ISC Class 12 Physics Sample Paper 2025 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) Figure 1 below shows a circuit containing an ammeter A, a galvanometer G and a plug key K. When the key is closed: [1 Mark]
(a) both G and A show deflections.
(b) neither G nor A shows deflection.
(c) G shows deflection but A does not.
(d) A shows deflection but G does not.

[Figure: A Wheatstone bridge circuit with resistors 2 ohms, 5 ohms, 4 ohms, and 10 ohms in arms P, Q, R, and the opposite arm, a galvanometer G connected across the diagonal, an ammeter A in the main circuit with a 2V cell and a plug key K. The bridge is balanced since 2/4 = 5/10 = 1/2.]

Answer: (d) A shows deflection but G does not.

The bridge is balanced because the ratio of resistances in opposite arms is equal (2/4 = 5/10), so no current flows through the galvanometer, but the main current flows through ammeter A.

Teacher's Note:
a) Recognise the Wheatstone bridge condition \( \frac{P}{Q} = \frac{R}{S} \) for zero current through the galvanometer.
b) Students often confuse bridge balance with total circuit current cutoff; ammeter A is in the main line and records current.

(ii) A circular coil having N turns of radius R carrying a current I is used to produce a magnetic field B at its centre O. If this coil is opened and rewound such that the radius of the newly formed coil is 2R, carrying the same current I, what will be the magnetic field at the centre O? [1 Mark]
(a) 2B
(b) B
(c) B/2
(d) B/4

Answer: (d) B/4

Magnetic field is given by \( B = \frac{\mu_{0}NI}{2R} \). When radius becomes \( 2R \), the number of turns becomes \( N/2 \). Thus \( B' = \frac{\mu_{0}(N/2)I}{2(2R)} = \frac{1}{4} \left( \frac{\mu_{0}NI}{2R} \right) = \frac{B}{4} \).

Teacher's Note:
a) Keep in mind that rewinding a wire of fixed length into a larger radius reduces the number of turns proportionally.
b) A common mistake is to keep \( N \) constant while changing \( R \), ignoring the total length constraint of the wire.

(iii) Magnetic susceptibility of a diamagnetic substance: [1 Mark]
(a) decreases with the increase in its temperature.
(b) is not affected by the change in its temperature.
(c) increases with increase in its temperature.
(d) first increases then decreases with increase in its with temperature.

Answer: (b) is not affected by the change in its temperature.

Diamagnetic susceptibility is independent of temperature as orbital magnetic moments do not change with thermal agitation.

Teacher's Note:
a) Remember Curie's law applies to paramagnetic substances, where susceptibility varies inversely with temperature.
b) Do not confuse diamagnetism with paramagnetism or ferromagnetism regarding thermal dependence.

(iv) The de Broglie wavelength of an electron in the first Bohr's orbit of hydrogen atom is equal to: [1 Mark]
(a) diameter of the first orbit.
(b) circumference of the first orbit.
(c) radius of the second orbit.
(d) h/2\( \pi \).

Answer: (b) circumference of the first orbit.

According to Bohr's quantization condition, \( mvr = \frac{nℎ}{2\pi} \implies 2\pi r = n\lambda \). For the first orbit (\( n = 1 \)), \( \lambda = 2\pi r \).

Teacher's Note:
a) Relate de Broglie wavelength directly to Bohr's stationary orbit condition \( n\lambda = 2\pi r \).
b) Students frequently confuse circumference with radius or diameter in this derivation.

(v) An ideal PN junction diode offers: [1 Mark]
(a) zero resistance in forward as well as reverse bias.
(b) infinite resistance in forward as well as reverse bias.
(c) zero resistance in forward, but infinite resistance in reverse bias.
(d) infinite resistance in forward, but zero resistance in reverse bias.

Answer: (c) zero resistance in forward, but infinite resistance in reverse bias.

An ideal diode acts as a closed switch (zero resistance) in forward bias and an open switch (infinite resistance) in reverse bias.

Teacher's Note:
a) Perfect rectifying action implies complete conduction in one direction and complete blocking in the other.
b) Avoid mixing up forward and reverse bias resistance characteristics.

(vi) Given below are two statements marked Assertion and Reason. Read the two statements and choose the correct option.
Assertion: An astronomical telescope has an objective lens having large focal length.
Reason: Magnifying power of an astronomical telescope varies directly with focal length of the objective lens. [1 Mark]
(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 false and Reason is true.
(d) Both Assertion and Reason are false.

Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.

The magnifying power of an astronomical telescope is given by \( m = \frac{f_{o}}{f_{e}} \), hence a large objective focal length directly increases magnification.

Teacher's Note:
a) Note that magnifying power depends directly on objective focal length and inversely on eyepiece focal length.
b) Ensure students read both statements carefully to establish the direct cause-effect link.

(vii) Given below are two statements marked, Assertion and Reason. Read the two statements and choose the correct option.
Assertion: If critical angle of glass-air pair (\( \mu_{g} = 3/2 \)) is \( \theta_{1} \) and that of water-air pair (\( \mu_{w} = 4/3 \)) is \( \theta_{2} \), then the critical angle for the water-glass pair will lie between \( \theta_{1} \) and \( \theta_{2} \).
Reason: A medium is optically denser if its refractive index is greater. [1 Mark]
(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 false and Reason is true.
(d) Both Assertion and Reason are false.

