of 10 Ω is……………………..
(a) 0.2 Ω
(b) 0.5 Ω
(c) 0.8 Ω
(d) 1.0 Ω
Answer:
(b) 0.5 Ω

Question 4.
An electron moves straight inside a charged parallel plate capacitor of uniform charge density σ. The time taken by the electron to cross the parallel plate capacitor when the plates of the capacitor are kept under constant magnetic field of induction \(\overrightarrow{\mathrm{B}}\) is ……….


Answer:
(d)

Question 5.
In a series resonant RLC circuit, the voltage across 100Ω resistor is 40 V. The resonant frequency co is 250 rad/s. If the value of C is 4 pF, then the voltage across L is……………….
(a) 600 V
(b) 4000 V
(c) 400 V
(d) IV
Answer:
(c) 400 V

Question 6.
During the propagation of electromagnetic waves in a medium:
(a) electric energy density is double of the magnetic energy density
(b) electric energy density is half of the magnetic energy density
(c) electric energy density is equal to the magnetic energy density
(d) both electric and magnetic energy densities are zero
Answer:
(c) electric energy density is equal to the magnetic energy density

Question 7.
First diffraction minimum due to a single slit of width 1.0 x 10^-5 cm is at 30°. Then wavelength of light used is, …………………
(a) 400 Å
(b) 500 Å
(c) 600 Å
(d) 700 Å
Answer:
(b) 500 Å
Hint. For diffraction minima, d sin θ = nλ
\(\lambda=\frac{d \sin \theta}{n}=\frac{1 \times 10^{-5} \times 10^{-2} \times \sin 30^{\circ}}{1}=0.5 \times 10^{-7}\)
λ = 500 Å

Question 8.
The sky would appear red instead of blue if
(a) atmospheric particles scatter blue light more than red light
(b) atmospheric particles scatter all colours equally
(c) atmospheric particle scatter red light more than blue light
(d) the sun was much hotter
Answer:
(c) atmospheric particle scatter red light more than blue light

Question 9.
Kinetic energy of emitted electron depends upon
(a) frequency
(b) intensity
(c) nature of atmosphere surrounding the electron
(d) none of these
Answer:
(a) frequency
Hint: Kinetic energy of emitted electron depends on the frequency of incident radiation.

Question 10.
The ratio between the first three orbits of hydrogen atom is…………….
(a) 1:2:3
(b) 2:4:6
(c) 1:4:9
(d) 1:3:5
Answer:
(c) 1:4:9
Hint :
E_n = \(\frac{-13.6 \times z^{2}}{n^{2}}\)
n = 1; E₁ =- 13.6 eV/ atom
n = 2; E₂ = – 3.4 eV/ atom
n = 3; E₃ = -1.51 eV/atom                             ’
The ratio of three orbits E₁ : E₂ : E₃ = 13.6 : 3.4 : 1.51 = 1 : 4 : 9

Question 11.
Bohr’s theory of hydrogen atom did not explain fully
(a) diameter of H-atom
(b) emission spectra
(c) ionisation energy
(d) the fine structure of even hydrogen spectrum
Answer:
(d) the fine structure of even hydrogen spectrum
Hint: Bohr theory could not explain the five structure of hydrogen spectrum.

Question 12.
If a half-wave rectified voltage is fed to a load resistor, which part of a cycle the load current will flow?
(a) 0° – 90°
(b) 90° – 180°
(c) 0° – 180°
(d) 0° – 360°
Answer:
(c) 0° – 180°

Question 13.
Diamond is very hard because ……………..
(a) it is covalent solid
(b) it has large cohesive energy
(c) high melting point
(d) insoluble in all solvents
Answer:
(b) it has large cohesive energy

Question 14.
The internationally accepted frequency deviation for the purpose of FM broadcasts.
(a) 75 kHz
(b) 68 kHz
(c) 80 kHz
(d) 70 kHz
Answer:
(a) 75 kHz

Question 15.
The blue print for making ultra durable synthetic material is mimicked from
(a) Lotus leaf
(b) Morpho butterfly
(c) Parrot fish
(d) Peacock feather
Answer:
(c) Parrot fish

Part – II

Answer any six questions in which Q. No 17 is compulsory.   [6 x 2 = 12]

Question 16.
Define ‘electric flux’.
Answer:
The number of electric field lines crossing a given area kept normal to the electric field lines  is called electric flux. Its unit is N m² C^-1. Electric flux is a scalar quantity.

Question 17.
Determine the number of electrons flowing per second through a conductor, when a current of 32 A flows through it.
Answer:
I = 32A, t= 1 s
Charge of an electron, e = 1.6 x 10^-19 C
The number of electrons flowing per second, n =?

Question 18.
Define magnetic flux.
Answer:
The number of magnetic field lines crossing per unit area is called magnetic flux φ_B
\(\phi_{\mathrm{B}}=\overrightarrow{\mathrm{B}} \cdot\overrightarrow{\mathrm{A}}=\mathrm{B} \mathrm{A} \cos \theta=\mathrm{B} \perp \mathrm{A}\)

Question 19.
Give any one definition of power factor.
Answer:
The power factor is defined as the ratio of true power to the apparent power of an a.c. circuit.
It is equal to the cosine of the phase angle between current and voltage in the a.c. circuit.

Question 20.
State the laws of reflection.
Answer:

• The incident ray, reflected ray and normal to the reflecting surface all are coplanar (ie. lie in the same plane).
• The angle of incidence i is equal to the angle of reflection r. i = r

Question 21.
Why do metals have a large number of free electrons?
Answer:
In metals, the electrons in the outer most shells are loosely bound to the nucleus. Even at room temperature, there are a large number of free electrons which are moving inside the metal in a random manner.

Question 22.
The radius of the 5th orbit of hydrogen atom is 13.25 A. Calculate the wavelength of the electron in the 5th orbit.
Answer:
2πr = nλ
2 x 3.14 x 13.25 Å = 5 x λ
.’. λ = 16.64 Å

Question 23.
Draw the output waveform of a full wave rectifier.
Answer:

Question 24.
Explain centre frequency or resting frequency in frequency modulation.
Answer:
When the frequency of the baseband signal is zero (no input signal), there is no change in the frequency of the carrier wave. It is at its normal frequency and is called as centre frequency or resting frequency.

Part – III

Answer any six questions in which Q.No. 28 is compulsory. [6 x 3 = 18]

Question 25.
What is corona discharge?
Answer:
The electric field near the edge is very high and it ionizes the surrounding air. The positive ions are repelled at the sharp edge and negative ions are attracted towards the sharper edge.
This reduces the total charge of the conductor near the sharp edge. This is called action at points or corona discharge.

Question 26.
What is electric power and electric energy?
Answer:
Electric power: It is the rate at which an electric appliance converts electric energy into other forms of energy. Or, it is the rate at which work is done by a source of emf in maintaining an electric current through a circuit.
\({ P }=\frac { { W } }{ t } ={ VI }={ I }^{ 2 }{ R }=\frac { { V }^{ 2 } }{ { R } } \)
Electric energy: It is the total workdone in maintaining an electric current in an electric circuit for a given time.
W = Vt = VIr joule = I²R? joule.

Question 27.
A bar magnet having a magnetic moment \(\overrightarrow{\mathrm{M}}\) is cut into four pieces i.e., first cut in two pieces along the axis of the magnet and each piece is further cut into two pieces. Compute the magnetic moment of each piece.
Answer:
Consider a bar magnet of magnetic moment \(\overrightarrow{\mathrm{M}}\). When a bar magnet first cut in two pieces along the axis, their magnetic moment is \(\frac{\overrightarrow{\mathrm{M}}}{2}\)

Question 28.
The current in an inductive circuit is given by 0.3 sin (200t – 40°) A. Write the equation for the voltage across it if the inductance is 40 mH.
Answer:
L = 40 x 10^-3H; i = 0.1 sin (200 t – 40°); X_L = ωL = 200 x 40 x 10^-3 = 8 Ω
V_m = I_m X_L = 0.3 x 8 = 2.4 V
In an inductive circuit, the voltage leads the current by 90° Therefore,
υ = V_m sin (ωt + 90°)
υ = 2.4 sin (200f – 40°+ 90°)
υ = 2.4 sin (200f +50°)volt

Question 29.
Writedown the integral form of modified Ampere’s circuital law.
Answer:
This law relates the magnetic field around any closed path to the conduction current and displacement current through that path.

Question 30.
Two light sources have intensity of light as I_o. 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 I_o
The resultant intensity is, I = 4 I_o cos² (φ/2)
Resultant intensity when, ϕ = π/ 3, is I = 4I₀ cos² (π / 6) I
= 4I₀(√3 / 2)² =3I₀

Question 31.
Write the properties of cathode rays.
Answer:

• Cathode rays possess energy and momentum and travel in a straight line with high speed of the order of 10⁷m s^-1
• It can be deflected by application of electric and magnetic fields.
• When the cathode rays are allowed to fall on matter, they produce heat.
• They affect the photographic plates and also produce fluorescence when they fall on   certain crystals and minerals.
• When the cathode rays fall on a material of high atomic weight, x-rays are produced.
• Cathode rays ionize the gas through which they pass.
• The speed of cathode rays is up \(\left( \frac { 1 }{ 10 } \right) ^{ th }\) of the speed of light.

Question 32.
Distinguish between wireline and wireless communication.
Answer:

Wireline communication	Wireless communication
It is a point-to-point communication.	It is a broadcast mode communication.
It uses mediums like wires, cable and optical fibres.	It uses free space as a communication medium.
These systems cannot be used for long distance transmission as they are connected.	These systems can be used for long distance transmission.
Ex. telephone, intercom and cable TV.	Ex. mobile, radio or TV broadcasting and satellite communication.

