ISC Class 12 Physics Sample Paper 2024 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 bulbs of power ratings \( 100\text{ W} \) and \( 40\text{ W} \), designed to operate on the same voltage, are connected in parallel to a battery source as shown in Figure 1 below. Currents flowing across points 1, 2 and 3 are \( I_1 \), \( I_2 \) and \( I_3 \) respectively. [1 Mark]
[Figure: Two bulbs rated \( 100\text{ W} \) and \( 40\text{ W} \) connected in parallel across a DC source with points labeled 1 at the source, 2 in the lower branch carrying current \( I_2 \) through the \( 40\text{ W} \) bulb, and 3 in the upper branch carrying current \( I_3 \) through the \( 100\text{ W} \) bulb.]
(a) \( I_1 \gt I_2 \gt I_3 \)
(b) \( I_1 \gt I_2 \lt I_3 \)
(c) \( I_1 \lt I_2 = I_3 \)
(d) \( I_1 \gt I_2 = I_3 \)

Answer: (b) \( I_1 \gt I_2 \lt I_3 \)

\( P = VI \), so for the same voltage, current is directly proportional to power (\( I \propto P \)). Thus, the \( 100\text{ W} \) bulb draws more current than the \( 40\text{ W} \) bulb (\( I_3 \gt I_2 \)). The total current \( I_1 \) is the sum of both branch currents (\( I_1 = I_2 + I_3 \)), making \( I_1 \) greater than both \( I_2 \) and \( I_3 \).

Teacher's Note:
a) Recall that in a parallel connection, higher wattage appliances have lower resistance and draw higher current.
b) Students often confuse series and parallel current properties; remember that total current equals the sum of branch currents.

(ii) A circular coil having \( N \) turns of radius \( R \) carries a current \( I \) and produces a magnetic field \( T \) at its centre O. If this coil is opened and rewound such that the radius of the newly formed coil is \( 2R \), and the same current \( I \) passed through it, the magnetic field at the centre O will be: [1 Mark]
(a) \( 2T \)
(b) \( T \)
(c) \( T / 2 \)
(d) \( T / 4 \)

Answer: (d) \( T / 4 \)

Magnetic field at the centre of a coil is \( B = \frac{\mu_0 NI}{2R} \). When the radius becomes \( 2R \), the number of turns becomes \( N/2 \). Thus, the new magnetic field becomes \( B' = \frac{\mu_0 (N/2) I}{2(2R)} = \frac{1}{4} \left(\frac{\mu_0 NI}{2R}\right) = \frac{T}{4} \).

Teacher's Note:
a) The total length of the wire remains constant, so when the radius is doubled, the number of turns is halved.
b) Be careful to account for both the change in radius and the change in the number of turns in such rewinding problems.

(iii) Assertion: Magnetic lines of force do not intersect each other.
Reason: At the point of intersection, the magnetic field will have two directions. [1 Mark]
(a) Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are correct, 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: (a) Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion.

A magnetic field vector has a unique direction at any given point. If two lines intersected, it would imply two tangent vectors at that point.

Teacher's Note:
a) This principle applies identically to electric field lines and streamlines in fluid dynamics.
b) Ensure students write out the complete justification rather than just the option letter in descriptive tests.

(iv) The focal length of a double convex lens is equal to the radius of curvature of either surface. The refractive index of its material is: [1 Mark]
(a) \( 3/2 \)
(b) \( 1 \)
(c) \( 4/3 \)
(d) Zero

Answer: (a) \( 3/2 \)

Using lens maker's formula: \( \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) \). Given \( f = R \), \( R_1 = R \), and \( R_2 = -R \), we get \( \frac{1}{R} = (\mu - 1)\left(\frac{1}{R} + \frac{1}{R}\right) \implies 1 = (\mu - 1)(2) \implies \mu - 1 = 0.5 \implies \mu = 1.5 \) or \( 3/2 \).

Teacher's Note:
a) Remember to apply proper sign conventions for radii of curvature of a symmetric double convex lens.
b) A refractive index of \( 1.5 \) is standard for optical glass.

(v) Which one of the following statements is correct in case of Fraunhofer's single slit diffraction experiment? [1 Mark]
(a) All bright fringes are of same intensity.
(b) Central bright fringe is the brightest.
(c) Central fringe is a dark fringe.
(d) Angular width of central fringe increases with an increase in wavelength of the incident monochromatic light.

Answer: (b) Central bright fringe is the brightest.

In single slit diffraction, intensity decreases rapidly as we move away from the central maximum to secondary maxima.

Teacher's Note:
a) Contrast this with Young's double slit interference where all bright fringes have nearly equal intensity.
b) Option (d) is incorrect because angular width is \( 2\lambda/a \), so it increases with wavelength, but statement (b) is the primary defining characteristic of single-nit intensity distribution.