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

Total internal reflection and critical angle occur only when light travels from an optically denser medium to a rarer medium. Water to glass is rarer to denser, so critical angle does not exist for this pair.

Teacher's Note:
a) Always verify that critical angle is defined only for light travelling from denser to rarer medium.
b) The reason is a factual statement about optical density, but does not validate the false assertion about water-glass pair.

(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:

[Figure: A parabolic curve on a Cartesian coordinate system with heating power P on the vertical axis and current I on the horizontal axis, starting from the origin and curving upwards representing \( P = I^{2}R \).]

Teacher's Note:
a) Use the relation \( P = I^{2}R \) to show that power is directly proportional to the square of current.
b) The graph must be clearly curved (parabolic) and not a straight line.

(ii) Explain why core of a transformer is always laminated. [1 Mark]

Answer:
The core of a transformer is laminated to reduce the loss of energy due to eddy currents by increasing the resistance paths.

Teacher's Note:
a) Eddy currents are induced circulating currents in bulk metal cores that cause heating and energy loss.
b) Mentioning lamination breaks the path of eddy currents and improves efficiency.

(iii) Why does a car driver use a convex mirror as a rear-view mirror? [1 Mark]

Answer:
A convex mirror always forms a virtual, erect, and diminished image of objects behind, thereby providing a wide field of view.

Teacher's Note:
a) Emphasize both characteristics: diminished size and wide field of view.
b) Students often miss stating that the image formed is virtual and erect.

(iv) Represent diagrammatically how the incident planar wavefronts of wavelength \( \lambda \) pass through an aperture of size d, when d is approximately equal to \( \lambda \). [1 Mark]

Answer:

[Figure: Diagram showing parallel incident planar wavefronts striking a barrier with an aperture of size comparable to wavelength, emerging as circular or diverging spherical wavefronts due to diffraction.]

Teacher's Note:
a) When aperture size is comparable to wavelength, pronounced diffraction bending occurs.
b) Draw clear curved wavefronts emerging from the aperture edges.

(v) What is meant by "Dual nature of matter"? [1 Mark]

Answer:
Matter exhibits both particle nature and wave nature depending on the experimental circumstances.

Teacher's Note:
a) De Broglie hypothesized that moving material particles possess an associated wave nature.
b) State clearly that particles like electrons show interference and diffraction under suitable conditions.

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

Answer:
They annihilate each other, converting their mass energy into gamma ray photons.

Teacher's Note:
a) This process is known as pair annihilation.
b) Mention energy release in the form of photons for full credit.

(vii) What energy conversion takes place in a solar cell? [1 Mark]

Answer:
Solar or light energy is converted into electrical energy.

Teacher's Note:
a) A solar cell is a photovoltaic device operating on the inner photoelectric effect.
b) Keep the answer concise and direct.

SECTION B - 14 MARKS

Question 2 [2 Marks]

(i) A dielectric slab of dielectric constant K and thickness t is introduced between the two plates of a capacitor of plate-separation d (\( > t \)) and common area A. The capacitance of this system is given as: C = \( \frac{\epsilon_{0}A}{(d-t)+(t/K)} \).
How does the capacitance C modify in each of the following cases?
(a) The dielectric slab covers half the distance of separation between the two plates. [1 Mark]
(b) The whole space between the plates is filled with the dielectric. [1 Mark]

Answer:
(a) Substituting \( t = d/2 \) into the formula:
\( C = \frac{\epsilon_{0}A}{(d - d/2) + \frac{d/2}{K}} = \frac{\epsilon_{0}A}{\frac{d}{2} + \frac{d}{2K}} = \frac{2\epsilon_{0}AK}{d(K+1)} \)
(b) When the whole space is filled, \( t = d \):
\( C = \frac{\epsilon_{0}A}{(d-d) + d/K} = \frac{K\epsilon_{0}A}{d} \)

Teacher's Note:
a) Substitute the correct thickness value corresponding to each condition into the general capacitor formula.
b) Ensure proper algebraic simplification of fractions.

OR

(ii) In Figure 2 below, the current-voltage graphs for a conductor are given at two different temperatures \( T_{1} \) and \( T_{2} \).

[Figure: Current-voltage (I-V) characteristic graph with two straight lines for temperatures \( T_{1} \) and \( T_{2} \), showing angle \( \theta_{1} \) and \( \theta_{2} \) with the voltage axis, where \( \theta_{1} > \theta_{2} \).]

(a) At which temperature \( T_{1} \) or \( T_{2} \) is the resistance higher? [1 Mark]
(b) Which temperature (\( T_{1} \) or \( T_{2} \)) is higher? [1 Mark]

Answer:
(a) The slope of the I-V graph gives conductance (\( 1/R = \tan\theta \)). Since \( \theta_{1} > \theta_{2} \), \( \tan\theta_{1} > \tan\theta_{2} \), meaning conductance at \( T_{1} \) is greater, so resistance at \( T_{1} \) is lower and resistance at \( T_{2} \) is higher.
(b) Since resistance of a conductor increases with temperature and resistance at \( T_{2} \) is higher than at \( T_{1} \), temperature \( T_{2} > T_{1} \).