Question 33.
What is the difference between Nano materials and Bulk materials?
Answer:

• The solids are made up of particles. Each of the particle has a definite number of atoms, which might differ from material to material. If the particle of a solid is of size less than 100 nm, it is said to be a ‘nano solid’.
• When the particle size exceeds 100 nm, it is a ‘bulk solid’. It is to be noted that nano and bulk solids may be of the same chemical composition.
• For example, ZnO can be both in bulk and nano form.
• Though chemical composition is the; same, nano form of the material shows strikingly different properties when compared to its bulk counterpart.

Part – IV

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

Question 34.
(a) Calculate the electric field due to a dipole on its equatorial plane.
Answer:
Electric field due to an electric dipole at a point on the equatorial plane. Consider a point C at a distance r from the midpoint O of the dipole on the equatorial plane as shown in Figure.

Since the point C is equi­distant from +q and -q, the magnitude of the electric fields of +q and -q are the same. The direction of \(\overrightarrow{\mathrm{E}}_{+}\) is along BC \(\overrightarrow{\mathrm{E}}_{-} \) and the direction of E is along CA. \(\overrightarrow{\mathrm{E}}_{+} \) and \(\overrightarrow{\mathrm{E}}_{-} \) and E are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it. The perpendicular components \(\left|\overrightarrow{\mathrm{E}}_{+}\right| \sin \theta \) and \(\left|\overrightarrow{\mathrm{E}}_{-}\right| \sin \theta \) are oppositely directed and cancel each other. The magnitude of the total electric field at point C is the sum of the parallel components of E₊ and E and its direction is \(\overrightarrow { { E } } _{ + }\quad and\quad \overrightarrow { { E } } _{ – }\) and its direction is along \(-\hat{p}\)

(b) How the emf of two cells are compared using potentiometer?
Answer:
Comparison of emf of two cells with a potentiometer: To compare the emf of two cells, the circuit connections are made as shown in figure. Potentiometer wire CD is connected to a battery Bf and a key K in series. This is the primary circuit. The end C of the wire is connected to the terminal M of a DPDT (Double Pole Double Throw) switch and the other terminal N is connected to a jockey through a galvanometer G and a high resistance HR. The cells whose emf ξ₁ and ξ₂ and to be compared are connected to the terminals M₁, N₁ and M₂, N₂, of the DPDT switch. The positive terminals of Bf ξ₁ and ξ₂ should be connected to the same end C.

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

Question 35.
(a) Discuss the working of cyclotron in detail.
Answer:
Cyclotron : Cyclotron is a device used 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 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.

\(\frac { mv^{ 2 } }{ r } =qv{ B }\Rightarrow r=\frac { m }{ q{ B } } v\Rightarrow r\propto v\)
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 Very important condition in cyclotron operation is the resonance condition. It happens when the frequency { 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
\(f_{\text {osc }}=\frac{q \mathrm{B}}{2 \pi m} \Rightarrow \mathrm{T}=\frac{1}{f_{\text {osc }}}\)
The time Period of oscillation is \(\mathrm{T}=\frac{2 \pi m}{q \mathrm{B}}\)
The kinetic energy of the charged particle is \(\mathrm{KE}=\frac{1}{2} m v^{2}=\frac{q^{2} \mathrm{B}^{2} r^{2}}{2 m}\)

Limitations of cyclotron

• the speed of the ion is limited
• electron cannot be accelerated
• uncharged particles cannot be accelerated

[OR]

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

(1) Induction stove : Induction stove is used to cook the food quickly and safely with less energy consumption. Below the cooking zone, there is a tightly wound coil of insulated wire. The cooking pan made of suitable material, is placed over the cooking zone. When the stove is switched on, an alternating current flowing in the coil produces high frequency alternating magnetic field which induces very strong eddy currents in the cooking pan. The eddy currents in the pan produce so much of heat due to Joule heating which is used to cook the food.

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

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

(3) Eddy current testing : It is one of the simple non-destructive testing methods to find defects like surface cracks and air bubbles present in a specimen. A coil of insulated wire is given an alternating electric current so that it produces an alternating magnetic field. When this coil is brought near the test surface, eddy current is induced in the test surface. The presence of defects causes the change in phase and amplitude of the eddy current that can be detected by some other means. In this way, the defects present in the specimen are identified.

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

Question 36.
(a) Write down the properties of electromagnetic waves.
Answer:
Properties of electromagnetic waves:

• Electromagnetic waves are produced by any accelerated charge.
• Electromagnetic waves do not require any medium for propagation. So electromagnetic wave is a non-mechanical wave.
• 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.
• Electromagnetic waves travel with speed which is equal to the speed of light in vacuum
  or free space, \(c=\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\mu _{ 0 } } } =3\times 10^{ 8 }{ ms }^{ -1 }\)
• The speed of electromagnetic wave is less than speed in free space or vacuum, that is, v < c. In a medium of refractive index,

[OR]

(b) Explain the Young’s double slit experimental setup and obtain the equation for path difference.
Answer:
Experimental setup
Wavefronts from S₁ and S₂ spread out and overlapping takes place to the right side of double slit. When a screen is placed at a distance of about 1 meter from the slits, alternate bright and dark fringes which are equally spaced appear on the screen. These are called interference fringes or bands.

S₁ and S₂ travel equal distances and arrive in-phase. These two waves constructively interfere and bright fringe is observed at O. This is called central bright fringe.

The fringes disappear and there is uniform illumination on the screen when one of the slits is covered. This shows clearly that the bands are due to interference.

Equation for path difference
Let d be the distance between the double slits S₁ and S₂ which act as coherent sources of wavelength λ. A screen is placed parallel to the double slit at a distance D from it. The mid-point of S₁ and S₂ is C and the midpoint of the screen O is equidistant from S₁ and S₂. P is any point at a distance y  from O.

The waves from S₁ and S₂ meet at P either inphase or out-of-phase depending upon the path difference between the two waves. The path difference S between the light waves from S₁ and S₂ to the point P is, δ = S₂P and S₁P
A perpendicular is dropped from the point S₁, to the line S₂P at M to find the path difference more precisely.
δ = S₂P – MP = S₂M
The angular position of the point P from C is θ. ∠OCP = θ.
From the geometry, the angles ∠OCP and ∠S₂S₁ M are equal.
∠OCP = ∠S₂S₁ M = θ
In right angle triangle ΔS₁S₂M, the path difference, S₂M = d sin θ
δ = d sin θ
If the angle θ is small, sin θ ≈ tan θ ≈ θ From the right angle triangle ΔOCP, tan θ = y/D. The path difference, δ =dy/D

Question  37.
(a) 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.

• 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.

[OR]

Question 37.
(b) Explain the J.J. Thomson experiment to determine the specific charge of electron.
Answer:
In 1887, J. J. Thomson made remarkable improvement in the scope of study of gases in discharge tubes. In the presence of electric and magnetic fields, the cathode rays are deflected. By the variation of electric and magnetic fields, mass normalized charge or the specific charge (charge per unit mass) of the cathode rays is measured.

A highly evacuated discharge tube is used and cathode rays (electron beam) produced at cathode are attracted towards anode disc A. Anode disc is made with pin hole in order to allow only a narrow beam of cathode rays. These cathode rays are now allowed to pass through the parallel metal plates, maintained at high voltage. Further, this gas discharge tube is kept in between pole pieces of magnet such that both electric and magnetic fields are perpendicular to each other. When the cathode rays strike the screen, they produce scintillation and hence bright spot is observed. This is achieved by coating the screen with zinc sulphide.

(i) Determination of velocity of cathode rays
For a fixed electric field between the plates, the magnetic field is adjusted such that the cathode rays (electron beam) strike at the original position O. This means that the magnitude of electric force is balanced by the magnitude of force due to magnetic field. Let e be the charge of the cathode rays, then eE = eBv
\(v=\frac{E}{B}\) …………….. (1)

(ii) Determination of specific charge
Since the cathode rays (electron beam) are accelerated from cathode to anode, the potential energy of the electron beam at the cathode is converted into kinetic energy of the electron beam at the anode. Let V be the potential difference between anode and cathode, then the potential energy is eV. Then from law of conservation of energy,
\(e \mathrm{V}=\frac{1}{2} m v^{2} \Rightarrow \frac{e}{m}=\frac{v^{2}}{2 \mathrm{V}}\)
Substituting the value of velocity from equation (1) , we get
\(\frac{e}{m}=\frac{1}{2 \mathrm{V}} \frac{\mathrm{E}^{2}}{\mathrm{B}^{2}}\) ………..(2)
Substituting the value of E ,B and V, the specific charge vam be determined as
\(\frac{e}{m}=1.7 \times 10^{11} \mathrm{Ckg}^{-1}\)

(iii) Deflection of charge only due to uniform electric field
When the magnetic field is turned off, the deflection is only due to electric field. The
deflection in vertical direction is due to the electric force.
F_e = eE ………………….. (3)
Let m be the mass of the electron and by applying Newton’s second law of motion,
acceleration of the electron is
\(a_{e}=\frac{1}{m} \mathrm{F}_{e}\) ………………… (4)
Substituting equation (4) in equation (3),
\(a_{e}=\frac{1}{m} e E=\frac{e}{m} E\)
Let y be the deviation produced from original position on the screen. Let the initial upward velocity of cathode ray be μ = 0 before entering the parallel electric plates. Let l be the time taken by the cathode rays to travel in electric field. Let l be the length of one of the plates, then the time taken is
\(t=\frac{l}{v}\) …………(5)

Hence, the deflection y’ of cathod rays is( note : u = 0 and \(a_{e}=\frac{e}{m} \mathrm{E}\))

Therefore, the deflection y on the screen is
y α y’ ⇒ y = Cy’
where C is proportionality constant which depends on the geometry of the discharge tube and substituting y’ value in equation (6), we get

Substituting the values on RHS, the value of specific charge is calculated as
e/m = 1.7 x 10¹¹Ckg^-1

Question 38.
(a) What is an LED? Give the principle of operation with a diagram.
Answer:
Light Emitting Diode (LED): LED is a p-n junction diode which emits visible or invisible light when it is forward biased. Since, electrical energy is converted into light energy, this process is also called electroluminescence. The cross-sectional view of a commercial LED is shown in figure (b). It consists of a p-layer, n-layer and a substrate. A transparent window is used to allow light to travel in the desired direction. An external resistance in series with the biasing source is required to limit the forward current through the LED. In addition, it has two leads; anode and cathode.