(vi) The de Broglie wavelength of a moving particle varies: [1 Mark]
(a) directly with its velocity.
(b) directly with its momentum.
(c) directly with its kinetic energy.
(d) inversely with its linear momentum.

Answer: (d) inversely with its linear momentum.

De Broglie wavelength is given by \( \lambda = \frac{h}{p} \), which shows an inverse proportionality with linear momentum \( p \).

Teacher's Note:
a) Planck's constant \( h \) is the proportionality factor.
b) Students must remember alternate forms such as \( \lambda = \frac{h}{\sqrt{2mE}} \).

(vii) Which one of the following statements is correct with reference to Semiconductor Physics? [1 Mark]
(a) N type semiconductor has a majority of holes.
(b) To convert a pure semiconductor into a P type semiconductor, Phosphorous is used as a dopant.
(c) In energy band diagram of a metal, forbidden band is absent.
(d) Resistance of a semiconductor increases on increasing its temperature.

Answer: (c) In energy band diagram of a metal, forbidden band is absent.

In metals, the valence band and conduction band overlap, meaning there is no forbidden energy gap.

Teacher's Note:
a) Phosphorous is a pentavalent impurity used to make N-type semiconductors, making option (b) false.
b) Semiconductor resistance decreases with an increase in temperature due to an increase in charge carrier concentration.

(B) Answer the following questions briefly:

(i) State any one advantage of a full wave rectifier over that of a half wave rectifier. [1 Mark]

Answer:
A full wave rectifier has a higher efficiency (double that of a half wave rectifier) as it utilizes both halves of the alternating current cycle.

Teacher's Note:
a) Efficiency of a half wave rectifier is about \( 40.6\% \), whereas for a full wave rectifier it is about \( 81.2\% \).
b) Mentioning ripple frequency or ease of filtering is also accepted as an advantage.

(ii) State the condition for a balanced Wheatstone bridge. [1 Mark]

Answer:
The condition for a balanced Wheatstone bridge is \( \frac{P}{Q} = \frac{R}{S} \), where no current flows through the galvanometer connected between the intermediate junctions.

Teacher's Note:
a) When balanced, the potential difference across the galvanometer terminals is zero.
b) This principle is utilized in practical instruments like the meter bridge.

(iii) Name the rule used to determine the direction of induced current in the case of a straight conductor moving in a magnetic field. [1 Mark]

Answer:
Fleming's Right Hand Rule.

Teacher's Note:
a) Fleming's Left Hand Rule is used for force on a current-carrying conductor, whereas the Right Hand Rule is for generators and motional emf.
b) Ensure students do not confuse the two rules in exams.

(iv) With what type of source of light are plane wavefronts associated? [1 Mark]

Answer:
A point source at infinity or a distant light source (such as a star).

Teacher's Note:
a) Spherical wavefronts are associated with a point source at a finite distance.
b) Cylindrical wavefronts are produced by a linear source like a slit.

(v) Write an expression for the torque acting on an electric dipole kept in a uniform electric field in vector form. [1 Mark]

Answer:
\( \vec{\tau} = \vec{p} \times \vec{E} \)

Teacher's Note:
a) Here \( \vec{p} \) is the electric dipole moment and \( \vec{E} \) is the uniform electric field.
b) Vector notation is mandatory since the question explicitly asks for vector form.

(vi) In Young's double slit experiment, what should be the path difference between two overlapping waves to form a bright band / fringe? [1 Mark]

Answer:
The path difference must be an integral multiple of wavelength (\( \Delta x = n\lambda \), where \( n = 0, 1, 2, 3, \dots \)).

Teacher's Note:
a) Constructive interference occurs under this condition, resulting in a bright fringe.
b) For dark fringes, the path difference is an odd multiple of half the wavelength.

(vii) What are isotones? [1 Mark]

Answer:
Isotones are nuclides that have the same number of neutrons (\( N \)) but different mass numbers (\( A \)) and atomic numbers (\( Z \)).

Teacher's Note:
a) Example: \( _6^{14}\text{C} \) and \( _8^{16}\text{O} \) both have \( 8 \) neutrons.
b) Do not confuse with isotopes (same \( Z \)) or isobars (same \( A \)).

SECTION B - 14 MARKS

Question 2 [2 Marks]

(i) (a) State Gauss' theorem.
(b) What is the SI unit of electric flux?

Answer:
(a) Gauss' theorem states that the total electric flux through any closed surface is equal to \( \frac{1}{\varepsilon_0} \) times the total charge enclosed by that surface: \( \oint \vec{E} \cdot d\vec{S} = \frac{q_{\text{enclosed}}}{\varepsilon_0} \).
(b) The SI unit of electric flux is Newton metre squared per Coulomb (\( \text{N m}^2 \text{C}^{-1} \)) or Volt metre (\( \text{V m} \)).

Teacher's Note:
a) The closed surface chosen for applying Gauss' law is called a Gaussian surface.
b) Both units for electric flux are equally accepted by examiners.

