CBSE Class 12 Physics Question Paper 2020 (Series: HMJ/5) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Physics with Solutions and CBSE Class 12 Physics Question Paper 2020 (Series: HMJ/5) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Physics Question Paper 2020 (Series: HMJ/5) with Solutions

Time Allowed: 3 hours
Maximum Marks: 70

General Instructions:

1. This questions paper comprises four sections-A, B, C and D.
2. There are 37 questions in the question paper. All questions are compulsory.
3. Section A : Q. no. 1 to 20 are very short-answer type questions carrying 1 mark each.
4. Section B : Q. no. 21 to 27 are short-answer type questions carrying 2 marks each.
5. Section C : Q. no. 28 to 34 are long-answer type questions carrying 3 marks each.
6. Section D : Q. no. 35 to 37 are also long answer type questions carrying 5 marks each.
7. There is no overall choice in the question paper. However, an internal choice has been provided in two questions of one mark, one question of two marks. You have to attempt only one of the choices in such questions.
8. However, separate instructions are given with each section and question, wherever necessary.
9. Use of calculators and log tables is not permitted.
10. You may use the following values of physical constants wherever necessary:
    c = 3 × 10 ⁸ m/s, h = 6.63 × 10 ^-34 Js, e = 1.6 × 10-19 C, µ0 = 4π × 10 ^-7 TmA ^-1
    ε ₀ = 8.854 × 10 ^-12 C ² N ^-1 m ^-2 , \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10 ⁹ Nm ² C ^-2
    Mass of electron (m) = 9.1 × 10 ^-31 kg, Mass of neutron = 1.675 × 10 ^-27 kg
    Mass of proton = 1.673 × 10 ^-27 Kg, Avogadro’s number = 6.023 × 10 ²³ per gram mole
    Boltzmann constant = 1.38 × 10 ^-23 JK ^-1
    * Modified as per Latest CBSE Curriculum.

SET I Code No. 55/5/1
Section-A

Select the most appropriate option from those given below each question.

Question 1.
A 20 volt AC is applied to a circuit consisting of a resistance and a coil with negligible resistance. If the voltage across the resistance is 12 volt, the voltage across the coil is —
(a) 16 V
(b) 10 V
(c) 8 V
(d) 6 V
Answer:
(a) 16 V

Given, V = 20 Volts and V _R = 12 volts
V _R = Effective Voltage across R
∴ V _R = I _eff R
V _L = Effective Voltage across L
∴ V _L = I _eff × L

Question 2.
The instantaneous values of emf and the current in a series ac circuit are —
E = E0 sin ωt and I = I ₀ sin(ωt + π/3) respectively, then it is
(a) Necessarily a RL circuit
(b) Necessarily a RC circuit
(c) Necessarily a LCR circuit
(d) Can be RC or LCR circuit
Answer:
(d) Can be RC or LCR circuit

Given, E = E ₀ sin ωt
and I = I ₀ sin (ωt + \(\frac{\pi}{3}\))
As I can lead the Voltage in RC and LCR circuit, so it can be RC or LCR circuit.

Question 3.
The selectivity of a series LCR ac. circuit is large, When [1]
(a) L is large and R is large
(b) L is small and R is small
(c) L is large and R is small
(d) L = R
Answer:
(c) L is large and R is small

[Hint: Q = \(\frac{1}{\mathrm{R}} \sqrt{\frac{\mathrm{L}}{\mathrm{C}}}\)

Question 4.
The graph showing the correct variation of linear momentum (p) of a charge particle with its de-Broglie wavelength (λ) is —

Answer:
(b)

[Hint: p = \(\frac{h}{\lambda}\)

Question 5.
The wavelength and intensity of light emitted by an LED depend upon [1]
(a) forward bias and energy gap of the semiconductor
(b) energy gap of the semiconductor and reverse bias
(c) energy gap only
(d) forward bias only
Answer:
(a) forward bias and energy gap of the semiconductor

Question 6.
A charge particle after being accelerated through a potential difference ‘V’ enters in a uniform magnetic field and moves in a circle of radius r. If V is doubled, the radius of the circle will become [1]
(a) 2r
(b) \(\sqrt{2}\)r
(c) 4r
(d) r/\(\sqrt{2}\)
Answer:
(b) \(\sqrt{2}\)r

