ISC Class 12 Physics Sample Paper 2026 with Solutions

SECTION A - 14 MARKS

Question 1

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

(i) Two point charges \( +50\text{ nC} \) and \( -50\text{ nC} \) separated a distance of \( 1\text{ mm} \) are kept well inside a large sphere of radium \( 1\text{ m} \). Electric flux emanating from the sphere is: [1 Mark]
(a) \( 50 \times 10^{-12}\text{ Vm} \)
(b) \( 50 \times 10^{-9}\text{ Vm} \)
(c) \( 50 \times 10^{-6}\text{ Vm} \)
(d) Zero

Answer: (d) Zero

According to Gauss's Law, the total electric flux through a closed surface depends only on the net charge enclosed by the surface. Since the system is an electric dipole with total net charge \( q_{\text{net}} = (+50\text{ nC}) + (-50\text{ nC}) = 0 \), the total electric flux emanating from the sphere is zero.

Teacher's Note:
a) Always check for the algebraic sum of all charges enclosed by the Gaussian surface to find the net enclosed charge.
b) Students often confuse individual charges with the net charge when applying Gauss's Law for electric dipoles.

(ii) Three straight, parallel wires are coplanar and perpendicular to the plane of the page. The currents \( I_{1} \) and \( I_{3} \) are directed out of the page. If wire 3 experiences no force due to the currents \( I_{1} \) and \( I_{2} \), then the current in the wire 2 is: [1 Mark]
[Figure: Three vertical parallel wires 1, 2 and 3 separated by distances \( r \) and \( 2r \), with current directions indicated.]
(a) \( I_{2} = 2I_{1} \) and directed into the page
(b) \( I_{2} = 0.5I_{1} \) and directed into the page
(c) \( I_{2} = 2I_{1} \) and directed out of the page
(d) \( I_{2} = 0.5I_{1} \) and directed out of the page

Answer: (c) \( I_{2} = 2I_{1} \) and directed out of the page

For wire 3 to experience zero net force, the magnetic force due to wire 1 and wire 2 on wire 3 must be equal in magnitude and opposite in direction. Since wire 1 and wire 3 carry currents in the same direction (out of the page) and wire 1 is further away, wire 2 must carry current in the same direction (out of the page) with double the magnitude because of the distance ratio.

Teacher's Note:
a) Parallel currents in the same direction attract each other, while parallel currents in opposite directions repel.
b) Ensure proper application of the force per unit length formula \( F/l = \frac{\mu_{0}I_{1}I_{2}}{2\pi d} \) considering relative distances.

(iii) The variation of magnetic susceptibility (\( \chi \)) with absolute temperature (\( T \)) for a diamagnetic substance is: [1 Mark]
[Figure: Four graphs showing variation of susceptibility \( \chi \) with absolute temperature \( T \).]
(a) graph showing constant negative susceptibility independent of temperature
(b) graph showing inverse variation
(c) graph showing inverse variation with critical temperature
(d) graph showing linear increase

Answer: (a)

The magnetic susceptibility of a diamagnetic substance is small, negative, and independent of absolute temperature \( T \).

Teacher's Note:
a) Remember that unlike paramagnetic and ferromagnetic substances which follow Curie's law, diamagnetic susceptibility is independent of temperature.
b) Students frequently mistake diamagnetic behavior for paramagnetic behavior.

(iv) The wavelength \( \lambda_{e} \) of an electron and \( \lambda_{p} \) of a photon of same energy \( E \) are related by: [1 Mark]
(a) \( \lambda_{p} \propto \sqrt{\lambda_{e}} \)
(b) \( \lambda_{p} \propto \frac{1}{\sqrt{\lambda_{e}}} \)
(c) \( \lambda_{p} \propto \lambda_{e}^{2} \)
(d) \( \lambda_{p} \propto \lambda_{e} \)

Answer: (c) \( \lambda_{p} \propto \lambda_{e}^{2} \)

For an electron, \( \lambda_{e} = \frac{h}{\sqrt{2mE}} \), so \( \lambda_{e}^{2} \propto \frac{1}{E} \). For a photon, \( \lambda_{p} = \frac{hc}{E} \), so \( \lambda_{p} \propto \frac{1}{E} \). Therefore, \( \lambda_{p} \propto \lambda_{e}^{2} \).

Teacher's Note:
a) Use De Broglie wavelength for the electron and Planck's energy relation for the photon.
b) Pay close attention to proportionality powers during derivation.