OR

(ii) Calculate the equivalent capacitance \( C' \) if \( n \) identical capacitors each of capacitance \( C \) are connected in series. [2 Marks]

Answer:
For \( n \) identical capacitors in series, the equivalent capacitance \( C' \) is given by:
\( \frac{1}{C'} = \frac{1}{C} + \frac{1}{C} + \dots + \text{n times} = \frac{n}{C} \)
\( C' = \frac{C}{n} \)

Teacher's Note:
a) Capacitors in series add inversely, unlike resistors which add directly.
b) This formula is frequently used in simplifying complex bridge networks.

Question 3 [2 Marks]

A hollow charged metallic sphere of radius \( R \) carries a charge \( Q \). How much work has to be done in moving a charge \( q \) on its surface through a distance \( d \)? Explain.

Answer:
Zero work is done (\( W = 0 \)).
Explanation: The surface of a charged conducting sphere is an equipotential surface. Since every point on the surface is at the same potential, the potential difference (\( \Delta V \)) between any two points on the surface is zero. Work done is given by \( W = q \Delta V = 0 \).

Teacher's Note:
a) The electric field is always perpendicular to the surface of a conductor, so no component of force acts along the surface.
b) Emphasize the concept of equipotential surfaces in electrostatic shielding.

Question 4 [2 Marks]

(i) What is meant by pair annihilation?
(ii) Give any one example where energy is converted to matter.

Answer:
(i) Pair annihilation is the process in which an elementary particle and its antiparticle collide and destroy each other, converting their mass energy into electromagnetic radiation (photons). Example: An electron and a positron annihilate to produce two gamma-ray photons.
(ii) Example: Pair production, where a high-energy gamma-ray photon interacts with a nucleus and converts into an electron-positron pair (\( \gamma \rightarrow e^- + e^+ \)).

Teacher's Note:
a) Pair annihilation obeys conservation of energy, momentum, and charge.
b) Pair production is a direct demonstration of Einstein's mass-energy equivalence equation \( E = mc^2 \).

Question 5 [2 Marks]

(i) Figure 2 given below shows a part of Wheatstone bridge circuit. With the help of Kirchhoff's loop rule, calculate the current flowing through the \( 4\,\Omega \) resistor. [2 Marks]
[Figure: Part of a Wheatstone bridge circuit with junction A splitting into upper branch through a \( 3\,\Omega \) resistor carrying \( 2\text{ A} \) current to node B, and lower branch through a \( 4\,\Omega \) resistor. From node B, \( 1\text{ A} \) flows down through a \( 6\,\Omega \) resistor to node C, where both branches rejoin.]

Answer:
Let the current through the \( 4\,\Omega \) resistor be \( I_4 \) entering node C.
At node A, applying Kirchhoff's Current Law: total current entering is split into \( 2\text{ A} \) going through the \( 3\,\Omega \) branch and current \( I_4 \) going through the \( 4\,\Omega \) branch.
At node B, incoming current from \( 3\,\Omega \) branch is \( 2\text{ A} \). Out of this, \( 1\text{ A} \) flows down through the \( 6\,\Omega \) resistor, so the remaining \( 1\text{ A} \) must flow towards node C.
Applying KCL at node C: current from \( 4\,\Omega \) branch (\( I_4 \)) plus \( 1\text{ A} \) from the \( 6\,\Omega \) branch combine. Alternatively, applying loop rule across the lower loop: current through \( 4\,\Omega \) resistor is \( 1\text{ A} \).

Teacher's Note:
a) Kirchhoff's current law states that the algebraic sum of currents meeting at a junction is zero.
b) Careful node analysis simplifies complex network problems without needing full loop equations.

OR

(ii) In an experiment conducted to determine the internal resistance of a cell, its emf is balanced against a p.d. across \( 110\text{ cm} \) of the potentiometer wire. The balancing length becomes \( 100\text{ cm} \) when the cell is shunted by a resistance of \( 10\,\Omega \). Calculate the internal resistance of the cell. [2 Marks]

Answer:
Given: \( l_1 = 110\text{ cm} \), \( l_2 = 100\text{ cm} \), \( R = 10\,\Omega \).
Internal resistance \( r \) is given by the formula:
\( r = R \left(\frac{l_1 - l_2}{l_2}\right) \)
\( r = 10 \left(\frac{110 - 100}{100}\right) = 10 \left(\frac{10}{100}\right) = 1\,\Omega \).

Teacher's Note:
a) The potentiometer method is accurate because no current is drawn from the cell at the null point.
b) Ensure units of balancing lengths are consistent in the ratio.

Question 6 [2 Marks]

(i) On which principle do optical fibres work?
(ii) What is a thin prism?

Answer:
(i) Total Internal Reflection (TIR).
(ii) A thin prism is a prism whose refracting angle (\( A \)) is very small, typically less than \( 10^{\circ} \).

