ISC Class 12 Physics Board Exam Question Paper 2026 with Solutions

ISC Class 12 Physics Board Exam Question Paper with Solutions

SECTION A - 14 MARKS

Question 1

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

(i) A fish called an electric eel can generate a voltage up to \( 800\text{ V} \). If it can supply a current of \( 5\text{ mA} \), the maximum power it can deliver is: [1 Mark]
(a) \( 4000\text{ W} \)
(b) \( 400\text{ W} \)
(c) \( 40\text{ W} \)
(d) \( 4\text{ W} \)

Answer: (d) \( 4\text{ W} \)

\( P = V \times I = 800\text{ V} \times 5 \times 10^{-3}\text{ A} = 4\text{ W} \)

Teacher's Note:
a) Use the standard electrical power formula \( P = VI \).
b) Remember to convert milliamperes into amperes before multiplying.

(ii) A collimated beam of protons is travelling with a constant velocity along X-axis. There is a current carrying solenoid whose axis is also X-axis. If the beam now enters the solenoid along its axis, it will: [1 Mark]
(a) describe a circular path.
(b) get deflected along Z-axis.
(c) get deflected along Y-axis.
(d) continue to travel undeflected.

Answer: (d) continue to travel undeflected.

The magnetic field inside a solenoid is directed along its axis. Since the protons are travelling parallel to the axis of the solenoid, the angle between velocity and magnetic field is zero, so magnetic force is zero.

Teacher's Note:
a) Magnetic force formula is \( \vec{F} = q(\vec{v} \times \vec{B}) \).
b) When velocity and magnetic field are parallel, force is zero irrespective of charge or speed.

(iii) An electron, a proton, a deuteron and an alpha particle are all accelerated from rest through the same potential difference. Which one will have the maximum de Broglie wavelength? [1 Mark]
(a) Proton
(b) Deuteron
(c) Electron
(d) Alpha particle

Answer: (c) Electron

de Broglie wavelength is given by \( \lambda = \frac{h}{\sqrt{2mqV}} \). Since mass of the electron is smallest, its de Broglie wavelength is maximum for the same potential difference.

Teacher's Note:
a) Wavelength is inversely proportional to the square root of the particle mass when accelerating potential and charge are compared carefully, but electron has the smallest mass by orders of magnitude.
b) Do not confuse this with velocity or momentum comparisons.

(iv) Magnetic susceptibility (\( \psi \)) of a material is found to decrease with the rise in its temperature. The material is: [1 Mark]
(a) diamagnetic.
(b) paramagnetic.
(c) ferromagnetic.
(d) not magnetic.

Answer: (b) paramagnetic.

Paramagnetic susceptibility varies inversely with absolute temperature according to Curie's Law.

Teacher's Note:
a) Diamagnetic susceptibility is practically independent of temperature.
b) Ferromagnetic susceptibility follows the Curie-Weiss law above the Curie temperature.

(v) When a p-n junction diode is forward-biased: [1 Mark]
(a) it acts as an insulator and no current flows through it.
(b) it offers a low resistance and a large current flows through it.
(c) it offers a high resistance and a small current flows through it.
(d) the width of depletion region increases.

Answer: (b) it offers a low resistance and a large current flows through it.

Forward biasing reduces the potential barrier, narrowing the depletion region and allowing substantial current flow.

Teacher's Note:
a) Forward bias decreases depletion layer width.
b) Reverse bias increases depletion layer width and offers high resistance.

(vi) When a biconvex lens of glass of refractive index \( 1\cdot6 \) was dipped in a certain transparent liquid, it was found to behave like a plane sheet of glass. So, the refractive index of the liquid is: [1 Mark]
(a) \( 1\cdot0 \)
(b) \( 1\cdot4 \)
(c) \( 1\cdot5 \)
(d) \( 1\cdot6 \)

Answer: (d) \( 1\cdot6 \)

When a lens behaves like a plane sheet of glass, its focal length becomes infinity, which happens when the refractive index of the lens equals the refractive index of the surrounding medium.

Teacher's Note:
a) Lens maker formula indicates focal length becomes infinite when \( n_{lens} = n_{medium} \).
b) This is a standard zero-refraction condition in liquid immersion experiments.

(vii) Given below are two statements marked, Assertion and Reason. Read the two statements and choose the correct option. [1 Mark]
Assertion: In Young's double slit experiment, the first bright band for blue light is closer to the centre than the first bright band for green light.
Reason: Blue light has a longer wavelength than green light.
(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: (c) Assertion is true and Reason is false.

Fringe width is \( \beta = \frac{\lambda D}{d} \). Since wavelength of blue light is less than that of green light, the fringe width and position of the first bright band for blue light are smaller, making the assertion true. However, blue light has a shorter wavelength than green light, making the reason false.

Teacher's Note:
a) Verify wavelengths in the visible spectrum where violet/blue have shorter wavelengths than yellow/green/red.
b) Read both statements carefully before linking them.

