CBSE Class 12 Physics Question Paper 2023 (Series: GEFH1/5) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Physics with Solutions and CBSE Class 12 Physics Question Paper 2023 (Series: GEFH1/5) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Physics Question Paper 2023 (Series: GEFH1/5) with Solutions

General Instructions:

Read the following instructions very carefully and follow them:

1. This question paper contains 35 questions. All questions are compulsory.
2. Question paper is divided into FIVE sections – Section A, B, C, D and E.
3. In Section-A: question number 1 to 18 are Multiple Choice (MCQ) type questions carrying 1 mark each.
4. In Section-B: question number 19 to 25 are Short Answer-1 (SA-1) type questions carrying 2 marks each.
5. In Section-C: question number 26 to 30 are Short Answer-2 (SA-2) type questions carrying 3 marks each.
6. In Section-D: question number 31 to 33 are Long Answer (LA) type questions carrying 5 marks each.
7. In Section-E: question number 34 and 35 are case-based questions carrying 4 marks each.
8. There is no overall choice. However, an internal choice has been provided in 2 questions in Section-B, 2 questions in Section-C, 3 questions in Section-D and 2 questions in Section-E.
9. Use of calculators is NOT allowed.
   c = 3 × 10 ⁸ m/s; h = 6.63 × 10 ^-34 Js
   e = 1.6 × 10 ^-19 ; µ ₀ = 4π × 10 ^-7 T m A ^-1
   \(\varepsilon_0\) = 8.854 × 10 ^-12 C ² N ^-1 m ^-2
   \(\frac{1}{4 \pi \varepsilon_0}\) = 9 × 10 ⁹ N m ² C ^-2
   Mass of electron (m _e ) = 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 × 1o ²³ per gram mole
   Boltzmann constant = 1.38 × 10 ^-23 JK ^-1

SET I Code No. 55/5/1

Question 1.
An electric dipole of length 2 cm iš placed at an angle of 30° with an electric field 2 × 10 ⁵ N / C. If the dipole experiences a torque of 8 × 10 ^-3 Nm, the magnitude of either charge
of the dipole, is [1]
(a) 4 µC
(b) 7 µC
(c) 8 mC
(d) 2 mC
Answer:
(a) 4 µC

Given. θ = 30°; E = 2 × 10 ⁵ N/C; \(\tau\) = 8 × 10 ^-3 Nm
dipole length, l = 2 cm or \(\frac{2}{100}\) m
So, the Dipole moment of dipole will be, P = ql = \(\left(\frac{2}{100}\right)\)q
⇒ P = 0.02q
Torque acting on dipole, \(\tau\) = PE sin θ
⇒ 8 × 10 ^-3 = (0.02q)(2 × 10 ⁵ ) sin 30°
⇒ 8 × 10 ^-3 = 0.02q × 2 × 10 ⁵ × \(\frac{1}{2}\)
⇒ 8 × 10 ^-3 = 2q × 10 ³
⇒ q = 4 × 10 ^-6
∴ q = 4µC

Question 2.
Two long parallel wires kept 2 m apart carry 3A current each, in the same direction. The force per unit length on one wire due to the other is [1]
(a) 4.5 × 10 ^-5 Nm ^-1 , attractive
(b) 4.5 × 10 ^-7 N/m, repulsive
(c) 9 × 10 ^-7 N/m, repulsive
(d) 9 × 10 ^-5 N/m, attractive
Answer:
No correct option is given
Given: I ₁ = I ₂ = 3A
Distance between the wires, d = 2 m
Force of attraction per unit length, f = \(\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2}{2 \pi d}\) N/m
∴ f = \(\frac{4 \pi \times 10^{-7} \times(3)(3)}{2 \pi \times 2}\) = 9 × 10 ^-7 N/m
As current are in the same direction so there would be an attractive force between them.

Question 3.
Which of the following has its permeability less than that of free space? [1]
(a) Copper
(b) Aluminium
(c) Copper chloride
(d) Nickel
Answer:
(a) Copper

Question 4.
A square shaped coil of side 10 cm, having 100 turns is placed perpendicular to a magnetic field which is increasing at 1 T/s. The induced emf in the coil is [1]
(a) 0.1 V
(b) 0.5 V
(c) 0.75 V
(d) 1.0 V
Answer:
(d) 1.0 V

Question 5.
Which one of the following electromagnetic radiation has the least wavelength? [1]
(a) Gamma rays
(b) Microwaves
(c) Visible light
(d) X-rays
Answer:
(a) Gamma rays

Question 6.
In a Young’s double-slit experiment, the screen is moved away from the plane of the slits. What will be its effect on the following? [1]
(i) Angular separation of the fringes.
(ii) Fringe-width.
(a) Both (i) and (ii) remain constant.
(b) (i) remains constant, but (ii) decreases.
(c) (i) remains constant, but (ii) increases
(d) Both (i) and (ii) increase.
Answer:
(c) (i) remains constant, but (ii) increases

Question 7.
The energy of a photon of wavelength λ is [1]
(a) he λ
(b) hc/λ
(c) λ/hc
(d) λh/c
Answer:
(b) hc/λ
E = hv
∴ E = \(\frac{h c}{\lambda}\) …. where[∴ v = \(\frac{c}{\lambda}\)

Question 8.
The ratio of the nuclear densities of two nuclei having mass numbers 64 and 125 is 1
(a) \(\frac{64}{125}\)
(b) \(\frac{4}{5}\)
(c) \(\frac{5}{4}\)
(d) 1
Answer:
(d) 1

Density of nuclear matter is independent of mass number, so the required ratio is 1 : 1.