(v) In which of the following figures, is the p-n diode forward biased? [1 Mark]
[Figure: Four circuit diagrams labeled (P), (Q), (R), (S) with various voltage configurations across p-n diodes with resistors \( R \).]
(a) Only (P), (Q) and (S)
(b) Only (R)
(c) Only (P) and (R)
(d) Only (Q) and (S)

Answer: (b) Only (R)

A p-n junction diode is forward biased when the p-region is at a higher potential relative to the n-region. In diagram (R), the p-side is connected to \( -5\text{ V} \) and the n-side to \( -12\text{ V} \), making \( -5\text{ V} \gt -12\text{ V} \).

Teacher's Note:
a) Forward bias requires the anode potential to be higher than the cathode potential, regardless of whether the potentials are positive or negative.
b) Always check the relative potential difference across the terminals carefully.

(vi) When a beam of white light is incident on a prism, the prism: [1 Mark]
(a) only disperses the incident light.
(b) only deviates the incident light.
(c) deviates as well as disperses incident light.
(d) neither deviates nor disperses incident light.

Answer: (c) deviates as well as disperses incident light.

A prism bends (deviates) light rays toward its base and also splits polychromatic white light into its component colors (disperses) due to wavelength-dependent refractive index.

Teacher's Note:
a) Deviation and dispersion occur simultaneously whenever white light passes through a refracting medium like a prism.
b) Do not confuse dispersion with pure refraction or deviation.

(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 the convex mirror will increase, if the mirror is placed in water.
Reason: The focal length of a convex mirror of radius \( R \) is equal to \( R/2 \).
(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. Therefore, placing a convex mirror in water does not change its focal length, making the assertion false while the reason is true.

Teacher's Note:
a) Unlike lenses, the focal length of a mirror does not depend on the refractive index of the surrounding medium.
b) Students often confuse mirror properties with lens maker's formula parameters.

(B) Answer the following questions briefly:

(i) Current \( I \) flowing through a metallic wire is gradually increased. Show graphically how heating power (\( P \)) developed in it varies with the current (\( I \)). [1 Mark]

Answer:

Teacher's Note:
a) Heating power is given by \( P = I^{2}R \), which shows a quadratic (parabolic) relationship with current.
b) Ensure axes are properly labeled with \( P \) on the y-axis and \( I \) on the x-axis.

(ii) State one method to minimise flux loss in a transformer. [1 Mark]

Answer:
By winding the primary and secondary coils close together, ideally one over the other (interleaving), and by using a core with a high permeability.

Teacher's Note:
a) Magnetic flux leakage reduces transformer efficiency.
b) Interleaving primary and secondary windings ensures maximum magnetic flux linkage.

(iii) Why are giant telescopes of reflecting type? Give any one scientific reason. [1 Mark]

Answer:
Images formed by a reflecting telescope are free from the defect of chromatic aberration (or spherical aberration, or they are brighter).

Teacher's Note:
a) Large lenses suffer from severe chromatic and spherical aberrations and are difficult to support mechanically.
b) Mirror-based systems reflect light from the surface, eliminating chromatic aberration entirely.

(iv) Give any one example where a ray of light travelling from one optical medium to another travels undeviated. [1 Mark]

Answer:
When a ray of light is incident normally on the surface of separation (or when the refractive index of medium 1 is equal to the refractive index of medium 2, i.e., \( \mu_{1} = \mu_{2} \)).

Teacher's Note:
a) Normal incidence corresponds to angle of incidence \( i = 0^{\circ} \), leading to angle of refraction \( r = 0^{\circ} \).
b) Equal refractive indices mean no optical boundary exists for refraction.

(v) Two charged particles having same charge but different masses, are passed through a potential difference \( V \). When \( V \) is varied, de-Broglie wavelength (\( \lambda \)) of the particles varies as shown in graphs below. Which graph is for heavier particles and why? [1 Mark]
[Figure: Graph showing linear relationship of \( \lambda \) against \( 1/\sqrt{V} \) with two straight lines labeled A and B.]

Answer:
Graph B is for heavier particles, because \( M_{B} \gt M_{A} \) as the slope of line B is greater than the slope of line A, and \( \lambda = \frac{h}{\sqrt{2mEV}} \).

Teacher's Note:
a) The de-Broglie wavelength is inversely proportional to the square root of mass: \( \lambda \propto \frac{1}{\sqrt{m}} \).
b) Steeper slope indicates a larger value of the constant factor involving mass.

(vi) What happens when an electron collides with a positron? [1 Mark]

Answer:
They annihilate each other producing gamma ray photons.

Teacher's Note:
a) Electron-positron annihilation converts mass entirely into energy in the form of photons.
b) Conserves both energy and momentum during the process.

(vii) What is the direction of flow of electrons in a solar cell? [1 Mark]

Answer:
Electrons move towards the n-type material, and holes move towards the p-type material, creating a flow of current when a circuit is connected.

Teacher's Note:
a) The built-in electric field at the p-n junction sweeps photogenerated electrons to the n-side.
b) This creates a potential difference that drives external current.