Teacher's Note:
a) Recall that slope represents conductance (\( I/V \)), not direct resistance.
b) Metallic resistance increases with an increase in temperature.

Question 3 [2 Marks]

Arrangement of an oxygen ion and two hydrogen ions in a water molecule is shown in Figure 3 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 structure showing oxygen ion at the vertex with negative charge and two hydrogen ions at distance l forming angle \( \theta \) between the two O-H bonds.]

Answer:
The dipole moment of each O-H bond is \( p_{1} = p_{2} = e \times l \).
The net dipole moment is the vector sum of the two bond dipole moments acting at angle \( \theta \):
\( P = \sqrt{p_{1}^{2} + p_{2}^{2} + 2p_{1}p_{2}\cos\theta} = \sqrt{2p^{2}(1 + \cos\theta)} = 2p\cos(\theta/2) = 2el\cos(\theta/2) \)

Teacher's Note:
a) Use vector addition formula for two equal vectors inclined at an angle \( \theta \).
b) Apply trigonometric identity \( 1 + \cos\theta = 2\cos^{2}(\theta/2) \) to simplify the expression.

Question 4 [2 Marks]

(i) Figure 4 below shows a part of an electric circuit.

[Figure: Circuit diagram showing junction A with incoming current 5A, branch currents \( I_{1} = 2\text{A} \), \( I_{2} \), and resistors 4 ohms, 5 ohms, 7 ohms with branch currents \( I_{3} \).]

(a) Apply Kirchhoff's junction rule to the junction A to find current flowing through the 5\( \Omega \) resistor. [1 Mark]
(b) Apply Kirchhoff's loop rule to the loop ABDA to find current flowing through the 7\( \Omega \) resistor. [1 Mark]

Answer:
(a) At junction A, total incoming current equals total outgoing current: \( I_{1} + I_{2} = I \)
\( 2 + I_{2} = 5 \implies I_{2} = 3\text{A} \) (current through 5\( \Omega \) resistor is 3 A).
(b) Applying loop rule to loop ABDA: \( I_{1} \times 4 + I_{3} \times 7 = I_{2} \times 5 \)
\( 2 \times 4 + 7I_{3} = 3 \times 5 \)
\( 8 + 7I_{3} = 15 \implies 7I_{3} = 7 \implies I_{3} = 1\text{A} \).

Teacher's Note:
a) Junction rule states algebraic sum of currents meeting at a junction is zero.
b) Loop rule accounts for potential drops and rises around a closed path correctly.

OR

(ii) In a potentiometer, a cell is balanced against 110 cm when the circuit is open. A cell is balanced at 100 cm when short-circuited through a resistance of 10 \( \Omega \). Find the internal resistance of the cell. [2 Marks]

Answer:
Given \( l_{1} = 110\text{ cm} \), \( l_{2} = 100\text{ cm} \), \( R = 10\text{ }\Omega \).
Internal resistance \( r = R\left(\frac{l_{1}}{l_{2}} - 1\right) \)
\( r = 10\left(\frac{110}{100} - 1\right) = 10(1.1 - 1) = 10(0.1) = 1\text{ }\Omega \).

Teacher's Note:
a) Use the standard potentiometer formula relating internal resistance to balancing lengths.
b) Ensure correct substitution and unit verification.

Question 5 [2 Marks]

Figure 5 below are two long, parallel wires carrying current in the same direction such that \( I_{1} < I_{2} \).

[Figure: Two parallel vertical wires carrying currents \( I_{1} \) and \( I_{2} \) in the same downward direction.]

(i) In which direction will wire \( I_{1} \) move? [1 Mark]
(ii) If the direction of the current \( I_{2} \) is reversed, in which direction will the wire \( I_{1} \) move now? [1 Mark]

Answer:
(a) It will move towards wire \( I_{1} \) (actually towards wire \( I_{2} \), as parallel currents attract each other).
(b) It will move away from wire \( I_{2} \) because parallel currents in opposite directions repel each other.

Teacher's Note:
a) Currents flowing in the same direction attract; currents in opposite directions repel.
b) Read the question carefully regarding which wire's motion is asked.

Question 6 [2 Marks]

(i) The focal length of a double convex lens is equal to the radius of curvature of either surface. What is the refractive index of its material? [1 Mark]
(ii) What is meant by a thin prism? [1 Mark]

Answer:
(a) Using lens maker's formula: \( \frac{1}{f} = (n-1)\left(\frac{1}{R_{1}} - \frac{1}{R_{2}}\right) \)
Given \( f = R \) and \( R_{1} = R, R_{2} = -R \):
\( \frac{1}{R} = (n-1)\left(\frac{1}{R} - \frac{1}{-R}\right) = (n-1)\frac{2}{R} \implies n - 1 = \frac{1}{2} \implies n = 1.5 \).
(b) A prism whose refracting angle is small, i.e., less than \( 4^{\circ} \), is called a thin prism.

Teacher's Note:
a) Apply sign conventions correctly for radii of curvature of a double convex lens.
b) Define thin prism precisely by its refracting angle limit.

Question 7 [2 Marks]

(i) Name the electromagnetic radiation that has been used in obtaining the image below. [1 Mark]

[Figure: X-ray radiograph showing human chest bones and lungs.]