Question 9.
During the formation of a p-n junction:
(a) diffusion current keeps increasing.
(b) drift current remains constant.
(c) both the diffusion current and drift current remain constant.
(d) diffusion current remains almost constant but drift current increases till both currents become equal.
Answer:
(d) diffusion current remains almost constant but drift current increases till both currents become equal.

Question 10.
The diagram shows four energy level of an electron in Bohr model of hydrogen atom. Identify the transition in which the emitted photon will have the highest energy. [1]

(a) I
(b) II
(c) III
(d) IV
Answer:
(a) I

Applying the formula, E = Rhc\(\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\)
……. where
[R = Rydberg constant,
h = Planck’s constant,
c = Speed of light,
n ₁ = final state
n ₂ = initial state.

By checking simultaneously from each energy level from the given diagram, Option I where n ₁ = 1 and n ₂ = 3 has the highest energy level possessed by the emitted photon.
Hence, Option (A) is correct.

Question 11.
Which of the following graphs correctly represents the variation of a particle momentum with its associated de-Broglie wavelength? [1]

Answer:
(d)

Question 12.
The capacitors, each of 4 μF are to be connected in such a way that the effective capacitance of the combination is 6 μF. This can be achieved by connecting [1]
(a) All three in parallel
(b) All three in series.
(c) Two of them connected in series and the combination in parallel to the third.
(d) Two of them connected in parallel and the combination in series to the third.
Answer:
(c) Two of them connected in series and the combination in parallel to the third.

To get equivalent capacitance of 6µF from the given set of three 4µF capacitors, two should be connected in series and the third one in parallel.
∴ C _eq = \(\frac{(4 \times 4)}{(4+4)}\) + 4 ⇒ C _eq = 2 + 4 = 6µF

Question 13.
Which of the following statements about a series LCR circuit connected to an ac source is correct? [1]
(a) If the frequency of the source is increased, the impedance of the circuit first decreases and then increases.
(b) If the net reactance (X _L – X _C ) of circuit becomes equal to its resistance, then the current leads the voltage by 45°.
(c) At resonance, the voltage drop across the inductor is more than that across the capacitor.
(d) At resonance, the voltage drop across the capacitor is more than that across the inductor.
Answer:
(a) If the frequency of the source is increased, the impedance of the circuit first decreases and then increases.

Question 14.
According to Huygens principle, the amplitude of secondary wavelets is [1]
(a) equal in both the forward and the backward directions.
(b) maximum in the forward direction and zero in the backward direction.
(c) large in the forward direction and small in the backward direction.
(d) small in the forward direction and large in the backward direction.
Answer:
(b) maximum in the forward direction and zero in the backward direction.

Question 15.
The radius of the nth orbit in Bohr model of hydrogen atom is proportional to [1]
(a) n ²
(b) \(\frac{1}{n^2}\)
(c) n
(d) \(\frac{1}{n}\)
Answer:
(a) n ²

Note: In question number 16 to 18 two statements are given one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
(a) Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true and (R) is NOT the correct explanation of (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also false.

Question 16.
Assertion (A): The resistance of an intrinsic semiconductor decreases with increase in its temperature. [1]
Reason (R): The number of conduction electrons as well as hole increase in an intrinsic semiconductor with rise in its temperature.
Answer:
(a) Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of (A).

Question 17.
Assertion (A): The equivalent resistance between points A and B in the given network is 2R. [1]
Reason (R): All the resistors are connected in parallel.

Answer:
(c) Assertion (A) is true and Reason (R) is false.

Question 18.
Assertion (A): The deflecting torque acting on a current carrying loop is zero when its plane is perpendicular to the direction of magnetic field. [1]
Reason (R): The deflecting torque acting on a loop of magnetic moment \(\vec{m}\) in a magnetic field \(\overrightarrow{\mathrm{B}}\) is given by the dot product of \(\vec{m}\) and \(\overrightarrow{\mathrm{B}}\)
Answer:
(c) Assertion (A) is true and Reason (R) is false.

Section – B

Question 19.
Draw a graph showing the variation of potential energy of a pair of nucleons as a function of their separation. Indicate the region in which the nuclear force is
(a) attractive and
(b) repulsive. [2]
Answer:

For a separation greater than r ₀ , the force is attractive.
For a separation less than r ₀ , the force is repulsive.

Question 20.
(a) How will the De-Broglie wavelength associated with an electron be affected when the
(i) velocity of the electron decreases? and
(ii) accelerating potential is increased? Justify your answer. [2]

Or

(b) How would the stopping potential for a given photosensitive surface change if
(i) the frequency of the incident radiation were increased? and
(ii) the intensity of incident radiation were decreased? Justify your answer. [2]
Answer:
(i) de-broglie wavelength, associated with an electron, λ = \(\frac{h}{m v}\) with decrease in velocity, de-Broglie wavelength increases.
(ii) λ = \(\frac{h}{(\sqrt{2 m e V})}\)
With increase in accelerating potential, de-Broglie wavelength decreases.

Or

(b) (i) with increase in frequency, stopping potential will increase V _α = \(\frac{h}{e} v-\frac{\phi_0}{e}\)
(ii) Stopping potential remains same.
Energy of the photo electron ejected is independent of intensity of radiation.

Question 21.
Identify the electromagnetic wave whose wavelengths range is from about
(a) 10 ^-12 m to about 10 ^-8 m.
(b) 10 ^-3 m to about 10 ^-1 m.
Write one use of each. [2]
Answer:
(a) X-rays:
Applications.

