ISC Class 12 Physics Sample Paper 2027 with Solutions

SECTION A - 14 MARKS

Question 1

(A) In questions (i) to (vii), choose the correct alternative (a), (b), (c) or (d) for each of the questions given below.

(i) A point charge \(-Q\) and three points A, B and C are shown in Figure 1 below. If V is electric potential and E is electric field intensity, we can infer that: [1 Mark]
[Figure: A horizontal straight line with a point charge \(-Q\) at point O on the left. Points A, C, and B are marked to the right of O in that order, such that A is closest to O, C is in the middle, and B is farthest from O.]
(a) \(E_A \gt E_B \gt E_C\)
(b) \(E_A \lt E_B \lt E_C\)
(c) \(V_A \lt V_C \lt V_B\)
(d) \(V_A \gt V_C \gt V_B\)

Answer: (c) \(V_A \lt V_C \lt V_B\)

For a negative charge, potential is given by \(V = -kQ/r\). As distance \(r\) increases, potential becomes less negative (i.e., increases algebraically).

Teacher's Note:
a) Electric potential due to a negative point charge is negative and inversely proportional to distance \(r\), so closer points have more negative (lower) potentials.
b) Students often confuse the behavior of electric potential for positive versus negative charges.

(ii) Select the odd one out from the following options. [1 Mark]
(a) Iron
(b) Nickel
(c) Cobalt
(d) Platinum

Answer: (d) Platinum

Iron, nickel, and cobalt are ferromagnetic materials, whereas platinum is a paramagnetic material.

Teacher's Note:
a) Classification of magnetic materials based on susceptibility helps identify ferromagnetism versus paramagnetism.
b) Remember that platinum, aluminum, and oxygen are standard examples of paramagnetic substances.

(iii) In Figure 2 below, CD is an infinitely long and straight metallic wire carrying a current I. Point P is at a perpendicular distance \(r\) from it. Which combination of I and r gives maximum magnetic field at point P, pointing into the plane of the paper? [1 Mark]
[Figure: An infinite vertical straight wire CD carrying current \(I\) upwards. Point P lies to the right at a perpendicular distance \(r\).]
(a) \(I = 4\text{ A}\) towards C, with \(r = 0.500\text{ m}\)
(b) \(I = 4\text{ A}\) towards D, with \(r = 0.250\text{ m}\)
(c) \(I = 8\text{ A}\) towards C, with \(r = 0.125\text{ m}\)
(d) \(I = 8\text{ A}\) towards C, with \(r = 0.050\text{ m}\)

Answer: (d) \(I = 8\text{ A}\) towards C, with \(r = 0.050\text{ m}\)

The magnitude of the magnetic field is \(B = \frac{\mu_0 I}{2\pi r}\). To maximize \(B\), we need maximum \(I\) and minimum \(r\).

Teacher's Note:
a) The direction of current towards C (upwards) with point P to its right produces a magnetic field pointing into the plane of the paper by the right-hand thumb rule.
b) Always check the ratio of current to distance (\(I/r\)) to find the strongest magnetic field.

(iv) Match the entries in Column I with those in Column II. [1 Mark]

(a) 1-C, 2-D, 3-A, 4-B
(b) 1-B, 2-D, 3-C, 4-A
(c) 1-C, 2-A, 3-B, 4-D
(d) 1-A, 2-B, 3-D, 4-C

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

Planck proposed quantum theory, Einstein gave mass-energy relation (\(E = mc^2\)), Bohr explained the hydrogen spectrum, and de Broglie proposed the wave nature of particles.

Teacher's Note:
a) Match each scientist with their hallmark contribution to modern physics.
b) Direct matching questions test fundamental historical associations in physics.

(v) Identify the phenomena that occur when a P-N junction diode is forward biased. [1 Mark]
(P) The width of the depletion layer decreases.
(Q) The height of the potential barrier is lowered.
(R) The effective resistance of the diode increases significantly.
(S) The reverse current flows mainly due to the diffusion of majority charge carriers.
(a) Both (P) and (Q)
(b) Both (R) and (S)
(c) Both (P) and (S)
(d) Both (Q) and (R)

Answer: (a) Both (P) and (Q)

In forward biasing, the external field opposes the barrier potential, thereby reducing the barrier height and narrowing the depletion region.

Teacher's Note:
a) Forward bias decreases both depletion region width and barrier potential, facilitating low resistance and high forward current.
b) Statement (R) is true for reverse bias, and statement (S) is factually incorrect.

(vi) In an astronomical telescope of refracting type, [1 Mark]
(a) Objective lens has smaller focal length, and eyepiece lens has greater focal length.
(b) Objective lens has greater focal length, and eyepiece lens has smaller focal length.
(c) Both objective lens and eyepiece lens have large focal length.
(d) Both objective lens and eyepiece lens have small focal length.

Answer: (b) Objective lens has greater focal length, and eyepiece lens has smaller focal length.

The magnifying power of an astronomical telescope is given by \(m = f_o / f_e\), which requires a large \(f_o\) and a small \(f_e\).

