# HBSE Class 12th Physics Solved Question Paper 2020

HBSE Class 12th Physics Solved Question Paper 2020

HBSE Class 12 Physics Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 12 Physics 2020. HBSE 12th Question Paper Download 2020. HBSE Class 12 Physics Paper Solution 2020. Haryana Board Class 12th Physics Question Paper 2020 Pdf Download with Answer.

SET-A

Q1.(i) Write SI unit of resistivity of a semiconductor.

Ans. Ohm-metre or -m

(ii) What is the value of power factor of a purely capacitive ac circuit ?

Ans. Zero

(iii) Which phenomenon illustrates the nature of light waves ?

Ans. Polarisation of Light

(iv) What is SI unit of activity of radioactive substance ?

Ans. Becquerel or Bq

(v) Draw logic symbol of NOR gate.

Ans.

(vi) How does the power radiated by an antenna vary with wavelength ?

Ans. Power radiated (P) ∝ 1/λ²

(vii) What is the maximum current that can be drawn from a storage battery of emf 10 V and internal resistance 0.2 Ω ?

Ans. Imax = e/r = 10V/0.2Ω = 50A

(viii) The force of attraction between a positively charged particle and a negatively charged particle is F. When distance between them is made one fourth, what will be the value of this force ?

Ans. New Force = 16 × Old Force

F’ = 16F (:. F = G×m1×m2/r²,  F ∝ 1/r² )

(ix) Write SI Unit of magnetic flux density.

Ans. Tesla

(x) Which dopant cannot be used to make silicon (Si) a p-type semi-conductor ?

(A) In

(B) Al

(C) B

(D) P (5th group element)

Ans. (D) P (5th group element)

(xi) The ionization energy of hydrogen atom is :

(A) 13.6 J

(B) 13.6 eV

(C) 1 eV

(D) 10.2 eV

Ans. (B) 13.6 eV

(xii) A thin convex lens of focal length 15 cm is placed in contact with a thin concave lens of focal length 15 cm. What is the power of the combination ?

(A) 0 D

(B) ∞ D

(C) 3 D

(D) 6 D

Ans. (A) 0 D

1/F = 1/F1 + 1/F2

1/F = 1/(+15) + 1/(-15)

P = 1/F = 0

(xiii) Which of the following has lowest wavelength ?

(A) Gamma rays

(B) X-rays

(C) Infrared rays

(D) Short radio waves

Ans. (A) Gamma rays

(xiv) In the magnetic meridian at a certain location, the horizontal component of earth’s magnetic field is 0.16 G and angle of dip is 60°. The earth’s magnetic field at this location is :

(A) 0.32 G

(B) 0.36 G

(C) 0.18 G

(D) 0.16 G

Ans. (A) 0.32 G

BH = B cosδ

0.16 = B cos60°

0.16 = B (1/2)

B = 0.16 × 2 = 0.32 G

Q2. What are electric field lines ? Draw these for a point charge -Q.

Ans. An electric field line is an imaginary line or curve drawn from a point of an electric field such that tangent to it (at any point) gives the direction of the electric field at that point. Electric field lines starting from positive charge and terminating at a negative charge.

Q3. Derive relation between electric field E and potential V at a point.

Ans. |E| dr = V – (V + dV) = – dV

|E| = – dV/dr

Q4. Calculate the equivalent resistance between the points A and B in the given Figure.

Ans. 1/RAB = 1/9 + 1/9 + 1/9 + 1/9 = 4/9

RAB = 9/4 Ω

Q5. What is demodulation ?

Ans. The process of recovering of audio frequency signal from the modulated carrier waves is called detection or demodulation. It is the reverse process of modulation.

Q6. Calculate the energy equivalent of 1.6 gram of a substance first in Joules and then in MeV.

Ans. E = mc² = (1.6×10-³ kg) × (3×10⁸ m/s)²

= (14.4×10¹³) ÷ (1.6×10-¹⁹) = 9×10³² eV

Q7. If the work function of Caesium is 2.14 eV then find its threshold frequency.

Ans. W = hv

v = W/h = (2.14×1.6×10-¹⁹) ÷ (6.6×10-³⁴) = 5.2×10¹⁴ Hz

Q8. Define Displacement current and write its mathematical form.

Ans. In electromagnetism, displacement current is a quantity appearing in Maxwell’s equations that is defined in terms of the rate of change of electric displacement field. Displacement current has the units of electric current density, and it has an associated magnetic field just as actual currents do.

It is mathematically represented as

Id = dQ/dt = ϵ dϕ/dt

Q9. State Lenz’s law.

Ans. Lenz’s law states that the current induced in the circuit which is due to the change in the magnetic field is directed which opposes the change in the flux and it exerts the mechanical force which opposes the motion.

Q10. A coil of inductance 0.5 H and a resistance of 100 Ω are connected to a 240V, 50Hz ac supply in series. If the rms current in the circuit is Irms = 1.29 A, calculate the rms value of potential drop across each element.

Ans. VR = Irms × R = 1.29×100 = 129 V

VL = Irms × XL = 1.29×2π×50×0.5 = 202.5 V

Q11. Using Biot-Savart’s law derive an expression for the magnetic field at the centre of a current carrying circular loop.

Ans. Biot Savart law states that If there is a current carrying wire and a point P close to it, then the intensity of the magnetic field produced at that point, due to a very small part of the wire is given by

d= μ/4π × IdLsinθ/r²

B = ∫ dB = μ/4π × I(2πr)/r² = μI/2r

Q12. Draw a circuit diagram of a full wave rectifier and explain its output waveform.

Ans. Rectifier is a circuit which convert ac to unidirectional pulsating output. In other words its convert ac to dc.

Full Wave Rectifier–

Waveforms of input and output voltages–

Q13. Draw the energy level diagram for hydrogen atom and show the transitions for Lyman and Balmer series in its spectrum.

Ans. When an electron makes a transition from one orbit to the other in an atom, it gains or loses energy due to which it emits wavelengths corresponding to the energy differences. These wavelengths are observed as lines called spectral lines. Different wavelengths are emitted for different transitions.

Q14. What is photoelectric effect ? Explain variation of photoelectric current with applied potential.

Ans. The photoelectric effect refers to the emission, or ejection of electrons from the surface of a metal in response to incident light. This takes place because of the energy of incident photons of light have energy more than the work potential of the metal surface, ejecting electrons with positive kinetic energy. These electrons are used as the current and it is known as photo electric current.

Q15. In Young’s double-slit experiment the distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of the light used if the slits are separated by 0.28 mm and the screen is placed 1.4 m away.

Ans. β = 1.2/4 = 0.3 cm = 3 mm

β = D/d × λ

Wavelength (λ) = dβ ÷ D = (0.28×10-³)(3×10-³) ÷ (1.4) = 600 nm

Q16. Explain sharpness of resonance in a series LCR circuit. Define its quality factor Q and write expression for it.

Ans. Sharpness of resonance in a series resonant circuit is defined by the  Q factor . This can be defined as how quickly the energy of the oscillating system decays.

Q factor or quality factor is a dimensionless parameter that is used to describe the underdamped resonator and characterizes the bandwidth and centre frequency of the resonator.

Q factor in a series circuit, Q = 1/R × √L/C

Q factor in a parallel circuit, Q = R × √C/L

Q17. Draw the circuit diagram of a Wheatstone bridge. Derive its balance condition to give null deflection in galvanometer.

