ISC Class 12 Physics Board Exam Question Paper 2025 with Solutions

ISC Class 12 Physics Board Exam Question Paper with Solutions

SECTION A

Question 1

(A) In questions (i) to (vii) given below, choose the correct alternative (a), (b), (c) or (d). [7 Marks]

(i) When a battery is connected between the terminals P and Q, as shown in Figure 1 below, it is found that no current flows through \(5\,\Omega\) resistor. Then, the value of resistor X is: [1 Mark]
[Figure: Wheatstone bridge circuit with four arms containing \(10\,\Omega\), \(10\,\Omega\), \(25\,\Omega\) and \(X\), with a central \(5\,\Omega\) resistor connected between terminals P and Q. Figure 1]
(A) \(10\,\Omega$
(B) \(20\,\Omega$
(C) \(25\,\Omega$
(D) \(45\,\Omega$

Answer: (C) \(25\,\Omega$

Using the balanced Wheatstone bridge condition, \(\frac{X}{25} = \frac{10}{10}\), which gives \(X = 25\,\Omega\).

Teacher's Note:
a) Recall that no current flows through the central resistor when the bridge is balanced.
b) Do not confuse the position of the resistors when applying the ratio of opposite arms.

(ii) The relative permeability of substance 'X' is slightly less than one and that of substance 'Y' is slightly more than one. Then: [1 Mark]
(A) 'X' is paramagnetic and 'Y' is ferromagnetic.
(B) 'X' is diamagnetic and 'Y' is ferromagnetic.
(C) 'X' is paramagnetic and 'Y' is diamagnetic.
(D) 'X' is diamagnetic and 'Y' is paramagnetic.

Answer: (D) 'X' is diamagnetic and 'Y' is paramagnetic.

Diamagnetic substances have relative permeability slightly less than one, whereas paramagnetic substances have it slightly greater than one.

Teacher's Note:
a) Remember that for diamagnetic materials \(\mu_r \lt 1\), for paramagnetic \(\mu_r \gt 1\), and for ferromagnetic \(\mu_r \gg 1$.
b) Pay close attention to the terms slightly less and slightly more.

(iii) If kinetic energy of moving electrons is made four times, then their de Broglie wavelength becomes: [1 Mark]
(A) eight times.
(B) four times.
(C) two times.
(D) half.

Answer: (D) half.

The de Broglie wavelength is given by \(\lambda = \frac{h}{\sqrt{2mK}}\). Since \(\lambda \propto \frac{1}{\sqrt{K}}\), making \(K\) four times makes wavelength half.

Teacher's Note:
a) Always use the relation between de Broglie wavelength and kinetic energy: \(\lambda = \frac{h}{\sqrt{2mK}}\).
b) A common mistake is relating wavelength inversely to \(K\) directly instead of its square root.

(iv) A student has made connections, as shown in Figure 2 below, so that the wires MN and ST repel each other. But it is observed that they are attracting each other. What change should the student make for the wires to repel each other? [1 Mark]
[Figure: Two parallel vertical wires MN and ST connected in separate circuits with power sources \(E_1\) and \(E_2\) and rheostats \(R_1\) and \(R_2\). Figure 2]
(A) Reverse the terminals of batteries \(E_1\) and \(E_2$.
(B) Reverse the terminals of battery \(E_1\) or \(E_2$.
(C) Choose supply voltage such that \(E_1 = E_2$.
(D) Add key and ammeter to the circuit.

Answer: (B) Reverse the terminals of battery \(E_1\) or \(E_2\).

Reversing one battery reverses the direction of current in one wire, changing the force from attractive to repulsive.

Teacher's Note:
a) Parallel wires carrying currents in opposite directions repel each other.
b) Reversing both currents keeps them in the same relative direction, maintaining attraction.

(v) N-type semiconductor is that which has: [1 Mark]
(A) majority of holes as charge carriers.
(B) majority of free electrons as charge carriers.
(C) trivalent element added as an impurity.
(D) an equal number of holes and free electrons.

Answer: (B) majority of free electrons as charge carriers.

N-type semiconductors are doped with pentavalent impurities, making free electrons the majority charge carriers.

Teacher's Note:
a) Recall that pentavalent impurities provide extra electrons.
b) Do not confuse N-type with P-type, which has holes as majority carriers.

(vi) Given below are two statements marked, Assertion and Reason. Read the two statements and choose the correct option. [1 Mark]
Assertion: When a convex lens made of glass is completely immersed in water, its focal length increases.
Reason: Refractive index of glass with respect to water is greater than that of glass with respect to air.
(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) Both Assertion and Reason are false.

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

The refractive index of glass with respect to water is smaller than that with respect to air, making the lens less bending and thus increasing focal length.

Teacher's Note:
a) Use Lens Maker's Formula to analyze the dependence of focal length on surrounding medium.
b) Note that the relative refractive index decreases when immersed in water, not increases.