1. These are used as a diagnostic tool in medicine.
2. These are also used as a treatment for certain forms of cancer.

or Gamma rays:
Application. These are used in medicine to destroy cancer cells.

(b) Microwave:
Applications:

1. It is used in radar communication.
2. It is used in microwave ovens.
3. It is also used in analysis of fine details of molecular and atomic structure.

Question 22.
Depict the orientation of an electric dipole in
(a) stable and
(b) unstable equilibrium in an external uniform electric field.
Write the potential energy of the dipole in each case. [2]
Answer:
(a) For stable equilibrium, the angle between p and E is 0°
U = -pE

(b) For unstable equilibrium, the angle between p and E is 180°
U = pE

Question 23.
(a) Write the expression for the Lorentz force on a particle of charge q moving with a velocity \(\vec{v}\) in a magnetic field \(\overrightarrow{\mathrm{B}}\). When is the magnitude of this force maximum? Show that no work is done by this force on the particle during its motion from a point \(\vec{r}_1\) to point \(\vec{r}_2\).
Or
(b) A long straight wire AB carries a current I. A particle (mass m and charge q) moves with a velocity \(\vec{v}\), parallel to the wire, at a distance d from it as shown in the figure. Obtain the expression for the force experienced by the particle and mention its directions.

Answer:
(a) Expression for Lorentz magnetic force on a particle of charge ‘q’ moving with velocity
\(\vec{v}\) in a magnetic field \(\vec{B}\) is \(\vec{F}\) = q(\(\vec{E}\) + \(\vec{v}\) × \(\vec{B}\))
Force is maximum when θ = 90°
Work done by a magnetic force on a charged particle:
The magnetic force \(\vec{F}\) = q(\(\vec{E}\) + \(\vec{v}\) × \(\vec{B}\)) always acts perpendicular to the velocity v on the direction of motion of charge q.
\(\vec{F}\) . \(\vec{v}\) = q(\(\vec{v}\) × \(\vec{B}\)). \(\vec{v}\) = 0

Or

(b) Force experienced by the particle,
F = q(\(\vec{v}\) × \(\vec{B}\))
As magnetic field due to tlw current carrying wire is directed into the plane of the paper (θ = 90°)
Here q = e; B = \(\frac{\mu_0 \mathrm{I}}{2 \pi d}\) ∴ F = \(\frac{\mu_0 e v \mathrm{I}}{2 \pi d}\)
Force is directed away from the current carrying wire or in the right direction of observer.

Question 24.
The potential difference applied across a given conductor is doubled. How will this affect
(i) the mobility of electrons and
(ii) the current density in the conductor? Justify your answers.
Answer:
(i) Mobility, µ = \(\frac{v_d}{E}=\frac{v_d}{\frac{V}{l}}\)
So, as potential is doubled, drift velocity algo gets doubled, therefore, there will be no change in mobility.
(ii) As J ∝ V, current density gets doubled
∴ J = \(\frac{\mathrm{I}}{\mathrm{A}}\) = \(\frac{\mathrm{V}}{\mathrm{RA}}\)

Question 25.
Two coils C ₁ and C ₂ are placed close to each other. The magnetic flux φ ₂ linked with the coil C ₂ , varies with the current I ₁ flowing in coil C ₁ , as shown in the figure. Find:

(i) the mutual inductance of the arrangement, and
(ii) the rate of change of current \(\left(\frac{d \mathrm{I}_1}{d t}\right)\) that will induce an emf of 100 V in coil C ₂ .
Answer:

Section – C

Question 26.
(a) A plane wave-front propagating in a medium of refractive index ‘µ ₁ ‘ is incident on a plane surface making an angle of incidence (i). It enters into a medium of refractive index µ ₂ (µ ₂ > µ ₁ ). Use Huygen’s construction of secondary wavelets to trace the refracted wave-front. Hence verify Snell’s law of refraction. [3]
Or
(b) Using Huygen’s construction, show how a plane wave is reflected from a surface. Hence verify the law of reflection. [3]
Answer:
(a) We take a plane wavefront AB incident at a plane surface XY. We use secondary wavelets starting at different times. We get refracted wavefront only when the time taken by light to travel along different rays from one wavefront to another is same. We take any arbitrary ray starting from point ‘P’ on incident wavefront to refracted wavefront at point ‘Q’. Let total time be ‘t’.

As time should be independent of the ray to be considered.
The coefficient of AO in the above equation should be zero
i.e., \(\frac{\sin i}{\sin r}\) = \(\frac{v_1}{v_2}\) = \({ }^1 \mu_2\) …where [\({ }^1 \mu_2\) is called refractive index of medium 2 w.r.t. medium 1.
This is Snell’s law of refraction.

Or

We take any point Q on the incident wavefront. When disturbance from point B on the incident wavefront reaches point B’, the disturbance from point Q reaches Q’ via point K on the reflecting surface. Since B’A’ represents the reflected wavefront, time by light to travel from any point on incident wavefront to the corresponding point on the reflected wavefront should always be same.