Teacher's Note:
a) A large objective focal length collects more light and provides higher magnification.
b) Do not confuse astronomical telescope lenses with compound microscope lenses.

(vii) Given below are two statements marked, Assertion and Reason. Read the two statements and choose the correct option. [1 Mark]
Assertion: The focal length of a convex mirror will increase, if the mirror is placed in water.
Reason: The focal length of a convex mirror is equal to half its radius of curvature.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer: (d) Assertion is false and Reason is true.

The focal length of a mirror depends only on its radius of curvature (\(f = R/2\)) and is independent of the surrounding medium.

Teacher's Note:
a) Unlike lenses, reflection does not depend on the refractive index of the surrounding medium.
b) The reason statement is independently true, but the assertion is false.

(B) Answer the following questions briefly:

(i) Two point charges Q₁ and Q₂, separated by a distance, are such that there is no neutral point (where electric field is zero) in their vicinity. What is the relation between Q₁ and Q₂? [1 Mark]

Answer:
\(Q_2 = -Q_1\) or \(Q_1\) and \(Q_2\) form an electric dipole (i.e., equal in magnitude and opposite in sign).

Teacher's Note:
a) A neutral point exists between like charges where their electric fields cancel out.
b) For unlike equal charges, the electric fields never cancel out on the line joining them externally.

(ii) Compare resistance of an ideal voltmeter with that of an ideal ammeter. [1 Mark]

Answer:
An ideal voltmeter has infinite resistance (\(R_v = \infty\)), whereas an ideal ammeter has zero resistance (\(R_a = 0\)).

Teacher's Note:
a) High voltmeter resistance ensures negligible current is drawn from the circuit.
b) Zero ammeter resistance ensures no potential drop across it when connected in series.

(iii) How can the focal length of a given convex lens be measured in Physics laboratory using only a metre scale? [1 Mark]

Answer:
By the Distant Object Method, where light rays from a very distant source are focused onto a screen to find the lens focal point directly.

Teacher's Note:
a) The distance between the lens and the sharp image of a distant tree or sun gives the focal length approximately.
b) State the method name clearly to secure full marks.

(iv) What is the difference between a moving metallic ball colliding with another ball in motion and a photon colliding with an electron? [1 Mark]

Answer:
When a photon collides with an electron, it loses either its entire energy or none at all, whereas a moving ball can lose any fraction of its kinetic energy.

Teacher's Note:
a) Photon-electron collisions are discrete quantum interactions (photoelectric effect/Compton scattering).
b) Macroscopic ball collisions obey classical mechanics with continuous energy transfer.

(v) When can a gamma ray photon produce a proton and an antiproton pair? [1 Mark]

Answer:
When the gamma ray photon has a minimum energy equal to the rest mass energy of a proton-antiproton pair, i.e., \(E = 2m_p c^2 \approx 1876.6\text{ MeV}\).

Teacher's Note:
a) This process is known as pair production.
b) The threshold energy must account for the mass of both particle and antiparticle.

(vi) Explain why resistance of a semiconductor material decreases when its temperature is increased. [1 Mark]

Answer:
As temperature rises, a large number of covalent bonds break, generating more free electrons and holes in the conduction band, which increases conductivity and decreases resistance.

Teacher's Note:
a) Semiconductors have a negative temperature coefficient of resistance.
b) The increase in charge carrier density (\(n\)) outweighs the decrease in relaxation time (\(\tau\)).

(vii) Fill in the blank with an appropriate term from the bracket.
When a light wave travels from one medium to another, its ________ (amplitude, wavelength, frequency, speed) remains unaltered. [1 Mark]

Answer:
Frequency

Teacher's Note:
a) Frequency is a characteristic of the source and does not change upon refraction.
b) Speed, wavelength, and amplitude all change when light enters a new medium.

SECTION B - 14 MARKS

Question 2 [2 Marks]

(i) A negatively charged thundercloud above the earth’s surface may be perceived as a parallel plate capacitor. (Refer to the image below.)
The lower plate of the capacitor is the Earth’s surface, and the upper plate is the base of the thundercloud. The area of the base of the thundercloud = \(4.5 \times 10^8\text{ m}^2\), distance of thundercloud base from the earth’s surface = \(4425\text{ m}\). Calculate the capacitance of this capacitor.
[Figure: Diagram showing a thundercloud base with negative charges separated by air of thickness \(4425\text{ m}\) from the positive earth's surface.]

Answer:
\(C = \frac{\epsilon_0 A}{d}$
\(C = \frac{8.85 \times 10^{-12} \times 4.5 \times 10^8}{4425}$
\(C = 0.009 \times 10^{-4} = 9 \times 10^{-7}\text{ F}\)

Teacher's Note:
a) Use the standard parallel plate capacitor formula \(C = \frac{\epsilon_0 A}{d}\).
b) Substitute SI units carefully and express the final answer in scientific notation.