[Hint: r = \(\frac{1}{\mathrm{~B}} \sqrt{\frac{2 m V}{q}}\)

Question 7.
The electric flux through a closed Gaussian surface depends upon [1]
(a) Net charge enclosed and permittivity of the medium
(b) Net charge enclosed, permittivity of the medium and the size of Gaussian surface
(c) Net charge enclosed only
(d) Permittivity of the medium only
Answer:
(a) Net charge enclosed and permittivity of the medium

Question 8.
If photons of frequency v are incident on the surfaces of metals A & B of threshold frequencies v/2 and v/3 respectively, the ratio of the maximum kinetic energy of electrons emitted from A to that from B is 1
(a) 2 : 3
(b) 3 : 4
(c) 1 : 3
(d) \(\sqrt{3}\) : \(\sqrt{2}\)
Answer:
(b) 3 : 4

3 : 4 [Hint: hv = hv ₀ + E _k

Question 9.
The power factor of a series LCR circuit at resonance will be 1
(a) 1
(b) 0
(c) 1/2
(d) 1/\(\sqrt{2}\)
Answer:
(a) 1
[Hint: Power factor R/Z becomes 1 at resonance when R = Z

Question 10.
A biconcave lens of power P vertically splits into two identical piano concave parts. The power of each part will be [1]
(a) 2P
(b) P/2
(c) P
(d) P/\(\sqrt{2}\)
Answer:
(b) P/2

[Hint: P = P/2 + P/2

Note: Fill in the blanks with appropriate answers:

Question 11.
The physical quantity having SI unit NC ^-1 m is _____. [1]
Answer:
Electric potential

Question 12.
A copper wire of non-uniform area of cross-section is connected to a d.c. battery. The physical quantity which remains constant along the wire is ____. [1]
Answer:
Electric current

Question 13.
A point charge is placed at the centre of a hollow conducting sphere of internal radius V and outer radius ‘2r’. The ratio of the surface charge density of the inner surface to that of the outer surface will be _____. [1]
Answer:
4 : 1 [Hint: Surface charge density = q/4πr ²

Question 14.
The ____, a property of materials C, Si and Ge depends upon the energy gap between their conduction and valence bands. [1]
Answer:
Conductivity

Question 15.
The ability of a junction diode to ____ an alternating voltage, is based on the fact that it allows current to pass only when it is forward biased. [1]
Answer:
Rectify

Note: Answer the following:

Question 16.
Define the term ‘current sensitivity’ of a moving coil galvanometer. [1]
Answer:
Current sensitivity is defined as the deflection produced in the galvanometer when unit current is passed through its coil.
I _s = \(\frac{\theta}{\mathrm{I}}\) = \(\frac{n \mathrm{AB}}{k}\) = radian/ ampere of division A ^-1 .
…where [n = No. of turns in galvanometer
K = Restoring couple per unit twist

Question 17.
Depict the fields diagram of an electromagnetic wave propagating along positive X-axis with its electric field along Y-axis. [1]
Answer:

Question 18.
Write the conditions on path difference under which (i) constructive (ii) destructive interference occur in Young’s double slit experiment. [1]
Answer:
(i) Path difference for constructive interference: ∆ _p = nλ
(ii) Path difference for destructive interference: ∆ _p = (2n + 1)λ/2 …where [n = 0, 1, 2, 3 ….
[Hint: Induced emf in a coil, ε = -L\(\frac{d l}{d t}\)

Question 19.
Plot a graph showing variation of induced e.m.f with the rate of change of current flowing through a coil. [1]
Or
A series combination of an inductor (L), capacitor (C) and a resistor (R) is connected across an ac source of emf of peak value Eu and angular frequency (ω). Plot a graph to show variation of impedance of the circuit with angular frequency (ω). [1]
Answer:

Or

Question 20.
An electron moves along +x direction. It enters into a region of uniform magnetic field \(\overrightarrow{\mathrm{B}}\) directed along -z direction as shown in figure. Draw the shape of trajectory followed by the electron after entering the field.

Or
A square shaped current carrying loop MNOP is placed near a straight long current carrying wire AB as shown in the figure. The wire and the loop lie in the same plane. If the loop experiences a net force F towards the wire, find the magnitude of the force on the side ‘NO’ of the loop.

Answer:
Circular path in x-y plane in clockwise direction.