(ii) What is the wavelength range of electromagnetic radiation used in radio broadcast? [1 Mark]

Answer:
(a) X-rays
(b) Greater than 0.1 m (or greater than 1 m depending on radio wave bands).

Teacher's Note:
a) Medical radiographs are produced using X-rays due to their high penetration power.
b) Radio waves occupy the longest wavelength region of the electromagnetic spectrum.

Question 8 [2 Marks]

How does stopping potential in photoelectric emission vary if
(i) the intensity of the incident radiation increases? [1 Mark]
(ii) the frequency of incident radiation decreases? [1 Mark]

Answer:
(a) Stopping potential remains independent of the intensity of incident radiation.
(b) Stopping potential decreases when the frequency of incident radiation decreases.

Teacher's Note:
a) Intensity affects photocurrent (number of emitted electrons), not their maximum kinetic energy.
b) Stopping potential depends linearly on frequency according to Einstein's photoelectric equation.

SECTION C - 27 MARKS

Question 9 [3 Marks]

"A uniformly charged conducting spherical shell for the points outside the shell behaves as if the entire charge of the shell is concentrated at its centre". Show this with the help of a proper diagram and verify this statement.

[Figure: A spherical shell of radius R with surface charge density \( \sigma \), surrounded by a concentric Gaussian sphere of radius \( r > R \) passing through point P.]

Answer:
Consider a spherical shell of radius \( R \) carrying total charge \( q \) uniformly distributed with surface charge density \( \sigma \).
For an external point P at a distance \( r \) (\( r > R \)) from the centre, construct a concentric spherical Gaussian surface of radius \( r \).
By Gauss' law:
\( \oint E \cdot ds = \frac{q}{\epsilon_{0}} \)
\( E \times 4\pi r^{2} = \frac{q}{\epsilon_{0}} \implies E = \frac{1}{4\pi\epsilon_{0}} \frac{q}{r^{2}} \)
This electric field expression is identical to that produced by a point charge \( q \) placed at the centre, thus verifying the statement.

Teacher's Note:
a) Apply Gauss' law on a spherical Gaussian surface enclosing the charged shell.
b) Conclude clearly that outside the shell, the field matches Coulomb's law for a point charge at the origin.

Question 10 [3 Marks]

(i) Study the two circuits shown in Figure 6 below. The cells in the two circuits are identical to each other. The resistance of the load resistor R is the same in both circuits.
If the same current flows through the resistor R in both circuits, calculate the internal resistance of each cell in terms of the resistance of resistor R. Show your calculations.

[Figure: Two circuits with load resistor R. Circuit 1 has two identical cells of emf E and internal resistance r connected in series. Circuit 2 has two identical cells connected in parallel.]

Answer:
For Circuit 1 (cells in series):
\( E_{eq} = 2E \), \( r_{eq} = 2r \)
Current \( I = \frac{2E}{R + 2r} \)
For Circuit 2 (cells in parallel):
\( E_{eq} = E \), \( r_{eq} = \frac{r}{2} \)
Current \( I = \frac{E}{R + r/2} \)
Since current is the same in both circuits:
\( \frac{2E}{R + 2r} = \frac{E}{R + r/2} \)
\( 2(R + r/2) = R + 2r \)
\( 2R + r = R + 2r \implies r = R \).

Teacher's Note:
a) Determine equivalent emf and equivalent internal resistance for both series and parallel cell combinations.
b) Equate the expressions for current through the external load resistor R to solve for r.

OR

(ii) The drift velocity of electrons in a conductor connected to a battery is given by \( v_{d} = -\frac{eE\tau}{m} \). Here, e is the charge of the electron, E is the electric field, \( \tau \) is the average time between collisions and m is the mass of the electron.
Based on this, answer the following:
(a) How does the drift velocity change with a change in the potential difference across the conductor? [1.5 Marks]
(b) A copper wire of length 'l' is connected to a source. If the copper wire is replaced by another copper wire of the same area of cross-section but of length '4l', how will the drift velocity change? Explain your answer. [1.5 Marks]

Answer:
(a) As the potential difference across the conductor is increased, the electric field inside increases (\( E = V/l \)). Since \( v_{d} \propto E \), drift velocity increases directly with potential difference.
(b) Since \( v_{d} \propto E \) and \( E = \frac{V}{l} \), we have \( v_{d} \propto \frac{1}{l} \). When length is increased to \( 4l \) keeping voltage constant, the electric field becomes one-fourth, so the drift velocity becomes one-fourth (\( 1/4 \text{th} \)) of its original value.

Teacher's Note:
a) Relate drift velocity to electric field and potential difference using the given formula.
b) Analyze the inverse proportionality with length when potential difference across the wire remains constant.