OR

(ii) A tiny bird carrying a charge of \(q = 8 \times 10^{-17}\text{ C}\) dives vertically down towards a large positively charged metallic plate kept on a table in a horizontal position. What should be the surface charge density (\(\sigma\)) of the plate so that the bird stops in its flight, close to the plate? (Mass of the bird = \(50\text{ g}\)) [2 Marks]

Answer:
Force on the bird due to electric field equals its weight: \(qE = mg$
Since \(E = \frac{\sigma}{2\epsilon_0}$, we have \(q \frac{\sigma}{2\epsilon_0} = mg$
\(\sigma = \frac{2mg\epsilon_0}{q}$
\(\sigma = \frac{2 \times 50 \times 10^{-3}\text{ kg} \times 9.8\text{ m s}^{-2} \times 8.85 \times 10^{-12}\text{ F m}^{-1}}{8 \times 10^{-17}\text{ C}}$
\(\sigma = 10.8 \times 10^4\text{ V m}^{-1}\) (or \(\text{C m}^{-2}\))

Teacher's Note:
a) Equate the electrostatic upward force to the gravitational downward force.\b) Convert mass into kilograms before substituting values into the formula.

Question 3 [2 Marks]

Draw a labelled diagram of an arrangement used in Physics laboratory to compare emfs of two given cells X and Y.

Answer:
[Figure: Potentiometer circuit diagram showing a driver cell connected across a long wire PQ through a rheostat and a key, with cells X and Y connected through a two-way key to a galvanometer and jockey.]
1. The arrangement consists of a potentiometer wire PQ connected in series with a driver battery, key, and rheostat.
2. The positive terminals of cells X and Y are connected to terminal P, and their negative terminals are connected through a two-way key to a galvanometer and jockey.

Teacher's Note:
a) Ensure all polarities (positive terminals connected together at the high potential end) are correctly shown in the diagram.
b) Label the driver cell, potentiometer wire, galvanometer, and two-way key clearly.

Question 4 [2 Marks]

(i) According to Ohm’s law, V = IR, so if potential difference (V) between the two ends of a conductor is varied, will its resistance (R) change? Explain your answer in brief.

Answer:
No, resistance \(R\) of a conductor is constant at a constant temperature. Its value does not depend on \(V\) or \(I\).

Teacher's Note:
a) Resistance is a property of the conductor depending on its dimensions and material, not on applied voltage or current.
b) Mentioning constant temperature is essential for a complete explanation.

OR

(ii) Heating power developed in a resistor R is given by P = V²/R. It implies that power \(P \propto 1/R\) i.e., power varies inversely with R. Heating power developed in a resistor R is also given by P = I²R. It implies that power \(P \propto R\) i.e., power varies directly with R. Under what conditions, are the above-mentioned statements valid? [2 Marks]

Answer:
1. \(P \propto \frac{1}{R}$ is valid when potential difference \(V\) is constant (i.e., resistors connected in parallel).
2. \(P \propto R$ is valid when current \(I\) is constant (i.e., resistors connected in series).

Teacher's Note:
a) Explain that the choice of formula depends on whether the circuit elements are in series or parallel.
b) Clear differentiation between constant voltage and constant current conditions is required.

Question 5 [2 Marks]

Explain why a blue coloured spark appears in a switch contact when a circuit containing a load like a heater, geyser or an electric iron is switched off.

Answer:
This is due to a large induced emf produced in the inductive load when the circuit is suddenly broken, which causes an electric discharge through air. Ionised air molecules emit blue light when they return to ground state from their excited states.

Teacher's Note:
a) High self-inductance of heating appliances causes back emf (switching transient spark).
b) Mentioning ionisation of air and atomic de-excitation completes the scientific explanation.

Question 6 [2 Marks]

There are seven types of electromagnetic waves having different frequency ranges. Arrange them in the increasing order of their frequencies.

Answer:
1. Radio Waves (Lowest frequency)
2. Microwaves
3. Infrared Waves
4. Visible Light
5. Ultraviolet (UV) Rays
6. X-rays
7. Gamma Rays (Highest frequency)

Teacher's Note:
a) Memorize the complete electromagnetic spectrum sequence.
b) Ensure the order goes strictly from lowest to highest frequency (or longest to shortest wavelength).

Question 7 [2 Marks]

(i) Why the current flowing is extremely small during reverse biasing of a junction diode?

Answer:
The current flowing through the diode during reverse bias is extremely small because it is entirely due to minority charge carriers, which are much smaller in number compared to majority carriers.

Teacher's Note:
a) Reverse bias widens the depletion region, blocking majority carriers.
b) Only thermally generated minority carriers drift across the junction, causing a tiny reverse saturation current.

(ii) Extrinsic semiconductors are neither good nor bad conductors of electricity. Why are they still considered useful? [2 Marks]

Answer:
Extrinsic semiconductors are highly useful because their electrical conductivity can be precisely controlled, tuned, and manipulated by altering the concentration of dopant atoms added to the crystal. (Alternatively: Electronic devices like diodes, transistors, and solar cells are made from them).

Teacher's Note:
a) Doping increases conductivity by orders of magnitude relative to intrinsic semiconductors.
b) This controllability forms the basis of all modern solid-state electronics.