(B) Answer the following questions briefly:

(i) Calculate the momentum of a photon having an energy of \( 4\cdot8 \times 10^{-19}\text{ J} \). [1 Mark]

Answer:
Momentum \( p = \frac{E}{c} = \frac{4\cdot8 \times 10^{-19}\text{ J}}{3 \times 10^8\text{ ms}^{-1}} = 1\cdot6 \times 10^{-27}\text{ kg ms}^{-1} \).

Teacher's Note:
a) Use the relation between energy and momentum of a photon: \( E = pc \).
b) Always include proper SI units in the final answer.

(ii) What is motional emf? [1 Mark]

Answer:
The potential difference induced across the ends of a conductor when it moves in a magnetic field is called motional emf.

Teacher's Note:
a) State the formula \( e = BvL \) if required for clarity.
b) Mention that it arises due to the magnetic Lorentz force acting on free electrons inside the moving conductor.

(iii) Calculate work done in moving a point charge 'q' through a distance 'd' along perpendicular bisector of an electric dipole, which consists of two-point charges -Q and +Q separated by a distance 'L'. [1 Mark]

Answer:
Zero work is done.

Teacher's Note:
a) The perpendicular bisector of an electric dipole is an equipotential surface where electric potential is zero at all points.
b) Since potential difference between any two points on the equipotential surface is zero, work done \( W = q \Delta V = 0 \).

(iv) Draw the symbol of a Zener diode. [1 Mark]

Answer:
[Figure: Standard Zener diode symbol showing a p-n junction diode triangle and bar, with the vertical ends of the cathode bar bent outwards in opposite Z-like shapes.]

Teacher's Note:
a) Ensure the bent tips on the cathode bar are clearly drawn to distinguish it from a normal junction diode.
b) Label anode and cathode terminals if necessary.

(v) A thin convex lens of power \( 5\text{D} \) is kept in contact with a thin concave lens of power \( -8\text{D} \). Calculate the focal length of the combination. [1 Mark]

Answer:
Net power \( P = P_1 + P_2 = 5\text{D} + (-8\text{D}) = -3\text{D} \).
Focal length \( f = \frac{1}{P} = \frac{1}{-3}\text{ m} = -0\cdot33\text{ m} \) or \( -33\cdot3\text{ cm} \).

Teacher's Note:
a) Power of lenses in contact is the algebraic sum of individual powers.
b) Keep track of signs for convex (positive) and concave (negative) lenses.

(vi) The image below shows a compound microscope. Which one of the two lenses \( L_1 \) or \( L_2 \) has a larger focal length? [1 Mark]
[Figure: A compound microscope line diagram showing eyepiece lens \( L_1 \) at the top near 'View' and objective lens \( L_2 \) at the bottom near 'Object'.]

Answer:
Lens \( L_1 \) (eyepiece) has a larger focal length than lens \( L_2 \) (objective).

Teacher's Note:
a) In a compound microscope, both objective and eyepiece are convex lenses, but objective has a very short focal length while eyepiece has a moderately larger focal length.
b) \( L_1 \) is the eyepiece located near the eye/view position.

(vii) Radius of the first Bohr orbit of hydrogen atom is \( 0\cdot05\text{nm} \). Calculate the radius of the third orbit. [1 Mark]

Answer:
Radius of orbit is given by \( r_n = r_1 n^2 \).
For the third orbit (\( n = 3 \)):
\( r_3 = 0\cdot05\text{ nm} \times (3)^2 = 0\cdot05 \times 9 = 0\cdot45\text{ nm} \).

Teacher's Note:
a) Radius is directly proportional to the square of the principal quantum number (\( n^2 \)).
b) Substitute \( n = 3 \) correctly.

SECTION B - 14 MARKS

Question 2 [2 Marks]
Four-point charges \( Q_1 = +17\cdot7\mu\text{C} \), \( Q_2 = -8\cdot85\mu\text{C} \), \( Q_3 = -17\cdot7\mu\text{C} \) and \( Q_4 = 35\cdot4\mu\text{C} \) are kept as shown in Figure 1 below.
[Figure: An elliptical closed surface S enclosing charges \( Q_1 \) and \( Q_2 \), while charges \( Q_3 \) and \( Q_4 \) are outside the closed surface S. Specifically, inside surface S: \( Q_1 = +17\cdot7\mu\text{C} \), \( Q_2 = -8\cdot85\mu\text{C} \); outside surface S: \( Q_3 = -17\cdot7\mu\text{C} \) and \( Q_4 = 35\cdot4\mu\text{C} \).]
Calculate electric flux emanating from the closed surface S.

Answer:
According to Gauss's Law, electric flux depends only on the total charge enclosed by the surface.
Enclosed charge \( q_{enclosed} = Q_1 + Q_2 = +17\cdot7\mu\text{C} - 8\cdot85\mu\text{C} = +8\cdot85\mu\text{C} = 8\cdot85 \times 10^{-6}\text{ C} \).
Electric flux \( \phi = \frac{q_{enclosed}}{\varepsilon_0} = \frac{8\cdot85 \times 10^{-6}}{8\cdot85 \times 10^{-12}} = 10^6\text{ N m}^2\text{C}^{-1} \).

Teacher's Note:
a) Charges outside the closed surface do not contribute to the electric flux through the surface.
b) Use the exact value of permittivity of free space \( \varepsilon_0 = 8\cdot85 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2} \).