Or, Force on the side NO, F = \(\frac{\left(\mu_0 \mathbf{I}_1 \mathbf{I}_2\right)}{2 \pi}\) [Hint: F = \(\frac{\left(\mu_0 \mathrm{I}_1 \mathrm{I}_2\right) \times \mathrm{L}}{2 \pi \mathrm{L}}\)

Section – B

Question 21.
Derive the expression for the torque acting on an electric dipole, when it is held in a uniform electric field. Identify the orientation of the dipole in the electric field, in which it attains a stable equilibrium. [2]

Or

Obtain the expression for the energy stored in a capacitor connected across a dc battery. Hence define energy density of the capacitor. [2]
Answer:
Torque on electric dipole. Consider an electric dipole consisting of two equal and opposite point charges separated by a small distance 2a having dipole moment | \(\vec{p}\) | = q (2 \(\vec{a}\))

Let the dipole held in a uniform external electric field at an angle θ
∴ Force on charge (+q) = q \(\overrightarrow{\mathrm{E}}\) along the direction of \(\overrightarrow{\mathrm{E}}\)
Force on charge (-q) = -q \(\overrightarrow{\mathrm{E}}\) along the opposite direction of \(\overrightarrow{\mathrm{E}}\)
∴ Net translatory force on the dipole = q \(\overrightarrow{\mathrm{E}}\) – q \(\overrightarrow{\mathrm{E}}\) = 0
So net force on the dipole is zero
Since \(\overrightarrow{\mathrm{E}}\) is uniform, hence the dipole does not undergo any translatory motion.
These forces being equal, unlike and parallel, form a couple, which rotates the dipole in clock-wise direction
∴ Magnitude of torque = Force × arm of couple
\(\tau\) = F. AC = qE. AB sin θ = (qE) 2a sin θ
or \(\tau\) = q(2a) E sin θ or \(\tau\) = pE sin θ [p = q(2\(\vec{a}\))]
∴ \(\vec{\tau}\) = \(\vec{p} \times \overrightarrow{\mathrm{E}}\) [The direction of \(\vec{\tau}\) is given by right hand screw rule and is normal to \(\vec{p}\) and \(\overrightarrow{\mathrm{E}}\).

Special cases

(i) When θ = 0 then \(\tau\) = PE sin θ = 0
∴ Torque is zero and the dipole is in stable equilibrium.
(ii) When θ = 90° then \(\tau\) = PE sin 90° = PE
∴ The torque is maximum

Or, Potential of capacitor = \(\frac{q}{C}\)
Small amount of work done in giving an additional charge dq to the capacitor,
dW = \(\frac{q}{C} \times d q\)
Total work done in giving a charge Q to the capacitor

As electrostatic force is conservative, thus work is stored in the form of potential energy (U) of the capacitor.
U = W = \(\frac{1}{2} \frac{Q^2}{C}\)
Put Q = CV
∴ U = \(\frac{1}{2}\) CV ²
Energy density is defined as the ‘energy stored per unit volume’.
Energy density of electric field, \(\frac{1}{2} \in E_0 E^2\)

Question 22.
Gamma rays and radio waves travel with the same velocity in free space. Distinguish between them in terms of their origin and the main application.
Answer:

1. Origin. Gamma Rays are emitted by radioactive nuclei produced in nuclear reactions; while Radio waves are produced by accelerated oscillating charges in L-C circuits.
2. Main application. Gamma Rays are used for treatment of cancer; while Radio waves are used in communication system (Radio, TV and Mobile phones).

Question 23.
Define the term ‘wave front of light’. A plane wave front AB propagating from rarer medium (1) into a denser medium (2) is incident on the surface P P, separating the two media as show in figure.

Using Huygen’s principle, draw the secondary wavelets and obtain the refracted wave front in the diagram.
Answer:
Wavefront: Wavefront is defined as the continuous locus of all such particles of the medium which are vibrating in the same phase at any instant.