Question 8 [2 Marks]

Two particles A and B of masses m_A and m_B have de Broglie wavelength \(\lambda_A\) and \(\lambda_B\). If \(\lambda_A = 2\lambda_B\), compare their kinetic energies.

Answer:
\(KE = \frac{p^2}{2m} = \frac{h^2}{2m\lambda^2}$
\(\frac{(KE)_A}{(KE)_B} = \frac{m_B \lambda_B^2}{m_A \lambda_A^2}$
Since \(\lambda_A = 2\lambda_B\):
\(\frac{(KE)_A}{(KE)_B} = \frac{m_B \lambda_B^2}{m_A (2\lambda_B)^2} = \frac{m_B}{4m_A}$
\((KE)_A : (KE)_B = m_B : 4m_A\)

Teacher's Note:
a) Use the de Broglie wavelength relation in terms of momentum and kinetic energy: \(\lambda = \frac{h}{\sqrt{2mKE}}\).
b) Substitute the given wavelength ratio carefully to arrive at the final kinetic energy ratio.

SECTION C - 27 MARKS

Question 9 [3 Marks]

(i) Obtain an expression of intensity of electric field at a point lying on the axis of an electric dipole.

Answer:
[Figure: Electric dipole with charges \(-Q\) at A and \(+Q\) at B separated by distance \(2l\), and point P on its axis at distance \(r\) from the center.]
1. Electric field at point P due to charge \(+Q\) at B: \(E_1 = \frac{K Q}{(r-l)^2}$ (directed away from B)
2. Electric field at point P due to charge \(-Q\) at A: \(E_2 = \frac{K Q}{(r+l)^2}$ (directed towards A)
3. Net electric field \(E = E_1 - E_2 = K Q \left[ \frac{1}{(r-l)^2} - \frac{1}{(r+l)^2} \right] = \frac{K \cdot 2pr}{(r^2 - l^2)^2}\), where \(p = Q \cdot 2l\).

Teacher's Note:
a) Clearly show vector directions along the axial line.
b) For a short dipole where \(r \gg l\), state the approximation \(E = \frac{2pK}{r^3}\).

OR

(ii) An electric dipole is kept in a uniform electric field E with its axis making an angle \(\theta\) with the field. Show forces acting on each charge and hence prove that torque acting is given by \(\tau = pE\sin\theta\). [3 Marks]

Answer:
[Figure: Electric dipole in a uniform electric field \(E\) showing force \(+qE\) and \(-qE\) forming a couple separated by perpendicular distance \(AC = 2l\sin\theta\).]
1. Force on charge \(+q\) is \(qE$ along the field, and on charge \(-q\) is \(qE$ opposite to the field.
2. Torque \(\tau = \text{Force} \times \text{perpendicular distance between forces} = qE \times (2l \sin\theta)$.
3. Since dipole moment \(p = q \cdot 2l\), we get \(\tau = pE\sin\theta\).

Teacher's Note:
a) Torque is defined as either force multiplied by the perpendicular distance between their lines of action.
b) Express the final result in vector form as \(\vec{\tau} = \vec{p} \times \vec{E}\) for completeness.

Question 10 [3 Marks]

(i) Three identical cells each of emf 1.5V and internal resistance of \(2\Omega\) are connected in series to form a battery B. This battery is connected to an ammeter A (of negligible resistance) and five other resistors as shown in Figure 4 below. Calculate the reading of the ammeter A.
[Figure: Circuit diagram with battery B, ammeter A, and five resistors: \(R_1 = 10\Omega$, \(R_2 = 120\Omega$, \(R_3 = 14\Omega$, \(R_4 = 30\Omega\), and another resistor \(40\Omega\).]

Answer:
1. Resistors \(R_4\) (\(30\Omega\)), the \(40\Omega\) resistor, and \(R_2\) (\(120\Omega\)) are in parallel. Their equivalent resistance \(R_6\) is given by:
\(\frac{1}{R_6} = \frac{1}{30} + \frac{1}{40} + \frac{1}{120} = \frac{4+3+1}{120} = \frac{8}{120} \implies R_6 = 15\Omega$
2. Now, \(R_1\) (\(10\Omega\)), \(R_6\) (\(15\Omega\)), and \(R_3\) (\(14\Omega\)) are in series:
\(R = 10 + 15 + 14 = 39\Omega$
3. Total emf of battery \(E_{total} = 3 \times 1.5 = 4.5\text{ V}$, and total internal resistance \(r = 3 \times 2 = 6\Omega$.
4. Current \(I = \frac{E_{total}}{R + r} = \frac{4.5}{39 + 6} = \frac{4.5}{45} = 0.1\text{ A}$.
Reading of ammeter = \(0.1\text{ A}\).

Teacher's Note:
a) Combine series and parallel resistor networks systematically before applying Ohm's law.
b) Remember to account for the internal resistance of all three series-connected cells.

OR

(ii) In the circuit shown in Figure 5 below, the central zero galvanometer G shows no deflection when the key K is closed. Calculate the value of resistor R₁. [3 Marks]
[Figure: Wheatstone bridge circuit with resistance wire AB of length \(100\text{ cm}$, galvanometer G connected at \(60\text{ cm}$, resistors \(R_1\), \(R_2 = 12\Omega$, \(R_3 = 24\Omega\), and a \(2\text{ V}\) cell with key K.]

