ISC Class 12 Physics Board Exam Question Paper 2023 with Solutions

ISC Class 12 Physics Board Exam Question Paper with Solutions

Question 1.
A. In questions (i) to (vii) given below, choose the correct alternative (a), (b), (c) or (d) for each of the questions.

(i) A hollow sphere of radius R has a point charge Q at its centre. Electric flux emanating from it is \(\varphi\). If both the charge and the radius of the sphere be doubled, electric flux emanating from the sphere will: [1 Mark]
(a) Remain the same.
(b) Become \(2\varphi$
(c) Become \(4\varphi$
(d) Become \(8\varphi$

Answer: (b) Become \(2\varphi$

According to Gauss's law, electric flux depends only on the charge enclosed by the surface and is independent of its radius. When the charge is doubled, the flux also becomes twice.

Teacher's Note:
a) Remember that Gauss's law formula is \(\Delta\phi = \frac{q_{in}}{\epsilon_0}\), meaning the size or radius of the Gaussian surface does not affect the total flux.
b) Students often make the mistake of considering the radius change in the inverse-square law for electric field, but flux depends linearly only on the enclosed charge.

(ii) An electric current (\(I\)) flowing through a metallic wire is gradually increased. The graph of heating power (\(P\)) developed in it versus the current (\(I\)) is: [1 Mark]
A. [Figure: A linear graph passing through the origin showing P increasing straight with I]
B. [Figure: A hyperbola curve showing P decreasing as I increases]
C. [Figure: A parabolic curve opening upwards showing P increasing non-linearly with I]
D. [Figure: A straight line sloping downwards]

Answer: (C)

The heating power is given by \(P = I^2R\), which represents a quadratic relation between power and current.

Teacher's Note:
a) The power dissipated as heat in a resistor varies directly as the square of the current (\(P \propto I^2\)), yielding a parabolic graph.
b) Do not confuse this with Ohm's law relationships where voltage and current are linearly proportional.

(iii) A circular coil has radius 'r', number of turns 'N' and carries a current 'l'. Magnetic flux density 'B' at its centre is: [1 Mark]
(a) \(B = \mu_0 NI$
(b) \(B = \frac{\mu_0 NI}{2r}$
(c) \(B = \frac{\mu_0 NI}{4\pi r}$
(d) \(B = \frac{\mu_0 NI}{4r}$

Answer: (b) \(B = \frac{\mu_0 NI}{2r}$

The magnetic field at the centre of a circular coil of \(N\) turns and radius \(r\) carrying current \(I\) is derived using Biot-Savart law as \(B = \frac{\mu_0 NI}{2r}\).

Teacher's Note:
a) Ensure you include the factor \(N\) for the number of turns and use \(2r\) in the denominator for a full circular loop.
b) A common error is writing \(4\pi r\) in the denominator which corresponds to a circular arc segment calculation.

(iv) If an object is placed at a distance of \(10\text{ cm}\) in front of a concave mirror of focal length \(20\text{ cm}\), the image formed will be: [1 Mark]
(a) real and \(20\text{ cm}\) in front of the mirror.
(b) real and \(6.67\text{ cm}\) in front of the mirror.
(c) virtual and \(20\text{ cm}\) behind the mirror
(d) virtual and \(6.67\text{ cm}\) behind the mirror.

Answer: (c) virtual and \(20\text{ cm}\) behind the mirror

Using mirror formula \(\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\) with \(f = -20\text{ cm}\) and \(u = -10\text{ cm}\), we get \(\frac{1}{-20} = \frac{1}{-10} + \frac{1}{v} \implies v = +20\text{ cm}\).

Teacher's Note:
a) When an object is placed between the pole and focus of a concave mirror, it always forms a virtual, enlarged, and erect image behind the mirror.
b) Always apply proper sign conventions carefully for \(u\), \(v\), and \(f\) during numerical substitution.

(v) What type of wavefronts are associated with a source at infinity? [1 Mark]
(a) Cylindrical wavefronts
(b) Plane wavefronts
(c) Spherical wavefronts
(d) All types of wavefronts

Answer: (b) Plane wavefronts

Light rays originating from a source at a very large distance (infinity) are parallel to each other, forming plane wavefronts.

Teacher's Note:
a) A point source at a finite distance produces spherical wavefronts, and a line source produces cylindrical wavefronts.
b) As the distance approaches infinity, the radius of curvature becomes infinite, flattening the wavefront into a plane.

(vi) Matter waves are: [1 Mark]
(a) Waves associated with moving particles.
(b) Waves associated with stationary particles.
(c) Waves associated with any charged particles.
(d) Waves associated with electrons only.

Answer: (a) Waves associated with moving particles.

According to De Broglie hypothesis, every moving particle possesses a wave characteristic known as matter waves or de Broglie waves.