OR

(ii) (a) Find the value of current 'I' in Figure 2 given below. [2 Marks]
[Figure: A junction network showing incoming currents \( 8\text{A} \), \( 7\text{A} \), current 'I' and outgoing currents \( 5\text{A} \), \( 3\text{A} \).]
(b) State the conservation principle on which your calculation is based.

Answer:
(a) Applying Kirchhoff's Current Law at the junction (sum of incoming currents = sum of outgoing currents):
\( 8 + 7 + I = 5 + 3 \)
\( 15 + I = 8 \)
\( I = 8 - 15 = -7\text{ A} \) (or magnitude of \( 7\text{ A} \) away from junction depending on arrow direction, treating incoming as positive: \( 8 + 7 + I = 5 + 3 \) gives \( I = -7\text{ A} \)).
(b) The calculation is based on the **Law of Conservation of Charge**.

Teacher's Note:
a) Carefully observe incoming and outgoing arrows at the junction node.
b) Kirchhoff's First Law is a direct consequence of charge conservation.

Question 3 [2 Marks]
You are provided with three identical capacitors, each of capacitance 'C'. If \( C_p \) is equivalent capacitance when they are connected in parallel and \( C_s \) is equivalent capacitance when they are connected in series, calculate the ratio \( \frac{C_s}{C_p} \).

Answer:
For three identical capacitors in parallel: \( C_p = C + C + C = 3C \).
For three identical capacitors in series: \( \frac{1}{C_s} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C} \Rightarrow C_s = \frac{C}{3} \).
Ratio \( \frac{C_s}{C_p} = \frac{C/3}{3C} = \frac{1}{9} \).

Teacher's Note:
a) Parallel combination formula is \( C_p = nC \) and series is \( C_s = C/n \).
b) Ensure the ratio order requested in the question (\( C_s/C_p \)) is strictly maintained.

Question 4 [2 Marks]
(i) What is the effect of doubling the current flowing through a metallic wire on:
(a) drift speed of free electrons?
(b) relaxation time of free electrons?

Answer:
(a) Drift speed (\( v_d \)) is directly proportional to current (\( I \)) since \( I = nAe v_d \). Therefore, doubling the current **doubles** the drift speed.
(b) Relaxation time depends on temperature and atomic structure of the material, not directly on current. Therefore, relaxation time **remains unchanged** (assuming temperature is constant).

Teacher's Note:
a) Relate drift velocity using macroscopic current formula \( I = nAv_de \).
b) Clarify that microscopic parameters like relaxation time depend on temperature.

OR

(ii) Draw a labelled circuit diagram of a potentiometer used to compare emfs of two cells X and Y. (Procedure and formula are NOT required.) [2 Marks]

Answer:
[Figure: A potentiometer circuit diagram showing a driver cell connected across a long uniform resistance wire AB through a rheostat and a plug key. The positive terminals of cells X and Y are connected to end A through a two-way key, and their negative terminals are connected through a galvanometer and a jockey to wire AB.]

Teacher's Note:
a) Ensure the positive terminals of the driver cell and experimental cells X and Y are connected to the same terminal A.
b) Label the driver cell, rheostat, key, galvanometer, and jockey clearly.

Question 5 [2 Marks]
(i) Write an expression of Biot-Savart law in vector form.
(ii) State Ampere's circuital law.

Answer:
(i) \( d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \hat{r})}{r^2} \) or \( d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \vec{r})}{r^3} \).
(ii) Ampere's circuital law states that the line integral of magnetic field \( \vec{B} \) around any closed loop is equal to \( \mu_0 \) times the total net current passing through the surface bounded by the loop: \( \oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enclosed} \).

Teacher's Note:
a) Vector form must clearly show the cross product between current element vector and position vector.
b) Mention the enclosed current term explicitly in Ampere's law statement.

Question 6 [2 Marks]
A bi-convex lens of focal length \( f_1 \) is placed in air as shown in **Figure 3(a)** below. The radii of curvature of its first and second surfaces are \( R \) and \( 2R \) respectively. The lens is cut along the plane CD. Compare the focal length \( f_2 \) of the resulting lens shown in **Figure 3(b)** with that of the original lens shown in **Figure 3(a)**.
[Figure 3(a): A symmetric/asymmetric biconvex lens with surfaces having radii of curvature R (left) and 2R (right), cut along vertical plane CD through the optical axis AB.]
[Figure 3(b): Resulting plano-convex lens with a flat curved surface on one side and radius 2R on the other side.]

Answer:
Using lens maker's formula for original biconvex lens:
\( \frac{1}{f_1} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (\mu - 1) \left(\frac{1}{R} - \frac{1}{-2R}\right) = (\mu - 1) \left(\frac{3}{2R}\right) \)
For the cut plano-convex lens (Figure 3(b)), first surface is flat (\( R_1 = \infty \)) and second surface has radius \( R_2 = -2R \):
\( \frac{1}{f_2} = (\mu - 1) \left(\frac{1}{\infty} - \frac{1}{-2R}\right) = (\mu - 1) \left(\frac{1}{2R}\right) \)
Comparing the two:
\( \frac{1}{f_1} = 3 \left(\frac{1}{f_2}\right) \Rightarrow f_2 = 3f_1 \).