Answer:
1. Resistors \(R_2\) (\(12\Omega\)) and \(R_3\) (\(24\Omega\)) are in parallel. Their equivalent resistance \(R_4\) is:
\(\frac{1}{R_4} = \frac{1}{12} + \frac{1}{24} = \frac{3}{24} \implies R_4 = 8\Omega$
2. The circuit forms a balanced Wheatstone bridge where galvanometer shows zero deflection.
\(\frac{R_1}{R_4} = \frac{l}{100-l}$
\(\frac{R_1}{8} = \frac{60}{40}$
\(R_1 = \frac{60 \times 8}{40} = 12\Omega$.

Teacher's Note:
a) Recognize that zero galvanometer deflection implies a balanced bridge condition.
b) Calculate the parallel combination of \(R_2\) and \(R_3\) correctly first.

Question 11 [3 Marks]

An alpha particle is moving with a velocity v. It enters a uniform magnetic field (B) as shown below. The magnetic field is perpendicular and into the plane of the paper. A uniform electric field is applied in the same region as the magnetic field so that the alpha particle passes undeviated through the combined fields.
(i) What should be the direction of the electric field?
[Figure: Uniform magnetic field into the plane denoted by crosses (\(\times\)), with an alpha particle entering horizontally.]

Answer:
\(\vec{E}$ should be in the plane of the paper, pointing vertically downwards.

Teacher's Note:
a) Magnetic force on a positively charged alpha particle moving right in an inward magnetic field is directed downwards.
b) An opposing upward magnetic force requires a downward electric field to balance the forces (\(qE = qvB\)).

(ii) Without any change in the electric and magnetic field, the alpha particle is replaced by a proton moving with a velocity v. Will there be any change in the path of the particle? Give a reason for your answer. [3 Marks]

Answer:
No, there will be no change in the path of the particle.
Reason: The condition for undeviated motion is \(v = E/B$. Since the proton has the same velocity \(v$ and experiences the same electric and magnetic fields, it also passes undeviated.

Teacher's Note:
a) Velocity filter condition depends only on the ratio of electric field to magnetic field strength, independent of particle mass and charge.
b) State both the conclusion and the supporting velocity selector formula clearly.

Question 12 [3 Marks]

(i) State any one difference between the nature of magnetic field inside a current carrying solenoid and that along the axis of a current carrying circular coil.

Answer:
Magnetic field inside a solenoid is uniform, whereas the magnetic field along the axis of a circular coil is non-uniform.

Teacher's Note:
a) Solenoids produce a uniform magnetic field over a long central region.
b) Circular coils exhibit a varying magnetic field gradient along their axis.

(ii) A current is passed through a circular coil A in a clockwise direction, as shown in Figure 7. B is another circular coil which does not carry any current. However, when it is moved towards the coil A, current is induced in it. What is the direction of this induced current? [3 Marks]
[Figure: Two circular coils A and B placed along a common axis with an observer looking at coil A carrying a clockwise current.]

Answer:
In anticlockwise direction (according to Lenz's law).

Teacher's Note:
a) As coil B approaches coil A, magnetic flux linked with B increases.
b) By Lenz's law, coil B opposes this approach by acquiring the same magnetic polarity (clockwise facing away, meaning anticlockwise for the observer).

(iii) Why is steel preferred to soft iron in making permanent magnets?

Answer:
Steel has a large coercive force, meaning it is much more difficult to demagnetize than soft iron, making it ideal for permanent magnets.

Teacher's Note:
a) Permanent magnets require high retentivity and high coercivity.
b) Soft iron has high permeability and low coercivity, making it suitable for electromagnets and transformer cores instead.

Question 13 [3 Marks]

A physics teacher drew a ray diagram on the board as shown below and asked the following questions to the students. State whether the answer of the student is correct or incorrect. Correct the incorrect response with an explanation in brief.
[Figure: Young's double slit experiment setup showing double slit S₁, S₂ illuminated by source S, creating an interference pattern on screen P.]
(i) Teacher: There is a point P on the screen where the path difference between two interfering light waves is exactly \(1.2 \times 10^{-6}\text{ m}$. If the light emitted by the source has a wavelength of 600 nm, can anyone tell me whether a bright or a dark fringe will form at the point P?
Student: Dark fringe will be obtained.

Answer:
Incorrect. A bright fringe will be obtained because path difference \(\Delta x = 1.2 \times 10^{-6}\text{ m} = 1200\text{ nm} = 2\lambda$ (which is an integral multiple of \(\lambda\)).

Teacher's Note:
a) Integral multiples of wavelength produce constructive interference (bright fringes).
b) Always convert units into consistent SI units before checking interference conditions.

(ii) Teacher: Two students performed the same Young’s double slit experiment. One student obtained wider fringes. What is the possible reason for this observation?
Student: The student who obtained wider fringes must have increased the distance between the two slits. [3 Marks]

Answer:
Incorrect, because fringe width \(\beta = \frac{\lambda D}{d}\). To obtain wider fringes, the slit separation \(d$ should have been decreased (or screen distance \(D$ increased).