When the p-n junction is forward biased, the conduction band electrons on n-side and valence band holes on p-side diffuse across the junction. When they cross the junction, they become excess minority carriers (electrons in p-side and holes in n-side). These excess minority carriers recombine with oppositely charged majority carriers in the respective regions, i.e. the. electrons in the conduction band recombine with holes in the valence band as shown in the figure (c). During recombination process, energy is released in the form of light (radiative) or heat (non-radiative). For radiative recombination, a photon of energy hv is emitted. For non- radiative recombination, energy is liberated in the form of heat.

The colour of the light is determined by the energy band gap of the material. Therefore, LEDs are available in a wide range of colours such as blue (SiC), green (AlGaP) and red (GaAsP). Now a days, LED which emits white light (GalnN) is also available.

[OR]

(b) Give the applications of ICT in mining and agriculture sectors.
Answer:
(i) Agriculture
The implementation of information and communication technology (ICT) in agriculture sector enhances the productivity, improves the living standards of farmers and overcomes the Challenges and risk factors.

• ICT is widely used in increasing food productivity and farm management.
• It helps to optimize the use of water, seeds and fertilizers etc.
• Sophisticated technologies that include robots, temperature and moisture sensors, aerial images, and GPS technology can be used.
• Geographic information systems are extensively used in farming to decide the suitable place for the species to be planted.

(ii) Mining

• ICT in mining improves operational efficiency, remote monitoring and disaster locating system.
• Information and communication technology provides audio-visual warning to the trapped underground miners.
• It helps to connect remote sites.

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, helps students complete homework assignments and to score high marks in board exams.

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 A is a 3 × 3 non-singular matrix such that AA^T = A^T A and B = A^-1 A^T, then BB^T = ________.
(a) A
(b) B
(c) I₃
(d) B^T
Answer:
(c) I₃

Question 2.
The rank of the matrix \(\left[\begin{array}{cc}
7 & -1 \\
2 & 1
\end{array}\right]\) is ________.
(a) 9
(b) 2
(c) 1
(d) 5
Answer:
(b) 2

Question 3.
The value of \(\sum_{i=1}^{13}\left(i^{n}+i^{n-1}\right)\) is ________.
(a) 1 + i
(b) i
(c) 1
(d) 0
Answer:
(a) 1 + i

Question 4.
Which of the following is incorrect?
(d) Re(z) ≤ |z|
(b) Im (z) ≤ |z|
(c) \(z \bar{z}=|z|^{2}\)
(d) Re(z) ≥ |z|
Answer:
(d) Re(z) ≥ |z|

Question 5.
According to the rational root theorem, which number is not possible rational zero of 4x⁷ + 2x⁴ – 10x³ – 5?
(a) -1
(b) \(\frac{5}{4}\)
(c) \(\frac{4}{5}\)
(d)5
Answer:
(c) \(\frac{4}{5}\)

Question 6.
If \(\cot ^{-1}(\sqrt{\sin \alpha})+\tan ^{-1}(\sqrt{\sin \alpha})=u\), then cos 2u is equal to ______.
(a) tan² α
(b) o
(c) -1
(d) tan 2α
Answer:
(c) -1

Question 7.
The domain of the function defined by f(x) = sin^-1 \(\sqrt{x-1}\) is _______.
(a) [1, 2]
(b) [-1, 1]
(c) [0, 1]
(d) [-1, 0]
Answer:
(a) [1, 2]

Question 8.
The area of quadrilateral formed with foci of the hyperbolas \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1\) is ________.
(a) 4(a² + b²)
(b) 2(a² + b²)
(c) a² + b²
(d) \(\frac{1}{2}\) (a² + b²)
Answer:
(b) 2(a² + b²)

Question 9.
The directrix of the parabola x² = -4y is ________.
(a) x = 1
(b) x = 0
(c) y = 1
(d) y = 0
Answer:
(c) y = 1

Question 10.
If the line \(\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}\) lies in the plane x + 3y – αz + b = β, then (α, β) is ______
(a) (-5, 5)
(b) (-6, 7)
(c) (5, -5)
(d) (6, -7)
Answer:
(b) (-6, 7)

Question 11.
If \(\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \times \vec{b}) \times \vec{c}\) for non-coplanar vectors \(\vec{a}, \vec{b}, \vec{c}\) then ________.
(a) \(\vec{a}\) parallel to \(\vec{b}\)
(b) \(\vec{b}\) parallel to \(\vec{c}\)
(c) \(\vec{c}\) parallel to \(\vec{a}\)
(d) \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)
Answer:
(c) \(\vec{c}\) parallel to \(\vec{a}\)

Question 12.
The maximum product of two positive numbers, when their sum of the squares is 200, is ________.
(a) 100
(b) \(25 \sqrt{7}\)
(c) 28
(d) \(24 \sqrt{14}\)
Answer:
(a) 100

Question 13.
If w(x,y, z) = x² (y – z) + y² (z – x) + z² (x – y), then \(\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}+\frac{\partial w}{\partial z}\) is ________.
(a) xy + yz + zx
(b) x (y + z)
(c) y (z + x)
(d) 0
Answer:
(d) 0

Question 14.
If u (x, y) = x² + 3xy + y – 2019, then \(\left.\frac{\partial u}{\partial x}\right|_{(4,-5)}\) is equal to ______.
(a) -4
(b) -3
(c) -7
(d) 13
Answer:
(c) -7

Question 15.
The volume of solid of revolution of the region bounded by y² = x (a – x) about x-axis is ________.

Answer:
(d) \(\frac{\pi a^{3}}{6}\)

Question 16.
\(\int_{0}^{a} f(x) d x+\int_{0}^{a} f(2 a-x) d x\) = _________.

Answer:
(c) \(\int_{0}^{2 a} f(x) d x\)

Question 17.
The differential equation of the family of curves y = Ae^x + Be^-x, where A and B are arbitrary constants is _______.

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

Question 18.
The differential equation corresponding to xy = c² where c is an arbitrary constant, is ______.
(a) xy”+ x = 0
(b) y” = 0
(c) xy’ + y = 0
(d) xy”- x = 0
Answer:
(c) xy’ + y = 0

Question 19.
If \(f(x)=\left\{\begin{array}{ll}
2 x, & 0 \leq x \leq a \\
0 & , \text { otherwise }
\end{array}\right.\)
is a probability density function of a random variable, then the value of a is _________.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1

Question 20.
The proposition \(p \wedge(\neg p \vee q)\) is
(a) a tautology
(b) a contradiction
(c) logically equivalent to p ∧ q
(d) logically equivalent to p ∨ q
Answer:
(c) logically equivalent to p ∧ q

Part – II

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

Question 21.
A 12 metre tall tree was broken into two parts. It was found that the height of the part which was left standing was the cube root of the length of the part that was cut away. Formulate this into a mathematical problem to find the height of the part which was cut away.
Answer:
Let the two parts be x and (12 – x)
Given that x = \(\sqrt[3]{12-x}\)
Cubing on both side, x³ = 12 – x
x³ + x – 12 = 0

Question 22.
Find the value of \(\sin ^{-1}\left(\sin \frac{5 \pi}{9} \cos \frac{\pi}{9}+\cos \frac{5 \pi}{9} \sin \frac{\pi}{9}\right)\)
Answer:

Question 23.
Obtain the equation of the circles with radius 5 cm and touching x-axis at the origin in general form.
Answer:

Given radius = 5 cm and the circle is touching x axis
So centre will be (0, ± 5) and radius = 5
The equation of the circle with centre (0, ± 5) and radius 5 units is
(x – 0)² + (y ± 5)² = 5²
(i.e) x² + y² ± 10y + 25 – 25 = 0
(i.e) x² + y² ± 10y = 0

Question 24.
Find the length of the perpendicular from the origin to the plane.
\(\bar{r} \cdot(3 \vec{i}+4 \bar{j}+12 \vec{k})=26\).
Answer:
Taking the equation of the plane in cartesian form we get,
\((x \vec{i}+y \vec{j}+z \vec{k}) \cdot(3 \vec{i}+4 \vec{j}+12 \vec{k})=26\)
i.e. 3x + 4y + 12z – 26 = 0
The length of the perpendicular from (0, 0, 0) to the above plane is

Question 25.
Evaluate \(\lim _{x \rightarrow \pi / 2} \frac{\log (\sin x)}{(\pi-2 x)^{2}}\)
Answer:

Note that here l’ Hospitals rule, applied twice yields the result.

Question 26.
Evaluate \(\int_{-1}^{1} e^{-\lambda x}\left(1-x^{2}\right) d x\)
Answer:
Taking u = 1 – x² and v= e^-λx, and applying the Bernoulli’s formula, we get

Question 27.
Solve: \(\frac{d y}{d x}+2 y=e^{-x}\)
Answer:
Given that \(\frac{d y}{d x}+2 y=e^{-x}\)
This is a linear differential equation.
Here P = 2; Q = e
∫P dx = ∫2 dx = 2x
Thus, I.F = e^∫Pdx = e^2x
Hence the solution of (1) is \(y e^{\int \mathrm{P} d x}=\int \mathrm{Q} e^{\int \mathrm{P} d x} d x+c\)
That is, ye^2x = ∫e^-x e^2x dx + c (or) ye^2x = e^x + c (or) y = e^-x + ce^-2x is the required solution.