Question 24.
A heavy nucleus P of mass number 240 and binding energy 7.6 MeV per nucleon splits into two nuclei Q and R of mass numbers 110, 130 and binding energy per nucleon 8.5 MeV and 8.4 MeV, respectively. Calculate the energy released in the fission. [2]
Answer:
Given.A _P = 240, (E) _P = 7.6 MV/nucleon
A _Q = 110, (E) _Q = 8.5 MeV/nucleon
A _R = 130, (E) _R = 8.4 MeV/nucleon
\({ }^{240} \mathrm{P}\) → \({ }^{110} Q+{ }^{130} R+E\)
Energy released, E = ([A _Q × E _Q ] + [A _R × E _R ]} — [A _P × E _P ]
= {[(110 × 8.5) + (130 × 8.4)) – (240 × 7.6)]
= [(935 + 1,092) — (1,824)]
= [2,027 – 1,8241 = 203 MeV

Question 25.
Figure shows the stopping potential (V ₀ ) for the photo electron versus (1/λ) graph, for two metals A and B, λ being the wavelength of incident light.

(a) How is the value of Planck’ constant determined from the graph?
(b) If the distance between the light source and the surface of metal
A is increased, how will the stopping potential for the electrons emitted from it be effected? Justify your answer. [2]
Answer:
(a) According to Einstein’s photoelectric equation, hv = φ0 + eV ₅

(b) Stopping potential will remain same, if the distance between the light source and the surface of metal A is increased.
Justification. Variation of distance of light source from the metal surface will alter the intensity, while the stopping potential depends only on the frequency and not on the intensity of incident light.

Question 26.
Use Bohr’s model of hydrogen atom to obtain the relationship between the angular momentum and the magnetic moment of the revolving electron. [2]
Answer:
Let us consider an electron revolving around in a circle of radius ‘r’ with a velocity ‘v’, the charge and mass of electron are ‘e’ and ‘m _e ’ respectively.
Time period, T of electron orbit \(=\frac{\text { Circumference of circle }}{\text { Velocity }}\) = \(\frac{2 \pi r}{v}\)
Current, I due to motion of electron is the charge flowing through the given time period.

It is the relationship between angular momentum and the magnetic moment of revolving electron.

Question 27.
In a single slit diffraction experiment, the width of the slit is doubled. How will the (i) size and (ii) intensity of central bright band be affected? Justify your answer. [2]
Answer:
According to Huygen’s principle “The net effect at any point due to a number of wavelets is equal to sum total of contribution of all wavelets with proper phase difference.”

The point O is maxima because contribution from each half of the slit S ₁ S ₂ is in phase, i.e., the path difference is zero.
At point P
(a) If S ₂ P = S ₁ P = nλ ⇒ the point P would be minima
(b) If S ₂ P – S ₁ P = (2n + 1)\(\frac{\lambda}{2}\)
⇒ the point would be maxima 1ut with decreasing intensity.
∴ The width of central maxima = \(\frac{2 \lambda \mathrm{D}}{a}\)
β = \(\frac{\lambda \mathrm{D}}{a}\) …when [β is the fringe width
If width is doubled, then d’ = 2d ⇒ β’ = \(\frac{\beta}{2}\)
When width of slit is doubled, contrast between maxima and minima decreases due to the overlapping of interference patterns formed by various narrow pairs of the slits.
Therefore, size of central maxima will be reduced to half and intensity of central maxima will be four times.

Section – C

Question 28.
(a) Differentiate between electrical resistance and resistivity of a conductor.
(b) Two metallic rods, each of length L, area of cross A ₁ and A ₂ , having resistivities ρ ₁ and ρ ₂ are connected in parallel across a d.c. battery. Obtain the expression for the effective resistivity of this combination. [3]
Answer:
(a) Resistance of a conductor. The resistance of a conductor of length l and cross-section area A is given by R = \(\rho \frac{l}{\mathrm{~A}}\) ..,where [ρ is the resistivity of the material of the conductor Resistivity of a conductor. It is defined as the resistance offered by a wire of this
material of unit length and unit cross-section area. It is also known as specific resistance of this material of the conductor. The SI unit of resistivity is Ohm metre ‘Ω m’.

(b)

Question 29.
Calculate the de-Broglie wavelength associated with the electron revolving in the first excited state of hydrogen atom. The ground state energy of the hydrogen atom is —13.6 eV. [3]
Answer:
Given : n = 2 of hydrogen atom 1, λ = ?
Kinetic energy for the second state,

Question 30.
Not in Syllabus.
Answer:
Not in Syllabus.