Teacher's Note:
a) Cutting a symmetric lens vertically produces two identical plano-convex lenses.
b) The focal length becomes three times the original focal length.

Question 7 [2 Marks]
(i) Name the electromagnetic wave that scans the contents of the luggage of a traveller during a security check.
(ii) Which physical quantity is the same for UV rays, red light and radio waves when they are travelling in vacuum?

Answer:
(i) X-rays.
(ii) Speed (or velocity of light in vacuum, \( c = 3 \times 10^8\text{ ms}^{-1} \)).

Teacher's Note:
a) X-rays have high penetrating power and are used in airport baggage scanners.
b) All electromagnetic waves travel with the exact same speed in vacuum.

Question 8 [2 Marks]
The work functions for metals \( M_1 \) and \( M_2 \) are \( 1\cdot9\text{ eV} \) and \( 5\cdot0\text{ eV} \) respectively. Perform necessary calculations to find out which metal emits photoelectrons, when monochromatic light of wavelength \( 410\text{nm} \) is incident on them.

Answer:
Energy of incident photon:
\( E = \frac{hc}{\lambda} = \frac{6\cdot63 \times 10^{-34} \times 3 \times 10^8}{410 \times 10^{-9}}\text{ J} = \frac{1\cdot989 \times 10^{-25}}{410 \times 10^{-9}} = 4\cdot85 \times 10^{-19}\text{ J} \)
Converting energy into electron-volts:
\( E = \frac{4\cdot85 \times 10^{-19}}{1\cdot6 \times 10^{-19}}\text{ eV} = 3\cdot03\text{ eV} \).
Since incident photon energy (\( 3\cdot03\text{ eV} \)) is greater than the work function of metal \( M_1 \) (\( 1\cdot9\text{ eV} \)) but less than the work function of metal \( M_2 \) (\( 5\cdot0\text{ eV} \)), **Metal \( M_1 \)** will emit photoelectrons.

Teacher's Note:
a) Compare incident photon energy directly with individual work functions.
b) Photoemission occurs only when \( E \gt \phi \).

SECTION C - 27 MARKS

Question 9 [3 Marks]
Show that the intensity of electric field at a point in end on position i.e., axial position of an electric dipole is given by:
\( E = \left(\frac{1}{4\pi\varepsilon_0}\right) \frac{2pr}{(r^2 - l^2)^2} \)
where the terms have their usual meaning.

Answer:
Consider an electric dipole consisting of charges \( -q \) and \( +q \) separated by distance \( 2l \). Let point P be at distance \( r \) from the center of the dipole on its axial line.
Electric field due to charge \( +q \) at P: \( E_1 = \frac{1}{4\pi\varepsilon_0} \frac{q}{(r - l)^2} \) (directed away from dipole).
Electric field due to charge \( -q \) at P: \( E_2 = \frac{1}{4\pi\varepsilon_0} \frac{q}{(r + l)^2} \) (directed towards dipole).
Net electric field \( E = E_1 - E_2 = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{(r - l)^2} - \frac{1}{(r + l)^2} \right] \)
\( E = \frac{q}{4\pi\varepsilon_0} \left[ \frac{(r + l)^2 - (r - l)^2}{(r^2 - l^2)^2} \right] = \frac{q}{4\pi\varepsilon_0} \left[ \frac{4rl}{(r^2 - l^2)^2} \right] \)
Since dipole moment \( p = q \times 2l \):
\( E = \frac{1}{4\pi\varepsilon_0} \frac{2pr}{(r^2 - l^2)^2} \).

Teacher's Note:
a) Clearly specify the directions of electric field vectors due to both charges.
b) Substitute dipole moment \( p = 2ql \) correctly in the final step.

Question 10 [3 Marks]
(i) In the circuit shown in Figure 4 below, how much resistance should be connected to a \( 10\Omega \) resistor so that the points M and N are the same potential? [3 Marks]
[Figure 4: A bridge circuit connected across a 2V source. Upper branch has resistors \( 3\Omega \) and \( 6\Omega \) with node M between them. Lower branch has resistors \( 4\Omega \) and \( 10\Omega \) with node N between them.]

Answer:
For points M and N to be at the same potential, the bridge must be balanced (Wheatstone bridge condition).
The ratio of resistances in the upper branch equals the ratio in the lower branch:
\( \frac{R_1}{R_2} = \frac{R_3}{R_4} \)
Here, \( R_1 = 3\Omega \), \( R_2 = 6\Omega \), \( R_3 = 4\Omega \), and let the required parallel or series connected resistance be \( R_x \).
If the \( 10\Omega \) resistor is replaced or adjusted to satisfy balance: \( \frac{3}{6} = \frac{4}{R_{new}} \Rightarrow R_{new} = \frac{24}{3} = 8\Omega \).
Since there is already a \( 10\Omega \) resistor, let a resistance \( R \) be connected in parallel across the \( 10\Omega \) resistor such that equivalent resistance of that arm becomes \( 8\Omega \):
\( \frac{10R}{10 + R} = 8 \Rightarrow 10R = 80 + 8R \Rightarrow 2R = 80 \Rightarrow R = 40\Omega \).