Question 28.
Three fair coins are tossed simultaneously. Find the probability mass function for number of heads occurred.
Answer:
When three coins are tossed, the sample space is
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
‘X’ is the random variable denotes the number of heads.
∴ ‘X’ can take the values of 0, 1, 2 and 3
Hence, the probabilities
P(X = 0) = P (No heads) = \(\frac{1}{8}\);
P(X = 1) = P (1 head) = \(\frac{3}{8}\);
P(X = 2) = P (2 heads) = \(\frac{3}{8}\);
P(X = 3) = P (3 heads)= \(\frac{1}{8}\);
∴ The probability mass function is
\(f(x)=\left\{\begin{array}{lll}
1 / 8 & \text { for } & x=0,3 \\
3 / 8 & \text { for } & x=1,2
\end{array}\right.\)

Question 29.
Construct the truth table for \((p \vee q) \vee \neg q\)
Answer:
truth table for \((p \vee q) \vee \neg q\)

Question 30.
Show that f(x, y) = \(\frac{x^{2}-y^{2}}{y^{2}+1}\) is continuous at every (x, y) ∈ R².
Answer:

Here, f satisfies all the three conditions of continuity at (a, b). Hence, f is continuous at every point of R² as (a, b) ∈ R².

Part – III

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

Question 31.
Form a polynomial equation with integer coefficients with \(\sqrt{\frac{\sqrt{2}}{\sqrt{3}}}\) as a root.

Question 32.
Find the equation of the tangents from the point (2, -3) to the parabola y² = 4x

Question 33.
Find the vector and cartesian equations of the straight line passing through the points (-5, 2, 3) and (4,-3, 6).

Question 34.
Find the points on the curve y = x³ – 6x² + x + 3 where the normal is parallel to the line x + y =1729.

Question 35.
Assuming log₁₀ e = 0.4343, find an approximate value of log₁₀ 1003.

Question 36.
Evaluate: \(\int_{0}^{\frac{\pi}{2}} \sin ^{2} x \cos ^{4} x d x\)

Question 37.
Solve the differential equation: \(\frac{d y}{d x}\) = e^x+y + x³ e^y

Question 38.
Using binomial distribution find the mean and variance of X for the following experiments
(i) A fair coin is tossed 100 times, and X denote the number of heads.
(ii) A fair die is tossed 240 times, and X denote the number of times that four appeared.

Question 39.
Let

be any three Boolean matrices of the same type. Find (A ∨ B) ∧C

Question 40.
Sketch the graph of y = sin\(\left(\frac{1}{3} x\right)\) for 0 ≤ x < 6π.

Part – IV

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

Question 41.
(a) A fish tank can be filled in 10 minutes using both pumps A and B simultaneously. However, pump B can pump water in or out at the same rate. If pump B is inadvertently run in reverse, then the tank will be filled in 30 minutes. How long would it take each pump to fill the tank by itself ? (Use Cramer’s rule to solve the problem).
[OR]
(b) Using elementary transformations find the inverse of the matrix \(\left[\begin{array}{ccc}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)

Question 42.
42. (a) Let z₁, z₂, and z₃ be complex numbers such that |z₁| = |z₂| = |z₃| = r > 0 and z₁ +z₂ + z₃ ≠ 0.
Prove that \(\left|\frac{z_{1} z_{2}+z_{2} z_{3}+z_{3} z_{1}}{z_{1}+z_{2}+z_{3}}\right|=r\)
[OR]
(b) Find all cube roots of \(\sqrt{3}\) + i.

Question 43.
(a) Find all zeros of the polynomial x⁶ – 3x⁵ – 5x⁴ + 22x³ – 39x² – 39x + 135, if it is known that 1 + 2i and \(\sqrt{3}\) are two of its zeros.
[OR]
(b) Let W(x, y, z) = x² – xy + 3 sin z, x, y, z∈R. Find the linear approximation at (2, -1, 0)

Question 44.
(a) Prove p → (q → r) ≡ (p ∧ q) → r without using truth table.
[OR]
(b) Evaluate \(\int_{0}^{\frac{\pi}{4}} \frac{1}{\sin x+\cos x} d x\)

Question 45.
(a) Prove that \(\text { (i) } \tan ^{-1} \frac{2}{11}+\tan ^{-1} \frac{7}{24}=\tan ^{-1} \frac{1}{2} \text { (ii) } \sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{12}{13}=\sin ^{-1} \frac{16}{65}\)
[OR]
(b) Find two positive numbers whose product is 100 and whose sum is minimum.

Question 46.
(a) If X is the random variable with distribution function F (x) given by,
\(\mathrm{F}(x)=\left\{\begin{array}{ll}
0, & x<0 \\
\frac{1}{2}\left(x^{2}+x\right) & 0 \leq x<1 \\
1, & x \geq 1
\end{array}\right.\)
then find (i) the probability density function f(x) (ii) P(0.3 ≤ X ≤ 0.6)
[OR]
(b) Find the vertex, focus, equation of directrix and length of the latus rectum of the following: y² – 4y – 8x + 12 = 0

Question 47.
(a) The rate at which the population of a city increases at any time is proportional to the population at that time. If there were 1,30,000 people in the city in 1960 and 1,60,000 in 1990, what population may be anticipated in 2020? [log_e \(\left(\frac{16}{3}\right)\) = 0.2070; e^-0.42 = 1.52]
[OR]
(b) Find the parametric vector, non-parametric vector and Cartesian form of the equation of the plane passing through the point (3, 6, -2), (-1, -2, 6) and (6, 4, -2).

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, helps students complete homework assignments and to score high marks in board exams.

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

General Instructions:

• The question paper comprises of four parts.
• You are to attempt all the parts. An internal choice of questions is provided wherever applicable.
• All questions of Part I, II, III, and IV are to be attempted separately.
• 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
• Question numbers 16 to 24 in Part II are two-mark questions. These are to be answered
  in about one or two sentences.
• Question numbers 25 to 33 in Part III are three-mark questions. These are to be answered
  in about three to five short sentences.
• Question numbers 34 to 38 in Part IV are five-mark questions. These are to be answered
  in detaiL Draw diagrams wherever necessary.

Time: 3 Hours
Max Marks: 70

Part – I

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

Question 1.
A thin conducting spherical shell of radius R has a charge Q which is uniformly distributed on its surface. the correct plot for electrostatic potential due to this spherical shell is ………

Answer:
(b)

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

Question 3.
n resistances, each of r Ω, when connected in parallel give an equivalent resistance of R Ω. If these resistances were connected in series, the combination would have a resistance in horns equal to ………..
(a) n²R
(b) \(\frac{R}{n^{2}}\)
(c) \(\frac{R}{n}\)
(d) nR
Answer:
(a) n²R
Hint :
Resistance in parallel combination, R= r/n = r = Rn
Resistance in series combination , R = nr = n²R

Question 4.
A thin insulated wire forms a plane spiral of N = 100 tight turns carrying a current 1 = 8 m A (milli ampere). The radii of inside and outside turns are a = 50 mm and b = 100 mm respectively. The magnetic induction at the center of the spiral is……………………….
(a) 5 μT
(b) 7 μT
(c) 8 μT
(d) 10 μT
Answer:
(b) 7 μT

Question 5.
In an oscillating LC circuit, the maximum charge on the capacitor is Q. The charge on the capacitor, when the energy is stored equally between the electric and magnetic fields, is……………………….
(a) \(\frac{Q}{2} \)
(b) \(\frac{Q}{\sqrt{3}}\)
(c) \(\frac{Q}{\sqrt{2}}\)
(d) Q
Answer:
(c) \(\frac{Q}{\sqrt{2}}\)

Question 6.
The direction of induced current during electro magnetic induction is given by …………………………………..
(a) Faraday’s law
(b) Lenz’s law
(c) Maxwell’s law
(d)  Ampere’s law
Answer:
(b) Lenz’s law

Question 7.
In an electromagnetic wave in free space the rms value of the electric field is 3 V m . The peak value of the magnetic field is………………………..
(a) 1.414 x 10^-8 T
(b) 1.0 x 10^-8 T
(c) 2.828 x 10^-8 T
(d) 2.0 x 10^-8 T
Answer:
(a) 1.414 x 10^-8 T

Question 8.
Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm approximately. The maximum distance at which these dots can be resolved by the eye is, [take wavelength of light, λ = 500 nm]
(a) 1m
(b) 5m
(c) 3m
(d) 6m
Answer:
(b) 5m
Hint:
Resolution limit sin θ =

Question 9.
Two mirrors are kept at 60° to each other and a body is placed at middle. The total number of images formed is ……………..
(a) six
(b) four
(c) five
(d) three
Answer:
(a) six
Hint:
Number of images formed \(n=\frac{360}{\theta}-1=\frac{360}{60}-1=5\)

Question 10.
The threshold wavelength for a metal surface whose photoelectric work function is 3.313 eV is
{a) 4125 Å
(b) 3750  Å
(c) 6000 Å
(d) 2062.5 Å
Answer:
(b) 3750  Å

Question 11.
Suppose an alpha particle accelerated by a potential of V volt is allowed to collide with a nucleus whose atomic number is Z, then the distance of closest approach of alpha particle to the nucleus is…..

Answer:
(c)

Question 12.
The speed of the particle, that can take discrete values is proportional to ……………
(a) n^-3/2
(b) n^-3
(c) n^1/2
(d) n
Answer:
(d) n
Hint:
\(P=m v=\frac{n h}{2 a} ; V \propto n\)

Question 13.
Doping semiconductor results in ………………………………
(a) The decrease in mobile charge carriers
(b) The change in chemical properties
(c) The change in the crystal structure
(d) The breaking of the covalent bond
Answer:
(c) The change in the crystal structure

Question 14.
The signal is affected by noise in a communication system …………………….
(a) At the transmitter
(b) At the modulator
(c) In the channel
(d) At the receiver
Answer:
(c) In the channel

Question 15.
The technology used for stopping the brain from processing pain is…………………………..
(a) Precision medicine
(b) Wireless brain sensor
(c) Virtual reality
(d) Radiology
Answer:
(c) Virtual reality

Part – II

Answer any six questions. Question No. 21 is compulsory. [6 x 2 = 12]

Question 16.
Define ‘Electric field’.
Answer:
The electric field at the point P at a distance r from the point charge q is the force experienced by a unit charge and is given by
\(\overrightarrow{\mathrm{E}}=\frac{\overrightarrow{\mathrm{F}}}{q_{0}}\)
The electric field is a vector quantity and its Sl unit is Newton per Coulomb (NC^-1).