Ans. It works on the principle of null deflection, which means the ratio of their resistances are equal and hence no current flows through the circuit. Under normal conditions, the bridge will be in the unbalanced condition where current flows through the galvanometer. The bridge will be in a balanced condition when no current flows through the galvanometer. One may achieve this condition by adjusting the known resistance and variable resistance.

The Wheatstone bridge principle states that if four resistances P, Q, R, and S are arranged to form a bridge with a cell and key between A and C, and a galvanometer between B and D then the bridge is said to be balanced when the galvanometer shows a zero deflection.

In balanced condition, Ig = 0

so VB = VD  or  P/Q = R/S .This is called condition of balance.

Q18. State and prove Gauss’s law in Electrostatics.

Ans. According to Gauss’s theorem the net-outward normal electric flux through any closed surface of any shape is equivalent to 1/∈ times the total amount of charge contained within that surface.

Proof of Gauss’s Theorem

Let’s say the charge is equal to q.

Let’s make a Gaussian sphere with radius = r

Now imagine surface A or area ds has a ds vector

At ds, the flux is:

dΦ  = E(vector) ds(vector) cos θ

But , θ = 0

Hence , Total flux Φ = E × 4πr²

Here, E = 1/4π∈ × q/r²

Hence,  Φ = 1/4π∈ × q/r² × 4πr²

Φ = q/∈

As per the Gauss law, the total flux associated with a sealed surface equals 1/∈ times the charge encompassed by the closed surface.

Q19. Draw a labelled ray diagram showing image formation in an astronomical telescope. Derive expression for its magnifying power.

Ans. Astronomical Telescope– Astronomical Telescope is used to observe objects which are very far from us. Telescopes produce magnified images of distant objects. It produces virtual and﻿ inverted image and is used to see heavenly bodies like﻿ sun, stars, planets etc. so the inverted image does not﻿ affect the observation.﻿

Principle– It is based on the principle that when rays﻿ of light are made to incident on an objective from a﻿ distant object, the objective forms the real and inverted﻿ image at its focal plane. The eye lens is so adjusted that﻿ the final image is formed at least distance of distinct﻿ vision.﻿

Construction– The refracting type astronomical﻿ telescope consists of two convex lenses one of which is﻿ called the objective and the other eye piece. The objective﻿ is a convex lens of large focal length and large aperture. It is generally a combination of two lenses in contact so﻿ as to reduce spherical and chromatic aberrations. The﻿ eye piece is also a convex lens but of short focal length﻿ and small aperture. The objective is mounted at one end﻿ of a brass tube and the eye piece at the other end in a﻿ smaller tube which can slide inside the bigger tube﻿ carrying the objective.

Magnifying Power– The magnifying power of a refracting type astronomical telescope is defined as the ratio of angle subtended by the final image at eye to the angle subtended by the object at eye.

M = fo/fe (1 + fe/D)

OR

Draw a labelled ray diagram showing image formation in a compound microscope. Define its magnifying power and write expression for it.

Ans.

Magnifying Power– Magnifying power of a compound microscope is defined as, the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the image and object are situated at the least distance of distinct vision from the eye.

Q20. Draw a Circuit diagram for studying V-I characteristics for a transistor in common emitter (CE) configuration. Draw and explain its typical output V-I characteristics.

Ans. Output characteristics– These characteristics are obtained by plotting collector current IC versus collector emitter voltage VCE at a fixed value of base current IB. The base current is changed to some other fixed value and the observations of IC versus VCE are repeated. The figure given below represents the output characteristic curves.

OR

Explain the working of common emitter transistor amplifier using circuit diagram.

Ans. The common emitter amplifier is a three basic single-stage bipolar junction transistor and is used as a voltage amplifier. The input of this amplifier is taken from the base terminal, the output is collected from the collector terminal and the emitter terminal is common for both the terminals.

Working– The below circuit diagram shows the working of the common emitter amplifier circuit and it consists of voltage divider biasing, used to supply the base bias voltage as per the necessity. The voltage divider biasing has a potential divider with two resistors are connected in a way that the midpoint is used for supplying base bias voltage. There are different types of electronic components in the common emitter amplifier which are R1 resistor is used for the forward bias, the R2 resistor is used for the development of bias, the RL resistor is used at the output it is called the load resistance. The RE resistor is used for thermal stability. The C1 capacitor is used to separate the AC signals from the DC biasing voltage and the capacitor is known as the coupling capacitor.

The figure shows that the bias vs gain common emitter amplifier transistor characteristics if the R2 resistor increases then there is an increase in the forward bias and R1 & bias are inversely proportional to each other. The alternating current is applied to the base of the transistor of the common emitter amplifier circuit then there is a flow of small base current. Hence there is a large amount of current flow through the collector with the help of the RC resistance. The voltage near the resistance RC will change because the value is very high and the values are from 4 to 10k ohm. Hence there is a huge amount of current present in the collector circuit which amplified from the weak signal, therefore common emitter transistors work as an amplifier circuit.

Q21. Explain the principle and working of a cyclotron using its schematic sketch. Show that time of revolution of an ion is independent of its speed or radius of its orbit.

Ans. Cyclotron– Cyclotron is a device used to accelerate charged particles to high energies.

Principle– Cyclotron works on the principle that a charged particle moving normal to a magnetic field experiences magnetic lorentz force due to which the particle moves in a circular path.

Construction– It consists of a hollow metal cylinder divided into two sections D1 and D2 called Dees, enclosed in an evacuated chamber (Figure). The Dees are kept separated and a source of ions is placed at the centre in the gap between the Dees. They are placed between the pole pieces of a strong electromagnet. The magnetic field acts perpendicular to the plane of the Dees. The Dees are connected to a high frequency oscillator.

Working– When a positive ion of charge q and mass m is emitted from the source, it is accelerated towards the Dee having a negative potential at that instant of time. Due to the normal magnetic field, the ion experiences magnetic lorentz force and moves in a circular path. By the time the ion arrives at the gap between the Dees, the polarity of the Dees gets reversed.

Hence the particles is once again accelerated and moves into the other Dee with a greater velocity along a circle of greater radius. Thus the particle moves in a spiral path of increasing radius and when it comes near the edge, it is taken out with the help of a deflection plate (D.P.). The particle with high energy is now allowed to hit the target T.

When the particle moves along a circle of radius r with a velocity v, the magnetic Lorentz force provides the necessary centripetal force.

Bqv = mv²/r

v = qBr/m

Period of revolution, T = 2πr/v

T = 2πrm/qBr = 2πm/qB

T = 2πm/qB = constant

Thus ‘T’ is independent of ‘v’.

OR

Derive an expression for the force between two long straight parallel conductors carrying current in same direction. Hence define one Ampere.

Ans.

Consider a small length L of the long straight conductor

B1 = μI1/2πd

B2 = μI2/2πd

F12 = I2B1

F21 = I1B2

F = F12 = F21 = μI1I2L/2πd

F/L = μI1I2/2πd

Definition of ampere– Fundamental unit of current ampere has been defined assuming the force between the two current carrying wires as standard. The force between two parallel current carrying conductor of separation r is 2×10-⁷ N/m.

Thus 1 ampere is the current which when flowing in each of parallel conductors placed at separation 1 m in vacuum exert a force of 2×10-⁷ N/m on 1 m length of either wire.

SET-B

Q1.(i) The force of repulsion between two positively charged particles is F. When distance between them is made one fourth, what will be the value of this force ?