SECTION B - 14 MARKS

Question 2 [2 Marks]

(i) Find the capacitance of the following combinations between terminals A and B. Area of each plate is \( A \) and separation between nearest two plates is ' \( d \) '.
(a)
(b)
[Figure: Diagrams of parallel plate multi-plate capacitor arrangements connected to terminals A and B, Figures 2 and 3.]

Answer:
(a) When terminals A and B are connected to plates, it is clear from the figure that two capacitors are formed in parallel. Thus, \( C_{AB} = 2C = \frac{2\varepsilon_{0}A}{d} \).
(b) There are three capacitors in parallel. Thus, \( C_{AB} = 3C = \frac{3\varepsilon_{0}A}{d} \).

Teacher's Note:
a) Count the number of active dielectric spaces (capacitors) formed between adjacent pairs of plates.
b) Parallel plate combinations share common potential differences across each adjacent layer.

OR

(ii) The current (\( I \))-voltage (\( V \)) graphs for a conductor are given at two different temperatures \( T_{1} \) and \( T_{2} \).
(a) At which temperature \( T_{1} \) or \( T_{2} \) is the resistance higher?
(b) Which temperature \( T_{1} \) or \( T_{2} \) is higher? [2 Marks]
[Figure: I-V characteristic curves at temperatures \( T_{1} \) and \( T_{2} \).]

Answer:
(a) For the I-V graph, the slope gives conductance: \( R = \frac{V}{I} = \frac{1}{\tan\theta} \). Since \( \theta_{1} \gt \theta_{2} \), \( \tan\theta_{1} \gt \tan\theta_{2} \), so resistance at \( T_{1} \) is less than resistance at \( T_{2} \). Thus, resistance is higher at \( T_{2} \).
(b) Since the resistance of a metallic conductor increases with temperature and resistance at \( T_{2} \) is higher than at \( T_{1} \), we conclude that \( T_{2} \gt T_{1} \).

Teacher's Note:
a) In an \( I \)-\( V \) graph, slope represents conductance (\( I/V \)), whereas in a \( V \)-\( I \) graph, slope represents resistance.
b) Metallic resistance increases directly with absolute temperature.

Question 3 [2 Marks]

Arrangement of an oxygen ion and two hydrogen ions in a water molecule is shown in Figure 4 below. Calculate electric dipole moment of water molecule. Express your answer in terms of \( e \) (charge on hydrogen ion), \( l \) and \( \theta \).
[Figure: Water molecule geometry showing bond length \( l \) and bond angle \( \theta \), Figure 4.]

Answer:
\( p_{1} = p_{2} = e \times l \)
\( P = 2p \cos\alpha = 2e \times l \cos(\theta / 2) \)

Teacher's Note:
a) The net dipole moment is the vector sum of the two individual bond dipole moments.
b) Use vector addition for angles to resolve components along the symmetry axis.

Question 4 [2 Marks]

(i) Two cells of same emf \( E \), but different internal resistance \( r_{1} \) and \( r_{2} \) are connected to an external resistance \( R \) as shown in Figure 5 given below. The voltmeter V reads zero. Obtain an expression for \( R \) in terms of \( r_{1} \) and \( r_{2} \). (Assume that the voltmeter V is of infinite resistance).
[Figure: Circuit diagram with two cells in opposition/series configuration connected to resistor \( R \) and voltmeter V, Figure 5.]

Answer:
Total emf in the circuit = \( E + E = 2E \)
Total resistance in the circuit = \( R + r_{1} + r_{2} \)
Current in the circuit: \( I = \frac{2E}{R + r_{1} + r_{2}} \)
Given terminal potential difference across first cell is \( V = E - Ir_{1} = 0 \)
\( \Rightarrow Ir_{1} = E \)
Substitute \( I \): \( \frac{2E}{R + r_{1} + r_{2}} \times r_{1} = E \)
\( R + r_{1} + r_{2} = 2r_{1} \)
\( R = r_{1} - r_{2} \)

Teacher's Note:
a) Terminal potential difference becomes zero when the internal drop equals the emf.
b) Check polarity connections of cells carefully before writing loop equations.

OR

(ii) Ramesh performed an experiment to determine an unknown resistance \( R \) using the circuit shown in Figure 6 below. X is a resistance box and PQ is a \( 100\text{ cm} \) potentiometer wire. He closed the key and inserted \( X = 1\Omega, 2\Omega \dots \) and recorded the null point (\( l \)) corresponding to different values of \( X \).
(a) Identify the principle involved in calculating \( R \).
(b) Write down a relation required to calculate the resistance \( R \) in terms of \( X \) and (\( l \)). [2 Marks]
[Figure: Potentiometer setup with resistance box \( X \) and unknown resistor \( R \), Figure 6.]