Question 17.
State the principle of potentiometer.
Answer:
The basic principle of a potentiometer is that when a constant current flows through a wire of uniform cross-sectional area and composition, the potential drop across any length of the wire is directly proportional to that length.

Question 18.
Compute the intensity of magnetisation of the bar magnet whose mass, magnetic moment and density are 200 g, 2 A m² and 8 g cm^-3, respectively.
Answer:

Question 19.
What is displacement current?
Answer:
The displacement current can be defined as the current which comes into play in the region in ‘ which the electric field and the electric flux are changing with time.

Question 20.
What is principle of reversibility?
Answer:
The principle of reversibility states that light will follow exactly the same path if its direction of travel is reversed.

Question 21.
Calculate the de Broglie wavelength of a proton whose kinetic energy is equal to. 81.9 x 10¹⁵ (Given: mass of proton is 1836 times that of electron).
Answer:

Question 22.
Give the symbolic representation of alpha decay, beta decay and gamma decay.
Answer:
Alpha decay: The alpha decay process symbolically in the following way
Beta decay: β decay is represented by \(_{ z }^{ A }X\rightarrow _{ z,-1 }^{ A }Y+e^{ + }+v\)
Gamma decay: The gamma decay is given \(\mathrm{z} \mathrm{X}^{*} \rightarrow_{Z}^{\mathrm{A}} \mathrm{X} \)+ gamma(y)rays

Question 23.
Explain 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 24.
Give the factors that are responsible for transmission impairments.
Answer:

• Attenuation
• Distortion (Harmonic)
• Noise

Part – III

Answer any six questions. Question No. 32 is compulsory. [6 x 3 = 18]

Question 25.
A water molecule has an electric dipole moment of 6.3 x 10 Cm. A sample contains 10 water molecules, with all the dipole moments aligned parallel to the external electric field of magnitude 3 x 10^s NC^-1. How much work is required to rotate all the water molecules from θ = 0° to 90°?
Answer:
When the water molecules are aligned in the direction of the electric field, it has minimum potential energy. The work done to rotate the dipole from θ = 0° to 90° is equal to the potential energy difference between these two configurations.
W = ΔU = U(90°) – U(0°)
As we know, U = -pE cos θ, Next we calculate the work done to rotate one water molecule
from θ = 0° to 90°,
For one water molecule, W = – pE cos 90° + pE cos 0° = pE
W = 6.3 x10^-30 x 3 x 10⁵ = 18.9 x 10^-25
For 10²² water molecules, the total work done is W_tot = 18.9x 10 ²⁵ x 10²² = 18.9 x 10 ³ J

Question 26.
What are the properties of an equipotential surface?
Answer:
Properties of equipotential surfaces
(i) The work done to move a charge q between any two points A and B,
W = q (V_B – V_A ). If the points A and B lie on the same equipotential surface, work done is zero because V_B – V_A

(ii) The electric field is normal to an equipotential surface. If it is not normal, then there is a component of the field parallel to the surface. Then work must be done to move a charge between two points on the same surface. This is a contradiction. Therefore the electric field must always be normal to an equipotential surface.

Question 27.
An electronics hobbyist is building a radio which requires 150 ft in her circuit, but she has only 220 Ω 79 Ω and 92 Ω resistors available. How can she connect the available resistors to get desired value of resistance?
Answer:
Required effective resistance = 150 ft
Given resistors of resistance, R = 220 ft, R = 79 ft, R = 92 ft
Parallel combination of R₁ and R₂

Question 28.
What is magnetic permeability?
Answer:
Magnetic permeability: The magnetic permeability can be defined as the measure of ability of the material to allow the passage of magnetic field lines through it or measure of the capacity of the substance to take magnetisation or the degree of penetration of magnetic field through the substance.

Question 29.
A circular metal of area 0.03 m² rotates in a uniform magnetic field of 0.4 T. The axis of rotation passes through the centre and perpendicular to its plane and is also parallel to the field. If the disc completes 20 revolutions in one second and the resistance of the disc is 4 Ω calculate the induced emf between the axis and the rim and induced current flowing in the disc.
Answer:
A = 0.03 m²; B = 0.4T; f = 20rps; R = 4Ω
Area covered in 1 sec = Area of the disc x frequency
= 0.03 x 20 = 0.6 m²

Question 30.
Give the characteristics of image formed by a plane mirror.
Answer:

• The image formed by a plane mirror is virtual, erect, and laterally inverted.
• The size of the image is equal to the size of the object.
• The image distance far behind the mirror is equal to the object distance in front of it.
• If an object is placed between two plane mirrors inclined at an angle 0, then the number of images n formed is as,\(n=\left(\frac{360}{\theta}-1\right)\)

Question 31.
An electron and an alpha particle have same kinetic energy. How are the de Broglie wavelengths associated with them related?
Answer:

Question 32.
Assuming V_CEsat = 0.2 V and β = 50, find the minimum base current (I_B) required to drive the transistor given in the figure to saturation.
Answer:

V_CEsat = 0.2 V and β = 50
V_CE = V_CC -I_C R_C
0.2= 3 – I_c(1 k)

Question 33.
Mention any two advantages and disadvantages of Robotics.
Answer:
(i) Advantages of Robotics:

• The robots are much cheaper than humans.
• Robots never get tired like humans. It can work for 24 x 7. Hence absenteeism in work place can be reduced.
• Robots are more precise and error free in performing the task.

(ii) Disadvantages of Robotics:

• Robots have no sense of emotions or conscience.
• They lack empathy and hence create an emotionless workplace.
• If ultimately robots would do all the work, and the humans will just sit and monitor them* health hazards will increase rapidly.

Part – IV

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

Question 34.
(a) Derive an expression for electrostatic potential energy of the dipole in a uniform electric field.
Answer:
Expression for electrostatic potential energy of the dipole in a uniform electric field:
Considera dipole placed in the uniform electric field \(\overrightarrow{\mathrm{E}}\) placed in the uniform electric field \(\overrightarrow{\mathrm{E}}\) . A dipole experiences a torque when kept in an uniform electric field E.

This torque rotates the dipole to align it with the direction of the electric field. To rotate the dipole (at constant angular velocity) from its initial angle θ’ to another angle θ against the torque exerted by the electric field, an equal and opposite external torque must be applied on the dipole. The work done by the external torque to rotate the dipole from angle θ’ to θ at constant angular velocity is

This work done is equal to the potential energy difference between the angular positions 0 and 0′. U(θ) – (Uθ’) = ΔU = -pE cos θ + pE cos θ’.

If the initial angle is = θ’ = 90° and is taken as reference point, then U(θ’) + pE cos 90° = 0. The potential energy stored in the system of dipole kept in the uniform electric field is given by.
U = -pE cos θ = -p-E  ……………. (3)

In addition to p and E, the potential energy also depends on the orientation θ of the electric dipole with respect to the external electric field.

The potential energy is maximum when the dipole is aligned anti-parallel (θ = Jt) to the external electric field and minimum when the dipole is aligned parallel (θ = 0) to the external electric field.

[OR]

(b) 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 \(d \vec{l} \text { and } \vec{r}\). Then, the magnetic field at P due to the element is
\(d \overrightarrow{\mathrm{B}}=\frac{\mu_{0} \mathrm{Id} l}{4 \pi r^{2}} \sin \theta\)(unit vector perpendicular to \(d \vec{l} \text { and } \vec{r}\)

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 \(d \vec{l} \text { and } \vec{r}\)
(let it be \(\hat{n}\)). The net magnetic field can be determined by integrating equation with proper limits. \(\overrightarrow{\mathrm{B}}=\int d \overrightarrow{\mathrm{B}}\) From the figure, in a right angle triangle PAO,

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 \(d \overrightarrow{\mathrm{B}} \) by varying the angle from θ = φ₁ to θ =φ₂ is

For a an infinitely long straight wire, 1 =0 and 2 =, the magnetic field is \(\overrightarrow{\mathrm{B}}=\frac{\mu_{0} I}{2 \pi a} \hat{n} \) ………… (3)
Note that here \(\hat{n} \) represents the unit vector from the point O to P.

Question 35.
(a) Obtain an expression for motional emf from Lorentz force.
Answer:
Motional emf from Lorentz force: Consider a straight conducting rod AB of length l in a uniform magnetic field \(\vec{B}\) which is directed perpendicularly into the plane of the paper. The length of the rod is normal to the magnetic field. Let the rod move with a constant velocity \(\vec{v}\) towards right side. When the rod moves, the free electrons present in it also move with same velocity \(\vec{v}\) in B. As a result, the Lorentz force acts on free electrons in the direction from B to A and is given by the relation  \(\overrightarrow{\mathrm{F}}_{\mathrm{B}}\)=-
\( e(\vec{v} \times \overrightarrow{\mathrm{B}}) \) …………. (1)

The action of this Lorentz force is to accumulate the free electrons at the end A. This accumulation of free electrons produces a potential difference across the rod which in turn establishes an electric field E directed along BA. Due to the electric field \(\overrightarrow{\mathrm{E}}\), the coulomb force starts acting on the free electrons along AB and is given by
\(\overrightarrow{\mathrm{F}}_{\mathrm{E}}=-e \overrightarrow{\mathrm{E}}\) ……….. (2)

The magnitude of the electric field E keeps on increasing as long as accumulation of electrons at the end A continues. The force \(\overrightarrow{\mathrm{F}}_{\mathrm{B}}\) also increases until equilibrium is reached. At equilibrium, the magnetic Lorentz force \(\overrightarrow{\mathrm{F}}_{\mathrm{B}}\) and the coulomb force F balance each other and no further accumulation of free electrons at the end A takes place,
i.e

The potential difference between two ends of the rod is
V= El
V= vBl
Thus the Lorentz force on the free electrons is responsible to maintain this potential difference and hence produces an emf

ε = B lv    ………….. (4)
As this emf is produced due to the movement of the rod, it is often called as motional emf.