Question 31.
Draw the circuit diagram of a full wave rectifier. Explain its working showing its input and output waveforms.
Answer:
Diagram of full wave rectifier:

Working of a full wave rectifier:

1. A full wave rectifier uses two diodes and gives the rectified output voltage corresponding to both the positive and negative half-cycle of alternating current.
2. The p-side of the two diodes are connected to the ends of the secondary of the transformer and, the n-sides of the diodes are connected together.
3. Output is taken from between the common- point of the two diodes and secondary of the transformer. Hence, the secondary of the transformer is provided with center tapping and is also called the centre-tap transformer.
4. Let, the input voltage to A with respect to the centre be positive and, at the same instant, voltage at B being out-of phase will be negative. Therefore, diode D ₁ isforward biased and starts conducting whereas, D ₂ being reverse biased does not conduct.
5. Thus, we get an output current and an output voltage across the load resistance R _L in the first positive half-cycle.
6. During the course of the negative half-cycle, that is, when voltage at A becomes negative and voltage at B becomes positive, we will have D ₁ as reverse biased and D ₂ forward biased.
7. In the negative part of the cycle, only diode D2 will conduct giving an output current and output voltage across R _L .
8. For both positive and negative half cycle we will get the output voltage. This rectified output voltage has the shape of half sinusoids.

Question 32.
An optical instrument uses a lens of power 100 D for objective lens and 50 D for its eyepiece. When the tube length is kept at 25 cm. the final image is formed at infinity.
(a) Identify the optical instrument.
(b) Calculate the magnification produced by the instrument. [3]
Answer:
(a) Theoptical instrument is th Compound microscope.
for objective lens
P ₀ = 100 D ∴ f ₀ = \(\frac{1}{100}\) m or 1 cm …..[f = 1/P
and for eye lens
P _e = 50 D ∴ f _e = \(\frac{1}{50}\) m or 2 cm ……[f = 1/P
As both the objective and eye lens are convex lenses of short focal length, hence instrument is compound microscope.

(b) Given.P ₀ = 100 D, P _e = 50 D, L = 25 cm, D = 25 cm
The magnifying power of compound microscope, (m) = \(\frac{\mathrm{L}}{f_0} \times \frac{\mathrm{D}}{f_e}\) = \(\frac{25}{1} \times \frac{25}{2}\) = 312.5

Question 33.
(a) Two point charges q ₁ and q ₂ are kept at a distance of r ₁₂ in air. Deduce the expression for the electrostatic potential energy of this system.
(b) If an external electric field (E) is applied on the system, write the expression for the total energy of this system. [3]
Answer:
(a) Let us consider two charges q ₁ and q ₂ kept at a distance r ₁₂ in air.

Hence, it is the expression for the electrostatic potential energy of the given system of two charges.

(b) Work done in bringing the charge q ₁ to a point against external electric field,
W ₁ = q ₁ V\(\left(\overrightarrow{r_1}\right)\)
Work done in bringing the charge q ₂ against the external electric field and the electric field produced due to charge q ₁

Question 34.
When a conducting loop of resistance 10 Ω and area 10 cm ² is removed from an external magnetic field acting normally, the variation of induced current in the loop with time is shown in the figure.

Find the
(i) total charge passed through the loop.
(ii) change in magnetic flux through the loop.
(iii) magnitude of the magnetic field applied.

Answer:
Given. Resistnce, R = 10 Ω, Area, A = 10 cm ² = 10 × 10 ^-4 m ² ,
change in current (∆I) = 0.4 A; change in time (∆t) = 1.0 s

(i) Total charge passed through the loop, (Q) = Area under I-t graph
= \(\frac{1}{2}\) × (0.4) × (1) = 0.2 C

(ii) Change in magnetic flux, (dφ ₁ ) = Q × R = (0.2) × (10) = 2 Wb

(iii) Initial magnetic flux, (φ ₁ ) = BA = B × (10 × 10 ^-4 )
Final magnetic flux, (φ ₂ ) = zero
Change is magnetic flux (dφ) = B × 10 ^-3 = 2 ∴ B = 2 × 10 ³ W _b m ^-2

Section – D

Question 35.
(a) Define the term ‘focal length of mirror’. With the help of a ray diagram, obtain the relation between its focal length and radius of curvature.
(b) Calculate the angle of emergence (e) of the ray of light incident normally on the face AC of a glass prism ABC of refractive index \(\sqrt{3}\). How will the angle of emergence change qualitatively, if the ray of light emerges from the prism into a liquid of refractive index 1.3 instead of air? [5]

Answer:
(a) (i) Focal length of a mirror is defined as the distance of point from the pole of a mirror through which ray of light moving parallel to its principle axis passes (or apper to come from). It is denoted by f and f = \(\frac{R}{2}\)
(ii) Relationship between the focal length and radius of curvature. The geometry of reflection of an incident ray is show in the figure.