Let total time be ‘t’.
t = \(\frac{\mathrm{QK}}{v}+\frac{\mathrm{KQ}^{\prime}}{v}\) …. where [v = C
or t = \(\frac{\mathrm{QK}}{C}+\frac{\mathrm{KQ}^{\prime}}{C}\) or t = \(\frac{\mathrm{AK}}{\mathrm{C}} \sin i+\frac{\left(\mathrm{AB}^{\prime}-\mathrm{AK}\right) \sin r}{\mathrm{C}}\)
∴ t = \(\frac{\mathrm{AB}^{\prime} \sin r}{\mathrm{C}}+\frac{\mathrm{AK}}{\mathrm{C}}(\sin i-\sin r)\)

As time of the ray to be considered should be independent, the coefficient of AK in the above equation should be zero.
That is, sin i = sin r or i = r.
Hence, angle of incidence is equal to angle of reflection.

Question 27.
An alternating voltage of 220 V is applied across a device X. A current of 0.22 A flows in the circuit and it lags behind the applied voltage in phase by π/2 radian. When the same voltage is applied across another device Y, the current in the circuit remains the same and it is in phase with the applied voltage.
(i) Name the devices X and Y and,
(ii) Calculate the current flowing in the circuit when the same voltage is applied across the series combination of X and Y. [3]
Answer:
(i) X is inductor and Y is resistor.
(ii)

Question 28.
State the basic principle behind the working of an ac generator. Briefly describe its working and obtain the expression for the instantaneous value of emf induced. [3]
Answer:
AC generator. A dynamo or generator is : device which converts mechanical energy into electrical energy. It is based on the principal of electromagnetic energy into electrical energy. It is based on the principle of electromagnetic induction.

Working. When the armature coil is rotated in the strong magnetic field, the magnetic flux linked with the coil changes and the current is induced in the coil, its direction being given by Fleming’s right hand rule. Considering the armature to be in vertical position and as it rotates in anticlockwise direction, the wire ab moves upward and cd downward, so that the direction of induced current is shown in fig. In the external circuit, the current flows along B ₁ RLB ₂ . The direction of current remains unchanged during the first half turn of armature. During the second half revolution, the wire ab moves downward and cd upward, so the direction of current is reversed and in external circuit it flows along B ₂ R _L B ₁ . Thus the direction of induced emf and current changes in the external circuit after each half revolution.
Expression from Induced emf.

N = number of turns in the coil
A = area enclosed by each turn of coil
\(\vec{B}\) = strength of magnetic field
θ = angle, which is normal to the coil, makes with \(\vec{B}\) at any instant t
∴ Magnetic flux linked with the coil in this position, φ = N \((\vec{B} \cdot \overrightarrow{\mathrm{A}})\) = NBA cos θ
If the coil rotates with an angular velocity ω and turns through an angle θ in time t

Question 29.
(a) Briefly describe how the current sensitivity of a moving coil galvanometer can be increased.
(b) A galvanometer shows full scale deflection for current I _g . A resistance R, is required to convert it into a voltmeter of range (0 – V) and a resistance R ₂ to convert it into a voltmeter of range (0 – 2V). Find the resistance of the galvanometer. [3]
Answer:
(a) Current sensitivity of a moving coil galvanometer can be increased by

1. increasing number of turns.
2. Increasing magnetic field strength.
3. Using a material having lesser value of torsional constant.
4. Increasing area of cross—section of the coil.

(b) As we know, Resistance, R = \(\frac{\mathrm{V}}{\mathrm{I}_8}\) – G

Question 30.
(a)
(i) Differentiate between ‘distance of closest approach’ and ‘impact parameter’. [3]
(ii) Determine the distance of closest approach when an alpha particle of kinetic energy 3.95 MeV approaches a nucleus of Z = 79, stops and reverses its directions.

Or

(b)
(i) State three postulates of Bohr’s theory of hydrogen atom.
(ii) Find the angular momentum of an electron revolving in the second orbit in Bohr’s hydrogen atom. [3]
Answer:
(a)

(i) Difference between Distance of Closet Approach and Impact Parameter.
Distance of closest approach. The distance from the nucleus at which the kinetic energy of alpha particle is zero is called the distance of closest approach. It is given by

Impact parameter. It is defined as the perpendicular distance of the velocity vector of the alpha particle from the central line of the nucleus when the particle is far away from the nucleus.
The impact parameter b for scattering angle θ is,
b = \(\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Z e^2 \cot \frac{\theta}{2}}{E}\)

(ii)

Or

(b) (i) Basic postulates of Bohr’s atomic model:

1. Every atom consists of a central core called nucleus in which entire positive charge and mass of the atom are concentrated. A suitable number of electrons revolve around the nucleus in circular orbit. The centripetal force required for revolution is provided by the electrostatic force of attraction between the electron and the nucleus.

2. Electron can resolve only in certain discrete non-radiating orbits, called stationary orbit. Total angular momentum of the revolving electron in an integral multiple of \(\frac{h}{2 \pi}\).
mvr = \(\frac{n h}{2 \pi}\) … where [h is plank constant

3. The radiation of energy occurs only when an electron jumps from one permitted orbit to another. The difference in the total energy of electron in the two permitted orbit is absorbed when the electron jumps from inner to the outer orbit and emitted when electron jumps from outer to inner orbit.