Teacher's Note:
a) Recognize the circuit as a Wheatstone bridge when points M and N are at the same potential (no current through potential-equalizing branch).
b) Calculate the required arm resistance and find the parallel combination value.

OR

(ii) Figure 5 below shows a battery consisting of three cells. Their emfs and internal resistances are also shown. [3 Marks]
[Figure 5: Three cells connected in parallel. Cell 1: \( 2\text{V}, 1\Omega \); Cell 2: \( 4\text{V}, 3\Omega \); Cell 3: \( 6\text{V}, 12\Omega \).]
Calculate:
(a) emf of the battery.
(b) internal resistance of the battery.

Answer:
(a) Equivalent emf of parallel combination of cells is given by:
\( E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2} + \frac{E_3}{r_3}}{\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}} \)
Substituting values (\( E_1 = 2, r_1 = 1; E_2 = 4, r_2 = 3; E_3 = 6, r_3 = 12 \)):
Numerator = \( \frac{2}{1} + \frac{4}{3} + \frac{6}{12} = 2 + 1\cdot33 + 0\cdot5 = 3\cdot83 \text{ A} \) (exact fraction: \( \frac{24 + 16 + 6}{12} = \frac{46}{12} \)).
Denominator = \( \frac{1}{1} + \frac{1}{3} + \frac{1}{12} = \frac{12 + 4 + 1}{12} = \frac{17}{12} \).
\( E_{eq} = \frac{46/12}{17/12} = \frac{46}{17}\text{ V} \approx 2\cdot71\text{ V} \).
(b) Equivalent internal resistance \( r_{eq} \):
\( \frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{17}{12} \Rightarrow r_{eq} = \frac{12}{17}\Omega \approx 0\cdot71\Omega \).

Teacher's Note:
a) Use standard formulas for parallel combination of multiple cells with internal resistances.
b) Keep fractions till the final calculation step for high accuracy.

Question 11 [3 Marks]
(i) Obtain an expression for magnetic field (B) at the centre of a circular coil having 'n' turns of radius 'R' when it is carrying a current 'I'.

Answer:
According to Biot-Savart law, magnetic field due to a small current element \( Id\vec{l} \) at distance \( R \) is:
\( dB = \frac{\mu_0}{4\pi} \frac{I dl \sin 90^{\circ}}{R^2} = \frac{\mu_0}{4\pi} \frac{I dl}{R^2} \)
Integrating around the complete circular loop of length \( 2\pi R \):
\( B = \int dB = \frac{\mu_0 I}{4\pi R^2} \int dl = \frac{\mu_0 I}{4\pi R^2} (2\pi R) = \frac{\mu_0 I}{2R} \)
For a coil with \( n \) turns:
\( B = \frac{\mu_0 n I}{2R} \).

Teacher's Note:
a) The angle between current element and position vector is \( 90^{\circ} \) everywhere on a circle.
b) Multiply by the number of turns \( n \) to get the total magnetic field.

OR

(ii) (a) Two moving coil galvanometers \( G_1 \) and \( G_2 \) are identical except for the following features: [3 Marks]

Perform calculations to determine which galvanometer has greater current sensitivity.
(b) In a moving coil galvanometer, how is magnetic field made radial?

Answer:
(a) Current sensitivity \( I_s = \frac{NBA}{k} \). Assuming magnetic field \( B \) and restoring torque per unit twist \( k \) are the same for both:
Sensitivity ratio is proportional to \( N \times A \).
For \( G_1 \): \( N_1 A_1 = 50 \times 30\text{ cm}^2 = 1500\text{ cm}^2 \).
For \( G_2 \): \( N_2 A_2 = 60 \times 40\text{ cm}^2 = 2400\text{ cm}^2 \).
Since \( N_2 A_2 \gt N_1 A_1 \), galvanometer \( G_2 \) has greater current sensitivity.
(b) The magnetic field is made radial by using a cylindrical soft iron core placed symmetrically inside the coil and using concave pole pieces of the magnet.

Teacher's Note:
a) Current sensitivity formula is \( I_s = \frac{NBA}{k} \). Resistance does not directly affect current sensitivity unless voltage sensitivity is asked.
b) Radial magnetic field ensures that the deflecting torque remains maximum for all orientations of the coil.

Question 12 [3 Marks]
Figure 6 below shows a long straight conductor X carrying a current \( I_1 \). A point P is at a perpendicular distance \( r_1 \) from it.
[Figure 6: Three parallel long straight conductors X, Y, Z carrying currents \( I_1 \), \( I \), \( I_2 \) respectively, with perpendicular distances \( r_1 \) and \( r_2 \).]
(i) How much force conductor X exerts on a short conductor Y of length 'l' which is carrying a current 'I' and is kept at point P parallel to the conductor X?
(ii) Another long wire Z carrying a current \( I_2 \) is now kept parallel to X and Y at a distance \( r_2 \) from Y such that the conductor Y remains at rest. Obtain the relation between the currents \( I_1 \) and \( I_2 \).