C is the centre of curvature of the mirror. A ray parallel to principal axis strikes the mirror at M. CM is the perpendicular to the mirror at M. Let θ be the angle of incidence and MD is the perpendicular to the mirror at M.
∠MCP = θ and ∠MFP = 2θ
Now, tan θ = \(\frac{M D}{C D}\) and tan 2θ = \(\frac{M D}{\overline{F D}}\) …….. (i)
For small values of θ, tan θ = θ and tan 2θ = 2θ, therefore equation (i) gives
\(\frac{\mathrm{MD}}{\mathrm{FD}}\) = 2\(\frac{M D}{C D}\) or FD = \(\frac{C D}{2}\) ……….. (ii)
Now, for small θ, the point D is very close to the point P.
∴ FD = f and CD = R. Thus equation (ii) gives f = \(\frac{R}{2}\).

(b) Given, μ of glass prism, \(\left(a_{\mu_g}\right)\) = \(\sqrt{3}\) ;
μ of liquid \(\left(a_{\mu_l}\right)\) = 1.3
It can be easily seen that the angle of incidence on the surface AB i.e., ∠A = 30°, because it is a right-angled prism with ∠B = 90° and also ∠C is 60°.

Applying Snell’s law at face AB when emerging out in air from glass prism, \(g_{\mu_a}\) = \(\frac{1}{\sqrt{3}}\)
\(g_{\mu_a}\) = \(\frac{\sin (i)}{\sin (e)}\) = \(\frac{\sin \left(30^{\circ}\right)}{\sin (e)}\) = \(\frac{1}{\sqrt{3}}\)
sin (e) = \(\sqrt{3}\) × sin(30°) = \(\sqrt{3} \times \frac{1}{2}=\frac{\sqrt{3}}{2}\) = sin(60°) ∴ ∠e = 60°
When the medium (air) in which the prism is kept, is replaced with a liquid of refractive index 1.3, the angle of emergence would decrease. It is because the bending in the ray of light would be lesser.
According to Snell’s law,
\(\sqrt{3}\) × sin(30°) = 1.3 × sin(e)
sin(e) = \(\frac{\sqrt{3}}{1.3 \times 2}\) = 0.66 ∴ ∠e = 41.83°

Question 36.
The variation of inductive reactance X _L of an inductor with the frequency (f) of the ac source of 100 V and variable frequency is shown in the figure.

(i) Calculate the self-inductance of the inductor.
(ii) When this inductor is used in series with a capacitor of unknown value and a resistor of 10 Ω at 300 s ^-1 , maximum power dissipation occurs in the circuit. Calculate the capacitance of the capacitor. [5]
Answer:
(i) From the graph, it is seen that inductive reactance (X _L ) of 20 Ω corresponds to the value of frequency (f) 100 Hz.
As we know, X _L = ωL …where [ω = 2πf
∴ X _L = 2πfL
⇒ L = \(\frac{X_L}{2 \pi f}\) = \(\frac{20}{2 \times 3.14 \times 100}\) = 0.03184 H
∴ L = 31.84 × 10 ^-3 ≈ 32 mH

(ii) As we know, maximum power dissipation takes place when X _L = X _C or ωL = \(\frac{1}{\omega \mathrm{C}}\)
C = \(\frac{1}{\omega^2 L}\) = \(\frac{1}{(2 \pi f)^2} \times \frac{1}{\mathrm{~L}}\) = \(\frac{1}{4 \times(3.14 \times 3.14)} \times \frac{1}{(300 \times 300)} \times \frac{1}{\left(32 \times 10^{-3}\right)}\)
∴ C = 8.8 × 10 ^-6 F = 8.8 µF

Question 37.
(a) Write two important characteristics of equipotential surfaces.
(b) The magnitude of electric field (in NC ^-1 ) in a region varies with the distance r(in m) as
E = 10 r + 5
By how much does the electric potential increase in moving from point at r = 1 m to a point at r = 10 m. [5]
Answer:
(a) Properties of equipotential surfaces:
(i) No work is done in moving a test charge over an equipotential surface.
(ii) No two equipotential surfaces can intersect each other.