Answer:
(a) Wheatstone Bridge Principle
(b) \( R = \frac{X(100 - l)}{l} \)

Teacher's Note:
a) A meter bridge or potentiometer setup operates on the balanced Wheatstone bridge principle.
b) The length ratio directly corresponds to the ratio of resistance arms.

Question 5 [2 Marks]

Two moving coil galvanometers \( G_{1} \) and \( G_{2} \) are identical except that they have \( 50 \) turns and \( 20 \) turns and resistance of \( 10\Omega \) and \( 1\Omega \) respectively. Perform necessary calculations to check which one has greater voltage sensitivity.

Answer:
Voltage sensitivity \( \beta = \frac{BAN}{CR} \)
\( \beta(G_{1}) = \frac{B \cdot A \cdot 50}{C \cdot 10} = 5\frac{BA}{C} \)
\( \beta(G_{2}) = \frac{B \cdot A \cdot 20}{C \cdot 1} = 20\frac{BA}{C} \)
Since \( 20 \gt 5 \), \( G_{2} \) has greater voltage sensitivity.

Teacher's Note:
a) Voltage sensitivity is defined as deflection per unit voltage: \( \frac{\theta}{V} = \frac{NAB}{CR} \).
b) Increasing turns increases both sensitivity and resistance proportionately, so careful ratio analysis is required.

Question 6 [2 Marks]

Two similar convex lenses are made up of two different materials as shown in Figures 7 and 8 below. Find the number of images formed in the following set ups:
(i)
(ii)
[Figure: Two convex lenses made of split/composite materials receiving parallel beam light rays, Figures 7 and 8.]

Answer:
(i) Two images
(ii) One image

Teacher's Note:
a) Composite lenses made of different refractive index materials have different focal lengths for each half.
b) Uniform material lenses focus all rays to a single point producing one image.

Question 7 [2 Marks]

Name the electromagnetic wave used in:
(a) radars
(b) crystallography

Answer:
(a) Microwaves
(b) X-rays

Teacher's Note:
a) Microwaves have short wavelengths suitable for radar tracking systems.
b) X-rays have wavelengths comparable to atomic spacings, making them ideal for crystal structure analysis.

Question 8 [2 Marks]

In the photoelectric effect, the maximum kinetic energy of the emitted photoelectron is ' \( a \) ' and the work function of the metal is \( W_{0} \). If the frequency of incident radiation is made ' \( K \) ' times, then calculate the change in maximum KE of the ejected electron.

Answer:
\( a_{1} = h\nu - W_{0} \)
\( a_{2} = h(K\nu) - W_{0} \)
Change in maximum KE: \( a_{2} - a_{1} = (hK\nu - W_{0}) - (h\nu - W_{0}) \)
\(= hK\nu - h\nu \)
\(= (K - 1)h\nu \)

Teacher's Note:
a) Use Einstein's photoelectric equation: \( K_{\max} = h\nu - \Phi_{0} \).
b) Work function remains constant for a given metal surface.

SECTION C - 27 MARKS

Question 9 [3 Marks]

Obtain an expression for electric potential (\( V \)) at a point near a point charge ' \( Q \) '.

Answer:
[Figure: Point charge Q at origin and test point P at distance r, with incremental displacement dx.]
Electrostatic force between charge \( Q \) and a unit test charge at distance \( x \):
\( F = \frac{1}{4\pi\varepsilon_{0}} \frac{Q \times 1}{x^{2}} \)
Work done in moving the test charge through a small displacement \( dx \):
\( dW = -F \, dx = -\frac{1}{4\pi\varepsilon_{0}} \frac{Q}{x^{2}} \, dx \)
Total work done in bringing the unit positive charge from infinity to point \( P \) at distance \( r \):
\( V = W = \int_{\infty}^{r} dW = \int_{\infty}^{r} -\left(\frac{1}{4\pi\varepsilon_{0}} \frac{Q}{x^{2}}\right) dx \)
\( V = \frac{1}{4\pi\varepsilon_{0}} \frac{Q}{r} \)
This is the desired expression.

Teacher's Note:
a) Electric potential is defined as the work done in bringing a unit positive test charge from infinity to that point.
b) Pay close attention to the negative sign indicating work done against electrostatic repulsion.

Question 10 [3 Marks]

(i) Five identical charges \( Q = 2\mu\text{C} \) are placed equidistant on a semicircle as shown in Figure 9. Another point charge \( q = 1\mu\text{C} \) is kept at the center of the circle of radius \( 2\text{ cm} \). Calculate the electrostatic force experienced by the charge \( q \).
[Figure: Five charges placed on a semicircle around central charge q, Figure 9.]