Ans. New Force = 16 × Old Force

F’ = 16F (:. F = G×m1×m2/r²,  F ∝ 1/r² )

(ii) The maximum current that can be drawn from a storage battery of emf 6V is 10A. What is its internal resistance ?

Ans. Imax = e/r

r = e/Imax = 6V/10A = 0.6Ω

(iii) Write SI Unit of electrical power.

Ans. Watt

(iv) How much activity is one Becqueral (Bq) ?

Ans. 1 Becquerel = 1 disintegration/second

1 Bq = 1 dis/sec or 1 decay/second

(v) Draw logic symbol of NAND gate.

Ans.

(vi) What do you mean by channel in a communication system ?

Ans. The channel is the physical medium that connects the transmitter and the receiver.

(vii) What is the value of power factor of a purely resistive ac circuit ?

Ans. One

(viii) Write SI Unit of magnetic flux.

Ans. Weber

(ix) Name any one effect of light which does not show its particle nature.

Ans. interference or Diffraction

(x) In the magnetic meridian at a certain location, earth’s magnetic field is 0.52 G and the angle of dip is 60°. The horizontal component of earth’s magnetic field at this location is :

(A) 0.36 G

(B) 0.52 G

(C) 0.13 G

(D) 0.26 G

Ans. (D) 0.26 G

B= B cosδ = 0.52 cos60° = 0.52 (1/2) = 0.26 G

(xi) Which of the following has highest   frequency ?

(A) Ultraviolet rays

(B) Long radio waves

(C) Gamma rays

(D) X-rays

Ans. (C) Gamma rays

(xii) The energy of an atom in ground state is -3.4 eV. The ionisation energy of this atom   is :

(A) 1.7 eV

(B) 6.8 eV

(C) 3.4 eV

(D) 5.1 eV

Ans. (C) 3.4 eV

(xiii) Which dopant cannot be used to make Germanium (Ge) an n-type semiconductor ?

(A) P

(B) In

(C) As

(D) Sb

Ans. (B) In (indium)

(xiv) A thin convex lens of focal length 15 cm is placed in contact with a thin concave lens of focal length 10 cm. What is the focal length of the combination ?

(A) +25 cm

(B) +5 cm

(C) -30 cm

(D) -5 cm

Ans. (C) -30 cm

1/F = 1/F1 + 1/F2

1/F = 1/(+15) + 1/(-10) = 1/(-30)

F = -30

Q2. Draw a labelled block diagram of a generalised communication system. What is the function of transmitter.

Ans.

It modifies the input signal for efficient transmission. The signal must pass through an electronic circuit containing noise filter, analog-to-digital converter, encoder, modulator and signal amplifier. The signal is amplified for transmission and then released as electromagnetic waves at the end of the circuit.

Q3. The ground state energy of hydrogen atom is -13.6 eV. Find the kinetic energy and the potential energy of the electron in this state.

Ans. K.E. = -Total Energy = -(-13.6) = 13.6 eV

Total Energy = K.E. + P.E.

P.E. = Total Energy – K.E. = -13.6-13.6 = -27.2 eV

Q4. Calculate the equivalent resistance between the points A and B in the given Figure.

Ans. 1/RAB = 1/7 + 1/7 + 1/7 + 1/7 = 4/7

RAB = 7/4 Ω

Q5. Find the energy of each photon in a monochromatic light beam of wavelength 632.8 nm.

Ans. E = hv = hc/λ = (6.63×10-³⁴)(3×10⁸) ÷ (632.8×10-⁹)

= 3.14×10-¹⁹ J = (3.14×10-¹⁹) ÷ (1.6×10-¹⁹) eV = 1.96 eV

Q6. Write any two important features of electric field E and magnetic field B of an electromagnetic wave.

Ans. E/B = C, E⊥B, E×B is direction of propagation.

Q7. State Faraday’s law of electromagnetic induction. Write mathematical form also.

Ans. First Law– Whenever magnetic flux linked with the circuit changes, an emf is induced in it.

Second Law– The magnitude of emf is directly proportional to the change in magnetic flux.

The mathematical expression for the Faraday’s law, E = – dϕ/dt

Q8. A 80 mH inductor and a 60 µF capacitor are connected to a 230 V, 50 Hz supply. If the rms current in the circuit is Irms = 8.24 A, calculate the rms value of potential drop across each element.

Ans. VL = Irms × XL = 8.24 × 2πvL = 8.24 × 2π × 50 × 80 × 10-³ = 207 V

VC = IrmsXC = 8.24 × (1/2πvC) = 8.24 × (1/2π×50×60×10-⁶) = 437.36 V

Q9. Using Ampere’s law find an expression for the magnetic field due to a straight long current carrying wire at a distance r outside it.

Ans. Using Ampere’s law,  ∮B.dL = μI

B.2πr = μI

B = μI/2πr

Q10. What are equipotential surfaces ? Draw equipotential surface due to point charge.

Ans. Equipotential surface is a surface which has equal potential at every Point on it. Equipotential surfaces due to single point charge are concentric sphere having charge at the centre.

Q11. State any two important properties of electrostatic field lines.

Ans. important properties of electrostatic field lines-

(i) Field lines never intersect each other.

(ii) They are perpendicular to the surface charge.

(iii) The field is strong when the lines are close together, and it is weak when the field lines move apart from each other.

(iv) The number of field lines is directly proportional to the magnitude of the charge.

Q12 Draw the circuit diagram of an half wave rectifier and explain its output waveform.

Ans.

Q13. What are the functions of the control rods and the moderators ? Explain Nuclear Fission with example.

Ans. Control Rods– The control rods are made up of cadmium or boron. The function of this control rod is to absorb the neutron efficiently. It can also be used to shut down the reactor.

Moderator– Heavy water and graphite are used as a moderator. The function of the moderator is to slow down the neutrons which are emitted in the fission reaction.

Nuclear Fission– Nuclear fission is a nuclear reaction in which the nucleus of an atom splits into smaller parts.  The fission process often produce free neutrons and photons and release a large amount of energy. e.g.

Q14. Write Einstein’s photoelectric equation. Define the threshold frequency and the stopping potential.

Ans. Einstein’s photoelectric equation is-

hv = hv0 = ½mv²

½mv² = h(v−v0)

Threshold frequency– The minimum value of frequency of incident radiations below which the photoelectric emission stops altogether is called threshold frequency.

Stopping potential– The value of retarding potential at which the photoelectric current becomes zero is called stopping potential for the given frequency of incident radiations.

Q15. Explain the principle and working of a transformer using a diagram.

Ans. Transformer– An electrical device that can change the A.C. current is known as a transformer.

Principle– A transformer works on the principle of mutual induction. Mutual induction is the phenomenon by which when the amount of magnetic flux linked with a coil changes, an E.M.F. is induced in the neighboring coil.

Construction– A transformer is made up of a rectangular iron core. Two coils, a primary (P) coil with two sides P1 and P2, and a secondary (S) coil with two sides S1 and S2. Both these coils are insulated from the Ferro-magnetic iron core. The source of the alternate current is connected to the primary winding and the output is obtained through the secondary winding which is connected in parallel to a resistance R.

Working– The transformer works on the principle of Faraday’s law of electromagnetic induction and mutual induction. There are usually two coils primary coil and secondary coil on the transformer core. The core laminations are joined in the form of strips. The two coils have high mutual inductance. When an alternating current pass through the primary coil it creates a varying magnetic flux. As per faraday’s law of electromagnetic induction, this change in magnetic flux induces an emf (electromotive force) in the secondary coil which is linked to the core having a primary coil. This is mutual induction.