Teacher's Note:
a) For total internal reflection to take place, light must travel from an optically denser to a rarer medium, and the angle of incidence must exceed the critical angle.
b) For a thin prism, the angle of deviation is given by \( \delta = (\mu - 1)A \), which is independent of the angle of incidence.

Question 7 [2 Marks]

(i) State any one application of microwaves.
(ii) Which one of the following rays/waves is NOT an electromagnetic wave?
X rays; UV rays; \( \gamma \) rays; matter waves.

Answer:
(i) Microwaves are used in radar systems for aircraft navigation / microwave ovens for cooking.
(ii) Matter waves.

Teacher's Note:
a) Matter waves (de Broglie waves) are associated with moving particles and are mechanical/probability waves, not electromagnetic waves.
b) X-rays, UV rays, and gamma rays are all parts of the electromagnetic spectrum.

Question 8 [2 Marks]

(i) What is motional emf?
(ii) A \( 220\text{ V} \) ac voltage is to be converted to \( 33,000\text{ V} \) ac voltage. What kind / type of transformer will you use?

Answer:
(i) Motional emf is the electromotive force induced across a conductor when it moves through a magnetic field, cutting magnetic field lines.
(ii) Step-up transformer.

Teacher's Note:
a) The magnitude of motional emf is given by \( e = Blv \).
b) A step-up transformer increases voltage while decreasing current, with turns ratio \( N_s \gt N_p \).

SECTION C - 27 MARKS

Question 9 [3 Marks]

Obtain an expression for intensity of electric field at a point in broadside position of an electric dipole.

Answer:
1. Consider an electric dipole consisting of charges \( -q \) and \( +q \) separated by distance \( 2l \). Let \( P \) be a point on the broadside (equatorial) position at a distance \( r \) from the center of the dipole.
2. The electric field at point \( P \) due to charge \( +q \) is \( E_1 = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + l^2} \) along \( BP \) produced. The electric field due to charge \( -q \) is \( E_2 = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + l^2} \) along \( PA \).
3. The magnitudes of \( E_1 \) and \( E_2 \) are equal (\( E_1 = E_2 = E \)). Resolving them into components, the vertical components \( E \sin\theta \) cancel out, and the horizontal components \( E \cos\theta \) add up in the direction parallel to the dipole axis (opposite to dipole moment direction).
4. Total electric field \( E_{\text{axial}} = 2E \cos\theta = 2 \left(\frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + l^2}\right) \left(\frac{l}{\sqrt{r^2 + l^2}}\right) = \frac{1}{4\pi\varepsilon_0} \frac{q(2l)}{(r^2 + l^2)^{3/2}} \).
5. Since dipole moment \( p = q(2l) \), \( E = \frac{1}{4\pi\varepsilon_0} \frac{p}{(r^2 + l^2)^{3/2}} \). For a short dipole where \( r \gg l \), \( E = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3} \).

Teacher's Note:
a) The electric field in the broadside position is directed opposite to the direction of the dipole moment vector.
b) Compare this with the end-on (axial) position where the field is twice as strong (\( E = \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3} \)).

Question 10 [3 Marks]

(i) In Figure 3 shown below, all the cells are identical. The external resistance \( R \) is also same in both the circuits. If the same current flows through the resistor \( R \) in both the circuits, calculate the relation between \( r \), the internal resistance of each cell and the external resistance \( R \). [3 Marks]
[Figure: Circuit 1 shows two cells of emf E and internal resistance r connected in series across external resistor R. Circuit 2 shows two identical cells connected in parallel across external resistor R.]

Answer:
1. For Circuit 1 (cells in series): Total emf = \( 2E \), total internal resistance = \( 2r \). Current \( I_1 = \frac{2E}{R + 2r} \).
2. For Circuit 2 (cells in parallel): Total emf = \( E \), total internal resistance = \( r/2 \). Current \( I_2 = \frac{E}{R + r/2} = \frac{2E}{2R + r} \).
3. Given that the currents are equal (\( I_1 = I_2 \)):
\( \frac{2E}{R + 2r} = \frac{2E}{2R + r} \)
\( R + 2r = 2R + r \)
\( r = R \).

Teacher's Note:
a) Students must correctly write equivalent emf and internal resistance formulas for series and parallel cell combinations.
b) The resulting relation \( r = R \) indicates that internal and external resistances are equal under these conditions.

OR

(ii) The drift velocity of electrons in a conductor connected to a battery is given by \( v_d = -eE\tau/m \). 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.
(a) How does the drift velocity change with an increase in the potential difference across the conductor?
(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. [3 Marks]

Answer:
(a) Drift velocity increases linearly with an increase in potential difference because \( v_d \propto E \) and \( E = V/l \).
(b) When length is increased to \( 4l \) keeping voltage constant, the electric field becomes \( E' = V/(4l) = E/4 \). Since drift velocity is directly proportional to electric field, the new drift velocity becomes one-fourth of its initial value (\( v_d' = v_d / 4 \)).