(ii) Angular momentum, L = \(\frac{2 h}{2 \pi}\) …[n = 2 (given)
= \(\frac{6.63 \times 10^{-34}}{3.14}\)
= 2.1 × 10 ^-34 kg m ² s ^-1

Section – D

Question 31.
(i) Explain how free electrons in a metal at constant temperature attain an average velocity under the action of an electric field. Hence obtain an expression for it. [3]
(ii) Consider two conducting wires A and B of the same diameter but made of different materials joined in series across a battery. The number density of electrons in A is 1.5 times that in B. Find the ratio of drift velocity of electrons in wire A to that in wire B. [2]
Or
(b) (i) A cell emf of (E) and internal resistance (r) is connected across a variable load resistance (R). Draw plots showing the variation of terminal voltage V with (a) R and (b) the current (I) in the load. 2
(ii) Three cells, each of emf E but internal resistances 2r, 3r and 6r are connected in parallel across a resistor R.
Obtain expressions for (a) current flowing in the circuit, and (b) the terminal potential difference across the equivalent cell. [3]
Answer:
(a)
(i) Drift velocity. Drift velocity is defined as the velocity of the free electrons with which they get drifted towards the positive terminal under the influence of the external electric field. The drift velocity of electron is of the order of 10 ^-5 m/ sec. Derivation. Let ‘m’ be the mass of an electron and V be the charge on it. When an external electric field ‘E’ is applied, the acceleration acquired by an electron is given by
F = ma ⇒ a = \(\frac{\mathrm{F}}{\mathrm{m}}\) ⇒ a = \(\frac{e \mathrm{E}}{m}\)
Let v ₁ , v ₂ , v ₃ ….. v _n be final velocities of electrons then average velocity of the electrons is given by

(b)

(b)
The terminal potential difference V = E _eq – Ir _eq
⇒ V = E – \(\frac{\mathrm{E}}{\mathrm{R}+\mathrm{r}}\) × r ∴ V = \(\frac{\mathrm{E}(\mathrm{R}+r)-\mathrm{E} r}{\mathrm{R}+r}\) = \(\frac{\mathrm{ER}}{(\mathrm{R}+r)}\)

Question 32.
(a) Draw the circuit arrangement for studying V-I characteristics of a p-n junction diode in
(i) forward biasing and
(ii) reverse biasing. Draw the typical V-I characteristics of a silicon diode.
Describe briefly the following terms:
(i) minority carrier injection in forward biasing and
(ii) breakdown voltage in reverse biasing. [5]
Or
(b) Name two important processes involved in the formation of a p-n junction diode. With the help of a circuit diagram, explain the working of junction diode as a full wave rectifier. Draw its input and output waveforms. State the characteristic property of a junction diode that makes it suitable for rectification. [5]
Answer:
(a)

(i) Minority carrier injection. Under forward bias electrons from n-side cross the depletion region and reach p-side. Similarly, holes from p-side cross the junction and reach the n-side.
(ii) Breakdown voltage. It is the voltage under reverse bias for which reverse current increases sharply.

Or

(b) Process involved in the formation of a p-n junction diode:
(i) Diffusion and
(ii) Drift
Working of p-n junction diode as full wave rectifier.
A full wave rectifier consists of two diodes and special type of transformer known as centre tap transformer as shown in the circuit. The secondary of transformer gives the desired a.c. voltage across A and B.

During the positive half cycle of a.c. input, the diode D ₁ is in forward bias and conducts current while D ₂ is in reverse biased and does not conduct current. So we get an output
voltage across the load resistor R _L . During the negative half cycle of a.c, input, the diode
D ₁ is in reverse biased and does not conduct current while diode D ₂ in forward biased and conducts current. So we get an output voltage across the load resistor R _L .

Let I = I ₀ sin ωt be the input current to be rectified
∴ The average current = \(\frac{2 I_0}{\pi}\)
Hence output voltage = \(\frac{2 \mathrm{I}_0}{\pi} \mathrm{R}_{\mathrm{L}}\)
Characteristic property of Junction Diode. Junction diode allows current to pass only when it is forward biased.

Question 33.
(a)
(i) Draw a ray diagram to show the working of a compound microscope. Obtain the expression for the total magnification for the final image to be formed at the near point. [3]
(ii) In a compound microscope an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye-piece has a focal length of 5 cm and the final image is formed at the near point, find the magnifying power of the microscope. [2]

Or

(i) Draw a ray diagram for the formation of image of an object by an astronomical telescope, in normal adjustment. Obtain the expression for its magnifying power.
(ii) The magnifying power of an astronomical telescope in normal adjustment is 2.9 and the objective and the eyepiece are separated by a distance of 150 cm. Find the focal lengths of the two lenses. [5]
Answer:
(a)
(i) Magnifying power. Thé magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final virtual image to the angle subtended at the eye by the object, when both are at the least distance of distinct vision from the eye.

Also image A’B is formed close to the eyelens whose focal length is short, therefore v ₀ = L = the length of the microscope tube or the distance between the two lenses
∴ m ₀ = \(\frac{v_0}{u_0}=-\frac{\mathrm{L}}{f_0}\) ∴ m = \(\frac{\mathrm{L}}{f_0}\left(1+\frac{\mathrm{D}}{f_e}\right)\) …… [For final image at D

(ii)

Or

(i) Ray Diagram of Astronomical Telescope:

(ii)

Section-E

Note: Questions number 34 and 35 are Case Study based questions. Read the following paragraph and answer the questions.

Question 34.
A lens is a transparent optical medium bounded by two surfaces; at least one of which should be spherical. Considering image formation by a single spherical surface successively at the two surfaces of a lens, lens maker’s formula is obtained. It is useful to design lenses of desired focal length using surfaces of suitable radii of curvature. This formula helps us obtain a relation between u, v and f for a lens. Lenses form images of objects and they are used in a number of optical devices, for example microscopes and telescopes. [4]

(i) An object AB is kept in front of a composite convex lens, as shown in figure. Will the lens produce one image? If not, explain.
(ii) A real image of an object formed by a convex lens is observed on a screen. If the screen is removed, will the image still be formed? Explain.
(iii) A double convex lens is made of glass of refractive index 1.55 with both faces of the same radius of curvature. Find the radius of curvature required if focal length is 20 cm.