(vii) Given below are two statements marked, Assertion and Reason. Read the two statements and choose the correct option. [1 Mark]
Assertion: Diffraction of light is difficult to observe in everyday situations but can be observed in laboratory conditions.
Reason: To produce diffraction of waves, size of an obstacle must be comparable to the wavelength of the waves.
(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) Both Assertion and Reason are false.

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

Everyday obstacles are much larger than the tiny wavelength of visible light, making diffraction negligible outside labs.

Teacher's Note:
a) Diffraction requires obstacle size to be comparable to wavelength.
b) Visible light has a very small wavelength, requiring specially arranged narrow slits.

(B) Answer the following questions briefly:

(i) How does the resistance of a semiconductor crystal vary with its temperature? [1 Mark]

Answer:
The resistance of a semiconductor crystal decreases with an increase in temperature due to an increase in the number of free charge carriers.

Teacher's Note:
a) Semiconductors have a negative temperature coefficient of resistance.
b) Higher temperature breaks more covalent bonds, increasing conductivity.

(ii) Figure 3 below shows an ideal transformer. Explain why current flowing through secondary coil is greater than that in primary coil. [1 Mark]
[Figure: Schematic diagram of an ideal transformer with primary coil input and secondary coil output. Figure 3]

Answer:
In a step-down transformer, since the output voltage is less than the input voltage, the secondary current must be greater to keep the input and output power constant (\(V_p I_p = V_s I_s\)).

Teacher's Note:
a) Apply the principle of conservation of energy for an ideal transformer.
b) State clearly that power remains constant in an ideal transformer.

(iii) State any one difference between a primary rainbow and a secondary rainbow. [1 Mark]

Answer:
A primary rainbow has red on the outer edge and violet on the inner edge, whereas a secondary rainbow has the order reversed (violet on the outer edge and red on the inner edge).

Teacher's Note:
a) Primary rainbow is formed due to one internal reflection and two refractions.
b) Secondary rainbow involves two internal reflections.

(iv) Hubble telescope employs a large parabolic mirror as an objective. State any one advantage of using a mirror in place of a lens in such a telescope. [1 Mark]

Answer:
Mirrors are completely free from chromatic aberration because they reflect all wavelengths of light equally.

Teacher's Note:
a) Mention absence of chromatic aberration as the primary optical advantage.
b) Other valid points include mechanical ease of supporting large mirrors from behind.

(v) Name any one phenomenon where moving particles behave like waves. [1 Mark]

Answer:
Electron diffraction.

Teacher's Note:
a) Davisson and Germer experiment demonstrated electron diffraction.
b) This confirms de Broglie's hypothesis of matter waves.

(vi) What is the minimum energy a gamma ray (\(\gamma\)) photon should possess to produce an electron - positron pair? [1 Mark]

Answer:
The minimum energy is \(1.022\,\text{MeV}\).

Teacher's Note:
a) This energy corresponds to the combined rest mass energy of an electron and a positron ($2 \times 0.511\,\text{MeV}$).
b) This process is known as pair production.

(vii) In an energy band diagram of a certain material, forbidden band is absent. Identify this material. [1 Mark]

Answer:
Conductor (or metal).

Teacher's Note:
a) In conductors, the valence band and conduction band overlap, so there is no forbidden gap.
b) This allows free movement of electrons even at room temperature.

SECTION B

Question 2 [2 Marks]

(i) (a) What is the effect on capacitance of a parallel plate capacitor if the distance between its plates is increased?

Answer:
The capacitance decreases because capacitance is inversely proportional to the distance between the plates (\(C = \frac{\varepsilon_0 A}{d}\)).

Teacher's Note:
a) State the formula \(C = \frac{\varepsilon_0 A}{d}\) clearly.
b) Explain the inverse proportionality between \(C\) and \(d\).

(b) How will capacitance of a capacitor change if a dielectric slab is introduced between its plates?

Answer:
The capacitance increases by a factor equal to the dielectric constant (\(k\)) of the slab.

Teacher's Note:
a) Mention that \(C' = k C\).
b) Mention that dielectric constant \(k\) is always greater than 1 for dielectric media.

OR

(ii) Figures 4 and 5 represent the combination of two identical cells having negligible internal resistance. [2 Marks]
[Figure: Two cells of emf E connected in series (Figure 4) and in parallel (Figure 5) with terminals P and Q. Figures 4 and 5]
(a) In which of the two combinations, emf of the battery is greater?

Answer:
The emf is greater in Figure 4 (series combination), where total emf is \(2E\).

Teacher's Note:
a) In series, emfs add up algebraically.
b) In parallel, the equivalent emf remains equal to that of a single cell.

(b) When a resistor 'R' is connected between the terminals P and Q, \(I_1\) and \(I_2\) are the currents flowing through 'R' in Figure 4 and Figure 5 respectively. Obtain the ratio \(\frac{I_1}{I_2}\).