Teacher's Note:
a) Emphasize that drift velocity depends on electric field intensity, not directly on the length of the conductor unless voltage is fixed.
b) Clear step-by-step reasoning is required for full credit in analytical parts.

Question 11 [3 Marks]

A galvanometer of resistance \( 100\,\Omega \) gives a full-scale deflection for a potential difference of \( 200\text{ mV} \).
(i) How will you convert this galvanometer into an ammeter of the range \( 0 - 2\text{ A} \)?
(ii) Define an ampere in terms of force between two current carrying conductors.

Answer:
(i) To convert the galvanometer into an ammeter, a very small resistance called a shunt (\( S \)) must be connected in parallel.
Given: \( G = 100\,\Omega \), full-scale voltage \( V_g = 200\text{ mV} = 0.2\text{ V} \).
Full-scale current \( I_g = \frac{V_g}{G} = \frac{0.2}{100} = 0.002\text{ A} \).
Range \( I = 2\text{ A} \).
Shunt resistance \( S = \frac{I_g G}{I - I_g} = \frac{0.002 \times 100}{2 - 0.002} = \frac{0.2}{1.998} \approx 0.1\,\Omega \) (connected in parallel).
(ii) One ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed \( 1\text{ metre} \) apart in vacuum, would produce between these conductors a force equal to \( 2 \times 10^{-7}\text{ newton per metre} \) of length.

Teacher's Note:
a) Always specify whether the shunt resistance is connected in series or parallel.
b) The SI definition of an ampere is standard and must be stated precisely with units and conditions.

Question 12 [3 Marks]

(i) (a) Draw the ray diagram of a compound microscope when the final image lies at the least distance of distinct vision, \( D \).
(b) Write an expression for magnifying power of an astronomical telescope when final image lies at infinity. (No derivation required) [3 Marks]

Answer:
(a) [Figure: Ray diagram showing objective lens forming a real, inverted, magnified image inside the focal point of the eyepiece, and eyepiece forming a virtual, magnified final image at least distance of distinct vision D.]
(b) Magnifying power of an astronomical telescope for final image at infinity: \( m = -\frac{f_o}{f_e} \).

Teacher's Note:
a) Ray diagrams must include arrowheads indicating direction of light rays and proper labeling of principal foci.
b) The negative sign in magnifying power indicates that the final image is inverted with respect to the distant object.

OR

(ii) For refraction at convex spherical surface of denser medium \( \eta_2 \) surrounded by a rarer medium of refractive index \( \eta_1 \), derive a formula connecting object distance \( u \), image distance \( v \), and radius of curvature \( R \) for the object in rarer medium and a real image formed inside the denser medium. [3 Marks]

Answer:
1. Let a point object \( O \) be placed in the rarer medium of refractive index \( \eta_1 \) in front of a spherical convex surface of radius of curvature \( R \) and refractive index \( \eta_2 \).
2. A ray starting from \( O \) is refracted at point \( N \) on the surface and forms a real image \( I \) in the denser medium.
3. From geometry of triangles, using Snell's law \( \eta_1 \sin i = \eta_2 \sin r \) and applying small angle approximations (\( \sin i \approx i \), \( \sin r \approx r \)):
\( \eta_1 (i) = \eta_2 (r) \)
4. Substituting angles in terms of exterior angles from normals: \( i = \alpha + \beta \) and \( r = \beta - \gamma \), where \( \alpha, \beta, \gamma \) are angles subtended by ray and normal at the center of curvature.
5. Substituting distances with sign convention (\( object\ distance = -u \), \( image\ distance = v \), \( radius = R \)):
\( \frac{\eta_2}{v} - \frac{\eta_1}{u} = \frac{\eta_2 - \eta_1}{R} \).

Teacher's Note:
a) This is a fundamental derivation in ray optics; students must practice sign conventions carefully.
b) The formula holds for both real and virtual images when proper sign conventions are applied.

Question 13 [3 Marks]

Using Huygen's wave theory, prove Snell's Law for refraction of light.

Answer:
1. Consider a plane wavefront \( AB \) incident obliquely on a refracting surface \( XY \) separating two media of refractive indices \( \mu_1 \) and \( \mu_2 \) with speeds of light \( v_1 \) and \( v_2 \).
2. According to Huygen's principle, every point on wavefront \( AB \) acts as a source of secondary wavelets. Let time taken by light to travel from \( B \) to \( B' \) be \( t \), so \( BB' = v_1 t \).
3. During this time, secondary wavelets from \( A \) travel a distance \( AA' = v_2 t \) in the second medium. Drawing a tangent from \( B' \) to sphere of radius \( v_2 t \) gives the refracted wavefront \( A'B' \).
4. From right-angled triangle \( ABB' \), \( \sin i = \frac{BB'}{AB'} = \frac{v_1 t}{AB'} \). From triangle \( AA'B' \), \( \sin r = \frac{AA'}{AB'} = \frac{v_2 t}{AB'} \).
5. Taking the ratio: \( \frac{\sin i}{\sin r} = \frac{v_1 t}{v_2 t} = \frac{v_1}{v_2} = \frac{c/\mu_1}{c/\mu_2} = \frac{\mu_2}{\mu_1} = { }^1\mu_2 \), which is Snell's Law.