Q16. Draw a circuit diagram of a meter bridge. Find the value of unknown resistance by using it.

Ans. The metre bridge, also known as the slide wire bridge consists of a one metre long wire of uniform cross sectional area, fixed on a wooden block. A scale is attached to the block. Two gaps are formed on it by using thick metal strips in order to make the Wheat stone’s bridge. The terminal B between the gaps is used to connect galvanometer and jockey.

A resistance wire is introduced in gap S and the resistance box is in gap R. One end of the galvanometer is connected to terminal D and its other end is connected to a jockey. As the jockey slides over the wire AC, it shows zero deflection at the balancing point (null point).

The unknown resistance can be calculated as :

X = R × l/(100-l)

Q17. State and prove Gauss law in electrostatics.

Ans. According to Gauss’s theorem the net-outward normal electric flux through any closed surface of any shape is equivalent to 1/ε times the total amount of charge contained within that surface.

Proof of Gauss’s Theorem

Let’s say the charge is equal to q.

Let’s make a Gaussian sphere with radius = r

Now imagine surface A or area ds has a ds vector

At ds, the flux is:

dΦ = E(vector) ds(vector) cos θ

But θ = 0

Hence, Total flux Φ = E × 4πr²

Here, E = 1/4π∈ × q/r²

Hence, Φ = 1/4π∈ × q/r² × 4πr²

Φ = q/∈

As per the Gauss law, the total flux associated with a sealed surface equals 1/∈ times the charge encompassed by the closed surface.

Q18. In Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. Calculate the fringe separation for light of wavelength 600 nm.

Ans. Fringe (β) = D/d × λ = 1.4/0.28×10-³ × (600×10-⁹) = 3×10-³ m = 3 mm

Q19. Explain the principle and working of a cyclotron using its schematic sketch. Show that time of revolution of an ion is independent of its speed or radius of its orbit.

Ans. Cyclotron– Cyclotron is a device used to accelerate charged particles to high energies.

Principle– Cyclotron works on the principle that a charged particle moving normal to a magnetic field experiences magnetic lorentz force due to which the particle moves in a circular path.

Construction– It consists of a hollow metal cylinder divided into two sections D1 and D2 called Dees, enclosed in an evacuated chamber (Figure). The Dees are kept separated and a source of ions is placed at the centre in the gap between the Dees. They are placed between the pole pieces of a strong electromagnet. The magnetic field acts perpendicular to the plane of the Dees. The Dees are connected to a high frequency oscillator.

Working– When a positive ion of charge q and mass m is emitted from the source, it is accelerated towards the Dee having a negative potential at that instant of time. Due to the normal magnetic field, the ion experiences magnetic lorentz force and moves in a circular path. By the time the ion arrives at the gap between the Dees, the polarity of the Dees gets reversed.

Hence the particles is once again accelerated and moves into the other Dee with a greater velocity along a circle of greater radius. Thus the particle moves in a spiral path of increasing radius and when it comes near the edge, it is taken out with the help of a deflection plate (D.P.). The particle with high energy is now allowed to hit the target T.

When the particle moves along a circle of radius r with a velocity v, the magnetic Lorentz force provides the necessary centripetal force.

Bqv = mv²/r

v = qBr/m

Period of revolution, T = 2πr/v

T = 2πrm/qBr = 2πm/qB = constant

Thus ‘T’ is independent of ‘v’.

OR

Derive an expression for the force between two long straight parallel conductors carrying current in same direction. Hence define one Ampere.

Ans.

Consider a small length L of the long straight conductor

B1 = μI1/2πd

B2 = μI2/2πd

F12 = I2B1

F21 = I1B2

F = F12 = F21 = μI1I2L/2πd

F/L = μI1I2/2πd

Definition of ampere : Fundamental unit of current ampere has been defined assuming the force between the two current carrying wires as standard. The force between two parallel current carrying conductor of separation r is 2×10-⁷ N/m.

Thus 1 ampere is the current which when flowing in each of parallel conductors placed at separation 1 m in vacuum exert a force of 2×10-⁷ N/m on 1 m length of either wire.

Q20. Draw a labelled ray diagram showing image formation in an astronomical telescope. Derive expression for its magnifying power.

Ans. Astronomical Telescope– Astronomical Telescope is used to observe objects which are very far from us. Telescopes produce magnified images of distant objects. It produces virtual and﻿ inverted image and is used to see heavenly bodies like﻿ sun, stars, planets etc. so the inverted image does not﻿ affect the observation.﻿

Principle– It is based on the principle that when rays﻿ of light are made to incident on an objective from a﻿ distant object, the objective forms the real and inverted﻿ image at its focal plane. The eye lens is so adjusted that﻿ the final image is formed at least distance of distinct﻿ vision.﻿

Construction– The refracting type astronomical﻿ telescope consists of two convex lenses one of which is﻿ called the objective and the other eye piece. The objective﻿ is a convex lens of large focal length and large aperture. It is generally a combination of two lenses in contact so﻿ as to reduce spherical and chromatic aberrations. The﻿ eye piece is also a convex lens but of short focal length﻿ and small aperture. The objective is mounted at one end﻿ of a brass tube and the eye piece at the other end in a﻿ smaller tube which can slide inside the bigger tube﻿ carrying the objective.

Magnifying Power– The magnifying power of a refracting type astronomical telescope is defined as the ratio of angle subtended by the final image at eye to the angle subtended by the object at eye.

M = fo/fe (1 + fe/D)

OR

Draw a labelled ray diagram showing image formation in a compound microscope. Define its magnifying power and write expression for it.

Ans.

Magnifying Power– Magnifying power of a compound microscope is defined as, the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the image and object are situated at the least distance of distinct vision from the eye.

Q21. Draw a Circuit diagram for studying V-I characteristics for a transistor in common emitter (CE) configuration. Draw and explain its typical output V-I characteristics.

Ans. Output characteristics– These characteristics are obtained by plotting collector current IC versus collector emitter voltage VCE at a fixed value of base current IB. The base current is changed to some other fixed value and the observations of IC versus VCE are repeated. The figure given below represents the output characteristic curves.

OR

Explain the working of transistor as a amplifier by using its circuit diagram.

Ans. Transistor Amplifier– A transistor acts as an amplifier by raising the strength of a weak signal. The DC bias voltage applied to the emitter base junction, makes it remain in forward biased condition. This forward bias is maintained regardless of the polarity of the signal. The below figure shows how a transistor looks like when connected as an amplifier.

The low resistance in input circuit, lets any small change in input signal to result in an appreciable change in the output. The emitter current caused by the input signal contributes the collector current, which when flows through the load resistor RL, results in a large voltage drop across it. Thus a small input voltage results in a large output voltage, which shows that the transistor works as an amplifier.

SET-C

Q1.(i) The force of repulsion between two negatively charged particles is F. When distance between them is halved, what will be the value of this force ?

Ans. New Force = 4 × Old Force

F’ = 4F (:. F = G×m1×m2/r²,  F ∝ 1/r² )

(ii) What is the maximum current that can be drawn from a storage battery of emf 12V and internal resistance 0.3Ω ?

Ans. Imax = e/r = 12V/0.3Ω = 40A

(iii) Write SI Unit of resistivity of a metal.

Ans. Ohm-metre or Ω-m

(iv) How much power is dissipated in a purely inductive ac circuit ?