[OR]

(b) Explain the Maxwell’s modification of Ampere’s circuital law.
Answer:
Maxwell argued that a changing electric field between the capacitor plates must induce a magnetic field. As currents are the usual sources of magnetic fields, a changing electric field must be associated with a current. Maxwell called this current as the displacement current.

If ‘A’ be the area of the capacitor plates and ‘cf be the charge on the plates at any instant ‘f during the charging process, then the electric field in the gap will be

But \(\frac{d q}{d t}\) is the rate of change of charge on the capacitor plates. It is called displacement current and given by \(\mathrm{I}_{d}=\frac{d q}{d t}=\varepsilon_{0} \frac{d \Phi_{\mathrm{E}}}{d t}\)

This is the missing term in Ampere’s Circuital Law. The total current must be the sum of the
conduction current I_c Hence, the modified form of the ampere’s law.
\(\oint_{c} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l}=\mu_{0}\left(\mathrm{I}_{c}+\varepsilon_{0} \frac{d \Phi_{\mathrm{E}}}{d t}\right)\)

Question 36.
(a) Obtain the equation for resolving power of microscope.
Answer:
Resolving power of microscope: The diagram related to the calculation of resolution of microscope. A microscope is used to see the details of the object under observation. The ability of microscope depends not only in magnifying the object objective but also in resolving two points on the object separated by a small distance d_min  Smaller the value of d_min better will be the resolving power of the microscope.

The radius of central maxima is, \(r_{0}=\frac{1.22 \lambda v}{a}\) . In the place of focal length f we have the image distance v. If the difference between the two points on the object to be resolved is d_min, then the magnification m is, \(m=\frac{r_{o}}{d_{\min }}\)


To further reduce the value of d_min the optical path of the light is increased by immersing the objective of the microscope into a bath containing oil of refractive index n.

Such an objective is called oil immersed objective. The term n sin β is called numerical aperture NA.

[OR]

(b) Derive an expression for de Broglie wavelength of electrons.
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
\(\frac{1}{2} m v^{2}=e \mathrm{V}\)
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 \(\lambda=\frac{h}{m v}=\frac{h}{\sqrt{2 e m \mathrm{V}}}\)
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
\(\lambda=\frac{h}{\sqrt{2 m \mathrm{K}}}\)

Question 37.
(a) Explain the variation of average binding energy with the mass number by graph and discuss its features.
Answer:
We can find the average binding energy per nucleon \(\overline{\mathrm{BE}}\).
It is given by

The average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus. \(\overline{\mathrm{BE}}\) is plotted against A of all known nuclei.

 

Important inferences from of the average binding energy curve:

• The value of BE rises as the mass number increases until it reaches a maximum value of 8.8 MeV for A= 56 (iron) and then it slowly decreases.
• The average binding energy per nucleon is about 8.5 MeV for nuclei having mass number between A = 40 and 120. These elements are comparatively more stable and not radioactive.
• For higher mass numbers, the curve reduces slowly and BE for uranium is about 7.6 MeV. They are unstable and radioactive.  If two light nuclei with A < 28 combine with a nucleus with A < 56, the binding energy per nucleon is more for final nucleus than the initial nuclei. Thus, if the lighter elements combine to produce a nucleus of medium value A, a large amount of energy will be released. This is the basis of nuclear fusion and is the principle of the hydrogen bomb.
• If a nucleus of heavy element is split (fission) into two or more nuclei of medium value A, the energy released would again be large. The atom bomb is based on this principle and huge energy of atom bombs comes from this fission when it is uncontrolled.

[OR]

(b) State Boolean laws. Elucidate how they are used to simplify Boolean expressions with suitable example.
Answer:
Laws of Boolean Algebra: The NOT, OR and AND operations are \(\bar{A}\), A + B, A.B are the Boolean operations. The results of these operations can be summarised as:

Complement law:

The complement law can be realised as \(\overline{\mathrm{A}}\) = A

Commutative laws
A + B = B +A
A . B = B . A

Associative laws
A + (B + C) = (A + B) + C
A . (B . C) = (A .B) . C

Distributive laws
A( B + C) = AB + AC
A + BC = (A + B) (A + C)
The above laws are used to simplify complicated expressions and to simplify the logic circuitry.

Question 38.
(a) Modulation helps to reduce the antenna size in wireless communication – Explain. Antenna size:
Answer:
Antenna is used at both transmitter and receiver end. Antenna height is an important parameter to be discussed. The height of the antenna must be a multiple of \(\frac{\lambda}{4}\)
\(h=\frac{\lambda}{4}\) ……………….. (1)
where X is wavelength \(\left(\lambda=\frac{c}{v}\right), c\) c is the velocity of light and v is the frequency of the signal v to be transmitted.

An example
Let us consider two baseband signals. One signal is modulated and the other is not modulated. The frequency of the original baseband signal is taken as v = 10 kHz while the modulated signal is v = 1 MHz. The height of the antenna required to transmit the original baseband signal of frequency
v = 10kHz is

The height of the antenna required to transmit the modulated signal of frequency v = 1 MHz is

Comparing equations (2) and (3), we can infer that it is practically feasible to construct an antenna of height 75 m while the one with 7.5 km is not possible. It clearly manifests that modulated signals reduce the antenna height and are required for long distance transmission.

[OR]

(b) What are the possible harmful effects of usage of Nanoparticles? Why?
Answer:
Possible harmful effects of usage of Nanoparticles:

The research on the harmful impact of application of nanotechnology is also equally important and fast developing. The major concern here is that the nanoparticles have the dimensions same as that of the biological molecules such as proteins. They may easily get absorbed onto the surface of living organisms and they might enter the tissues and fluids of the body.

The adsorbing nature depends on the surface of the nanoparticle. Indeed, it is possible to deliver a drug directly to a specific cell in the body by designing the surface of a nanoparticle so that it adsorbs specifically onto the surface of the target cell.

The interaction with living systems is also affected by the dimensions of the nanoparticles. For instance, nanoparticles of a few nanometers size may reach well inside biomolecules, which is not possible for larger nanoparticles. Nanoparticles can also cross cell membranes.

It is also possible for the inhaled nanoparticles to reach the blood, to reach other sites such as the liver, heart or blood cells. Researchers are trying to understand the response of living organisms to the presence of nanoparticles of varying size, shape, chemical composition and surface characteristics.

Students can Download Tamil Nadu 12th Chemistry Model Question Paper 2 English Medium Pdf, Tamil Nadu 12th Chemistry Model Question Papers helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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

Time: 3 Hours
Maximum Marks: 70

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-marks questions. These are to be answered in about three to five short sentences
7. Question numbers 34 to 38 in Part IV are five-mark Questions These are to answered in detail. Draw diagrams wherever necessary

Part – 1

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

Question 1.
Blistered copper is .
(a) 98% pure copper
(b) 96% pure copper
(c) 97% pure copper
(d) 88% pure copper
Answer:
(a) 98% pure copper

Question 2.
Which of the following statements is not correct?
(a) Beryl is a cyclic silicate
(b) Mg₂SiO₄ is an orthosilicate
(c) SiO₄ is the basic structural unit of silicates
(d) Feldspar is not aluminosilicate
Answer:
(d) Feldspar is not aluminosilicate

Question 3.
Consider the following statements.
(i) phosphine is the most important hydride of phosphorous
(ii) phosphine is a poisonous gas with rotten egg smell.
(iii) phosphine is a powerful reducing agent
Which of the above statement(s) is / are correct?
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (ii) only
Answer:
(c) (i) and (iii)

Question 4.
The correct order of increasing oxidizing power in the series


Answer:

Hint: VO₂⁺ < Cr₂O₇^2- < MnO₄^– greater the oxidation state, higher is the oxidising power.

Question 5.
The geometry possible in [Fe F₆]^4- and [COF₊₆]^4- is ……………
(a) Trigonal bipyramidal
(b) Square planar
(c) Octahedral
(d) Tetrahedral
Answer:
(c) Octahedral

Question 6.
The number of carbon atoms per unit cell of diamond is
(a) 8
(b) 6
(c) 1
(d) 4
Answer:
(a) 8

Hint: In diamond carbon forming fee. Carbon occupies comers and face centres and also occupying half of the tetrahedral voids.

Question 7.
In a homogeneous reaction
A → B + C + D, the initial pressure was P₀ and after time t it was P. Expression for rate constant in terms of P₀ , P and t will be ……..

Answer:

Solution:

Question 8.
The solubility of an aqueous solution of Mg(OH)₂ be x then its K_sp is
(a) 4x³
(b) 108x⁵
(c) 27x⁴
(d) 9x
Answer:
(a) 4x³
Solution:
Mg(OH)₂ Mg²⁺(x) + 2OH^– ((2x)²)
K_sp = 4x³

Question 9.
During electrolysis of molten sodium chloride, the time required to produce 0. lmol of chlorine gas using a current of 3 A is ……………..
(a) 55 minutes
(b) 107.2 minutes
(c) 220 minutes
(d) 330 minutes
Hint: m = ZIt (mass of 1 mole of Cl₂ gas = 71)
Answer:
(b) 107.2 minutes

Question 10:
The coagulation values in millimoles per litre of the electrolytes used for the coagulation of AS₂S₃ are given below
(I) (NaCl) = 52
(II) (BaCl) = 0.69
(III) (MgSO₄) = 0.22
The correct order of their coagulating power is
(a) III > II > I
(b) I > II > III
(c) I > III > II
(d) II > III> I
coagulating power ± \(\frac{1}{\text { coagulation value }}\)
Answer:
(a) III > II > I

Question 10.
Assertion : Coagulation power of Al³⁺ is more than Na⁺ .
Reason : greater the valency of the flocculating ion added, greater is its power to cause precipitation
(a) if both assertion and reason are true and reason is the correct explanation of assertion. (Hardy-Sechulze rule)
(b) if both assertion and reason are true but reason is not the correct explanation of assertion.
(c) assertion is true but reason is false
(d) both assertion and reason are false.
Answer:
(a) if both assertion and reason are true and reason is the correct explanation of assertion. (Hardy-Sechulze rule)

Question 11.