Teacher's Note:
a) Fringe width is inversely proportional to slit separation \(d\).
b) Point out the exact relationship from the fringe width formula.

(iii) Teacher: If one of the slits is closed with a finger while performing the experiment, what change will be noticed on the screen?
Student: The screen will get illuminated uniformly.

Answer:
Incorrect. A diffraction pattern (or general illumination without interference fringes) is seen on the screen.

Teacher's Note:
a) Closing one slit turns the double-slit experiment into single-slit diffraction.
b) Interference fringes disappear, leaving a broad central maximum.

Question 14 [3 Marks]

A student uses two prisms, A and B, made of different materials but having the same prism angle (60°).

(i) Calculate the refractive index of the material of prism A.

Answer:
\(n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{60^{\circ} + 34^{\circ}}{2}\right)}{\sin\left(\frac{60^{\circ}}{2}\right)} = \frac{\sin(47^{\circ})}{\sin(30^{\circ})} = \frac{0.7314}{0.5} = 1.463 \approx 1.47\)

Teacher's Note:
a) Use prism formula with minimum deviation and prism angle.\b) Substitute values carefully and evaluate trigonometric ratios.

(ii) ‘Since prism B has larger minimum deviation, the student claims it must also have the greater dispersive power.’ Is this statement always correct? Justify your answer. [3 Marks]

Answer:
The student's statement is not always correct. A larger minimum deviation indicates a higher mean refractive index, whereas dispersive power depends on the difference in refractive indices for violet and red light relative to mean refractive index.

Teacher's Note:
a) Dispersive power depends on angular dispersion divided by mean deviation, not simply on minimum deviation alone.
b) Emphasize that refractive index and dispersive power are independent material properties.

Question 15 [3 Marks]

(i) An equiconvex lens forms a full size magnified image of a candle flame on a screen. How will this image change, if at all, when the lens is cut horizontally into two equal parts and only one of them is used? Explain.

Answer:
The nature, size, and position of the image will not change at all, but the brightness (intensity) of the image will be reduced to half because the light-gathering area is halved.

Teacher's Note:
a) Each half of the lens retains the same focal length and forms a complete image.
b) Cutting the lens horizontally reduces aperture area and image intensity, but not image dimensions.

(ii) State what happens to magnifying power of a simple microscope if its focal length is decreased. [3 Marks]

Answer:
Magnifying power increases when the focal length is decreased.

Teacher's Note:
a) Magnifying power of a simple microscope is given by \(m = 1 + \frac{D}{f}\) (when image is formed at least distance of distinct vision).
b) Smaller focal length gives higher angular magnification.

Question 16 [3 Marks]

Study the images shown in Figure 9 below carefully and answer the questions that follow. (All three surfaces have same work function = 2.3 eV)
(i) Why are no photoelectrons emitted in image ‘A’?
[Figure: Three metal plates A, B, and C illuminated by different light sources with varying photon energies causing photoelectron emission in B and C, but none in A.]

Answer:
No photoelectrons are emitted because the energy of the incident photons in image A is less than the work function (\(2.3\text{ eV}\)) of the material.

Teacher's Note:
a) Photoelectric emission requires incident photon energy to exceed threshold work function (\(E \ge \phi\)).
b) Below threshold frequency or energy, no electrons are ejected regardless of light intensity.

(ii) Which image corresponds to the highest stopping potential? [3 Marks]

Answer:
C

Teacher's Note:
a) Higher stopping potential corresponds to higher maximum kinetic energy of emitted photoelectrons.
b) Image C represents the highest incident photon energy among the emitting surfaces.

(iii) If the photon energy in image B = 2.8 eV, calculate the maximum kinetic energy of the emitted photo electrons.

Answer:
Using Einstein's photoelectric equation:
\(K_{max} = E - \phi$
\(K_{max} = 2.8\text{ eV} - 2.3\text{ eV} = 0.5\text{ eV}\).

Teacher's Note:
a) Apply Einstein's photoelectric energy conservation equation directly.
b) Ensure proper unit conversion if required, though expressing in electron-volts is standard.

Question 17 [3 Marks]

(i) What is a solar cell?

Answer:
A solar cell is a semiconductor P-N junction diode that directly converts solar energy (light energy) into electrical energy.

Teacher's Note:
a) Solar cells operate on the photovoltaic effect.
b) They do not require an external bias voltage.

(ii) Explain its working in brief. [3 Marks]

Answer:
Incoming photons break covalent bonds near the junction, generating free electron-hole pairs. The electric field at the junction separates these charge carriers, creating a potential difference. When an external circuit is connected, a direct electric current flows.

Teacher's Note:
a) Mention photon absorption, electron-hole pair generation, and charge separation by the depletion region field.
b) Keep the explanation concise and structured in bullet points.

(iii) In an energy band diagram of a certain material, forbidden band is absent. Identify the material.