Or

(iii) Two convex lenses A and B of focal lengths 15 cm and 10 cm respectively are placed coaxially ‘d! distance apart. A point object is kept at a distance of 30 cm in front of lens A. Find the value of ‘d! so that the rays emerging from lens B are parallel to its principal axis.
Answer:
(i) No, The lens is made up of two materials of different refractive indices. It has two focal lengths.

(ii) Yes, Rays are still intersecting/converging at the location of image.

(iii) Given, μ = 1.55, f = 20 cm

Or

For lens A:
Given: f = 15 cm, u = -30 cm
Usingg Lens formula, \(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\) ⇒ \(\frac{1}{v_1}\) =\(\frac{1}{15}-\frac{1}{30}\) = \(\frac{1}{30}\) = 30 ∴ v ₁ = 30 cm
For lens B. Object is kept at 2f so image will be formed at 2f on the other side of the lens, i.e., at 30 cm. Now the final image is to be formed at infinity so the image formed must lie at the focus of the second lens (B).
So, d = 30 + 10 = 40 cm

Question 35.
A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference between them. The capacitance of a capacitor depends on the geometrical configuration (shape, size and separation) of the system and also on the nature of the insulator separating the two conductors. They are used to store charges. Like resistors, capacitors can be arranged in series or parallel or a combination of both to obtain desired value of capacitance. [4]

(i) Find the equivalent capacitance between points A and B in the given diagram.
(ii) A dielectric slab is inserted between the plates of a parallel plate Capacitor. The electric field between the plates decreases. Explain.
(iii) A capacitor A of capacitance C, having charge Q is connected across another uncharged capacitor B of capacitance 2C. Find an expression for (a) the potential difference across the combination and (b) the charge lost by capacitor A.

Or

(iii) Two slabs of dielectric constants 2K and K fill the space between the plates of a parallel plate capacitor of plate area A and plate separation d as shown in figure. Find an expression for capacitance of the system.

Answer:
(i)

(ii) Within the dielectric slab, induced electric field due to polarization, decreases the electric field.
∴ E = E ₀ – Ep

(iii)

SET II Code No. 55/5/2
Section-A

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

Question 1.
A charge Q is placed at the centre of a cube. The electric flux through one if its face is [1]
(a) \(\frac{\mathrm{Q}}{\varepsilon_0}\)
(b) \(\frac{\mathrm{Q}}{6 \varepsilon_0}\)
(c) \(\frac{\mathrm{Q}}{8 \varepsilon_0}\)
(d) \(\frac{\mathrm{Q}}{3 \varepsilon_0}\)
Answer:
(b) \(\frac{\mathrm{Q}}{6 \varepsilon_0}\)

As the charge is at the centre of the cube, the flux through each surface is sanie.
Using Gauss’s law, 6φ = \(\frac{\mathrm{Q}}{\varepsilon_0}\) ∴ φ = \(\frac{Q}{6 \varepsilon_0}\)

Question 5.
Choose the correct option related to wavelengths (λ) of different parts of electromagnetic spectrum. [1]

Answer:
(c) λ _{radio wave} > λ _{micro waves} > λ _x-rays

Question 11.
Figure shows a plot of stopping potential (V ₀ ) versus \(\frac{1}{\lambda}\), where λ is the wavelength of the radiation causing photoelectric emission from a surface. The slope of the line is equal to

(a) φ ₀
(b) \(\frac{h}{e}\)
(c) \(\frac{h c}{e}\)
(d) \(\frac{h^2 c}{e^2}\)
Answer:
(c) \(\frac{h c}{e}\)

Question 13.
An ideal inductor is connected across an AC source of voltage. The current in the circuit [1]
(a) is ahead of the voltage in phase by π.
(b) lags voltage in phase by π.
(c) is ahead of voltage in phase by π/2.
(d) lags voltage in phase by π/2.
Answer:
(d) lags voltage in phase by π/2.

Question 16.
In the following question two statement are given—one labelled two statements are given one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
Assertion (A): In insulators, the forbidden gap is very large. [1]
Reason (R): The valence electrons in an atom of an insulator are very tightly bound to the nucleus.
(a) Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true and (R) is NOT the correct explanation of (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also false.
Answer:
(a) Both assertion (A) and reason (R) are true and (R) is the correct explanation of (A).

Section – B

Question 21.
Identify the electromagnetic radiation and write its wavelength range, which is used to kill germs in water purifier. Name the two sources of these radiations. [2]
Answer:
Electromagnetic Radiation. Ultraviolet rays
Wavelength range—4 × 10 ^-7 m – 6 × 10 ^-10 m.
Source:
1. The sun
2. Any device in which inner shell electrons in atoms move from one energy level to a lower energy level.