Answer:
Force acting on \( q \) due to \( Q_{1} \) and \( Q_{5} \) are in opposite directions, so they cancel each other.
Force acting on \( q \) due to \( Q_{3} \) is \( F_{3} = \frac{qQ_{3}}{4\pi\varepsilon_{0}R^{2}} \).
Forces due to \( Q_{2} \) and \( Q_{4} \):
Resolving in two-component method:
1. Vertical Component: \( Q_{2}\sin\theta \) and \( Q_{4}\sin\theta \) are equal and opposite, so they cancel each other.
2. Horizontal Component: \( Q_{2}\cos\theta \) and \( Q_{4}\cos\theta \) are equal and in the same direction, so they add up.
\( F_{24} = F_{2q} + F_{4q} = F_{2}\cos 45^{\circ} + F_{4}\cos 45^{\circ} \)
\( F_{24} = \frac{qQ_{2}}{4\pi\varepsilon_{0}R^{2}}\cos 45^{\circ} + \frac{qQ_{4}}{4\pi\varepsilon_{0}R^{2}}\cos 45^{\circ} \)
Resultant net force \( F = \frac{1}{4\pi\varepsilon_{0}} \frac{qQ}{R^{2}}[1 + \sqrt{2}]\text{ N} \)

Teacher's Note:
a) Symmetry plays a crucial role in canceling opposing vector components in electrostatic equilibrium problems.
b) Substitute numerical values carefully using standard SI units for distance and charge.

OR

(ii) Using Kirchhoff's laws of electrical networks, calculate the current \( I_{3} \). [3 Marks]
[Figure: Multi-loop resistive network with voltage sources and branch currents, Figure 10.]

Answer:
\( I_{3} = I_{1} + I_{2} \)
Applying KVL to loop ABCDEFA:
\( 2I_{1} + 8(I_{1} + I_{2}) + 1I_{1} = 33 \)
\( 11I_{1} + 8I_{2} = 33 \)
Applying KVL to loop BCDEB:
\( 6I_{2} + 8(I_{1} + I_{2}) + 2I_{2} = 26 \)
\( 8I_{1} + 9I_{2} = 24 \)
Solving these simultaneous equations, we get:
\( I_{1} = 3\text{ A},\quad I_{2} = 0 \)
\( \therefore I_{3} = 3\text{ A} \)

Teacher's Note:
a) Apply Kirchhoff's Current Law (KCL) at junctions first to reduce the number of independent variables.
b) Follow sign conventions strictly when traversing loops in KVL equations.

Question 11 [3 Marks]

Using Ampere circuital law, obtain an expression for magnetic field ' \( B \) ' at a point at a perpendicular distance ' \( r \) ' from a long current carrying conductor.

Answer:
[Figure: Long straight current-carrying wire with circular Amperian loop of radius r and infinitesimal element dl.]
By Ampere's Circuital Law:
\( \oint \vec{B} \cdot d\vec{l} = \mu_{0}I \)
\( \oint B \, dl \cos 0^{\circ} = \mu_{0}I \)
\( B \oint dl = \mu_{0}I \)
\( B(2\pi r) = \mu_{0}I \)
\( B = \frac{\mu_{0}I}{2\pi r} \)

Teacher's Note:
a) Choose a circular Amperian loop concentric with the conductor where magnetic field magnitude is uniform.
b) State the angle between magnetic field vector and length element explicitly as zero degrees.

Question 12 [3 Marks]

(i) A student records the following data for the magnitudes (\( B \)) of the magnetic field at axial points at different distances \( x \) (See Figure 11 given below) from the centre O of a circular coil of radius \( a \) carrying a current \( I \). Verify (for any two) that these observations are in good agreement with the expected theoretical values of \( B \).
[Figure: Circular coil of radius a with axial distance x and observation point P, Figure 11.]

Answer:
The expression for \( B \) at an axial point of a circular coil carrying current is:
\( B = \frac{\mu_{0}}{4\pi} \frac{2NI(\pi a^{2})}{(a^{2} + x^{2})^{3/2}} \)
At the center, \( x = 0 \): \( B_{0} = \frac{\mu_{0}I}{2a} \)
At \( x = a \): \( B = \frac{\mu_{0}I a^{2}}{2(a^{2} + a^{2})^{3/2}} = \frac{\mu_{0}Ia^{2}}{2(2a^{2})^{3/2}} = \frac{\mu_{0}I}{2a(2\sqrt{2})} = \frac{B_{0}}{2\sqrt{2}} \)
At \( x = 2a \): \( B = \frac{\mu_{0}Ia^{2}}{2(a^{2} + 4a^{2})^{3/2}} = \frac{\mu_{0}Ia^{2}}{2(5a^{2})^{3/2}} = \frac{B_{0}}{5\sqrt{5}} \)
At \( x = 3a \): \( B = \frac{\mu_{0}Ia^{2}}{2(a^{2} + 9a^{2})^{3/2}} = \frac{\mu_{0}Ia^{2}}{2(10a^{2})^{3/2}} = \frac{B_{0}}{10\sqrt{10}} \)

Teacher's Note:
a) Use the standard Biot-Savart law derivation for the magnetic field on the axis of a current loop.
b) Substitute \( x = a, 2a, 3a \) systematically to verify the denominator scaling factors.