Answer:
For Figure 4, \(I_1 = \frac{2E}{R}\).
For Figure 5, \(I_2 = \frac{E}{R}\).
Therefore, the ratio \(\frac{I_1}{I_2} = 2:1\).

Teacher's Note:
a) Assume negligible internal resistance as given in the problem.
b) Apply Ohm's law directly for both circuit configurations.

Question 3 [2 Marks]

In case of a short electric dipole:
(i) What is the locus of a point having zero potential?

Answer:
The locus is the equatorial plane passing through the center of the dipole, perpendicular to the dipole axis.

Teacher's Note:
a) Electric potential due to a dipole is zero at all points on the equatorial line/plane.
b) Specify that it is perpendicular to the dipole axis.

(ii) If electric field intensity at a point in axial position is \(E_1\) and at an equidistant point in equatorial position \(E_2\), what is the ratio \(\frac{E_1}{E_2}\)?

Answer:
\(E_{\text{axial}} = \frac{1}{4\pi\varepsilon_0} \frac{2p}{r^3}\) and \(E_{\text{equatorial}} = \frac{1}{4\pi\varepsilon_0} \frac{p}{r^3}\).
Therefore, the ratio \(\frac{E_1}{E_2} = 2:1\).

Teacher's Note:
a) Recall standard formulas for axial and equatorial electric fields of a short dipole.
b) Note that the axial field is twice the magnitude of the equatorial field at the same distance.

Question 4 [2 Marks]

(i) Figure 6 below shows an electric circuit. [2 Marks]
[Figure: Resistor network connected to a 10 V source with nodes P, Q, R, S and E, showing resistors of \(3\,\Omega\), \(4\,\Omega\), \(1\,\Omega\), \(4\,\Omega\) and \(2\,\Omega\). Figure 6]
Apply Kirchhoff's laws to calculate:
(a) current 'I' flowing through the \(2\,\Omega\) resistor.

Answer:
Applying Kirchhoff's Current Law (KCL) at junction Q: \(I + 3\text{A} = 4\text{A}\), which gives \(I = 1\text{A}\).

Teacher's Note:
a) Use KCL at junction Q where total entering current equals total leaving current.
b) Check the node currents carefully from the given diagram.

(b) emf of the cell 'E'.

Answer:
Applying Kirchhoff's Voltage Law (KVL) in loop PQRSP: \((3\text{A} \times 4\,\Omega) + (4\text{A} \times 1\,\Omega) - E - 10\text{V} = 0\).
\(12\text{V} + 4\text{V} - 10\text{V} = E\), so \(E = 6\text{V}\).

Teacher's Note:
a) Follow a closed loop and sum up potential drops and rises correctly.
b) Ensure proper sign conventions for resistor voltage drops and battery emfs.

OR

(ii) In a potentiometer experiment, a cell of emf \(1.25\,\text{V}\) gives a balance point at \(35\,\text{cm}\) mark of the wire. If this cell is replaced by another cell, the balance point is at \(63\,\text{cm}\) mark. Calculate the emf of the second cell. [2 Marks]

Answer:
Using potentiometer relation: \(\frac{E_1}{E_2} = \frac{L_1}{L_2}\).
\(\frac{1.25}{E_2} = \frac{35}{63}\), which gives \(E_2 = 1.25 \times \frac{63}{35} = 2.25\,\text{V}\).

Teacher's Note:
a) Emf is directly proportional to the balancing length in a potentiometer.
b) Ensure units of length are consistent on both sides.

Question 5 [2 Marks]

Magnetic field at the centre of a circular coil is B. Calculate the magnetic field at the same point when each of the current, number of turns of the coil and its radius is doubled.

Answer:
Initial magnetic field \(B = \frac{\mu_0 NI}{2R}\).
When \(N' = 2N\), \(I' = 2I\), and \(R' = 2R\):
\(B' = \frac{\mu_0 (2N)(2I)}{2(2R)} = 2 \left(\frac{\mu_0 NI}{2R}\right) = 2B\).

Teacher's Note:
a) Write the original formula for the magnetic field at the center of a circular coil.
b) Substitute scaled variables carefully to find the new field.

Question 6 [2 Marks]

The objective of a telescope consists of two lenses kept in contact. One lens has an optical power of \(+2.0\,\text{D}\) whereas the other has an optical power of \(-1.5\,\text{D}\). Calculate focal length of the objective.

Answer:
Total power of lenses in contact \(P_{\text{total}} = P_1 + P_2 = +2.0\,\text{D} - 1.5\,\text{D} = +0.5\,\text{D}\).
Focal length \(f = \frac{1}{P_{\text{total}}} = \frac{1}{0.5} = 2.0\,\text{m}\).

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

Question 7 [2 Marks]

(i) Name the electromagnetic wave travelling from the satellite to the dish antenna in the image above. [1 Mark]
[Figure: Satellite transmitting signals to a ground dish antenna. Figure in Q7]

Answer:
Microwaves.