Question 22.
An electric dipole of dipole moment (\(\vec{p}\)) is kept in a uniform electric field \(\vec{E}\). Show graphically the variation of torque acting on the dipole \((\tau)\) with its orientation (θ) in the field. Find the orientation in which torque is (i) zero and (ii) maximum. [2]
Answer:
(i) Torque is zero for orientation corresponding to θ = 0° and θ = 180°.
(ii) Torque is maximum for orientation corresponding to θ = \(\frac{\pi}{2}\) and 3 \(\frac{\pi}{2}\)

Section – C

Question 27.
A current of 1A flows through a coil when it is connected across a DC battery of 100 V. If DC battery is replaced by an AC source of 100 V and angular frequency 100 rad s ^-1 , the current reduces to 0.5 A. Find
(i) impedance of the circuit.
(ii) self-inductance of coil.
(iii) phase difference between the voltage and the current. [3]
Answer:

Question 28.
The primary coil having N _p , turns of an ideal transformer is supplied with an alternating voltage V _p . Obtain an expression for the voltage vs induced in its secondary coil having N _s , turns. Mention two main sources of power loss in real transformers. [3]
Answer:
Working. As the alternating current flows through the primary, it generates an alternating magnetic flux in the core which also passes through the secondary. This changing flux sets up an induced emf in the secondary, also a self-induced emf in the primary. If there is no leakage of magnetic flux, then flux linked with each turn of the primary will be equal to that linked with each turn of the secondary.
V _P = -N _P \(\frac{d \phi}{d t}\) and V _s = -N _s \(\frac{d \phi}{d t}\)
…………. where
[N _P and N _S are number of turns in the primary and secondary respectively, V _P and V _S are their respective voltages
∴ \(\frac{V_S}{V_P}\) = \(\frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}}\) ……… (i)
This ratio \(\frac{\mathrm{N}_{\mathrm{S}}}{\mathrm{N}_{\mathrm{P}}}\) is called the turns ratio.
Assuming the transformer to be ideal one, so that there are no energy tosses, then
Input power = output power
V _P I _P = V _S I _S … where [I _P and I _S are the current in the primary and secondary respectively
\(\frac{\mathrm{V}_{\mathrm{S}}}{\mathrm{V}_{\mathrm{P}}}\) = \(\frac{I_P}{I_S}\)
From (i) and (ii), we get \(\frac{\mathrm{I}_{\mathrm{P}}}{\mathrm{I}_{\mathrm{S}}}\) = \(\frac{V_S}{V_P}=\) = \(\frac{N_S}{N_P}\)

In a step up transformer, N _S > N _P , i.e., the turns ratio is greater than 1 and therefore V _S > V _P . The output voltage is greater than the input voltage.

Two sources of energy loss in a transformer
1. Copper loss. Some energy is lost due to heating of copper wires used in the primary and
secondary windings. This power loss (= I ² R) can be minimised by using thick copper
wires of low resistance.
2. Eddy current loss. The alternating magnetic flux induces eddy currents in the iron core which leads to some energy loss in the form of heat. This loss can be reduced by using laminated iron core.

Question 30.
(a) (i) Write the limitations of Rutherford’s model of atom.
(ii) The wavelength of the second line of the Balmer series in the hydrogen spectrum is 4861 A. Calculate the wavelength of the first line of the same series. [3]

Or

(b) (i) Increase in the intensity of he radiation causing photo-electric emission from a surface, does not affect the maximum K.E. of the photo electrons. Explain.
(ii) The photon emitted during the de-excitation from the first excited level to the ground state of hydrogen atOm is used to irradiate a photo cathode in which stopping potential is 5 V. Calculate the work function of the cathode used. [3]
Answer:
(a) (i) Important limitations of Rutherford Model:
1. According to Rutherford model, electron orbiting around the nucleus, continuously radiates energy due to the acceleration; hence the atom will not remain stable.
2. As electron spirals inwards; its angular velocity and frequency change continuously; therefore it will emit a continuous spectrum.
(ii)

Or

(b) (i) hv = φ ₀ + K _max
It is evident from the above equation that the maximum kinetic energy depends on the frequency of incident radiation and is independent of the intensity of the radiations.
(ii) Given: n ₁ = 2, n ₂ = 1, V ₀ = 5V, eV ₀ = 5eV,
(φ _o ) =?
Energy of photon (E) = (13.6) – (3.4) eV = 10.2 eV
According to photoequation,
E = eV ₀ + φ ₀ ⇒ φ ₀ = E – (eV ₀ ) = (10.2) – (5)
∴ φ ₀ = 5.2eV

SET I Code No. 55/5/3
Section – A

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

Question 1.
Two horizontal thin long parallel wires, separated by a distance r carry current I each in the opposite directions. The net magnetic field at a point midway between them, will be [1]
(a) Zero
(b) \(\left(\frac{\mu_0 \mathrm{I}}{2 \pi r}\right)\) vertically downward
(c) \(\left(\frac{2 \mu_0 \mathrm{I}}{r}\right)\) vertically upward
(d) \(\left(\frac{\mu_0 \mathrm{I}}{\pi r}\right)\) vertically downward
Answer:
No option is correct
Net magnetic field = \(\frac{\mu_0 \mathrm{I}}{2 \pi \frac{r}{2}}+\frac{\mu_0 \mathrm{I}}{2 \pi \frac{r}{2}}\) = \(\frac{2 \mu_0 I}{\pi r}\)

Question 3.
Which of the following cannot modify an external magnetic field as shown in the figure? [1]

(a) Nickel
(b) Silicon
(c) Sodium Chloride
(d) Copper
Answer:
(a) Nickel

Question 7.
E, c and y represent the energy, velocity and frequency of a photon. Which of the following represents its wavelength? [1]
(a) \(\frac{h \nu}{c^2}\)
(b) hv
(c) \(\frac{h c}{\mathrm{E}}\)
(d) \(\frac{h v}{c}\)
Answer:
(c) \(\frac{h c}{\mathrm{E}}\)

Question 9.
The energy required by an electron to jump the forbidden band in silicon at room temperature is about [1]
(a) 0.01 eV
(b) 0.05 eV
(c) 0.7 eV
(d) 1.1 eV
Answer:
(d) 1.1 eV