OR

(ii) An electron moving along positive X axis with a velocity of \( 8 \times 10^{7}\text{ ms}^{-1} \) enters a region having uniform magnetic field \( B = 1.3 \times 10^{-3}\text{ T} \) along positive Y axis.
(a) Explain why the electron describes a circular path.
(b) Calculate the radius of the circular path described by the electron. [3 Marks]

Answer:
(a) Because the force exerted by the magnetic field is always perpendicular to its velocity vector (\( \vec{v} \times \vec{B} \)), it provides the necessary centripetal force for circular motion.
(b) Using \( \frac{e}{m} = \frac{v}{Br} \):
\( \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}} = \frac{8 \times 10^{7}}{1.3 \times 10^{-3} \times r} \)
Solving for \( r \): \( r \approx 0.35\text{ m} \)

Teacher's Note:
a) A magnetic force does no work on a charged particle, changing only its direction of motion.
b) Substitute standard values of electron charge and mass carefully during numerical calculation.

Question 13 [3 Marks]

Study the diagram shown in Figure 12 given below. Identify the following in Figure 12:
(i) A primary wavefront
(ii) A secondary wavefront
(iii) A wave normal
[Figure: Huygens construction diagram showing wavefront propagation, Figure 12.]

Answer:
(i) ABCD
(ii) \( A'B'C'D' \)
(iii) SDD' or SCC' or SBB' or SAA'

Teacher's Note:
a) Huygens' principle states that every point on a wavefront acts as a source of secondary spherical wavelets.
b) Wave normals are drawn perpendicular to the wavefronts indicating the direction of wave propagation.

Question 14 [3 Marks]

In Young's double slit experiment, show that fringe width \( \omega \) (fringe separation) is given by \( \omega = \frac{\lambda D}{d} \) where the terms have their usual meaning.

Answer:
[Figure: Young's double slit experimental geometry showing slits separated by d, screen at distance D, and fringe position \( y_{m} \).]
Because \( \theta \) is small, \( \tan\theta \approx \sin\theta \)
\( \frac{y_{m}}{D} = \frac{BN}{AB} = \frac{m\lambda}{d} \)
\( y_{m} = \frac{m\lambda D}{d} \)
Fringe width \( \omega = y_{m} - y_{m-1} \)
\( \omega = \frac{m\lambda D}{d} - (m - 1)\frac{\lambda D}{d} \)
\( \omega = \frac{\lambda D}{d} \) (Proved)

Teacher's Note:
a) Fringe width is defined as the separation between two consecutive bright or dark fringes.
b) Clearly state the small angle approximation assumption during path difference calculation.

Question 15 [3 Marks]

With reference to the lens maker's formula, answer the following questions:
[Figure: Ray refraction through a thin lens of refractive index bounded by radii of curvature \( R_{1} \) and \( R_{2} \), Figure 13.]
(i) Apply the formula (expression) of refraction at a single spherical surface to:
(a) refraction at first spherical surface.
(b) refraction at second spherical surface.
(ii) Combine these two expressions / equations to obtain an expression for focal length of the lens.

Answer:
(i) (a) Refraction at first surface: \( \frac{\mu_{1}}{u} + \frac{\mu_{2}}{\vartheta'} = \frac{\mu_{2} - \mu_{1}}{R_{1}} \)
\( \frac{1}{\mu}u + \frac{\mu_{2}}{\vartheta'} = \frac{\mu_{2} - 1}{R_{1}} \)
(b) Refraction at second surface: \( \frac{-\mu_{2}}{\vartheta'} + \frac{\mu_{2}}{\vartheta} = \frac{\mu_{2} - 1}{R_{2}} \)
(ii) Adding both equations:
\( \frac{1}{u} + \frac{1}{\vartheta} = (\mu_{2} - 1)\left(\frac{1}{R_{1}} + \frac{1}{R_{2}}\right) \)
When \( u = \infty \), \( \vartheta = f \):
\( \frac{1}{f} = (\mu_{2} - 1)\left(\frac{1}{R_{1}} + \frac{1}{R_{2}}\right) \)

Teacher's Note:
a) Apply single spherical surface refraction formula sequentially, treating the image of the first surface as a virtual object for the second surface.
b) Maintain proper sign convention for the radii of curvature.