Teacher's Note:
a) Microwaves are used for satellite communication because they can easily penetrate the atmosphere.
b) Ensure correct spelling of the wave type.

(ii) If the wavelength of the dish antenna shown in the image is \(6\,\text{nm}\), what is its frequency? [1 Mark]

Answer:
Using \(f = \frac{c}{\lambda} = \frac{3 \times 10^8}{6 \times 10^{-9}} = 5.0 \times 10^{16}\,\text{Hz}\).

Teacher's Note:
a) Convert nanometers to meters using \(1\,\text{nm} = 10^{-9}\,\text{m}\).
b) Use standard speed of light \(c = 3 \times 10^8\,\text{m/s}\).

Question 8 [2 Marks]

With reference to photoelectric effect, define the terms:
(i) Threshold frequency. [1 Mark]

Answer:
Threshold frequency is the minimum frequency of incident light required to eject electrons from the surface of a metal.

Teacher's Note:
a) Denoted usually by \(\nu_0\).
b) Below this frequency, no photoelectric emission takes place regardless of intensity.

(ii) Work function. [1 Mark]

Answer:
Work function is the minimum energy required to eject an electron from the surface of a metal.

Teacher's Note:
a) Denoted usually by \(\phi_0\).
b) It is characteristic of the metal surface.

SECTION C

Question 9 [3 Marks]

An infinite plane metallic sheet having surface charge density \(+\sigma\) is placed in vacuum. P is a point at a small distance \(r\) to its right.
(i) Write an expression for intensity of electric field at point P. [1 Mark]

Answer:
\(E = \frac{\sigma}{2\varepsilon_0}\), directed away from the sheet.

Teacher's Note:
a) Derived using Gauss's law for a thin infinite sheet of charge.
b) Note that the electric field is independent of distance \(r\).

(ii) Now, an identical charged sheet having surface charge density \(-\sigma\) is placed parallel to the first sheet such that \(-\sigma\) is placed parallel to the first sheet such that \(+\sigma\) is placed parallel to the first sheet such that \(-\sigma\) is placed to its left at the same distance \(r\). (The point P lies between the two plates.)
(a) What is the resultant intensity of electric field at point P? [1 Mark]

Answer:
\(E_{\text{resultant}} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}\).

Teacher's Note:
a) Electric fields due to both sheets point in the same direction between the plates.
b) Add the individual field magnitudes directly.

(b) What is its direction? [1 Mark]

Answer:
Directed towards the right (from the positively charged sheet to the negatively charged sheet).

Teacher's Note:
a) Field lines originate on positive charge and terminate on negative charge.
b) Both field vectors point towards the right.

Question 10 [3 Marks]

(i) Harry sets up a circuit as shown in Figure 7 below. He measures potential difference \(V\) across the variable resistor \(R\) with an instrument Y. He also measures current \(I\) flowing through \(R\) with another instrument X. [1 Mark]
[Figure: Circuit diagram showing battery with internal resistance, variable resistor R, and two meters X and Y. Figure 7]
(a) Identify the instruments X and Y.

Answer:
Instrument X is an Ammeter and Instrument Y is a Voltmeter.

Teacher's Note:
a) Ammeters are connected in series to measure current.
b) Voltmeters are connected in parallel to measure potential difference.

(b) Using the graph of V against I shown below, calculate: [2 Marks]
[Figure: V-I graph with voltage intercept at 2.5 V and current intercept at 0.5 A. Figure in Q10(ii)]
(1) emf (E) of the cell.

Answer:
From the graph, when \(I = 0\), \(V = 2.5\,\text{V}\). Thus, emf \(E = 2.5\,\text{V}\).

Teacher's Note:
a) The intercept on the voltage axis represents the emf of the cell.
b) Read the intercept value carefully from the given axes.

(2) Internal resistance (r) of the cell.

Answer:
Using \(V = E - Ir\), rearranging gives \(r = \frac{E - V}{I}\).
From the graph at \(I = 0.5\,\text{A}\), \(V = 0\,\text{V}\).
\(r = \frac{2.5 - 0}{0.5} = 5\,\Omega\).

Teacher's Note:
a) Internal resistance is equal to the magnitude of the slope of the V-I graph.
b) Alternatively, use any coordinate point on the line where current and voltage are known.

OR

(ii) (a) The graph below shows the variation of current density (j) with electric field (E) applied to two different metallic wires A and B. [2 Marks]
[Figure: j-E graph with two lines A and B of different slopes. Figure in Q10(ii) OR]
Which one of the wires, A or B, has higher resistivity?

Answer:
Wire A has a lower slope, meaning lower conductivity and thus a higher resistivity \(\rho\).

Teacher's Note:
a) Ohm's law in microscopic form is \(j = \sigma E\), where slope gives conductivity \(\sigma\).
b) Resistivity is the reciprocal of conductivity (\(\rho = \frac{1}{\sigma}\)).