Question 13.
What is the ratio of inductive and capacitive reactance in an ac circuit? [1]
(a) ω ² LC
(b) LC ²
(c) \(\frac{L C}{\omega^2}\)
(d) ω ² L
Answer:
(a) ω ² LC

Inductive reactant, X _L = ωL; Capacitive reactant, X _C = \(\frac{1}{\omega C}\)
∴ Ratio = \(\frac{X_L}{X_{\mathrm{C}}}\) = \(\frac{\omega \mathrm{L}}{\frac{1}{\omega \mathrm{C}}}\) = ω ² LC

Question 14.
In an interference experiment, third bright fringe is obtained at a point on the screen with a light of 700 nm. What should be the wavelength of the light source in order to obtain the fifth bright fringe at the same point? [1]
(a) 420 nm
(b) 750 nm
(c) 630 nm
(d) 500 nm
Answer:
(a) 420 nm

Question 17.
In the following question two statement are given —one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
Assertion (A): The given figure does not show a balanced Wheatstone bridge. [1]

Answer:
(c) Assertion (A) is true and Reason (R) is false.

Reason (R): For a balanced bridge small current should flow through the galvanometer. ‘
(a) Both Assertion (A) and Reason (R) are true and (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true and (R) is NOT the correct explanation of (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also false.

Section – B

Question 19.
Plot a graph showing the variation of photo electric current, as a function of anode potential for two light beams having the same frequency but different intensities I ₁ and I ₂ (I ₁ > I ₂ ). Mention its important features. [2]
Answer:

Important Features:

• Increase in intensity does not change the cut-off potential.
• The increase in intensity increase the value of saturation current.

Question 21.
How are electromagnetic waves produced? Write their two characteristics. [2]
Answer:
Methods to produce Electromagnetic waves:

1. Charged particle moving with varying speed.
2. Oscillating charge
3. By using LC circuit.

Characteris tics of Electromagnetic waves

1. Travels with speed of light in vacuum.
2. Transverse in nature.
3. No medium required for propagation

Question 22.
Three point charges Q, q and -q are kept at the vertices of an equilateral triangle of side L as shown in figure. What is [2]

(i) the electrostatic potential energy of the arrangement? and
(ii) the potential at point D?
Answer:

Question 23.
(a) Two identical circular loops P and Q, each of radius R carrying current I are kept in perpendicular planes such that they have a common centre O as shown in the figure. Find the magnitude and direction of the net magnetic field at point O. [2]

Or
(b) A long straight conductor kept along X’ X axis, carries a steady current I along +x direction. At an instant t, a particle of mass m and charge q at point (x, y) moves with a velocity \(\vec{v}\) along + y direction. Find the magnitude and direction of the force on the particle due to the conductor. [2]
Answer:

Note: Since direction of current is shown clockwise as well as anticlockwise in the vertical loop, so, no marks is assigned for the direction of the magnetic field.

Or

(b) Magnetic force, \(\overrightarrow{\mathrm{F}}\) = q\((\vec{V} \times \vec{B})\) sin θ
⇒ F = q\(\left(\mathrm{V} \times \frac{\mu_0 \mathrm{I}}{2 \pi y}\right)\) sin 90°
∴ F = \(\frac{\mu_0 q \mathrm{VI}}{2 \pi y}\)
Direction. Force is along + X axis.

Question 24.
Two conductors, made of the same material have equal lengths but different cross-sectional areas A ₁ and A ₂ (A ₁ > A ₂ ). They are connected in parallel across a cell. Show that the drift velocities of electrons in two conductors are equal. [2]
Answer:
Given. Length of two conductors are equal, l ₁ = l ₂ = l
Area of cross-sections of two conductors = A ₁ and A ₂
Let Drift velocities of wire V _d and V _d ,
Since two conductors are connected in parallel, the potential difference across the two conductors be same

Section – C

Question 27.
An alternating current I = 14 sin (100 π) A passes through a series combination of a resistor of 30 Ω and an inductor of \(\left(\frac{2}{5 \pi}\right)\)H. Taking \(\sqrt{2}\) = 1.4, clculate the—
(i) rms value of the voltage drops across the resistor and the inductor, and
(ii) power factor of the circuit.
Answer:

Question 30.
(a) Calculate the binding energy of an alpha particle in MeV.
Given: . [3]
mass of a proton 1.007825 u
mass of a neutron 1.008665 u
mass of He nucleus = 4.002800 u
1u = 931 MeV/c ²
Or
(b) 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 number 110 and 130 and binding energy per nucleon 8.5 MeV and 8.4 MeV respectively. Calculate the energy released in the fission. [3]
Answer:
(a) Number of Protons = 2; Number of neutrons = 2
BE = [(ZM _P + (A – Z) M _n ) – m _N )] C ²
⇒ ∆ _m = [(2 × 1.007825 + 2 × 1.008665) – (4.002800)]u
⇒ BE = 0.03018 × 931 meV
∴ BE = 28.097 MeV

Or

(b) Binding energy of nucleus, P = 7.6 × 240 = 1824 meV.
Binding energy of nucleus, Q = 8.5 × 11o = 935 meV
Binding energy of nucleus, R = 8.4 × 130 = 1092 MeV.
Binding energy of (Q + R) = 935 + 1092 = 2027 MeV.
∴ Energy released = Binding energy of (Q + R) – Binding energy of P
∴ Energy released = 2027 – 18 = 203 MeV