Question 11 [3 Marks]

A galvanometer of resistance 100 \( \Omega \) gives a full-scale deflection for a potential difference of 200 mV.
(i) What must be the resistance connected to convert the galvanometer into an ammeter of the range 0-200 mA? [2 Marks]
(ii) Determine resistance of the ammeter. [1 Mark]

Answer:
Given \( G = 100\text{ }\Omega \), \( V_{g} = 200\text{ mV} = 0.2\text{ V} \), range \( I = 200\text{ mA} = 0.2\text{ A} \).
Full scale deflection current of galvanometer \( I_{g} = \frac{V_{g}}{G} = \frac{0.2\text{ V}}{100\text{ }\Omega} = 0.002\text{ A} = 2\text{ mA} \).
(i) Shunt resistance required to convert into ammeter:
\( S = \frac{I_{g}G}{I - I_{g}} = \frac{2 \times 10^{-3} \times 100}{0.2 - 0.002} = \frac{0.2}{0.198} = 1.01\text{ }\Omega \).
(ii) Resistance of the ammeter \( R_{A} = \frac{G \times S}{G + S} = \frac{100 \times 1.01}{100 + 1.01} = \frac{101}{101.01} \approx 0.999\text{ }\Omega \).

Teacher's Note:
a) Calculate galvanometer current \( I_{g} \) first using given voltage and resistance.
b) Use the shunt resistance formula for ammeter conversion and parallel combination formula for total ammeter resistance.

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 7 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 carrying current I with an axis showing axial point P at distance x from centre O.]

Answer:
The theoretical magnetic field at an axial point at distance \( x \) is given by:
\( B = \frac{\mu_{0}Ia^{2}}{2(a^{2} + x^{2})^{3/2}} \)
At the centre (\( x = 0 \)), \( B_{0} = \frac{\mu_{0}I}{2a} \).
For \( x = a \):
\( B = \frac{\mu_{0}Ia^{2}}{2(a^{2} + a^{2})^{3/2}} = \frac{\mu_{0}Ia^{2}}{2(2a^{2})^{3/2}} = \frac{\mu_{0}I}{2a} \frac{1}{2\sqrt{2}} = \frac{B_{0}}{2\sqrt{2}} \).\br For \( x = 2a \):
\( B = \frac{\mu_{0}Ia^{2}}{2(a^{2} + (2a)^{2})^{3/2}} = \frac{\mu_{0}Ia^{2}}{2(5a^{2})^{3/2}} = \frac{\mu_{0}I}{2a} \frac{1}{5\sqrt{5}} = \frac{B_{0}}{5\sqrt{5}} \).
Both evaluated points match the recorded data values, verifying the theoretical formula.

Teacher's Note:
a) State the standard formula for magnetic field on the axis of a current-carrying circular coil.
b) Substitute \( x = a \) and \( x = 2a \) into the formula to verify the given fractional values.

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. [3 Marks]
(a) Explain why the electron describes a circular path. [1 Mark]
(b) Calculate the radius of the circular path described by the electron. [2 Marks]

Answer:
(a) The magnetic force acting on the moving electron is always perpendicular to its velocity vector and the magnetic field vector (\( \vec{F} = q(\vec{v} \times \vec{B}) \)). This perpendicular force acts as the centripetal force, causing the electron to move in a circular path.
(b) Radius \( r = \frac{mv}{eB} \)
Given \( m = 9.1 \times 10^{-31}\text{ kg} \), \( v = 8 \times 10^{7}\text{ ms}^{-1} \), \( e = 1.6 \times 10^{-19}\text{ C} \), \( B = 1.3 \times 10^{-3}\text{ T} \):
\( r = \frac{(9.1 \times 10^{-31}) \times (8 \times 10^{7})}{(1.6 \times 10^{-19}) \times (1.3 \times 10^{-3})} = \frac{72.8 \times 10^{-24}}{2.08 \times 10^{-22}} = 0.35\text{ m} \).

Teacher's Note:
a) Explain the perpendicular nature of magnetic Lorentz force providing centripetal acceleration.
b) Substitute standard values of electron mass and charge into the radius formula carefully.

Question 13 [3 Marks]

With the help of a neatly drawn labelled diagram, prove the law of reflection on the basis of Huygen's wave theory.

Answer:

[Figure: Diagram showing incident plane wavefront AB striking a reflecting surface at angle of incidence i, secondary wavelets originating from surface, and reflected wavefront A'B' at angle of reflection r, with normals and directional arrows.]

Consider a plane wavefront AB incident obliquely on a reflecting surface XY. Let \( v \) be the speed of light in the medium and \( t \) be the time taken by the wave disturbance to travel from B to B'.
Draw perpendiculars representing incident and reflected rays.
In triangles \( \text{ABB}' \) and \( \text{A'B'B} \):
- \( BB' = vt \) (distance travelled by wave from B to B')
- \( AA' = vt \) (distance travelled by secondary wavelet from A to A')
- \( AB' \) is common to both triangles.
- Angle \( \angle BAB' = \angle B'A'A = 90^{\circ} \).
Therefore, triangle \( \text{ABB}' \) is congruent to triangle \( \text{A'B'B} \).
Hence, angle of incidence \( i \) equals angle of reflection \( r \) (\( \angle i = \angle r \)). This proves the law of reflection.

Teacher's Note:
a) Draw clear wavefronts and normals with angle indicators \( i \) and \( r \).
b) Use triangle congruence to show that angle of incidence equals angle of reflection.

Question 14 [3 Marks]

A lens of focal length f is divided into two equal parts and then these parts are put in a combination as shown in Figure 8 below.