(b) For a certain value of electric field (E), in which wire 'A' or 'B' is drift velocity greater? Give a reason for your answer. [1 Mark]

Answer:
Drift velocity is greater in wire B because it has a higher current density for the same electric field.

Teacher's Note:
a) Use the relation \(j = nq v_d\).
b) Higher current density at the same electric field implies higher drift velocity for a constant charge carrier density \(n\).

Question 11 [3 Marks]

A galvanometer having a resistance of \(20\,\Omega\) shows a full scale deflection with a current of \(1\,\text{mA}\). How can it be converted into a voltmeter with a range of \(0-10\,\text{V}\)?

Answer:
Given \(G = 20\,\Omega\), \(I_g = 1\,\text{mA} = 10^{-3}\,\text{A}\), \(V = 10\,\text{V}\).
To convert a galvanometer into a voltmeter, a high resistance \(R\) is connected in series.
\(V = I_g(G + R)\)
\(R = \frac{V}{I_g} - G = \frac{10}{10^{-3}} - 20 = 10,000 - 20 = 9,980\,\Omega\).

Teacher's Note:
a) A voltmeter requires a very high resistance in series to limit current.
b) Remember to convert milliamperes into amperes before calculation.

Question 12 [3 Marks]

(i) Two infinitely long straight wires PQ and RS carrying currents \(I_1\) and \(I_2\) respectively are kept \(10\,\text{cm}\) apart in vacuum. Calculate magnetic field (B) at the point X shown in Figure 8 below. [3 Marks]
[Figure: Two parallel vertical wires PQ carrying 4A and RS carrying 16A, separated by 10 cm, with point X located 2 cm from PQ. Figure 8]

Answer:
Given \(I_1 = 4\,\text{A}\), \(I_2 = 16\,\text{A}\), distance between wires \(d = 10\,\text{cm} = 0.1\,\text{m}\).
Distance of point X from PQ: \(d_1 = 2\,\text{cm} = 0.02\,\text{m}\).
Distance of point X from RS: \(d_2 = 10 - 2 = 8\,\text{cm} = 0.08\,\text{m}\).
Magnetic field due to PQ at X: \(B_1 = \frac{\mu_0 I_1}{2\pi d_1} = \frac{4\pi \times 10^{-7} \times 4}{2\pi \times 0.02} = 4 \times 10^{-5}\,\text{T}\) (into the page).
Magnetic field due to RS at X: \(B_2 = \frac{\mu_0 I_2}{2\pi d_2} = \frac{4\pi \times 10^{-7} \times 16}{2\pi \times 0.08} = 4 \times 10^{-5}\,\text{T}\) (out of the page).
Net magnetic field \(B_{\text{net}} = B_1 - B_2 = 4 \times 10^{-5} - 4 \times 10^{-5} = 0\,\text{T}\).

Teacher's Note:
a) Use right-hand rule to find the directions of magnetic fields from both current-carrying wires.
b) Since fields are in opposite directions at point X, their magnitudes subtract.

OR

(ii) An electron and a proton are moving along the +X axis. If an external magnetic field \(B = 0.314\,\text{T}\) is applied along -Z axis: [3 Marks]
(a) What is the path followed by the electron due to the magnetic field?

Answer:
Circular path in the XY plane.

Teacher's Note:
a) Apply Fleming's left hand rule or Lorentz force formula to determine the direction of force.
b) A charge moving perpendicular to a uniform magnetic field traces a circular trajectory.

(b) Calculate the frequency of revolution of the proton.

Answer:
Cyclotron frequency formula: \(f = \frac{qB}{2\pi m_p}\).
\(f = \frac{1.6 \times 10^{-19} \times 0.314}{2 \times 3.14 \times (1.67 \times 10^{-27})} = \frac{5.024 \times 10^{-20}}{10.486 \times 10^{-27}} = 4.79 \times 10^6\,\text{Hz}\).

Teacher's Note:
a) Frequency of revolution is independent of particle velocity for non-relativistic speeds.
b) Use mass of proton \(m_p = 1.67 \times 10^{-27}\,\text{kg}\).

Question 13 [2 Marks]

A convex lens having small focal length is to be used as a magnifying glass (simple microscope) to obtain an image of a small diamond. If the image lies at least distance of distinct vision (D):
(i) Where will you keep the diamond to obtain its image at D? [1 Mark]

Answer:
Between the optical center and the principal focus of the convex lens.

Teacher's Note:
a) Object distance must be less than the focal length of the convex lens.
b) This produces a virtual and magnified image.

(ii) State any two characteristics of the image formed by the magnifying glass. [1 Mark]

Answer:
1. Virtual and erect.
2. Magnified (enlarged).

Teacher's Note:
a) The image is formed on the same side of the lens as the object.
b) It cannot be caught on a screen.