Teacher's Note:
a) Wave theory successfully explains refraction by showing that speed of light decreases in a denser medium (\( v \propto 1/\mu \)).
b) Neat geometrical diagrams showing incident and refracted wavefronts are essential for full marks.

Question 14 [3 Marks]

(i) A certain metal emits photoelectrons with a green light but not with a yellow light. If this surface is illuminated with blue light and red light, which one of the two will cause photoelectric emission? Give a reason for your answer.
(ii) What conclusion can be drawn from this phenomenon regarding the nature of light?

Answer:
(i) Blue light will cause photoelectric emission. Reason: Green light causes emission, meaning its frequency is greater than or equal to the threshold frequency. Since frequency order is red \( \lt \) yellow \( \lt \) green \( \lt \) blue, blue light has higher frequency and shorter wavelength than green light, exceeding the threshold frequency.
(ii) This phenomenon demonstrates the particle nature of light (photons interacting with electrons in discrete packets of energy \( E = h\nu \)).

Teacher's Note:
a) Threshold frequency is the minimum frequency required for photoelectric emission to occur.
b) Wave theory failed to explain why red light (regardless of high intensity) could not eject electrons, which photoelectric equations resolved.

Question 15 [3 Marks]

A lens of focal length \( f \) is divided into two equal parts as shown in Figure 4A below. These parts are then arranged as shown in Figure 4B below. In terms of the original focal length \( f \),
(i) What is the focal length of the lens \( L_1 \)?
(ii) What is the focal length of the final combination? [3 Marks]
[Figure: Figure 4A shows a convex lens split vertically into two identical halves \( L_1 \) and \( L_2 \). Figure 4B shows the two halves arranged with their curved surfaces facing each other / joined as a combination.]

Answer:
(i) When a symmetric convex lens is cut along its principal axis into two identical halves, the focal length of each half remains unchanged. Therefore, the focal length of lens \( L_1 \) is \( f \).
(ii) In the arrangement shown in Figure 4B, two lenses of focal length \( f \) and \( f \) are placed in contact. The equivalent focal length \( F \) is given by: \( \frac{1}{F} = \frac{1}{f} + \frac{1}{f} = \frac{2}{f} \implies F = \frac{f}{2} \).

Teacher's Note:
a) Cutting a lens perpendicular to its principal axis doubles the focal length of each part, whereas cutting it along the axis leaves focal length unchanged.
b) Lenses in contact add up their optical powers directly.

Question 16 [3 Marks]

Using Ampere circuital law, obtain an expression for magnetic field \( B \) near a long current carrying straight conductor.

Answer:
1. Consider a long straight conductor carrying current \( I \). By symmetry, the magnetic field lines are concentric circles centered on the conductor.
2. Choose a circular Amperian loop of radius \( r \) centered on the conductor in a plane perpendicular to it.
3. Applying Ampere's circuital law: \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}} \).
4. Since \( B \) is uniform at all points on the circular loop and parallel to the infinitesimal element \( d\vec{l} \): \( \oint B \, dl \cos 0^{\circ} = B \oint dl = B(2\pi r) \).\br />5. Equating to \( \mu_0 I \): \( B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r} \).

Teacher's Note:
a) Ampere's circuital law is particularly useful for highly symmetric current distributions, analogous to Gauss' law in electrostatics.
b) The magnetic field is inversely proportional to the perpendicular distance \( r \) from the wire.

Question 17 [3 Marks]

Young's double-slit experiment is performed by a student. The examiner gives the following instructions to the student. State whether the responses of the student, in each case, are correct or incorrect. Give a reason for your answer.
(i) EXAMINER: The fringes on the screen are too crowded. Increase the distance between the fringes.
STUDENT increases the distance between the two slits (thinking that distance between the fringes increases on increasing the distance between slits).
(ii) EXAMINER: Bright fringes are not very bright. Increase their brightness.
STUDENT moves the source of light towards the two slits.
(iii) EXAMINER: Compare the distance between central and first maxima with the distance between second and third maxima.
STUDENT measures and states that the distance between central and first maxima is MORE than that between second and third maxima. [3 Marks]

Answer:
(i) Incorrect. Fringe width \( \beta = \frac{\lambda D}{d} \) is inversely proportional to slit separation \( d \). Increasing \( d \) decreases fringe width, making fringes more crowded.
(ii) Correct. Moving the light source closer increases intensity of light falling on the slits, thereby increasing the brightness of the bright fringes.
(iii) Incorrect. All bright fringes in Young's double slit experiment have equal width (\( \beta = \frac{\lambda D}{d} \)), so the distance between any two consecutive maxima is constant.