Answer:
Metal (Conductor)

Teacher's Note:
a) In metals, valence and conduction bands overlap, making the forbidden energy gap zero.
b) This allows free movement of electrons, resulting in high electrical conductivity.

SECTION D - 15 MARKS

Question 18 [5 Marks]

(i) (a) The graph below shows variation of current (I) flowing through an electrical device with time (t), when an ac source \(e = 240\sin(\omega t)\text{ V}\) is connected to it.
(1) Identify the device. (resistor, capacitor or an inductor)
(2) Calculate the frequency of the current flowing through the device.
(3) How much is the opposition offered by this device?
[Figure: Sinusoidal alternating current graph showing peak current \(I_0 = 12\text{ A}\) and time period corresponding to \(80\text{ ms}\).]

Answer:
(1) Capacitor (current leads applied emf by \(90^{\circ}$)
(2) Frequency \(f = \frac{1}{T} = \frac{1}{80 \times 10^{-3}\text{ s}} = 12.5\text{ Hz}$
(3) Capacitive reactance \(X_c = \frac{e_0}{i_0} = \frac{240\text{ V}}{12\text{ A}} = 20\Omega$ (Note: using peak values from graph where \(i_0 = 12\text{ A}\) or as per key calculation values).

Teacher's Note:
a) Identify phase relationship: current leading voltage indicates a capacitive circuit.
b) Use peak voltage and peak current to find reactance (\(X_c = V_0 / I_0\)).

(b) A train is moving on horizontal tracks that are 1.2 m apart. Its displacement (\(x\)) depends on time (\(t\)) as \(x(\text{m}) = 3t^2 + 6\). Vertical component of earth’s magnetic field in its region of motion is \(2 \times 10^{-5}\text{ T}\). Calculate the emf induced between two ends of its axle rod 10s after the train starts moving. [5 Marks]

Answer:
1. Velocity \(v = \frac{dx}{dt} = \frac{d}{dt}(3t^2 + 6) = 6t$
2. At \(t = 10\text{ s}\), \(v = 6 \times 10 = 60\text{ m s}^{-1}$
3. Induced emf \(E = Blv = 2 \times 10^{-5}\text{ T} \times 1.2\text{ m} \times 60\text{ m s}^{-1} = 1.44 \times 10^{-3}\text{ V}\).

Teacher's Note:
a) Differentiate displacement with respect to time to obtain velocity as a function of time.
b) Apply motional emf formula \(E = Blv$ where \(l\) is the track separation (axle length).

OR

(ii) An ac generator generating an emf \(e = 300\sin(100\pi t)\text{ V}\) is connected to a \(100\Omega\) resistor.
(a) Draw a labelled graph of current (I) flowing through it versus time (t). Draw one cycle only.
(b) Calculate the average power consumed by the resistor.
(c) Why ac is preferred to dc, though it has varying values whereas the latter has a fixed value?
(d) State any one difference between a transformer used at the power generating station and the transformer in a city which is used to supply power to households. [5 Marks]

Answer:
(a) [Figure: Sine wave graph of alternating current with peak value \(I_0 = 3\text{ A}\) over one full time period \(T = 0.02\text{ s}\).]
(b) Peak current \(I_0 = \frac{e_0}{R} = \frac{300}{100} = 3\text{ A}$
Average power \(\langle P \rangle = \frac{I_0^2 R}{2} = \frac{3^2 \times 100}{2} = \frac{900}{2} = 450\text{ W}$
(c) AC can be easily stepped up or down using a transformer with minimal energy loss, whereas DC cannot be stepped up or down easily.
(d) The transformer at the power generating station is a step-up transformer, whereas the transformer used for household supply in a city is a step-down transformer.

Teacher's Note:
a) For a purely resistive circuit, power is calculated using RMS or peak values (\(P = I_{rms}^2 R = \frac{1}{2} I_0^2 R\)).
b) Clearly state transformer classifications for generation versus distribution.

Question 19 [5 Marks]

(i) (a) Write Bohr’s equations for circular motion of an electron around the nucleus and conservation of angular momentum. Using them, obtain an expression for orbital velocity of an electron.

Answer:
1. Electrostatic force provides centripetal force: \(\frac{m v^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{(Ze)(e)}{r^2}$
2. Quantization of angular momentum: \(mvr = \frac{nh}{2\pi}$
3. From angular momentum, \(r = \frac{nh}{2\pi mv}$. Substituting \(r\) into the first equation:
\(m v^2 = \frac{Ze^2}{4\pi\epsilon_0} \cdot \frac{2\pi mv}{nh} \implies v = \frac{Ze^2}{2\epsilon_0 nh}\).

Teacher's Note:
a) Combine Coulomb's law force balance with Bohr's quantization condition step by step.
b) Cancel common terms carefully to derive the final velocity expression.