[Figure: Diagram showing a double convex lens divided vertically into two symmetrical plano-convex lenses \( L_{1} \) and \( L_{2} \), arranged in two different combinations.]

(i) What is the focal length of \( L_{1} \)? [1.5 Marks]
(ii) What is the focal length of the final combination? [1.5 Marks]

Answer:
(a) For the original lens of focal length \( f \), \( \frac{1}{f} = (n-1)\left(\frac{2}{R}\right) \). When cut vertically into two equal plano-convex lenses \( L_{1} \) and \( L_{2} \), each has one flat surface (\( R_{2} = \infty \)) and one curved surface (\( R_{1} = R \)).
\( \frac{1}{f_{1}} = (n-1)\left(\frac{1}{R} - \frac{1}{\infty}\right) = (n-1)\frac{1}{R} = \frac{2}{f} \implies f_{1} = 2f \).
(b) In the final combination shown (two plano-convex lenses placed together symmetrically), the equivalent focal length \( F \) is given by:
\( \frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1}{2f} + \frac{1}{2f} = \frac{2}{2f} = \frac{1}{f} \implies F = f \).

Teacher's Note:
a) Cutting a lens vertically doubles the focal length of each individual plano-convex part.
b) Combine the individual focal lengths using the lens combination formula \( \frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \).

Question 15 [3 Marks]

In Young's double slit experiment, how is interference pattern affected when the following changes are made:
(i) Slits are brought closer to each other. [1 Mark]
(ii) Screen is moved away from the slits. [1 Mark]
(iii) Red coloured light is replaced with blue coloured light. [1 Mark]

Answer:
(a) Fringe width \( \beta = \frac{\lambda D}{d} \). When slits are brought closer (\( d \) decreases), fringe width increases.
(b) When screen is moved away (\( D \) increases), fringe width increases.
(c) When red light (\( \lambda_{\text{red}} \)) is replaced with blue light (\( \lambda_{\text{blue}} \)), wavelength decreases, so fringe width decreases.

Teacher's Note:
a) Base all answers on the fringe width formula \( \beta = \frac{\lambda D}{d} \).
b) Remember that wavelength of blue light is shorter than that of red light.

Question 16 [3 Marks]

(i) A virologist studies details of a virus with the help of an instrument. Name the instrument used by him. [1 Mark]
(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. [2 Marks]

Answer:
(a) Compound microscope.
(b)

[Figure: Ray diagram of a compound microscope showing objective lens forming a real, inverted, magnified intermediate image, and eyepiece acting as a magnifier to form a final virtual, magnified image at least distance of distinct vision D, with all principal rays and labels \( u_{o}, v_{o}, u_{e}, f_{o}, f_{e} \).]

Teacher's Note:
a) Viruses are extremely minute biological structures requiring high magnification provided by a compound microscope.
b) Ensure correct labelling of objective, eyepiece, and ray paths in the diagram.

Question 17 [3 Marks]

The graphs below show the variation of the stopping potential \( V_{s} \) with the frequency (\( \nu \)) of the incident radiations for two different photosensitive materials \( M_{1} \) and \( M_{2} \).

[Figure: Graph of stopping potential \( V_{s} \) versus frequency \( \nu \) showing two parallel straight lines for materials \( M_{1} \) and \( M_{2} \) with threshold frequencies \( \nu_{01} \) and \( \nu_{02} \).]

Express work function for \( M_{1} \) and \( M_{2} \) in terms of Planck's constant (h) and threshold frequency and charge of the electron (e).
If the values of stopping potential for \( M_{1} \) and \( M_{2} \) are \( V_{1} \) and \( V_{2} \) respectively then show that the slope of the lines equals to \( \frac{V_{1} - V_{2}}{\nu_{01} - \nu_{02}} \) for a frequency, \( \nu > \nu_{02} \) and also \( \nu > \nu_{01} \).

Answer:
Work functions are given by \( W_{01} = h\nu_{01} \) and \( W_{02} = h\nu_{02} \), or in terms of stopping potential units, \( W_{01}/e \) and \( W_{02}/e \).
According to Einstein's photoelectric equation: \( eV = h\nu - h\nu_{0} \implies V = \left(\frac{h}{e}\right)\nu - \frac{h\nu_{0}}{e} \)
For material \( M_{1} \): \( V_{1} = \left(\frac{h}{e}\right)\nu - \frac{h\nu_{01}}{e} \)
For material \( M_{2} \): \( V_{2} = \left(\frac{h}{e}\right)\nu - \frac{h\nu_{02}}{e} \)
Subtracting the two equations:
\( V_{1} - V_{2} = \frac{h}{e}(\nu_{02} - \nu_{01}) \)
Rearranging terms for slope: \( \frac{h}{e} = \frac{V_{1} - V_{2}}{\nu_{01} - \nu_{02}} \).

Teacher's Note:
a) Start with Einstein's photoelectric equation relating stopping potential to frequency and threshold frequency.
b) Subtract equations for the two materials to eliminate frequency \( \nu \) and express the slope \( h/e \).

SECTION D - 15 MARKS

Question 18 [5 Marks]

(i) The magnetic field through a single loop of wire, 12 cm in radius and 8.5 \( \Omega \) resistance, changes with time as shown in graph below (Figure 9). The magnetic field is perpendicular to the plane of the loop.