Question 14 [3 Marks]

A parallel beam of light is travelling obliquely from an optically rarer medium to an optically denser medium.
(i) Draw a labelled diagram showing incident and refracted wavefronts. Mark angle of incidence as 'i' and angle of refraction as 'r'. [1 Mark]

Answer:
[Figure: Diagram showing incident wavefront AB in rarer medium 1 and refracted wavefront CD in denser medium 2, with angle of incidence i and angle of refraction r marked.]

Teacher's Note:
a) Ensure wavefronts are drawn perpendicular to corresponding light rays.
b) Label all boundary lines, media, and angles clearly.

(ii) Use Huygen's wave theory to prove Snell's law. [2 Marks]

Answer:
Let \(v_1\) and \(v_2\) be the speeds of light in medium 1 and medium 2 respectively. In time \(t\), distance travelled by secondary wavelets is \(BC = v_1 t\) and \(AD = v_2 t$.
From triangle ABC: \(\sin i = \frac{BC}{AC} = \frac{v_1 t}{AC}\).
From triangle ADC: \(\sin r = \frac{AD}{AC} = \frac{v_2 t}{AC}\).
Dividing both: \(\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = {^1}\mu_2\) (a constant), which proves Snell's law.

Teacher's Note:
a) Geometry of wavefront construction is crucial for this proof.
b) State clearly that refractive index equals the ratio of speeds in the two media.

Question 15 [3 Marks]

For any prism, show that refractive index 'n' of its material is given by:
\(n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\)
where the terms have their usual meaning.

Answer:
At minimum deviation, \(\angle r_1 = \angle r_2 = \angle r\).
We know \(A = r_1 + r_2 = 2r\), so \(r = \frac{A}{2}\).
Also, angle of deviation \(\delta = i_1 + i_2 - A\). At minimum deviation \(\delta_m = 2i - A\), giving \(i = \frac{A + \delta_m}{2}\).
Applying Snell's law, refractive index \(n = \frac{\sin i}{\sin r} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\).

Teacher's Note:
a) Remember the condition for minimum deviation: angle of incidence equals angle of emergence.
b) Step-by-step substitution of \(i\) and \(r\) into Snell's law completes the derivation.

Question 16 [3 Marks]

Two identical rectangular slits \(5\,\text{mm}\) apart are illuminated with a monochromatic light of wavelength \(600\,\text{nm}\). The screen is kept \(1.2\,\text{m}\) away from the slits.
(i) Calculate the distance between the \(5^{\text{th}}\) bright fringe (band) on one side and the \(3^{\text{rd}}\) bright fringe on the other side of the central bright band. [2 Marks]

Answer:
Given \(d = 5\,\text{mm} = 5 \times 10^{-3}\,\text{m}\), \(\lambda = 600\,\text{nm} = 600 \times 10^{-9}\,\text{m}\), \(D = 1.2\,\text{m}\).
Fringe width \(\beta = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1.2}{5 \times 10^{-3}} = 1.44 \times 10^{-4}\,\text{m} = 0.144\,\text{mm}\).
Position of \(5^{\text{th}}\) bright fringe: \(y_5 = 5\beta = 5 \times 0.144 = 0.72\,\text{mm}\).
Position of \(3^{\text{rd}}\) bright fringe on opposite side: \(y_3 = 3\beta = 3 \times 0.144 = 0.432\,\text{mm}\).
Total distance = \(y_5 + y_3 = 0.72 + 0.432 = 1.152\,\text{mm}\).

Teacher's Note:
a) Fringe width formula is \(\beta = \frac{\lambda D}{d}\).
b) Add the distances on opposite sides of the central maximum to get the total separation.

(ii) What will be the change in the interference pattern if the given light is replaced with monochromatic light of wavelength \(500\,\text{nm}\)? [1 Mark]

Answer:
Since fringe width is directly proportional to wavelength (\(\beta \propto \lambda\)), reducing the wavelength to \(500\,\text{nm}\) will decrease the fringe width, making the fringes closer together.

Teacher's Note:
a) State the proportional dependence between fringe width and wavelength.
b) Conclude that the overall interference pattern becomes more compact.

Question 17 [3 Marks]

(i) Plot a labelled graph of stopping potential (\(V_s\)) versus frequency (\(\nu\)) of incident UV radiation. [2 Marks]

Answer:
[Figure: Graph of stopping potential \(V_s\) on Y-axis versus frequency \(\nu\) on X-axis, showing a straight line with a positive slope and an intercept \(-\frac{\phi_0}{e}\) on the negative Y-axis.]

Teacher's Note:
a) The graph is a straight line starting above the threshold frequency \(\nu_0\).
b) Label axes, intercept, and threshold frequency clearly.

(ii) State how the value of Planck's constant can be determined from this graph. [1 Mark]

Answer:
The Planck's constant \(h\) is determined by multiplying the slope of the \(V_s\) versus \(\nu\) graph by the electronic charge \(e\) (\(h = \text{slope} \times e\)).