Teacher's Note:
a) Formula familiarity (\( \beta = \lambda D / d \)) is crucial for analyzing changes in interference patterns.
b) Students must remember that fringe width is uniform across the entire interference pattern.

SECTION D - 15 MARKS

Question 18 [5 Marks]

(i) (a) A closed loop of area \( 1\text{ m}^2 \) and resistance of \( 10\,\Omega \) is free to rotate about an axis passing through its plane and the centre. If the loop is placed at right angles to a magnetic field \( B = 0.2\text{ Wb/m}^2 \) and turned through \( 180^{\circ} \) in \( 0.02\text{ second} \), then determine (I) induced emf and (II) induced current in the loop.
(b) A wire is wound into a solenoid of length \( l \) and radius \( r \). It has a self-inductance of \( L \). By what factor does the self-inductance change, if the same wire is wound into a solenoid of half the length and half the radius?

Answer:
(a) (I) Initial flux \( \Phi_1 = BA \cos 0^{\circ} = 0.2 \times 1 = 0.2\text{ Wb} \).
Final flux after \( 180^{\circ} \) rotation \( \Phi_2 = BA \cos 180^{\circ} = -0.2\text{ Wb} \).
Change in magnetic flux \( \Delta\Phi = \Phi_2 - \Phi_1 = -0.2 - 0.2 = -0.4\text{ Wb} \).
Induced emf \( e = -\frac{\Delta\Phi}{\Delta t} = -\frac{-0.4}{0.02} = 20\text{ V} \).
(II) Induced current \( I = \frac{e}{R} = \frac{20\text{ V}}{10\,\Omega} = 2\text{ A} \).
(b) Self-inductance of a solenoid is given by \( L = \frac{\mu_0 N^2 A}{l} = \frac{\mu_0 N^2 (\pi r^2)}{l} \).
Let total wire length be \( S \approx N(2\pi r) \), so \( N \propto \frac{1}{r} \).
Substituting \( N \): \( L \propto \frac{(1/r)^2 r^2}{l} \propto \frac{1}{l} \).
When length is halved (\( l' = l/2 \)), the self-inductance becomes twice the original value (\( L' = 2L \)).

Teacher's Note:
a) Pay close attention to initial and final flux signs when a coil is turned through \( 180^{\circ} \).
b) Solenoid self-inductance scaling problems require careful substitution of turns in terms of wire dimensions.

OR

(ii) (a) In a series LCR circuit connected to an alternating voltage source of \( 220\text{ V} - 50\text{ Hz} \), the readings of voltmeters across resistor, capacitor and inductor are \( 70\text{ V} \), \( 415\text{ V} \) and \( 210\text{ V} \) respectively. If the value of the resistance \( R = 100\,\Omega \), calculate
(1) current in the circuit.
(2) value of \( L \).
(3) value of \( C \).
(b) If the capacitance in an LC series circuit is doubled, by what value should the inductance be changed to keep its resonant frequency constant? [5 Marks]

Answer:
(a) (1) Current in the circuit: \( I = \frac{V_R}{R} = \frac{70\text{ V}}{100\,\Omega} = 0.7\text{ A} \).
(2) Inductive reactance \( X_L = \frac{V_L}{I} = \frac{210}{0.7} = 300\,\Omega \).
Since \( X_L = 2\pi f L \), \( L = \frac{X_L}{2\pi f} = \frac{300}{2 \times \frac{22}{7} \times 50} = \frac{300}{314.16} \approx 0.955\text{ H} \).
(3) Capacitive reactance \( X_C = \frac{V_C}{I} = \frac{415}{0.7} \approx 592.86\,\Omega \).
Since \( X_C = \frac{1}{2\pi f C} \), \( C = \frac{1}{2\pi f X_C} = \frac{1}{2 \times \frac{22}{7} \times 50 \times 592.86} \approx 5.37 \times 10^{-6}\text{ F} \) or \( 5.37\,\mu\text{F} \).
(b) Resonant frequency is \( f = \frac{1}{2\pi\sqrt{LC}} \). For \( f \) to remain constant, \( LC = \text{constant} \implies L_1 C_1 = L_2 C_2 \).
If capacitance is doubled (\( C_2 = 2C_1 \)), inductance must be halved (\( L_2 = \frac{L_1}{2} \)).

Teacher's Note:
a) Voltmeter readings in AC circuits represent rms values.
b) Resonance condition requires \( X_L = X_C \), and resonant frequency depends inversely on the square root of \( LC \).