Ans. Zero

(v) The bluish colour of sky is due to which effect of light ?

Ans. Scattering of Light

(vi) Who discovered radioactivity ?

Ans. Henry Becquerel

(vii) Draw logic symbol of OR gate.

Ans.

(viii) What is the relation between least size of antenna and wavelength of the signal transmitted  ?

Ans. Minimum size, Lmin λ/4

(ix) Write SI Unit of magnetic field.

Ans. Tesla

(x) A thin convex lens of focal length 10 cm is placed in contact with a thin concave lens of focal length 10 cm. What is the power of the combination ?

(A) ∞ D

(B) +5 D

(C) 0 D

(D) -5 D

Ans. (C) 0 D

1/F = 1/F1 + 1/F2

1/F = 1/(+10) + 1/(-10)

P = 1/F = 0

(xi) Which spectral series of hydrogen atom lies in the ultraviolet region ?

(A) Lyman

(B) Balmer

(C) Paschen

(D) Brackett

Ans. (A) Lyman

(xii) Which dopant cannot be used to make silicon (Si) an n-type semi-conductor ?

(A) B

(B) As

(C) P

(D) Sb

Ans. (A) B (Boron)

(xiii) Which of the following has highest wavelength  ?

(A) Gamma rays

(B) X-rays

(C) Ultraviolet rays

(D) Radio waves

Ans. (D) Radio waves

(xiv) In the magnetic meridian at a certain place, earth’s magnetic field is 0.32 G and its horizontal component is 0.16 G. The angle of dip of earth’s magnetic field at this location is :

(A) 30°

(B) 60°

(C) 45°

(D) 22°

Ans. (B) 60°

B= B cosδ

0.16 = 0.32 cosδ

cosδ = 0.16/0.32 = 1/2 = cos60°

δ = 60°

Q2. Explain amplitude modulation.

Ans. Amplitude modulation is a process by which the wave signal is transmitted by modulating the amplitude of the signal. It is often called AM and is commonly used in transmitting a piece of information through a radio carrier wave. Amplitude modulation is mostly used in the form of electronic communication.

Q3. Calculate the energy equivalent of 3.2 gram of a substance first in Joules and then in eV.

Ans. E = mc² = (3.2×10-³ kg) × (3×10⁸ m/s)² = 28.8×10¹³ J

= (28.8×10¹³) ÷ (1.6×10-¹⁹) = 1.8×10³³ eV

Q4. Calculate the equivalent resistance between the points A and B in the given Figure.

Ans. 1/RAB = 1/5 + 1/5 + 1/5 + 1/5 = 4/5

RAB = 5/4 Ω

Q5. What are electric field lines ? Draw these for a point charge -Q.

Ans. An electric field line is an imaginary line or curve drawn from a point of an electric field such that tangent to it (at any point) gives the direction of the electric field at that point. Electric field lines starting from positive charge and terminating at a negative charge.

Q6. Derive the equation E = – dV/dr.

Ans. As you know, standard formula of electric field intensity and potential are given by

E = KQ/r² , V = KQ/r , here Q is the charge , r is the distance between charge and observation point .

V = KQ/r differentiate with respect to r

dV/dr = – KQ/r² = – E

dV = – E.dr

E = – dV/dr

Hence, relation between Electric field intensity and potential is E = -∆V/∆r

Here negative sign shows that on uniform electric field intensity if distance increases then potential difference decreases.

Q7. Derive an expression for the magnetic field at the centre of a circular current carrying loop.

Ans. Biot Savart law states that If there is a current carrying wire and a point P close to it, then the intensity of the magnetic field produced at that point, due to a very small part of the wire is given by

dB= μ/4π × IdLsinθ/r²

B = ∫dB = μ/4π × I(2πr)/r² = μI/2r

Q8. State Lenz’s law.

Ans. Lenz’s law states that the current induced in the circuit which is due to the change in the magnetic field is directed which opposes the change in the flux and it exerts the mechanical force which opposes the motion.

Q9. Define displacement current and write its mathematical form.

Ans. In electromagnetism, displacement current is a quantity appearing in Maxwell’s equations that is defined in terms of the rate of change of electric displacement field. Displacement current has the units of electric current density, and it has an associated magnetic field just as actual currents do.

It is mathematically represented as

Id = dQ/dt = ϵ dϕ/dt

Q10. If the work function of potasium is 2.30 eV then find its threshold frequency.

Ans. W = hv

Threshold Frequency (v) = W/h = (2.30×1.6×10-¹⁹) ÷ (6.6×10-³⁴) = 5.2×10¹⁴ Hz

Q11. A 100 µF capacitor and a resistance of 40 Ω are connected to a 110 V, 60 Hz ac supply in series. If the rms current in the circuit is Irms = 2.28 A, calculate the rms value of potential drop across each element.

Ans. VC = Irms × XC = 2.28 × (1/2×3.14×60×100×10-⁶ = 60.5 V

VR = Irms × R = 2.28×40 = 91.2 V

Q12. State and prove Gauss’s law in Electrostatics.

Ans. According to Gauss’s theorem the net-outward normal electric flux through any closed surface of any shape is equivalent to 1/ε times the total amount of charge contained within that surface.

Proof of Gauss’s Theorem

Let’s say the charge is equal to q.

Let’s make a Gaussian sphere with radius = r

Now imagine surface A or area ds has a ds vector

At ds, the flux is:

dΦ = E(vector) ds(vector) cos θ

But θ = 0

Hence, Total flux Φ = E × 4πr²

Here, E = 1/4π∈ × q/r²

Hence,  Φ = 1/4π∈ × q/r² × 4πr²

Φ = q/∈

As per the Gauss law, the total flux associated with a sealed surface equals 1/∈ times the charge encompassed by the closed surface.

Q13. How will you measure the internal resistance of a cell using a potentiometer ? Explain using circuit diagram.

Ans. Circuit diagram to measure internal resistance of a cell by potentiometer.

We know that a potentiometer is used to determine the internal resistance of a cell. Besides this, we will have a determination of potential difference using a potentiometer.The internal resistance of a cell using the potentiometer formula is given by :

r = (E/I – R) ……..(1)

And, Determination of Potential Difference Using Potentiometer, the formula is :

E = I (R + r) ……..(2)

We know by Ohm’s law :

V = IR = E – Ir ….(3)

Equation (3) indicates that the value of V is less than E by an amount equal to the fall of potential inside the cell due to its internal resistance.

From equation (3), we have :

r/R = (E – V)/V

Therefore, the internal resistance of the cell is :

r  = R (E – V)/V  ……..(4)

So, equation (4) is the Formula for Internal Resistance In Potentiometer.

Then, E = k L1   and   V = k L2  ; where k is the potential gradient along the wire.

r = R (L1-L2)/L2

Q14. Explain the phenomenon of resonance in a series LCR circuit. Calculate its resonant frequency.

Ans. Resonance is the phenomenon in the circuit when the output of that electric circuit is maximum at one particular frequency. In an LCR circuit, this frequency is determined by the values of inductance, conductance, and resistance. If the inductor, capacitor and resistor are connected in series then the circuit is called a Series resonance circuit. But, if they are connected in parallel, then the circuit is called Parallel Resonance circuit.

In LCR series circuits, resonance occurs when the value of inductive and capacitive reactances have equal magnitude but have a phase difference of 180°. Thus, they cancel each other.