[Figure: Graph of magnetic field B (in Tesla) versus time t (in seconds), showing B = 0 at t = 0, increasing linearly to 1.0 T at t = 2.0 s, remaining constant at 1.0 T till t = 4.0 s, and decreasing linearly to 0 at t = 6.0 s.]

(a) Find the induced emf for the time intervals 0 to 2.0 s, 2.0 to 4.0 s, and 4.0 to 6.0 s. [3 Marks]
(b) Hence, plot induced current as a function of time. [2 Marks]

Answer:
Radius \( r = 12\text{ cm} = 0.12\text{ m} \), Area \( A = \pi r^{2} = \pi (0.12)^{2} = 0.04524\text{ m}^{2} \), Resistance \( R = 8.5\text{ }\Omega \).
(a) Induced emf \( e = -\frac{d\Phi}{dt} = -A\frac{dB}{dt} \):
- For interval 0 to 2.0 s: \( \frac{dB}{dt} = \frac{1.0 - 0}{2.0 - 0} = 0.5\text{ T/s} \).
\( e_{1} = -0.04524 \times 0.5 = -0.0226\text{ V} \approx -0.023\text{ V} \). Current \( i_{1} = \frac{-0.023}{8.5} = -2.7 \times 10^{-3}\text{ A} \).
- For interval 2.0 to 4.0 s: \( \frac{dB}{dt} = 0 \), so \( e_{2} = 0 \) and \( i_{2} = 0 \).
- For interval 4.0 to 6.0 s: \( \frac{dB}{dt} = \frac{0 - 1.0}{6.0 - 4.0} = -0.5\text{ T/s} \).
\( e_{3} = -0.04524 \times (-0.5) = +0.0226\text{ V} \). Current \( i_{3} = \frac{+0.023}{8.5} = +2.7 \times 10^{-3}\text{ A} \).
(b)

[Figure: Step graph of induced current i (in mA) versus time t (in s), showing current at -2.7 mA from t = 0 to 2 s, 0 from t = 2 to 4 s, and +2.7 mA from t = 4 to 6 s.]

Teacher's Note:
a) Calculate rate of change of magnetic field \( \frac{dB}{dt} \) from the slope of each segment in the B-t graph.
b) Use Faraday's law and Ohm's law to find induced emf and current for each time interval.

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.

[Figure: Resonance curves (current versus frequency) for three circuits X, Y, Z showing different sharpness and peak currents at resonant frequency \( f_{0} \).]

(a) Who used minimum resistance? [1 Mark]
(b) In which case will the quality Q factor be maximum? [1 Mark]
(c) What did the students conclude about the nature of impedance at resonant frequency (\( f_{0} \))? [1 Mark]
(d) An ideal capacitor is connected across 220V, 50Hz, and 220V, 100Hz supplies. Find the ratio of current flowing through it in the two cases. [2 Marks]

Answer:
(a) Student X used minimum resistance (indicated by the highest peak current at resonance).
(b) Quality Q factor is maximum for case X (sharpest resonance curve).
(c) At resonant frequency, impedance is purely resistive and attains its minimum value equal to the ohmic resistance R.
(d) Capacitive reactance \( X_{c} = \frac{1}{2\pi f C} \). Current \( I = \frac{V}{X_{c}} = 2\pi f C V \).
Since voltage is same, \( I \propto f \).
Ratio of currents: \( \frac{I_{1}}{I_{2}} = \frac{f_{1}}{f_{2}} = \frac{50}{100} = \frac{1}{2} \).

Teacher's Note:
a) Peak current at resonance is inversely proportional to circuit resistance (\( I_{m} = E/R \)).
b) Capacitive reactance is inversely proportional to frequency, making capacitive current directly proportional to frequency.

Question 19 [5 Marks]

(i) (a) Find the binding energy per nucleon of \( ^{235}_{\ 92}\text{U} \) based on the information given below. [2 Marks]

(b) Find the energy released during the following fission reaction: \( ^{235}_{\ 92}\text{U} + \,^{1}_{0}\text{n} \rightarrow \,^{236}_{\ 92}\text{U} \rightarrow \,^{90}_{\ 36}\text{Kr} + \,^{143}_{\ 56}\text{Ba} + 3\,^{1}_{0}\text{n} \). [2 Marks]

(c) During Rutherford's gold foil experiment, it was observed that most of the \( \alpha \)-particles did not deflect. However, some showed a deflection of \( 180^{\circ} \). What hypothesis was made to justify the deflection of \( \alpha \)-particle by \( 180^{\circ} \)? [1 Mark]

Answer:
(a) Number of protons = 92, number of neutrons = \( 235 - 92 = 143 \).
Total mass of nucleons = \( (92 \times 1.0073) + (143 \times 1.0087) = 92.6716 + 144.2441 = 236.9157\text{ u} \).
Mass defect \( \Delta m = 236.9157 - 235.0439 = 1.8718\text{ u} \).
Total binding energy = \( 1.8718 \times 931.5\text{ MeV} = 1743.6\text{ MeV} \).
Binding energy per nucleon = \( \frac{1743.6\text{ MeV}}{235} = 7.42\text{ MeV/nucleon} \).
(b) Mass of reactants = Mass of \( ^{235}\text{U} \) + Mass of neutron = \( 235.0439 + 1.0087 = 236.0526\text{ u} \).
Mass of products = Mass of \( ^{90}\text{Kr} \) + Mass of \( ^{143}\text{Ba} \) + 3(Mass of neutron)
= \( 89.9195 + 142.9206 + 3(1.0087) = 232.8401 + 3.0261 = 235.8662\text{ u} \).
Mass defect \( \Delta m = 236.0526 - 235.8662 = 0.1864\text{ u} \).
Energy released \( Q = 0.1864 \times 931.5\text{ MeV} = 173.63\text{ MeV} \).
(c) The entire positive charge and almost the entire mass of the atom are concentrated in a very tiny central core called the nucleus.