Answer:
(a) A- 2, B – 3, C – 4, D – 1

Question 12.
Which of the following represents the correct order of acidity in the given compounds (
(a) FCH₂COOH > CH₃COOH > BrCH₂COOH > ClCH₃COOH
(b) FCH₂COOH > ClCH2COOH > BrCH COOH > CH COOH
(c) CH₃COOH > ClCH₂COOH > FCH₂COOH > Br – CH₂COOH
(d) Cl CH₂COOH > CH₃COOH > BrCH₂COOH > ICH₂COOH
Hint. -I effect increases the acidity. If electronegativity is high, -I effect is also high.
Answer:
(b) FCH₂COOH > ClCH2COOH > BrCH COOH > CH COOH

Question 13.
1 -nitrobutane and 2-methyl-1 -nitropropane are belong to
(a) position isomerism
(b) functional isomerism
(c) Tautomerism
(d) chain isomerism
Answer:
(d) chain isomerism

Question 14.
Haemoglobin is
(a) an enzyme
(b) a globular protein
(c) a vitamin
(d) carbohydrate
Answer:
(b) a globular protein

Question 15.
An example of antifertility drug is
(a) novestrol
(b) seldane
(c) salvarsan
(d) Chloramphenicol
Answer:
(a) novestrol

Part – II

Answer any six questions. Question No. 20 is compulsory. [6 x 2 = 12]

Question 16.
Name the method used for the refining of (i) Nickel (ii) Zirconium
Answer:
(i) Mond’s process
(ii) Van Arkel’s method

Question 17.
Complete the following reactions:
(a) B(OH)₃ + NH₃ →
(b) Na₂B₄O₇ + H₂SO₄ + H₂O
Answer:

Question 18.
KMnO₄ does not act as oxidising agent in the presence of HC1. Why?
Answer:
HCl cannot be used for making acidified KMnO₄ as oxidising agent, since it reacts with KMnO₄ as follows.
2MnO₄^– + 10 Cl^– + 16H⁺ → 2Mn²⁺ + 5Cl₂ + 8H₂O

Question 19.
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled? ,
Answer:
(i) \(\frac{d x}{d t}=k[\mathrm{A}]^{1}[\mathrm{B}]^{2}\)
(ii) If concentration of ‘B ’ is tripled, then the rate will become 9 times.
(iii) When concentration of both A and B are doubled, then the rate will become 8 times.

Question 20.
Calculate the pH of 0.04 M HNO₃ Solution.
Answer:
Concentration of HNO₃ = 0.04M
[H₃O⁺] = 0.04 mol dm^-3
pH = -log[H₃O⁺]
= – log(0.04)
= – log(4 x 10^-2)
= 2 -log 4 = 2-0.6021
= 1.3979= 1.40

Question 21.
Write a note about Helmoholtz double layer.
Answer:
The surface of a colloidal particle adsorbs one type of ion due to preferential adsorption. This layer attracts the oppositely charged ions in the medium and hence the boundary separating the two electrical double layers are set up. This is called Helmholtz electrical double layer. As the particles nearby are having similar charges, they cannot come close and condense. Hence this helps to explain the stability of the colloid.

Question 22.
Why HCOOH does not give Hell-Volhard Zelinsky reaction but CH₃3COOH does?
Answer:
CH₃COOH contains α – hydrogen atom and hence gives HVZ reaction but HCOOH does not contain an α – hydrogen atom and hence does not give HVZ reaction.

Question 23.
What happens when
Oxidation of acetoneoxime with trifluoroperoxy acetic acid.
Answer:
Oxidation of acetoneoxime with trifluoroperoxy acetic acid gives 2-nitropropane.

Question 24.
What are food preservatives?
Answer:
Food preservatives are chemical substances are capable of inhibiting, retarding or arresting the process of fermentation, acidification or other decomposition of food by growth of micro organisms.
Examples:

• Acetic acid is used mainly as a preservative for preparation of pickles.
• Sodium meta sulphite is used as preservative for fresh vegetables and fruits.
• Sodium benzoate is used as preservative for juices.

Part – III

Answer any six questions. Question No. 31 is compulsory. [6 x 3 = 18]

Question 25.
Explain the principle of electrolytic refining with an example.
Answer:
The crude metal is refined by electrolysis. It is carried out in an electrolytic cell containing aqueous solution of the salts of the metal of interest. The rods of impure metal are used as anode and thin strips of pure metal are used as cathode. The metal of interest dissolves from the anode, pass into the solution while the same amount of metal ions from the solution will be deposited at the cathode. During electrolysis, the less electropositive impurities in the anode, settle down at the bottom and are removed as anode mud.
Let us understand this process by considering electrolytic refining of silver as an example.
Cathode: Pure silver
Anode: Impure silver rods

Electrolyte: Acidified aqueous solution of silver nitrate.
When a current is passed through the electrodes the following reactions will take place
Reaction at anode: 2Ag(s) → Ag⁺ (aq) + le^–
Reaction at cathode: Ag⁺(aq)+ le^– → Ag (s)

During electrolysis, at the anode the silver atoms lose electrons and enter the solution. The positively charged silver cations migrate towards the cathode and get discharged by gaining electrons and deposited on the cathode. Other metals such as copper, zinc etc.,can also be refined by this process in a similar manner.

Question 26.
How will you prepare phosphine and explain the purification of phosphine?
Phosphine’is prepared by action of sodium hydroxide with white phosphorous in an inert atmosphere of carbon dioxide or hydrogen.

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

Phosphine is prepared in pure form by heating phosphorous acid.

A pure sample of phosphine is prepared by heating phosphonium iodide with caustic soda S0lution.

Question 27.
Justify the position of lanthanides and actinides in the periodic table.
Answer:
(i) In sixth period after lanthanum, the electrons are preferentially filled in inner 4f sub shell and these 14 elements following lanthanum show similar chemical properties. Therefore these elements are grouped together and placed at the bottom of the periodic table. This position can be justified as follows.
(a) Lanthanoids have general electronic configuration [Xe] 4f2 14 5d° 1 6s2
(b) The common oxidation state of lanthanoids is +3
(c) All these elements have similar physical and chemical properties.

(ii) Similarly the fourteen elements following actinium resemble in their physical and chemical properties.

(iii) If we place these elements after Lanthanum in the periodic table below 4d series and actinides below 5d series, the properties of the elements belongs to a group would be different and it would affect the proper structure of the periodic table.

(iv) Hence a separate position is provided to the inner transition elements at the bottom of the periodic table.

Question 28.
Distinguish between isotropy and anisotropy?
Answer:
Isotropy:

1. Isotropy means uniformity in all directions.
2. Isotropy means having identical values of physical properties such as refractive index, electrical conductance in all directions.
3. Isotropy is the property of amorphous solids.

Anisotropy :

1. Anisotropy means non-uniformity in all directions.
2. Anisotropy is the property which depends on the direction of measurement. They show different values of physical properties when measured along different directions.
3. Anisotropy is the properly of crystalline solids.

Question 29.
What are the merits and limitations of the intermediate compound theory?
Answer:
Merits:
(i) The specificity of a catalyst
(ii) The increase in the rate of the reaction with increase in the concentration of catalyst.

Limitations:
(i) This theory fails to explain the action of catalytic poison and promoters. .
(ii) This theory is unable to explain the mechanism of heterogeneous catalysed reactions.

Question 30.
Write a note on sacrificial protection.
Answer:
In this method, the metallic structure to be protected is made cathode by connecting it with more active metal (anodic metal). So that all the corrosion will concentrate only on the active metal. The artificially made anode thus gradually gets corroded protecting the original metallic structure. Hence this process is otherwise known as sacrificial anodic protection.
Al, Zn and Mg are used as sacrificial anodes.

Question 31.
1mole of Hlis allowed to react with t-butyl methylether. Identify the product and write down the mechanism of the reaction.
Answer:

Question 32.
How will you prove the presence of aldehyde group in glucose?
Answer:
(i) Glucose is oxidised to gluconic acid with ammoniacal silver nitrate (Tollen’s reagent) and alkaline copper sulphate (Fehling’s solution). Tollen’s reagent is reduced to metallic silver and Fehling’s solution to cuprous oxide (red precipitate). These reactions confirm the presence of an aldehye group in glucose.

Question 33.
Explain about anaesthetics with their types.
(i) Local anaesthetics: It causes loss of sensation in the area in which it is applied without losing consciousness. They block pain perception that is transmitted via peripheral nerve fibres to the brain.
Example: Procaine, Li do caine.
They are often used during minor surgical procedures.

(ii) General anaesthetics: They cause a controlled and reversible loss of consciousness by affecting central nervous system.
Example: Propofol, Isoflurane.
They are often used for major surgical procedures.

Part – IV

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

Question 34.
(a) (i) CO is more stable at higher temperature. Why? (2)
(ii) How will you prepare potash alum? (3)
[OR]
(b) (i) Give the balanced equation for the reaction between chlorine with cold NaOH. (3)
(ii) Nitrogen exists as diatomic molecule and Phosphrus as P₄ Why? (2)
Answer:
(a) (i) In the Ellingham diagram, the formation of carbon monoxide is a straight line with negative slope. In this case AS is positive as 2 moles of CO gas is formed by the consumption of one mole of oxygen. Hence, CO is more stable at higher temperature.