Answer:
(i) Magnetic field at P due to conductor X: \( B_1 = \frac{\mu_0 I_1}{2\pi r_1} \).
Force on conductor Y of length \( l \) carrying current \( I \): \( F = I l B_1 = \frac{\mu_0 I I_1 l}{2\pi r_1} \).
(ii) For conductor Y to remain at rest, the magnetic force due to conductor Z must be equal and opposite to the force due to conductor X.
\( \frac{\mu_0 I I_1 l}{2\pi r_1} = \frac{\mu_0 I I_2 l}{2\pi r_2} \Rightarrow \frac{I_1}{r_1} = \frac{I_2}{r_2} \Rightarrow I_2 = I_1 \left(\frac{r_2}{r_1}\right) \).

Teacher's Note:
a) Use the standard formula for force between two parallel current-carrying conductors: \( F = \frac{\mu_0 I_1 I_2 l}{2\pi r} \).
b) For equilibrium, opposing magnetic forces must balance out.

Question 13 [3 Marks]
A monochromatic ray of light incident on one refracting surface of an equilateral prism, suffers a deviation as shown in the **Figure 7** below:
[Figure 7: An equilateral prism mounted on a prism table with collimator and telescope positions. Angle of prism \( A = 60^{\circ} \), angle of deviation \( \delta = 40^{\circ} \).]
(i) Calculate the refractive index of HB material of the prism.
(ii) What is meant by dispersive power of a transparent material?

Answer:
(i) For an equilateral prism, angle of prism \( A = 60^{\circ} \). Given deviation \( \delta = 40^{\circ} \). Assuming minimum deviation condition or standard prism formula where angle of incidence/deviation data corresponds to angle of minimum deviation \( \delta_m = 40^{\circ} \):
Refractive index \( \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{60^{\circ} + 40^{\circ}}{2}\right)}{\sin\left(\frac{60^{\circ}}{2}\right)} = \frac{\sin(50^{\circ})}{\sin(30^{\circ})} = \frac{0\cdot7660}{0\cdot5} = 1\cdot532 \).
(ii) Dispersive power of a transparent material is defined as the ratio of angular dispersion between the extreme colours (violet and red) to the mean deviation (for yellow colour) produced by the prism.

Teacher's Note:
a) Apply Snell's prism formula with \( A = 60^{\circ} \).
b) Dispersive power formula is \( \omega = \frac{\delta_v - \delta_r}{\delta_y} = \frac{n_v - n_r}{n_y - 1} \).

Question 14 [3 Marks]
Draw a labelled diagram of an astronomical telescope when its final image lies at infinity.

Answer:
[Figure: Ray diagram of an astronomical telescope in normal adjustment (image at infinity). Shows a large aperture objective lens and a small aperture eyepiece lens with parallel rays entering the objective and emerging parallel from the eyepiece, forming an inverted final image at infinity, with tube length \( L = f_o + f_e \).]

Teacher's Note:
a) Label both objective and eyepiece lenses clearly with their respective focal lengths \( f_o \) and \( f_e \).
b) Show arrows on light rays to indicate direction of propagation.

Question 15 [3 Marks]
Kabir and Neha conducted experiments on light as part of their school project. Both of them worked with the same source of monochromatic yellow light. Both obtained patterns on the screen which consisted of alternate bright and dark bands.
The images below depict their results.
[Figure: Kabir's result shows many closely spaced narrow alternative bright and dark fringes. Neha's result shows very few, wide alternative bright and dark fringes with central maximum significantly broader.]
(i) Identify the phenomenon observed by Neha.
(ii) The teacher asked Kabir to replace monochromatic yellow light with white light. What would be Kabir's observation now?
(iii) What conclusion would be drawn by Kabir and Neha about the nature of light?

Answer:
(i) Diffraction of light.
(ii) Kabir will observe a central white fringe surrounded by colored (rainbow-colored) interference bands because fringe width depends on wavelength.
(iii) They would conclude that light exhibits wave nature (interference and diffraction).

Teacher's Note:
a) Wide central maximum with alternating secondary maxima is characteristic of single-slit diffraction.
b) White light illumination produces a central white fringe since all wavelengths overlap at the center.

Question 16 [3 Marks]
Using Huygen's wave theory and a labelled diagram, show that angle of reflection (r) is equal to angle of incidence (i).

Asset:
Answer:
[Figure: Huygen's construction for reflection of a plane wavefront from a reflecting surface. Shows incident wavefront AB striking a plane mirror at angle of incidence i, and reflected wavefront CD making angle of reflection r.]
Proof: Consider a plane wavefront AB incident on a reflecting surface MM'. Let \( v \) be the speed of wave in the medium. Time taken by wave to travel from B to C is \( t = \frac{BC}{v} \). In the same time \( t \), secondary wavefronts from A spread out to D, such that \( AD = vt = BC \).
In triangles ABC and ADC:
- \( BC = AD = vt \) (since sound/light covers equal distance in equal time)
- \( AC = AC \) (common hypotenuse)
- \( \angle ABC = \angle ADC = 90^{\circ} \)
Therefore, \( \Delta ABC \cong \Delta ADC \) by RHS congruence.
Consequently, angle of incidence \( \angle i = \angle BAC = \angle DCA \) and angle of reflection \( r = \angle ACD = \angle DAC \), proving that angle of incidence equals angle of reflection (\( i = r \)).