Teacher's Note:
a) Calculate mass defect carefully by subtracting total nuclear rest mass from constituent nucleon masses.
b) Multiply mass defect by 931.5 MeV to obtain binding energy or fission energy release.

OR

(ii) (a) Define unified atomic mass unit. [1 Mark]
(b) Calculate its energy equivalent. [1 Mark]
(c) 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 10 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 eV for reference. The transition of electrons takes place between levels \( E_{1} \) and \( E_{2} \). [3 Marks]

[Figure: Energy level diagram showing three levels \( E_{1} = 0\text{ eV} \), \( E_{2} = 1.78\text{ eV} \), and \( E_{3} = 2.95\text{ eV} \).]

(1) What wavelength of radiation is needed to excite the atom to energy level \( E_{2} \) from \( E_{1} \)?
(2) Suppose the external source has a power of 100 W. What would be the rate of photon emission?

Answer:
(a) One unified atomic mass unit (1 u) is defined as exactly \( 1/12\text{th} \) of the mass of an unbound carbon-12 atom at rest in its ground state.
(b) Energy equivalent of 1 u is \( 1\text{ u} \times c^{2} = 931.5\text{ MeV} \).
(c)(1) Energy difference \( \Delta E = E_{2} - E_{1} = 1.78\text{ eV} = 1.78 \times 1.6 \times 10^{-19}\text{ J} = 2.848 \times 10^{-19}\text{ J} \).
Wavelength \( \lambda = \frac{hc}{\Delta E} = \frac{(6.6 \times 10^{-34}) \times (3 \times 10^{8})}{2.848 \times 10^{-19}} = \frac{1.98 \times 10^{-25}}{2.848 \times 10^{-19}} \approx 6.95 \times 10^{-7}\text{ m} \) (or 696.6 nm).
(c)(2) Energy of each photon \( E_{p} = 1.78\text{ eV} = 2.85 \times 10^{-19}\text{ J} \).
Rate of photon emission \( \frac{N}{t} = \frac{\text{Power}}{E_{p}} = \frac{100\text{ W}}{2.85 \times 10^{-19}\text{ J}} \approx 3.51 \times 10^{20}\text{ photons/second} \).

Teacher's Note:
a) Know the standard definition and energy equivalent of 1 u.
b) Use \( \lambda = hc/\Delta E \) for wavelength and divide total power by individual photon energy to find emission rate.

Question 20 [5 Marks]

Sanya performed an experiment of obtaining characteristic curves of a junction diode. When she forward biased it, she found that beyond forward voltage \( V = V_{k} \), the conductivity is very high. When she reverse biased the diode she found that a very small current (of about a few microamperes) flows in the diode. It remained constant even though she varied the voltage.

(i) In Figure 11, which one of the diodes is forward biased? [1 Mark]

[Figure: Circuit diagram showing two diodes \( D_{1} \) and \( D_{2} \) in parallel branches connected across voltage rails -10V and -5V.]

(ii) What is meant by a saturation current? [1 Mark]
(iii) When applied voltage during forward bias is small, why does no current flow in the diode? [1 Mark]
(iv) The circuit shown in the Figure 12 below contains two diodes, each with a forward resistance of 50 \( \Omega \) and with infinite resistance during reverse bias. If the battery voltage is 6V, then calculate the current through the 100 \( \Omega \) resistance. [2 Marks]

[Figure: Circuit diagram with a 6V battery and a bridge network containing resistors 150 ohms, 50 ohms, and 100 ohms along with two diodes.]

Answer:
(a) Diode \( D_{2} \) is forward biased (due to appropriate potential difference across its terminals).
(b) Saturation current is the constant minimum reverse current that flows due to minority charge carriers, independent of the applied reverse voltage.
(c) At small forward voltages, the external voltage is insufficient to overcome the potential barrier of the depletion region.
(d) Analyzing Figure 12: Based on diode polarities and resistances, one diode is reverse biased (infinite resistance) and blocks current in its branch. Ignoring that branch, the total resistance in the active path is \( R_{f} + R_{\text{other}} + 100\text{ }\Omega = 50 + 150 + 100 = 300\text{ }\Omega \).
Current \( I = \frac{V}{R_{total}} = \frac{6\text{ V}}{300\text{ }\Omega} = 0.02\text{ A} \).

Teacher's Note:
a) Check potential polarities at the terminals of each diode to determine forward or reverse bias.
b) Sum up series resistances in the conducting branch to calculate total current using Ohm's law.