Teacher's Note:
a) Einstein's photoelectric equation gives \(eV_s = h\nu - \phi_0\), hence slope is \(\frac{h}{e}\).
b) Calculate slope using any two points on the linear graph.

SECTION D

Question 18 [5 Marks]

(i) (a) Figure 9 shows a metallic rod AB of length \(l = 3\,\text{m}\) moving in an (external uniform magnetic field, \(B = \frac{1}{\pi}\,\text{T}\)) which is directed into the plane of this paper. [2 Marks]
[Figure: Metallic rod AB of length l moving with velocity v perpendicular to uniform magnetic field directed into the page. Figure 9]
If the position of the rod changes with time as \(x = \pi t\), where \(x\) is in metre and \(t\) is in second, then:
(1) Calculate the motional emf developed in the rod.

Answer:
Velocity \(v = \frac{dx}{dt} = \frac{d}{dt}(\pi t) = \pi\,\text{m/s}\).
Motional emf \(\varepsilon = Blv = \left(\frac{1}{\pi}\right) \times 3 \times \pi = 3\,\text{V}\).

Teacher's Note:
a) Velocity is the time derivative of position \(x\).
b) Use the standard motional emf formula \(\varepsilon = Blv\).

(2) Name the law used to find the direction of induced current. [1 Mark]

Answer:
Fleming's Right Hand Rule (or Lenz's law).

Teacher's Note:
a) Fleming's Right Hand Rule relates magnetic field, motion, and induced current direction.
b) Lenz's law is based on energy conservation.

(b) When current flowing through a solenoid decreases from \(15\,\text{A}\) to \(0\) in \(0.2\,\text{s}\), an emf of \(30\,\text{V}\) is induced in it. Calculate the coefficient of self-inductance of the solenoid. [2 Marks]

Answer:
Using Faraday's law of electromagnetic induction: \(\varepsilon = L \frac{dI}{dt}\).
\(30 = L \left(\frac{15 - 0}{0.2}\right)\)
\(30 = L \times 75\)
\(L = \frac{30}{75} = 0.4\,\text{H}\).

Teacher's Note:
a) Self-inductance measures opposition to a change in current in the same coil.
b) Substitute magnitude of rate of change of current properly.

OR

(ii) An alternating emf \(E = 5.0\sin(314\,t)\,\text{V}\) is applied to a circuit containing a resistor connected in series with an unknown component X. The current in the circuit is found to be \(I = 3.0\sin\left(314\,t - \frac{\pi}{3}\right)\,\text{A}\). [5 Marks]
(a) Identify the component X. [1 Mark]

Answer:
Since the current lags the voltage by \(\frac{\pi}{3}\), component X must be an inductor (L).

Teacher's Note:
a) In a purely inductive or RL circuit, current lags voltage.
b) Compare phase angles of given voltage and current equations.

(b) Calculate rms value (\(I_{\text{rms}}\)) of the current flowing through the circuit. [1 Mark]

Answer:
Peak current \(I_0 = 3.0\,\text{A}\).
\(I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{3.0}{1.414} = 2.12\,\text{A}\).

Teacher's Note:
a) RMS value is peak value divided by \(\sqrt{2}\).
b) Take peak current directly from the coefficient of the sine term.

(c) Find the frequency of the source. [1 Mark]

Answer:
Angular frequency \(\omega = 314\,\text{rad/s}\).
\(\omega = 2\pi f \implies 314 = 2 \times 3.14 \times f \implies f = 50\,\text{Hz}\).

Teacher's Note:
a) Standard frequency of AC supply in India is \(50\,\text{Hz}\).
b) Use \(\omega = 2\pi f\) for conversion.

(d) Calculate power factor. [1 Mark]

Answer:
Phase angle \(\phi = \frac{\pi}{3}\).
Power factor \(\cos\phi = \cos\left(\frac{\pi}{3}\right) = \cos(60^{\circ}) = 0.5\).

Teacher's Note:
a) Power factor is defined as \(\cos\phi\).
b) Phase difference \(\phi\) is obtained from the current equation.

(e) Find the impedance (Z) of the circuit. [1 Mark]

Answer:
Peak voltage \(E_0 = 5.0\,\text{V}\), peak current \(I_0 = 3.0\,\text{A}\).
\(Z = \frac{E_0}{I_0} = \frac{5.0}{3.0} = 1.67\,\Omega\).

Teacher's Note:
a) Impedance in AC circuits is the effective opposition given by \(Z = \frac{E_0}{I_0}\) or \(\frac{E_{\text{rms}}}{I_{\text{rms}}}\).
b) Keep calculations to two decimal places.