Teacher's Note:
a) Draw clear wavefronts and normal lines in the geometrical construction diagram.
b) State the congruence of triangles clearly to complete the proof.

Question 17 [3 Marks]
(i) The graphs below show variation of stopping potential versus frequency of incident radiation for metals A and B. [3 Marks]
[Figure: Graph of Stopping potential versus Frequency showing two parallel straight lines for Metal A and Metal B with threshold frequencies \( f_{oA} \) and \( f_{oB} \).]
UV radiation of appropriate wavelength is allowed to fall on both the metals. Which metal will emit photoelectrons with higher maximum kinetic energy (\( E_{max} \))? Give a reason.
(ii) State the conclusion that was drawn from Davisson-Germer's experiment.

Answer:
(i) Metal A will emit photoelectrons with higher maximum kinetic energy.
Reason: According to Einstein's photoelectric equation \( eV_s = hf - \phi \), stopping potential \( V_s \) is higher for Metal A at any given frequency because its threshold frequency (\( f_{oA} \)) and work function are smaller than that of Metal B.
(ii) Davisson-Germer experiment conclusively proved the wave nature of electrons (de Broglie hypothesis of matter waves) by demonstrating electron diffraction.

Teacher's Note:
a) Maximum kinetic energy is given by \( K_{max} = h(f - f_o) \).
b) Lower threshold frequency implies higher kinetic energy for a fixed incident frequency.

SECTION D - 15 MARKS

Question 18 [5 Marks]
(i) (a) When a coil is connected to a \( 200\text{V} \) dc supply, the current flowing through it is found to be \( 1\text{A} \). However, when it is connected to the \( 200\text{V}, 50\text{Hz} \) ac supply, the current is found to be \( 0\cdot5\text{A} \).
(1) Explain why current flowing is less when the coil is connected to an ac supply.
(2) Calculate coefficient of self-inductance (L) of the coil.
(b) If the power factor in an ac circuit is \( 0\cdot5 \), what is the phase difference between the voltage and the current in the circuit?

Answer:
(a)(1) In a dc supply, frequency is zero, so inductive reactance is zero and only the ohmic resistance of the coil limits the current. In an ac supply, the inductive reactance also offers opposition, increasing the total impedance of the circuit, which reduces the current.
(a)(2) Resistance of coil \( R = \frac{V}{I_{dc}} = \frac{200}{1} = 200\Omega \).
Impedance with ac supply \( Z = \frac{V}{I_{ac}} = \frac{200}{0\cdot5} = 400\Omega \).
Since \( Z = \sqrt{R^2 + X_L^2} \):
\( 400 = \sqrt{(200)^2 + X_L^2} \Rightarrow 160000 = 40000 + X_L^2 \Rightarrow X_L^2 = 120000 \Rightarrow X_L = \sqrt{120000} = 346\cdot4\Omega \).
Since \( X_L = 2\pi f L \):
\( L = \frac{346\cdot4}{2 \times 3\cdot14 \times 50} = \frac{346\cdot4}{314} = 1\cdot1\text{ H} \).
(b) Power factor \( \cos\phi = 0\cdot5 \Rightarrow \phi = \cos^{-1}(0\cdot5) = 60^{\circ} \) (or \( \frac{\pi}{3}\text{ radians} \)).

Teacher's Note:
a) Differentiate between dc resistance and ac impedance.
b) Calculate inductive reactance using \( X_L = \sqrt{Z^2 - R^2} \).

OR

(ii) (a) A \( 70\Omega \) resistor is connected to an ac source generating an emf 'e' given by \( e(V) = 495\sin(100\pi t) \). Calculate rms value of the current (\( I_{rms} \)) flowing through the resistor. [5 Marks]
(b) A \( 20\text{cm} \) length of a long iron cored solenoid has \( 25 \) turns. If the area of cross section of the solenoid is \( 1 \times 10^{-4}\text{ m}^2 \), calculate its coefficient of self-inductance (L).
(c) State why soft iron is preferred to steel as a material for the core of a transformer.

Answer:
(a) Peak emf \( E_0 = 495\text{ V} \).
RMS emf \( E_{rms} = \frac{E_0}{\sqrt{2}} = \frac{495}{1\cdot414} = 350\text{ V} \).
RMS current \( I_{rms} = \frac{E_{rms}}{R} = \frac{350}{70} = 5\text{ A} \).
(b) Self-inductance of a solenoid \( L = \frac{\mu_0 \mu_r N^2 A}{l} \).
Given \( l = 0\cdot2\text{ m} \), \( N = 25 \), \( A = 1 \times 10^{-4}\text{ m}^2 \), relative permeability of iron \( \mu_r = 3000 \) (from table).
\( L = \frac{(4\pi \times 10^{-7}) \times 3000 \times (25)^2 \times (1 \times 10^{-4})}{0\cdot2} \)
\( L = \frac{1\cdot256 \times 10^{-6} \times 3000 \times 625 \times 10^{-4}}{0\cdot2} = \frac{0\cdot2355}{0\cdot2} = 1\cdot18 \times 10^{-3}\text{ H} \) (or \( 1\cdot18\text{ mH} \)).
(c) Soft iron has high magnetic permeability, low retentivity, and low hysteresis loss compared to steel.