Z = √R²+(XL-XC)²

At resonance,  Z = R and XL = XC

ωL = 1/ωC

ω² = 1/LC

ω = 1/√LC

Q15. What is photoelectric effect ? Explain effect of intensity of light on photoelectric current.

Ans. The photoelectric effect refers to the emission, or ejection of electrons from the surface of a metal in response to incident light. This takes place because of the energy of incident photons of light have energy more than the work potential of the metal surface, ejecting electrons with positive kinetic energy. These electrons are used as the current and it is known as photo electric current.

Intensity has nothing to do with the energy of the photon. Therefore, the intensity of the radiation is increased, the rate of emission increases but there will be no change in the kinetic energy of electrons. With an increasing number of emitted electrons, the value of the photoelectric current increases.

Q16. Discuss the three postulates of Bohr’s model for hydrogen atom.

Ans. First Postulate– Electron revolves round the nucleus in discrete circular orbits called stationary orbits without emission of radiant energy. These orbits are called stable orbits or non-radiating orbits.

Second Postulate– Electrons revolve around the nucleus only in orbits in which their angular momentum is an integral multiple of h/2π.

Third Postulate– When an electron makes a transition from one of its non-radiating orbits to another of lower energy, a photon is emitted having energy equal to the energy difference between the two states. The frequency of the emitted photon is then given by, v = (Ei-Ef)/h.

Q17. In Young’s double-slit experiment the distance between the central bright fringe and the second bright fringe is measured to be 6 mm. Determine the wavelength of the light used if the slits are separated by 0.28 mm and the screen is placed 1.4 m away.

Ans. β = 6/2 = 3 mm

β = D/d × λ

Wavelength (λ) = dβ ÷ D = (0.28×10-³ × 3×10-³) ÷ (1.4) = 600 nm

Q18. Draw the circuit diagram of a full wave rectifier and explain its output waveform.

Ans. Rectifier is a circuit which convert ac to unidirectional pulsating output. In other words its convert ac to dc.

Full Wave Rectifier–

Waveforms of input and output voltages–

Q19. Draw a labelled ray diagram showing image formation in an astronomical telescope. Derive expression for its magnifying power.

Ans. Astronomical Telescope– Astronomical Telescope is used to observe objects which are very far from us. Telescopes produce magnified images of distant objects. It produces virtual and﻿ inverted image and is used to see heavenly bodies like﻿ sun, stars, planets etc. so the inverted image does not﻿ affect the observation.﻿

Principle– It is based on the principle that when rays﻿ of light are made to incident on an objective from a﻿ distant object, the objective forms the real and inverted﻿ image at its focal plane. The eye lens is so adjusted that﻿ the final image is formed at least distance of distinct﻿ vision.﻿

Construction– The refracting type astronomical﻿ telescope consists of two convex lenses one of which is﻿ called the objective and the other eye piece. The objective﻿ is a convex lens of large focal length and large aperture. It is generally a combination of two lenses in contact so﻿ as to reduce spherical and chromatic aberrations. The﻿ eye piece is also a convex lens but of short focal length﻿ and small aperture. The objective is mounted at one end﻿ of a brass tube and the eye piece at the other end in a﻿ smaller tube which can slide inside the bigger tube﻿ carrying the objective.

Magnifying Power– The magnifying power of a refracting type astronomical telescope is defined as the ratio of angle subtended by the final image at eye to the angle subtended by the object at eye.

M= fo/fe (1 + fe/D)

OR

Draw a labelled ray diagram showing image formation in a compound microscope. Define its magnifying power and write expression for it.

Ans.

Magnifying Power– Magnifying power of a compound microscope is defined as, the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the image and object are situated at the least distance of distinct vision from the eye.

Q20. Draw a Circuit diagram for studying V-I characteristics for a transistor in common emitter (CE) configuration. Draw and explain its typical output V-I characteristics.

Ans. Output characteristics– These characteristics are obtained by plotting collector current IC versus collector emitter voltage VCE at a fixed value of base current IB. The base current is changed to some other fixed value and the observations of IC versus VCE are repeated. The figure given below represents the output characteristic curves.

OR

Explain the working of transistor as a amplifier by using its circuit diagram.

Ans. Transistor Amplifier– A transistor acts as an amplifier by raising the strength of a weak signal. The DC bias voltage applied to the emitter base junction, makes it remain in forward biased condition. This forward bias is maintained regardless of the polarity of the signal. The below figure shows how a transistor looks like when connected as an amplifier.

The low resistance in input circuit, lets any small change in input signal to result in an appreciable change in the output. The emitter current caused by the input signal contributes the collector current, which when flows through the load resistor RL, results in a large voltage drop across it. Thus a small input voltage results in a large output voltage, which shows that the transistor works as an amplifier.

Q21. Explain the principle and working of a cyclotron using its schematic sketch. Show that time of revolution of an ion is independent of its speed or radius of its orbit.

Ans. Cyclotron– Cyclotron is a device used to accelerate charged particles to high energies.

Principle– Cyclotron works on the principle that a charged particle moving normal to a magnetic field experiences magnetic lorentz force due to which the particle moves in a circular path.

Construction– It consists of a hollow metal cylinder divided into two sections D1 and D2 called Dees, enclosed in an evacuated chamber (Figure). The Dees are kept separated and a source of ions is placed at the centre in the gap between the Dees. They are placed between the pole pieces of a strong electromagnet. The magnetic field acts perpendicular to the plane of the Dees. The Dees are connected to a high frequency oscillator.

Working– When a positive ion of charge q and mass m is emitted from the source, it is accelerated towards the Dee having a negative potential at that instant of time. Due to the normal magnetic field, the ion experiences magnetic lorentz force and moves in a circular path. By the time the ion arrives at the gap between the Dees, the polarity of the Dees gets reversed.

Hence the particles is once again accelerated and moves into the other Dee with a greater velocity along a circle of greater radius. Thus the particle moves in a spiral path of increasing radius and when it comes near the edge, it is taken out with the help of a deflection plate (D.P.). The particle with high energy is now allowed to hit the target T.

When the particle moves along a circle of radius r with a velocity v, the magnetic Lorentz force provides the necessary centripetal force.

Bqv = mv²/r

v = qBr/m

Period of revolution, T = 2πr/v

T = 2πrm/qBr = 2πm/qB = constant

Thus ‘T’ is independent of ‘v’.

OR

Derive an expression for the force between two long straight parallel conductors carrying current in same direction. Hence define one Ampere.

Ans.

Consider a small length L of the long straight conductor

B1 = μI1/2πd

B2 = μI2/2πd

F12 = I2B1

F21 = I1B2

F = F12 = F21 = μI1I2L/2πd

F/L = μI1I2/2πd

Definition of ampere– Fundamental unit of current ampere has been defined assuming the force between the two current carrying wires as standard. The force between two parallel current carrying conductor of separation r is 2×10-⁷ N/m.

Thus 1 ampere is the current which when flowing in each of parallel conductors placed at separation 1 m in vacuum exert a force of 2×10-⁷ N/m on 1 m length of either wire.

SET-D

Q1.(i) What is a transducer ?

Ans. Any device that converts one form of energy into another is called transducer.

(ii) Draw logic symbol of AND gate.

Ans.

(iii) How much activity is one Becquerel (1 Bq) ?

Ans. 1 Becquerel = 1 disintegration/second

1 Bq = 1 dis/sec or 1 decay/second

(iv) Write SI Unit of magnetic intensity.