Question 19 [5 Marks]

(i) (a) Show that radius (\(r_n\)) of the \(n^{\text{th}}\) Bohr orbit varies directly with square of the principal quantum number (\(n\)) of the orbit. [3 Marks]

Answer:
Electrostatic force provides the necessary centripetal force for an electron in orbit:
\(\frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{Ke^2}{r^2} \implies v^2 = \frac{ke^2}{mr}\) ----(1)
Using Bohr's quantization condition for angular momentum:
\(L = mvr = \frac{nh}{2\pi} \implies v = \frac{nh}{2\pi mr}\) ----(2)
Substituting equation (2) into equation (1):
\(\left(\frac{nh}{2\pi mr}\right)^2 = \frac{ke^2}{mr} \implies \frac{n^2 h^2}{4\pi^2 m^2 r^2} = \frac{ke^2}{mr}\)
Solving for \(r\):
\(r_n = \frac{n^2 h^2}{4\pi^2 m k e^2}\).
Since \(h\), \(\pi\), \(m\), \(k\), and \(e\) are constants, \(r_n \propto n^2\).

Teacher's Note:
a) Combine Coulomb's law and Bohr's angular momentum quantization condition.
b) Clearly show the cancellation of terms to arrive at the final proportional relation.

(b) (1) Where does nuclear fusion reaction take place continuously in the Universe? [1 Mark]

Answer:
In stars, including our Sun.

Teacher's Note:
a) Stellar energy is powered by hydrogen fusion into helium.
b) High temperature and pressure are required for fusion.

(2) What is meant by "Mass defect" of a nucleus? [1 Mark]

Answer:
Mass defect is the difference between the sum of the masses of individual nucleons and the actual mass of the nucleus.

Teacher's Note:
a) Formula: \(\Delta m = [Zm_p + (A - Z)m_n] - M\).
b) This missing mass is converted into binding energy.

OR

(ii) (a) A group of students went on an educational tour to Bhabha Atomic Research Centre, Mumbai. They visited various nuclear reactors like Apsara, Cirus, Zerlina and Dhruv and observed that Apsara was a swimming pool type reactor. The teacher explained how water slows down fast moving neutrons and absorbs the heat produced in the reactor. [3 Marks]
(1) What is the use of a nuclear reactor?

Answer:
To generate electricity by controlling nuclear fission reactions, produce radioactive isotopes, and conduct scientific research.

Teacher's Note:
a) Nuclear reactors harness controlled chain reactions.
b) Mention both power generation and research/medical applications.

(2) What is the name given to a material which slows down fast moving neutrons?

Answer:
Moderator (e.g., heavy water, graphite).

Teacher's Note:
a) Moderators slow down neutrons to thermal energies to increase fission probability.
b) Common moderators are graphite and heavy water.

(3) State how a nuclear reactor can be shut down in case of an emergency.

Answer:
By inserting control rods (made of neutron-absorbing materials like boron or cadmium) completely into the reactor core.

Teacher's Note:
a) Control rods absorb excess neutrons to halt the chain reaction.
b) Boron and cadmium are standard control rod materials.

(b) (1) What is meant by the statement: 'Angular momentum of an orbiting electron is quantised'? [1 Mark]

Answer:
It means that an electron cannot have any arbitrary value of angular momentum; instead, its angular momentum is restricted to integral multiples of \(\frac{h}{2\pi}\).

Teacher's Note:
a) Expressed as \(mvr = \frac{nh}{2\pi}\).
b) This was a fundamental postulate of Bohr's atomic model.

(2) What is the physical significance of the fact that total energy of an orbiting electron is negative? [1 Mark]

Answer:
It implies that the electron is bound to the nucleus by attractive forces and requires external energy to escape the atom.

Teacher's Note:
a) Negative total energy signifies a bound system.
b) Zero energy corresponds to complete ionization.

Question 20 [5 Marks]

When a television set (scientifically known as a television receiver) is opened, many components like semiconductor diodes, transistors, capacitors, resistors, etc. can be observed on its motherboard. There are different types of diodes like photo diode, Zener diode, LED, etc. found in the T.V. set. With the help of these components, a television receives audio as well as video signals.
(i) A semiconductor diode has two types of semiconducting materials: 'P' type and 'N' type. What is the difference between them? [2 Marks]

Answer:

Teacher's Note:
a) Tabular format is best suited for differences.
b) Ensure correct dopant types and majority carriers are mentioned.

(ii) Draw a labelled diagram of a full wave rectifier. Show graphically how its output voltage varies with time. [2 Marks]

Answer:
[Figure: Labelled circuit diagram of a full wave rectifier showing center-tap transformer, two diodes \(D_1\) and \(D_2\), load resistor \(R_L\), and corresponding input/output voltage waveforms varying with time.]

Teacher's Note:
a) Label input alternating wave, diodes, center-tap transformer, and pulsating output waveform.
b) Ensure output shows continuous positive half-cycles.

(iii) What type of diode can be used as a voltage regulator? [1 Mark]

Answer:
Zener diode.

Teacher's Note:
a) A Zener diode operates in the reverse breakdown region to maintain constant voltage.
b) It provides stable output voltage despite fluctuations in input voltage.