Teacher's Note:
a) Use peak voltage coefficient from the sine equation to find rms voltage.
b) Include relative permeability of iron from the constants table given at the end of the paper.

Question 19 [5 Marks]
Answer either subparts (i) to (iii) or (iv) to (vii).
(i) Draw a labelled graph showing the variation of binding energy per nucleon with the mass number (A) of the nucleus. On it, mark the nucleus that is most stable.
(ii) What is meant by the following statement? 'Angular momentum of an orbiting electron is quantised.'
(iii) Calculate the shortest wavelength of Balmer series.

Answer:
(i) [Figure: Graph of binding energy per nucleon versus mass number A showing a sharp rise for light nuclei, a broad peak around \( A = 56 \) (Iron, \( ^{56}\text{Fe} \)), and a gradual decrease for heavy nuclei.]
(ii) It means that the orbital angular momentum of an electron can only take specific discrete values that are integral multiples of \( \frac{h}{2\pi} \): \( L = mvr = \frac{nh}{2\pi} \).
(iii) For Balmer series, shortest wavelength corresponds to transition from \( n = \infty \) to \( n = 2 \):
\( \frac{1}{\lambda} = R_H \left(\frac{1}{2^2} - \frac{1}{\infty^2}\right) = R_H \left(\frac{1}{4}\right) \Rightarrow \lambda = \frac{4}{R_H} \).
Given Rydberg constant \( R_H = 1\cdot097 \times 10^7\text{ m}^{-1} \):
\( \lambda = \frac{4}{1\cdot097 \times 10^7} = 3\cdot646 \times 10^{-7}\text{ m} \) (or \( 364\cdot6\text{ nm} \)).

Teacher's Note:
a) Iron (\( ^{56}\text{Fe} \)) is the most stable nucleus with maximum binding energy per nucleon (\( \approx 8\cdot8\text{ MeV} \)).
b) Shortest wavelength in any spectral series corresponds to the series limit transition from infinity.

OR

(iv) Calculate the minimum amount of energy which a gamma ray photon should possess in order to produce a proton and an anti-proton pair. [5 Marks]
(v) Why is a nuclear fusion reaction also known as a thermonuclear reaction?
(vi) What is the physical significance of the fact that total energy of an orbiting electron is negative?
(vii) State any one limitation of Bohr's theory of hydrogen atom.

Answer:
(iv) Mass of proton \( m_p = 1\cdot67 \times 10^{-27}\text{ kg} \). Mass of anti-proton is the same.
Total rest mass energy required: \( E = 2 m_p c^2 \).
In terms of energy equivalent: rest mass energy of one nucleon \( \approx 931\text{ MeV} \), so for proton and anti-proton pair: \( 2 \times 938\text{ MeV} = 1876\text{ MeV} \) (or \( 1\cdot88\text{ GeV} \)).
(v) Nuclear fusion requires extremely high temperatures (of the order of \( 10^8\text{ K} \)) to overcome electrostatic repulsion between nuclei, hence it is called thermonuclear reaction.
(vi) Negative total energy signifies that the electron is bound to the nucleus by attractive electrostatic forces and cannot escape to infinity without external energy.

(vii) Bohr's theory fails to explain the spectra of multi-electron atoms (it works only for hydrogen-like single electron systems).

Teacher's Note:
a) Pair production requires energy equal to at least twice the rest mass energy of the particle.
b) High temperature provides thermal kinetic energy needed to initiate nuclear fusion.

Question 20 [5 Marks]
Read the scenario given below and answer the questions that follow.
Annie, a student of Physics, began her experiment by fabricating a p-n junction diode from an extrinsic semiconductor crystal, leading to the formation of a depletion region and a potential barrier at the interface. She then measured resistance of the diode in the linear region of the V-I characteristic curve. She also studied its characteristics during forward bias and reverse bias. Finally, she designed a circuit to convert ac voltage to dc voltage.
(i) Which instrument was used by Annie to measure current flowing through the diode during its reverse bias?
(ii) How did Annie obtain an extrinsic semiconductor from an intrinsic semiconductor?
(iii) What is meant by potential barrier?
(iv) Draw a labelled circuit diagram of a half wave rectifier.

Answer:
(i) Microammeter (or milliammeter sensitive to small reverse currents).
(ii) By adding a small, suitable impurity (dopant) to an intrinsic semiconductor (such as adding group V or group III elements to silicon/germanium).
(iii) The potential difference built across the p-n junction due to the diffusion of charge carriers and formation of an immobile space-charge region is called the potential barrier.
(iv) [Figure: Circuit diagram of a half wave rectifier showing an ac input source connected to a step-down transformer, a p-n junction diode in series with a load resistance \( R_L \), and output dc voltage across \( R_L \).]

Teacher's Note:
a) Reverse saturation current in diodes is very small, hence sensitive microammeters are used.
b) Doping creates n-type or p-type extrinsic semiconductors.