Ans. Ampere per metre or Am-¹ or Ampere/metre

(v) What is the value of power factor of a purely inductive ac circuit ?

Ans. Zero

(vi) The maximum current that can be drawn from a storage battery of emf 18 V is 30 A. What is its internal resistance ?

Ans. Imax = e/r

r = e/Imax = 18V/30A = 0.6Ω

(vii) Write SI Unit of electrical energy.

Ans. Joule

(viii) The force of attraction between a positively charged particle and a negatively charged particle is F. When distance between them is made half, what will be the value of this force ?

Ans. New Force = 4 × Old Force

F’ = 4F (:. F = G×m1×m2/r²,  F ∝ 1/r² )

(ix) Working of an optical fibre is based on which effect of light ?

Ans. Total internal reflection

(x) In the magnetic meridian of a certain place, earth’s magnetic field is 0.38 G and the angle of dip is 30°. The horizontal component of earth’s magnetic field at this location is :

(A) 0.19 G

(B) 0.38 G

(C) 0.33 G

(D) 0.57 G

Ans. (C) 0.33 G

B= B cosδ = 0.38 cos30° = 0.38 (√3/2) = 0.33 G

(xi) The energy of an atom in ground state is –4.9 eV. The ionisation energy of this atom  is :

(A) 2.45 eV

(B) 13.6 eV

(C) 9.8 eV

(D) 4.9 eV

Ans. (D) 4.9 eV

(xii) A thin convex lens of focal length 30 cm is placed in contact with a thin concave lens of focal length 20 cm. What is the focal length of the combination ?

(A) +50 cm

(B) -60 cm

(C) +10 cm

(D) -10 cm

Ans. (B) -60 cm

1/F = 1/F1 + 1/F2

1/F = 1/(+30) + 1/(-20) = 1/(-60)

F = -60 cm

(xiii) Which dopant cannot be used to make Germanium (Ge) a p-type semiconductor ?

(A) P

(B) In

(C) B

(D) Al

Ans. (A) P (phosphorus)

(xiv) Which of the following has lowest frequency ?

(A) Ultraviolet rays

(B) Long radio waves

(C) Gamma rays

(D) X-rays

Ans. (B) Long radio waves

Q2. Two electrostatic field lines never cross each other. Why ?

Ans. Electric lines of force never intersect each other because at the point of intersection, two tangents can be drawn to the two lines of force. This means two direction of electric field at the point of intersection, which is not possible.

Q3. What is Electrostatic Potential ? Write its unit.

Ans. It is defined as, An electric potential  is the amount of work needed to move a unit positive charge from a reference point to a specific point inside the field without producing any acceleration. Its SI unit is Volts.

Q4. Calculate the equivalent resistance between the points A and B in the given Figure :

Ans. 1/RAB = 1/3 + 1/3 + 1/3 + 1/3 = 4/3

RAB = 3/4 Ω

Q5. Find expression for magnetic field due to a straight infinite current carrying wire at a distance r outside it.

Ans. Using Ampere’s law,  ∮B.dL = μI

B.2πr = μI

B = μI/2πr

Q6. State Faraday’s law of electromagnetic induction and mathematical form.

Ans. First Law– Whenever magnetic flux linked with the circuit changes, an emf is induced in it.

Second Law– The magnitude of emf is directly proportional to the change in magnetic flux.

The mathematical expression for the Faraday’s law, E = – dϕ/dt

Q7. A resistor of 200 Ω and a capacitor of 15.0 µF are connected in series to a 220 V, 50 Hz ac supply. If the current in the circuit is Irms = 0.755 A, calculate the voltage across each element.

Ans. VR = Irms × R = 0.755 × 200 = 151 V

VC = Irms × XC = 0.755 × (1/2×3.14×50×15×10-⁶ = 160.3 V

Q8. Write any two important features of electric field E and magnetic field B of an electromagnetic wave.

Ans. important features of electric field E and magnetic field B of an electromagnetic wave-

(i) E/B = C

(ii) E⊥B

(iii) E×B is direction of propagation.

Q9. Find the energy of each photon in a monochromatic light beam of wavelength 550 nm.

Ans. E = hv = hc/λ = (6.63×10-³⁴ × 3×10⁸) ÷ (550×10-⁹)

= 3.6×10-¹⁹ J = (3.6×10-¹⁹) ÷ (1.6×10-¹⁹) = 2.26 eV

Q10. The ground state energy of hydrogen atom is -13.6 eV. Find the K.E. and potential energy of the electron in this state.

Ans. K.E. = -Total Energy = -(-13.6) = 13.6 eV

Total Energy = K.E. + P.E.

P.E. = Total Energy – K.E. = -13.6-13.6 = -27.2 eV

Q11. Draw a labelled block diagram of a generalised communication system. What is the function of a receiver ?

Ans.

Receiver– The transmitted signal is received at the receiver. The purpose of the receiver is to receive and de-modulate the message signal using a demodulator inside the receiver. The message is decoded and then sent to the user of information.

The function of receiver is to process the received signal to recover the appropriate message signal.

Q12. Draw a circuit diagram of a voltage regulator using a Zener diode and explain it.

Ans.

There is a series resistor connected to the circuit in order to limit the current into the diode. It is connected to the positive terminal of the d.c. It works in such a way the reverse-biased can also work in breakdown conditions. We do not use ordinary junction diode because the low power rating diode can get damaged when we apply reverse bias above its breakdown voltage. When the minimum input voltage and the maximum load current is applied, the Zener diode current should always be minimum.

Since the input voltage and the required output voltage is known, it is easier to choose a Zener diode with a voltage approximately equal to the load voltage, i.e. VZ  = VL.

Q13. Derive the relation N(t) = N0 e^−λt for radioactive decay. Sketch a graph of this relation. Show half life time (T½) in this graph.

Ans. When a radioactive material undergoes α, β or γ-decay, the number of nuclei undergoing the decay, per unit time, is proportional to the total number of nuclei in the sample material. So,

If N = total number of nuclei in the sample and ΔN = number of nuclei that undergo decay in time Δt then,

ΔN/ Δt ∝ N

Or, ΔN/ Δt = λN ……..(1)

where λ = radioactive decay constant or disintegration constant. Now, the change in the number of nuclei in the sample is, dN = – ΔN in time Δt. Hence, the rate of change of N (in the limit Δt → 0) is,

dN/dt = – λN

Or, dN/N = – λ dt

Now, integrating both the sides of the above equation, we get,

NN0∫ dN/N = λ tt0∫ dt ………(2)

Or,  ln N – ln N0 = – λ (t – t0) ……..(3)

Where, N0 is the number of radioactive nuclei in the sample at some arbitrary time t0 and N is the number of radioactive nuclei at any subsequent time t. Next, we set t0 = 0 and rearrange the above equation (3) to get,

ln (N/N0) = – λt

Or, N(t) = N0 e^-λt ….….(4)

Equation (4) is the Law of Radioactive Decay.

Q14. What is a photon ? Write its important properties.

Ans. Photon– A photon is a quantum or packet of energy which has zero rest mass and energy equal to product of the frequency of the radiation and Planck’s constant. i.e, E = hv

Properties– Photon always travel in straight lines. Photons are electricity neutral. They cannot be deflected by electric or magnetic field. They have no electric charge. In the empty space, they travel at the speed of light. Energy of photon is E = hv.

Q15. In Young’s double-slit experiment, the slits are separated by 1 mm and the screen is placed 1 m away. Calculate the fringe separation for light of wavelength 500 nm.