Question 19 [5 Marks]

(i) (a) Uranium undergoes nuclear fission as shown in the following nuclear reaction:
\( _{92}^{235}\text{U} + _0^1\text{n} \rightarrow _{92}^{236}\text{U} \rightarrow _{36}^{90}\text{Kr} + _{56}^{143}\text{Ba} + 3(_0^1\text{n}) \)
Calculate the energy released during the fission reaction.
(b) Write any one balanced equation to represent nuclear fusion.
(c) In Rutherford's \( \alpha \)-particle scattering experiment, it was observed that most of the \( \alpha \)-particles did not deflect; some showed a deflection of more than \( 90^{\circ} \) and some were deflected by even \( 180^{\circ} \).
(1) What hypothesis was made to justify the deflection of \( \alpha \)-particle by more than \( 90^{\circ} \)?
(2) State any one postulate of Bohr's theory.

Answer:
(a) Mass of reactants: Mass of \( _{92}^{235}\text{U} \) = \( 235.0439\text{ u} \); Mass of neutron = \( 1.0087\text{ u} \). Total reactant mass = \( 235.0439 + 1.0087 = 236.0526\text{ u} \).
Mass of products: Mass of \( _{36}^{90}\text{Kr} \) = \( 89.9195\text{ u} \); Mass of \( _{56}^{143}\text{Ba} \) = \( 142.9206\text{ u} \); Mass of 3 neutrons = \( 3 \times 1.0087 = 3.0261\text{ u} \). Total product mass = \( 89.9195 + 142.9206 + 3.0261 = 235.8662\text{ u} \).
Mass defect \( \Delta m = 236.0526 - 235.8662 = 0.1864\text{ u} \).
Energy released \( E = \Delta m \times 931.5\text{ MeV} = 0.1864 \times 931.5 \approx 173.6\text{ MeV} \).
(b) Example of nuclear fusion: \( _1^2\text{H} + _1^2\text{H} \rightarrow _2^3\text{He} + _0^1\text{n} + 3.27\text{ MeV} \).
(c) (1) The entire positive charge and almost all the mass of the atom are concentrated in a tiny central core called the nucleus.
(2) Electrons revolve around the nucleus only in certain stable, non-radiating circular orbits called stationary orbits where angular momentum is an integral multiple of \( h / 2\pi \).

Teacher's Note:
a) Always carry mass values to at least 4 decimal places for accurate mass defect calculations.
b) Nuclear fusion requires extremely high temperatures and pressures to overcome Coulomb repulsion between nuclei.

OR

(ii) (a)
(1) Define unified atomic mass unit.
(2) Write its energy equivalent.
(3) What is the physical significance of binding energy per nucleon of a nucleus?
(b) The wavelength of the first line of Balmer series (\( H_{\alpha} \)) is \( 656.3\text{ nm} \). Calculate the value of the Rydberg's constant. [5 Marks]

Answer:
(a) (1) One unified atomic mass unit (\( 1\text{ u} \)) is defined as \( 1/12^\text{th} \) of the mass of an unbound neutral carbon-12 atom at rest in its ground state.
(2) Energy equivalent of \( 1\text{ u} \) is \( 931.5\text{ MeV} \).
(3) Binding energy per nucleon is a measure of the stability of a nucleus; higher binding energy per nucleon means greater stability.
(b) For the first line of the Balmer series (\( H_{\alpha} \)), transition is from \( n_2 = 3 \) to \( n_1 = 2 \).
Using Rydberg formula: \( \frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \)
\( \frac{1}{656.3 \times 10^{-9}\text{ m}} = R \left(\frac{1}{2^2} - \frac{1}{3^2}\right) \)
\( \frac{1}{656.3 \times 10^{-9}} = R \left(\frac{1}{4} - \frac{1}{9}\right) = R \left(\frac{5}{36}\right) \)
\( R = \frac{36}{5 \times 656.3 \times 10^{-9}} \approx 1.097 \times 10^7\text{ m}^{-1} \).

Teacher's Note:
a) Binding energy per curve peaks around iron (\( A \approx 56 \)), explaining why fission and fusion both release energy.
b) Ensure proper unit conversion (nanometres to metres) when calculating Rydberg's constant.

Question 20 [5 Marks]

Read the passage given below and answer the questions that follow.
A rain sensor is a device used for sensing rain. They are used to automatically activate windscreen wipers to remove water from the windshields. An IR LED is used to shine infrared light onto the windshield at an angle of \( 45^{\circ} \), which is then detected using an IR photodiode.
(i) Name the two semiconductor diodes used in the rain sensor.
(ii) What kind of biasing is used in each of these diodes?
(iii) Draw the symbol of LED.

Answer:
(i) The two semiconductor diodes used are an Infrared Light Emitting Diode (IR LED) and an Infrared Photodiode.
(ii) The IR LED is forward biased (to emit infrared radiation), and the IR photodiode is reverse biased (to detect light and generate photocurrent).
(iii) Symbol of LED: A standard p-n junction diode symbol enclosed in a circle with two outward-pointing arrows representing emitted light.

Teacher's Note:
a) LEDs operate under forward bias where electrons and holes recombine, releasing energy in the form of photons.
b) Photodiodes operate under reverse bias where incident photons generate electron-hole pairs, increasing reverse saturation current.