(b) Prove that \(1\text{ u} = 931.5\text{ MeV}\) where the terms have their usual meaning. [5 Marks]

Answer:
1. \(1\text{ u} = 1.66 \times 10^{-27}\text{ kg}$
2. Using Einstein's mass-energy equivalence \(E = mc^2$, where \(c = 3 \times 10^8\text{ m s}^{-1}$:
\(E = 1.66 \times 10^{-27}\text{ kg} \times (3 \times 10^8\text{ m s}^{-1})^2 = 1.4929 \times 10^{-10}\text{ J}$
3. Converting joules to electron-volts: \(E = \frac{1.4929 \times 10^{-10}\text{ J}}{1.6 \times 10^{-19}\text{ J/eV}} = 9.33 \times 10^8\text{ eV} = 931.5\text{ MeV}\) (using exact constant \(1.6605 \times 10^{-27}\text{ kg}\)).

Teacher's Note:
a) Show the complete conversion from kilograms to joules, and then from joules to mega electron-volts.
b) Memorize the standard numerical value of \(1\text{ u}\) in MeV for quick verification.

OR

(ii) (a) Assuming that nuclei are spherical in shape, whose radius R is given by \(R = R_0 A^{1/3}\) where R₀ is a constant, show that the nuclear density (i.e. density of nuclear matter) is independent of mass number (A) of the nucleus / atom.

Answer:
1. Let average nucleon mass be \(m$. Mass of nucleus = \(mA$.
2. Volume of nucleus \(V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (R_0 A^{1/3})^3 = \frac{4}{3}\pi R_0^3 A$
3. Nuclear density \(\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{mA}{\frac{4}{3}\pi R_0^3 A} = \frac{3m}{4\pi R_0^3}\).
Since mass number \(A\) cancels out, nuclear density is independent of \(A\).

Teacher's Note:
a) This derivation is a board examination favorite. Show mass as \(mA$ and volume proportional to \(A\).
b) Conclude clearly that nuclear density is constant for all nuclei.

(b) What are isotones? [5 Marks]

Answer:
Isotones are atoms or nuclei that have the same number of neutrons (\(N\)) but different numbers of protons (\(Z\)).

Teacher's Note:
a) Differentiate clearly between isotopes, isobars, and isotones.
b) Give an example such as \(^{13}_{\ 6}\text{C}$ and \(^{14}_{\ 7}\text{N}\) if helpful.

(c) Name the series of lines in hydrogen spectrum which lies in:
(1) Visible region
(2) UV region

Answer:
(1) Visible region: Balmer series
(2) UV region: Lyman series

Teacher's Note:
a) Recall the complete spectral series: Lyman (UV), Balmer (Visible), Paschen, Brackett, and Pfund (Infrared).
b) Direct factual recall question.

Question 20 [5 Marks]

An innovative optical sensor system for a miniature drone incorporates a combination of various optical elements. A beam of light from an LED source first passes through a convex spherical refracting surface (radius of curvature R = 20 cm) separating air \(\mu_1 = 1.0\) from a glass medium of refractive index \(\mu_2 = 1.5\). The light then enters a thin biconvex lens of focal length 10 cm. Finally, the beam strikes a concave spherical mirror of radius of curvature 30 cm, which acts as a collector, focusing the rays onto a photodetector (optical instrument application) to measure light intensity.
(i) If the drone operates in a highly humid or misty environment where a film of water completely coats the lens surfaces, how will the refractive power of this lens change?

Answer:
The refractive power of the lens will decrease (focal length will increase) because the relative refractive index of the lens material with respect to the surrounding medium decreases when water coats the surface.

Teacher's Note:
a) Lens power depends on relative refractive index: \(P = (\mu_{rel} - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\).
b) Water coating reduces the refractive index contrast between lens and medium, reducing power.

(ii) A point object is placed in air at a distance of 60 cm in front of the convex spherical refracting surface. What is the position of the image formed by this surface? [5 Marks]

Answer:
Using refraction formula at spherical surface: \(\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$
\(\frac{1.5}{v} - \frac{1.0}{-60} = \frac{1.5 - 1.0}{20}$
\(\frac{1.5}{v} + \frac{1}{60} = \frac{0.5}{20} = \frac{1}{40}$
\(\frac{1.5}{v} = \frac{1}{40} - \frac{1}{60} = \frac{3 - 2}{120} = \frac{1}{120}$
\(v = 1.5 \times 120 = +180\text{ cm}$.
The image is formed at a distance of \(180\text{ cm}$ inside the denser medium (glass), measured from the pole of the refracting surface.

Teacher's Note:
a) Apply sign conventions strictly (\(u = -60\text{ cm}$, \(R = +20\text{ cm}\)).
b) Solve the refraction formula carefully for image distance \(v\).

(iii) If this setup is to be modified to function as a reflecting astronomical telescope, what roles would be played by the mirror and the lens?

Answer:
Objective: The concave spherical mirror must act as the objective.
Eyepiece: The thin biconvex lens must act as the eyepiece.

Teacher's Note:
a) Reflecting telescopes use a large concave mirror as the objective to eliminate chromatic aberration.
b) The eyepiece lens magnifies the final image for the observer.

(iv) Which mirror collects light from a distant object most efficiently?

Answer:
A parabolic mirror

Teacher's Note:
a) Parabolic mirrors bring parallel rays from distant stars to a sharp common focus without spherical aberration.
b) They provide maximum light-collecting efficiency in astronomical telescopes.