Ans. Fringe (β) = D/d × λ = 1/1×10-³ × (500×10-⁹) = 5×10-⁴ m

Q16. Explain the principle and working of a step-down transformer using a diagram.

Ans.

Step-down Transformer– A transformer that converts the high output voltage using less current to low output voltage through high current is known as a step-down transformer. There are two types of windings in this transformer namely primary as well as secondary. Primary winding includes more turns as compared with secondary.

Principle– A transformer works on the principle of mutual induction. Whenever the amount of magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.

Working– When an alternating current source is connected to the ends of primary coil, the current changes continuously in the primary coil, due to which magnetic flux linked with the secondary coil changes continuously. Therefore, the alternating emf of same frequency is developed across the secondary terminals. According to Faraday’s laws, the e.m.f. induced in the primary coil, EP = -NP × ∆ϕ/∆t

Q17. Name the practical device based on the principle of Wheatstone bridge. Determine the value of unknown resistance by this device.

Ans. Meter Bridge is based on the principle of Wheatstone bridge. The metre bridge, also known as the slide wire bridge consists of a one metre long wire of uniform cross sectional area, fixed on a wooden block. A scale is attached to the block. Two gaps are formed on it by using thick metal strips in order to make the Wheat stone’s bridge. The terminal B between the gaps is used to connect galvanometer and jockey.

A resistance wire is introduced in gap S and the resistance box is in gap R. One end of the galvanometer is connected to terminal D and its other end is connected to a jockey. As the jockey slides over the wire AC, it shows zero deflection at the balancing point (null point).

The unknown resistance can be calculated as :

X = R × l/(100-l)

Q18. State and prove Gauss’s law in Electrostatics.

Ans. According to Gauss’s theorem the net-outward normal electric flux through any closed surface of any shape is equivalent to 1/∈ times the total amount of charge contained within that surface.

Proof of Gauss’s Theorem

Let’s say the charge is equal to q.

Let’s make a Gaussian sphere with radius = r

Now imagine surface A or area ds has a ds vector

At ds, the flux is:

dΦ = E(vector) ds(vector) cos θ

But θ = 0

Hence, Total flux Φ = E × 4πr²

Here, E = 1/4π∈ × q/r²

Hence,  Φ = 1/4π∈ × q/r² × 4πr²

Φ = q/∈

As per the Gauss law, the total flux associated with a sealed surface equals 1/∈ times the charge encompassed by the closed surface.

Q19. Explain the principle and working of a cyclotron using its schematic sketch. Show that time of revolution of an ion is independent of its speed or radius of its orbit.

Ans. Cyclotron– Cyclotron is a device used to accelerate charged particles to high energies.

Principle– Cyclotron works on the principle that a charged particle moving normal to a magnetic field experiences magnetic lorentz force due to which the particle moves in a circular path.

Construction– It consists of a hollow metal cylinder divided into two sections D1 and D2 called Dees, enclosed in an evacuated chamber (Figure). The Dees are kept separated and a source of ions is placed at the centre in the gap between the Dees. They are placed between the pole pieces of a strong electromagnet. The magnetic field acts perpendicular to the plane of the Dees. The Dees are connected to a high frequency oscillator.

Working– When a positive ion of charge q and mass m is emitted from the source, it is accelerated towards the Dee having a negative potential at that instant of time. Due to the normal magnetic field, the ion experiences magnetic lorentz force and moves in a circular path. By the time the ion arrives at the gap between the Dees, the polarity of the Dees gets reversed.

Hence the particles is once again accelerated and moves into the other Dee with a greater velocity along a circle of greater radius. Thus the particle moves in a spiral path of increasing radius and when it comes near the edge, it is taken out with the help of a deflection plate (D.P.). The particle with high energy is now allowed to hit the target T.

When the particle moves along a circle of radius r with a velocity v, the magnetic Lorentz force provides the necessary centripetal force.

Bqv = mv²/r

v = qBr/m

Period of revolution, T = 2πr/v

T = 2πrm/qBr = 2πm/qB = constant

Thus ‘T’ is independent of ‘v’.

OR

Derive an expression for the force between two long straight parallel conductors carrying current in same direction. Hence define one Ampere.

Ans.

Consider a small length L of the long straight conductor

B1 = μI1/2πd

B2 = μI2/2πd

F12 = I2B1

F21 = I1B2

F = F12 = F21 = μI1I2L/2πd

F/L = μI1I2/2πd

Definition of ampere : Fundamental unit of current ampere has been defined assuming the force between the two current carrying wires as standard. The force between two parallel current carrying conductor of separation r is 2×10-⁷ N/m.

Thus 1 ampere is the current which when flowing in each of parallel conductors placed at separation 1 m in vacuum exert a force of 2×10-⁷ N/m on 1 m length of either wire.

Q20. Draw a labelled ray diagram showing image formation in an astronomical telescope. Derive expression for its magnifying power.

Ans. Astronomical Telescope– Astronomical Telescope is used to observe objects which are very far from us. Telescopes produce magnified images of distant objects. It produces virtual and﻿ inverted image and is used to see heavenly bodies like﻿ sun, stars, planets etc. so the inverted image does not﻿ affect the observation.﻿

Principle– It is based on the principle that when rays﻿ of light are made to incident on an objective from a﻿ distant object, the objective forms the real and inverted﻿ image at its focal plane. The eye lens is so adjusted that﻿ the final image is formed at least distance of distinct﻿ vision.﻿

Construction– The refracting type astronomical﻿ telescope consists of two convex lenses one of which is﻿ called the objective and the other eye piece. The objective﻿ is a convex lens of large focal length and large aperture. It is generally a combination of two lenses in contact so﻿ as to reduce spherical and chromatic aberrations. The﻿ eye piece is also a convex lens but of short focal length﻿ and small aperture. The objective is mounted at one end﻿ of a brass tube and the eye piece at the other end in a﻿ smaller tube which can slide inside the bigger tube﻿ carrying the objective.

Magnifying Power– The magnifying power of a refracting type astronomical telescope is defined as the ratio of angle subtended by the final image at eye to the angle subtended by the object at eye.

M= fo/fe (1 + fe/D)

OR

Draw a labelled ray diagram showing image formation in a compound microscope. Define its magnifying power and write expression for it.

Ans.

Magnifying Power– Magnifying power of a compound microscope is defined as, the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the image and object are situated at the least distance of distinct vision from the eye.

Q21. Draw a Circuit diagram for studying V-I characteristics for a transistor in common emitter (CE) configuration. Draw and explain its typical output V-I characteristics.

Ans. Output characteristics– These characteristics are obtained by plotting collector current IC versus collector emitter voltage VCE at a fixed value of base current IB. The base current is changed to some other fixed value and the observations of IC versus VCE are repeated. The figure given below represents the output characteristic curves.

OR

Explain the working of transistor as a amplifier by using its circuit diagram.

Ans. Transistor Amplifier– A transistor acts as an amplifier by raising the strength of a weak signal. The DC bias voltage applied to the emitter base junction, makes it remain in forward biased condition. This forward bias is maintained regardless of the polarity of the signal. The below figure shows how a transistor looks like when connected as an amplifier.

The low resistance in input circuit, lets any small change in input signal to result in an appreciable change in the output. The emitter current caused by the input signal contributes the collector current, which when flows through the load resistor RL, results in a large voltage drop across it. Thus a small input voltage results in a large output voltage, which shows that the transistor works as an amplifier.

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