(ii) The alunite the alum stone is the naturally occurring form and it is K₂SO₄ . Al₂(S0₄)₃.4Al(0H)₃. When alum stone is treated with excess of sulphuric acid, the aluminium hydroxide is converted to aluminium sulphate. A calculated quality of – potassium sulphate is added and the solution is crystallised to generate potash alum. It is purified by recrystallisation.
(2)

(b)(i) Reaction between chlorine with cold NaoH

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

(ii) Nitrogen has a triple bond between its two atoms because of its small size and high electronegativity. Phosphorus P₄ has single bond, that is why it is tetra-atomic.

Question 35.
(a) (i) Discuss the ortho and pyro silicates. (2)
(ii) Compare lanthanides and actinides. (3)
[OR]
(b) (i) Ni²⁺ is identified using alcoholic solution of dimethyl glyoxime. Write the structural formula for the rosy red precipitate of a complex formed in the reaction. (3)
Cu⁺, Zn²⁺, Sc³⁺, Ti⁴⁺ are colourless. Prove this statement. (2)
Answer:
Ortho silicates: The simplest silicates which contain discrete [SiO₄]^4- tetrahedral units are called ortho silicates or neso silicates.

Examples: Phenacite – Be₂SiO₄(Be²⁺ ions are tetrahedrally surrounded by O^2- ions), Olivine – (Fe/Mg)₂ SiO₄ (‘Fe²⁺ and Mg²⁺ cations are octahedrally surrounded by O

Pyro silicates: Silicates which contain [Si₂O₇]^6- ions are called pyro silicates (or) Soro silicates. They are formed by joining two [SiO₄]^4-

tetrahedral units by sharing one oxygen atom at one comer, (one oxygen is removed while joining). Example: Thortveitite – Sc₂Si₂O₇

(ii)
Lanthanoids

1. Differentiating electron enters in 4f orbital
2. Binding energy of 4f orbitals are higher
3. They show less tendency to form complexes
4. Most of the lanthanoids are colourless
5. They do not form oxo cations
6. Besides +3 oxidation states lanthanoids show +2 and +4 oxidation states in few cases.

Actinoids

1. Differentiating electron enters in 5f orbital
2. Binding energy of 5f orbitals are lower
3. They show greater tendency to form complexes
4. Most of the actinoids are coloured.
5. E.g : U³⁺ (red), U⁴⁺ (green), UO₂²⁺ (yellow)
6. They do form oxo cations such as UO₂²⁺ NpO₂²⁺ etc.
7. Besides +3 oxidation states actinoids show higher oxidation states such as +4, +5, +6 and +7.

[OR]

(b) (i) Ni²⁺ ions present in Nickel chloride solution is estimated accurately for forming an insoluble complex called [Ni(DMG)₂]
Nickel ion reacts with alcoholic solution of DMG in the presence of ammonical medium, to give rosy red precipitate of [Ni(DMG)₂] complex.

(ii) 1. Cu⁺, Zn²⁺ have d10 configuration and Sc³⁺, Ti⁴⁺have d¹ configuration.
2. d-d transition is not possible in the above complexes. So they are colourless.

Question 36.
(a) i)0 What is meant by the term “coordination number”? What is the coordination number of atoms in a bcc structure? (2)
(ii) Rate constant k of a reaction varies with temperature T according to the following Arrhenius equation
\(\log \mathbf{k}=\log \mathbf{A} \frac{\mathbf{E}_{\mathbf{a}}}{2.303 \mathbf{R}}\left(\frac{1}{\mathbf{T}}\right)\)

Where E_a is the activation energy. When a graph is plotted for logk Vs \(\frac{1}{T}\) a straight line
with a slope of- 4000K Is obtained. Calculate the activation energy. (3)

(OR)

(b) (i) Explain about the hydrolysis of salt of strong base and weak acid. Derive the value of K_h for that reaction. (3)
(ii) Identify the Lewis acid and the Lewis base in the following reactions. (2)

Answer:
(a) (i) 1. Let us consider the reaction between sodium hydroxide and acetic acid to give sodium acetate and water

NaOH_(aq) + CH₃COOH_(aq) ⇌ CH₃COONa_(aq) + H₂O

2. Coordination number of atoms in a bcc structure is 8

(ii) logk = logA – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{1}{\mathrm{T}}\right)\)
y = c + mx
m = \(-\frac{E_{a}}{2.303 R}\)
E_a =—2.303Rm
E_a = – 2.303 x 8.314 J K^-1 mol^-1 x (— 4000 K)
E_a = 76,589 J mol^-1
E_a =76589kJmol^-1

[OR]

(b) (i) 1. Coordination number: The number of nearest neighbours that surrounding a particle in a crystal is called the coordination number of that particle.
2. In aqueous solution, CH₃COONa is completely dissociated as follows.
CH₃COONa_(aq), →  CH₃COO^–_(aq) + Na⁺_(aq)

3. CH₃COO^– is a conjugate base of the weak acid CHCOOH and it has a tendency to react with H⁺ from water to produce unionised acid. But there is no such tendency
for Na⁺ to react with OH^–

4.

such cases, the solution is basic due to the hydrolysis and pH is greater than 7.

5. Relationship between equilibrium constant, hydrolysis constant and the dissociation constant of acid is derived as follows:

K_h value in terms of degree of hydrolysis (h) and the concentration of salt (c) for the equilibrium can be obtained as in the case of Ostwald’s dilution law K_h = h²C and \(\left[\mathrm{OH}^{-}\right]=\sqrt{\mathrm{K}_{\mathrm{h}} \cdot \mathrm{c}}\)

Question 37.
(a) (i) What are lyophilic and lyophobic sols? Give one example of each type. Why are
Answer:
(a) (i) Lyophilic Sols: Colloidal sols directly formed by mixing substances like gums, gelatin, starch, rubber, etc. with a suitable liquid (The dispersion medium) are lyophilic sols.
An important characteristic of these sols is that if the dispersion medium is separated from the dispersed phase (say by evaporation) the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols. These sols are quite stable and cannot be easily coagulated.

Lyophobic sols: These colloidal sols can only be prepared by some special methods. These sols are readily precipitated on the addition of small amount of electrolytes, by heating or by shaking and hence are not stable.
Hydrophobic sols are water hating. They are formed by indirect method. These sols are irreversible sols. These sols are readily precipitated by the addition of small amount of electrolytes, by heating or by shaking and hence are not stable.

(ii) 1. The substances when added to a catalysed reaction decreases or completely destroys the activity of a catalyst are often known as catalytic poisons.
2. In the reaction 2SO₂ + O₂ → 2SO₃ with Pt catalyst, the catalyst poison is AS₂O₃

[OR]

(b) (i) 1. Substances like AgCl, PbS04 are sparingly soluble in water. The solubility product can be determined using conductivity experiments.
2. Let us consider AgCl as an example
Agci_(s)) ⇌ Ag⁺ + Cl^–
K_Sp = [Ag ⁺] [Cl^–]

3. Let the concentration of [Ag⁺] be ‘C’ mol L^-1
If [Ag⁺] = C, then [Cl^–] is also equal to C mol L^-1.
.’. K_sp = C.C
K_sp = C²

4. The relationship between molar conductance and equivalent conductance is

(ii) 1. Formic acid contains both an aldehyde as well as an acid group. Hence, like other aldehydes, formic acid can easily be oxidised and therefore acts as a strong reducing agent.

2. Formic acid reduces Tollen’s reagent (ammonical silver nitrate solution) to metallic silver.

3. Formic acid reduces Fehling’s solution. It reduces blue coloured cupric ions to red coloured cuprous ions.

Question 38.
(a) (i) What are the uses of nitrobenzene? (2)
(ii) Write a note on formation of -helix. (3)
[OR]
(b) (i) Define TFM value. (2)
(ii) Differentiate thermoplastic and thermosetting. (3)
Answer:
(a) (i) 1. Nitro benzene is used to produce lubricating oils in motors and machinery.
2. It is used in the manufacture of dyes, drugs, pesticides, synthelic rubber, aniline and explosives like TNT, TNB.
(ii) 1. In the a-helix sub-structure, the aminoacids are arranged in a right handed helical (spiral) structure and are stabilised by the hydrogen bond between the carbonyl oxygen one aminoacid (nth residue) with amino hydrogen of the fifth residue (n+4th residue).

2. The side chains of the residues protrude outside of the helix. Each turn of an a-helix contains about 3.6 residues and is about 5.4 A long.

3. The amino acid pro line produces a line in the helical structure and often called as a helical breaker due to its rigid cyclic structure.

4. Many fibrous proteins such as a-Keratin in hair, nails, wool, skin and myosin in muscles have a-helix structure. Stretching property of human hair is due to the helical structure of a-keratin in hair.

5. Structure of α – helix

[OR]

(b) (i) 1. The quality of a soap is described in terms of total fatty matter (TFM value). It is defined as the total amount of fatty matter that can be separated from a sample after spliting with mineral acids. Higher the TFM value in the soap, better is its quality.
2. As per BIS standards, Grade I soaps should have 76% minimum TFM value (ii) Difference between thermoplastic and thermosetting:

Thermoplastic :

1. They soften on heating and harden on cooling, and they can be remoulded.
2. They consists of linear long chain polymers and low molecular weights polymers.
3. All the polymer chains are held together by weak Vanderwaals forces.
4. They are weak, soft and less brittle.
5. They are formed by addition polymerisation.
6. They are soluble in organic solvents.
7.  Example: PVC, polythene, polystrene etc.

Thermosetting

1. They do not soften on heating and they cannot be remoulded.
2. The consist of three dimensional network structure and high molecular weight polymers.
3. All the polymer chains are linked by strong covalent.
4. They are strong, hard and more brittle.
5. They are formed by condensation polymerisation.
6. They are insoluble in organic solvents.
7. Example: Bakelite, melamine etc.