(iii) Equipotential surface due to an isolated point charge is spherical.
(iv) The electric field at every point is normal to the equipotential surface passing through that point. As E = \(\frac{-d \mathrm{~V}}{d r}\)
or dr = \(\frac{-d \mathrm{~V}}{\mathrm{E}}\)
For the same charge in the value, V, i.e., when dV = constant, we have dr ∝ \(\frac{1}{\mathrm{E}}\)
Hence, equipotential surface gets closer as the distance between the equipotential surface and the source charge decreases.

(b) Given. E = lOr + 5
We know, dv = –\(\overrightarrow{\mathrm{E}} \cdot \overrightarrow{d r}\)
Integrating both sides

Set II Code No. 55/5/2

Except for the following questions, all the remaining questions have been asked in Set-I.

Question 6.
A biconvex lens of focal length f is cut into two identical piano convex lenses. The focal length of each part will be [1]
(a) f
(b) f/2
(c) 2f
(d) 4f
Answer:
(c) 2f

Question 7.
The phase difference between the current and the voltage in series LCR circuit at resonance is [1]
(a) π
(b) π/2
(c) π/3
(d) zero
Answer:
(d) zero

Question 8.
Photons of frequency v are incident on the surface of two metals A and B of threshold frequencies 3/4 v and 2/3v, respectively. The ratio of maximum kinetic energy of electrons emitted from A to that from B is [1]
(a) 2 : 3
(b) 4 : 3
(c) 3 : 4
(d) 3 : 2
Answer:
(c) 3 : 4

Question 23.
In case of photo electric effect experiment, explain the following facts, giving reasons.
(a) The wave theory of light could not explain the existence of the threshold frequency.
(b) The photo electric current increases with increase of intensity of incident light. [2]
Answer:
(a) According to wave theory, threshold frequency should not exist. Light of all frequencies should emit electrons, provided intensity of light is sufficient for electrons to eject.
(b) When intensity of incident light increases, the number of incident photons increases, as one photon ejects one electron, the increase in intensity will increase the number of ejected electrons. In other words, photo current will increase with increase of intensity.

Set III Code No. 55/5/3

Note: Except for the following questions, all the remaining questions have been asked in Set-I and II.

Question 24.
Calculate for how many years will the fusion of 2.0 kg deuterium keep 800 W electric lamp glowing. Take the fusion reaction as
\({ }_1^2 \mathrm{H}\) + \({ }_1^2 \mathrm{H}\) → \({ }_2^3 \mathrm{He}\) + \({ }_0^1 \mathrm{n}\) + 3.27 MeV [2]
Answer:
Given. Fusion reaction = \({ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H}\) → \({ }_2^3 \mathrm{He}+{ }_0^1 \mathrm{n}+3.27\) MeV
Mass of deuterium, m = 2.0 kg
Power of electric lamp, P = 800 W or 800 J/s
Number of years (y) = ?
Avagadro number (N) = 6.023 × 10 ²³
1 mole i.e., 2g of deuterium contains = 6.023 × 10 ²³ atoms
So, 2.0 kg of deuterium contains = \(\frac{\left(6.023 \times 10^{23}\right) \times(2,000)}{2}\) = 6.023 × 10 ²⁶ atoms
Since 2 atoms of deuterium fuse to release 3.27 MeV energy,

Question 30.
Calculate the de-Broglie wavelength associated with the electron in the 2 ^nd excited state of hydrogen atom. The ground state energy of the hydrogen atom is 13.6eV. [3]
Answer:

Question 32.
(a) Two point charges +Q ₁ and -Q ₂ are placed r distance apart. Obtain the expression for the amount of work done to place a third charge Q ₃ at the midpoint of the line joining the two charges.
(b) At what distance from charge + Q ₁ on the line joining the two charges (in terms of Q ₁ , Q ₂ and r) will this work done be zero. [3]
Answer:
(a) Work done in placing the charge Q ₃
= (Q ₃ ) × (net potential at the mid point)

(b) Let the required distance be x, to the left of Q ₁ . We then have