Question 16 [3 Marks]

(i) A student studies details of a microorganism with the help of an instrument. Name the instrument used by him.
(ii) Draw a labelled ray diagram of an image formed by this instrument, assuming
(a) a small upright object.
(b) image lies at least distance of distinct vision.

Answer:
(i) Compound microscope
(ii) [Figure: Ray diagram of a compound microscope showing objective and eyepiece lenses, intermediate real inverted image, and final virtual magnified image at least distance of distinct vision D.]

Teacher's Note:
a) A compound microscope uses two converging lens systems (objective and eyepiece) for high magnification.
b) Ensure all principal rays, focal points, and eye positions are accurately labeled in the ray diagram.

Question 17 [3 Marks]

In a hydrogen atom, an electron jumps from the first excited state to the ground state, and a photon is emitted. This photon is incident on a metal surface having a work function of \( 2\text{ eV} \). Calculate the stopping potential of the electron emitted from the metal surface.

Answer:
Energy of photon emitted during transition from \( n = 2 \) to \( n = 1 \):
\( E = E_{2} - E_{1} = -3.4\text{ eV} - (-13.6\text{ eV}) = 10.2\text{ eV} \)
Using Einstein's photoelectric equation:
\( eV_{0} = E - W_{0} \)
\( eV_{0} = 10.2\text{ eV} - 2\text{ eV} = 8.2\text{ eV} \)
\( V_{0} = 8.2\text{ V} \)

Teacher's Note:
a) Energy levels in a hydrogen atom are given by \( E_{n} = \frac{-13.6}{n^{2}}\text{ eV} \).
b) Stopping potential numerically equals the maximum kinetic energy in electron-volts divided by the electronic charge.

SECTION D - 15 MARKS

Question 18 [5 Marks]

(i) (a) Define the co-efficient of self-inductance.
(b) (1) Consider an A.C. source of frequency \( \left(\frac{200}{\pi}\right)\text{ Hz} \) applied across a coil. For each value of V-I in the tabulation, evaluate Inductive reactance and Self-Inductance of the coil.
(2) If a D.C. source be connected to the same coil, what would be the value of inductive reactance?

Answer:
(a) It is defined as magnetic flux linked with the solenoid when unit current flows through it (or the ratio of magnetic flux linked with the solenoid to the current flowing through it).
(b) (1)

Teacher's Note:
a) Inductive reactance \( X_{L} = \frac{V}{I} = 2\pi f L \). For DC sources, frequency \( f = 0 \), hence inductive reactance is zero.
b) Ensure proper unit conversions for frequency and inductance values.

OR

(ii) Three students, X, Y and Z performed an experiment for studying the variation of A.C. with frequency in a series LCR circuit and obtained the graphs as shown below. They all used an AC source of the same emf and inductance of the same value.
(a) Who used minimum resistance?
(b) In which case will the quality Q factor be maximum?
(c) What did the students conclude about the nature of impedance at resonant frequency (\( f_{0} \))?
(d) An ideal capacitor is connected across \( 220\text{ V}, 50\text{ Hz} \), and \( 220\text{ V}, 100\text{ Hz} \) supplies. Find the ratio of current flowing through it in the two cases. [5 Marks]
[Figure: Resonance curves for series LCR circuits for students X, Y, Z, Figure 14.]

Answer:
(a) Resistance used by X is the least and resistance used by Z is the maximum.
(b) Q factor will be maximum for X.
(c) At resonance, impedance is equal to ohmic resistance (\( Z = R \)).
(d) In a capacitor, current is directly proportional to frequency (\( I \propto f \)). Therefore, \( \frac{I_{1}}{I_{2}} = \frac{50}{100} = \frac{1}{2} \).

Teacher's Note:
a) Sharpness of resonance (Q-factor) is inversely proportional to circuit resistance: \( Q = \frac{1}{R}\sqrt{\frac{L}{C}} \).
b) Capacitive reactance decreases with an increase in AC frequency, causing current to scale linearly with frequency.

Question 19 [5 Marks]

(i) (a) Study the graph shown below and answer the questions that follow.
Indicate which region corresponds to:
(1) Nuclei prone to fission
(2) Nuclei prone to fusion
(3) Most stable nuclei
[Figure: Binding energy per nucleon versus mass number curve showing Regions A, B, and C.]
(b) In Rutherford's scattering experiment when an alpha particle (charge = \( +2e \), mass = \( 4m_{p} \)) approaches a gold nucleus (\( Z = 79 \)), it is continuously repelled, so it loses its kinetic energy (\( K \)) and its potential energy increases. Finally, \( \alpha \)-particle comes to rest momentarily when whole of the kinetic energy is change into the potential energy of the charge at that distance from the nucleus. Let this distance be \( r_{0} \) after which \( \alpha \)-particle returns back again due to electrostatic repulsion. Using the above information, derive an expression for \( r_{0} \).

Answer:
(a) (1) Region C
(2) Region A
(3) Region B
(b) At the distance of closest approach \( r_{0} \), kinetic energy is entirely converted into electrostatic potential energy \( U \) of the \( \alpha \)-particle:
\( U = \frac{1}{4\pi\varepsilon_{0}} \frac{(Ze \cdot 2e)}{r_{0}} \)
where \( Z = 79 \) is the atomic number of the gold nucleus.
Thus we have:
\( K = \frac{1}{4\pi\varepsilon_{0}} \frac{2(79)e^{2}}{r_{0}} \)
\( r_{0} = \frac{1}{4\pi\varepsilon_{0}} \frac{158e^{2}}{K} \)

Teacher's Note:
a) The distance of closest approach gives an upper limit estimate for the nuclear radius.
b) Conservation of energy equates initial kinetic energy directly to electric potential energy at point of zero velocity.

OR

(ii) (a) In an atom X, electrons absorb the energy from an external source. This energy "excites" the electrons from a lower-energy level to a higher-energy level around the nucleus of the atom. When electrons return to the ground state, they emit photons. Figure 15 below is the energy level diagram of atom X with three energy levels, \( E_{1} = 0.00\text{ eV} \), \( E_{2} = 1.78\text{ eV} \) and \( E_{3} = 2.95\text{ eV} \). The ground state is considered \( 0\text{ eV} \) for reference.
What wavelength of radiation is needed to excite the atom to energy level \( E_{2} \) from \( E_{1} \)?
(b) According to Bohr's theory of hydrogen atom, calculate:
(1) angular momentum of the electron in second Bohr orbit.
(2) radius of the third Bohr orbit. [5 Marks]
[Figure: Energy level diagram with levels \( E_{1}, E_{2}, E_{3} \), Figure 15.]

Answer:
(a) \( E = E_{2} - E_{1} \)
\( \frac{hc}{\lambda} = E_{2} - E_{1} \)
\( \lambda = \frac{hc}{E_{2} - E_{1}} = \frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{(1.78 - 0) \times 1.6 \times 10^{-19}} \)
\( \lambda = \frac{19.8 \times 10^{-26}}{2.848 \times 10^{-19}} \approx 6.95 \times 10^{-7}\text{ m} \)
(b) (1) For second Bohr orbit (\( n = 2 \)):
\( l_{2} = 2\hbar = \frac{2h}{2\pi} = \frac{6.6 \times 10^{-34}}{3.14} \approx 2.1 \times 10^{-34}\text{ J s} \)
(2) Radius of \( n \)-th Bohr orbit: \( r_{n} = n^{2}a_{0} \)
For \( n = 3 \): \( r_{3} = 3^{2} \times 5.3 \times 10^{-11}\text{ m} = 9 \times 5.3 \times 10^{-11} = 47.7 \times 10^{-11}\text{ m} \)

Teacher's Note:
a) Use Planck's energy-wavelength relation with proper joule-electron volt conversions.
b) Apply Bohr's quantization condition \( mvr = \frac{nh}{2\pi} \) and radius formula \( r_{n} \propto n^{2} \).

Question 20 [5 Marks]

(i) A band gap is the distance between the valence band of electrons and the conduction band. Essentially, the band gap represents the minimum energy that is required to excite an electron up to a state in the conduction band where it can participate in conduction. The lower energy level is the valence band, and thus if a gap exists between this level and the higher energy conduction band, energy must be input for electrons to become free.
An LED is made of a p-type semiconductor material (which has a higher concentration of "holes" or positive charge carriers) and an n-type semiconductor material (which has a higher concentration of electrons or negative charge carriers). This recombination process releases energy in the form of light and heat. The specific wavelength (and therefore the colour) of the emitted light depends on the energy band gap of the semiconductor material used.
I-V characteristic of LED bulb is given below.
Identify the wavelength that has:
(a) The maximum energy gap
(b) The minimum energy gap
(ii) \( E \) is the energy of the incident photon and \( E_{g} \) is energy gap, which is produced across the depletion layer. What will happen in the following cases:
(a) \( E \gt E_{g} \)
(b) \( E = E_{g} \)
(c) \( E \lt E_{g} \)?
[Figure: I-V characteristic curves labeled 1, 2, 3 for LED bulbs.]

Answer:
(i) (a) 3
(b) 1
(ii) (a) Emission of photon with energy
(b) Emission of photon with no energy
(c) No emission of photon

Teacher's Note:
a) Photon energy is inversely proportional to wavelength (\( E = \frac{hc}{\lambda} \)), so maximum band gap corresponds to the shortest wavelength.
b) Photons with energy less than the band gap cannot excite electrons across the depletion layer, resulting in no photon emission or absorption.