Teacher's Note:
a) The wavelength is given by \(\lambda = \frac{h}{mv}\), which applies to all material particles in motion.
b) Stationary particles have zero momentum, leading to undefined de Broglie wavelength.

(vii) With an increase in the temperature, electrical conductivity of a semiconductor: [1 Mark]
(a) Decreases.
(b) Increases.
(c) Does not change.
(d) First increases and then decreases.

Answer: (b) Increases.

As temperature increases, more covalent bonds break, generating additional electron-hole pairs and thereby increasing electrical conductivity.

Teacher's Note:
a) Semiconductors have a negative temperature coefficient of resistance.
b) Contrast this with metals, where conductivity decreases with temperature due to increased lattice vibrations.

B. Answer the following questions briefly.

(i) What is meant by an equipotential surface? [1 Mark]

Answer:
An equipotential surface is any surface defined by a constant value of electric potential at all points on its surface.

Teacher's Note:
a) The electric field lines are always directed normal to the equipotential surface at every point.
b) No work is done in moving a test charge from one point to another along an equipotential surface.

(ii) In case of metals, what is the relation between current density (J), electrical conductivity (\(\sigma\)) and electric field intensity (E)? [1 Mark]

Answer:
The vector relation is given by \(J = \sigma E\), where \(J\) is current density, \(\sigma\) is electrical conductivity, and \(E\) is electric field intensity.

Teacher's Note:
a) This relation represents the microscopic form of Ohm's law.
b) Ensure proper vector notation or statement of magnitudes when writing definitions in board exams.

(iii) What is meant by “Motional emf”? [1 Mark]

Answer:
Motional emf is the electromotive force induced across a conductor when it moves through a magnetic field, given by \(\varepsilon = vBl\).

Teacher's Note:
a) It arises due to the magnetic Lorentz force acting on the free electrons inside the moving conductor.
b) Mention all terms clearly: \(v\) for velocity, \(B\) for magnetic field, and \(l\) for length of the conductor perpendicular to motion.

(iv) What is meant by a microscope in normal use? [1 Mark]

Answer:
A microscope in normal use is an optical instrument adjusted so that the final virtual image is formed at the least distance of distinct vision (\(D\)).

Teacher's Note:
a) Normal adjustment sometimes also refers to the final image formed at infinity for relaxed eye viewing.
b) Clearly specify the eye condition when defining normal use in optical instruments.

(v) In a single slit Fraunhofer diffraction experiment, how does the angular width of central maximum change when the slit width is increased? [1 Mark]

Answer:
The angular width of the central maximum decreases when the slit width is increased, as angular width is inversely proportional to slit width (\(\theta = \frac{2\lambda}{b}\)).

Teacher's Note:
a) The formula for angular width of central maximum is \(2\theta = \frac{2\lambda}{b}\) where \(b\) is the slit width.
b) Students must explicitly state the inverse proportionality to score full marks.

(vi) Name the type of nuclear reaction that takes place in the core of the Sun. [1 Mark]

Answer:
Nuclear fusion reaction (specifically proton-proton chain reaction) takes place in the core of the Sun.

Teacher's Note:
a) Nuclear fusion involves lighter nuclei combining to form a heavier nucleus with a release of enormous energy.
b) Do not confuse fusion with nuclear fission, which is used in nuclear power reactors.

(vii) What type of semiconductor is obtained when a crystal of silicon is doped with a trivalent element? [1 Mark]

Answer:
A p-type semiconductor is obtained when silicon is doped with a trivalent element (such as boron, indium, or aluminium).

Teacher's Note:
a) Trivalent impurities create electron vacancies or holes, which act as majority charge carriers.
b) Doping with pentavalent elements like phosphorus yields an n-type semiconductor.

Question 2.

(i) Calculate equivalent capacitance of the circuit shown in Figure 1 given below: [2 Marks]
[Figure: Capacitors C1 (50 \(\mu\)F) and C2 (50 \(\mu\)F) are in parallel, and their combination is in series with C3 (25 \(\mu\)F)]

Answer:
1. Capacitors \(C_1\) and \(C_2\) are connected in parallel, so their equivalent capacitance is \(C' = C_1 + C_2 = 50\mu\text{F} + 50\mu\text{F} = 100\mu\text{F}\).
2. This equivalent capacitance \(C'\) is in series with \(C_3 = 25\mu\text{F}\).
3. The total equivalent capacitance \(C\) is given by \(\frac{1}{C} = \frac{1}{C'} + \frac{1}{C_3} = \frac{1}{100} + \frac{1}{25} = \frac{1 + 4}{100} = \frac{5}{100}\), which gives \(C = 20\mu\text{F}\).

Teacher's Note:
a) Always combine parallel branches first before solving series configurations.
b) Check units carefully; all values are given in microfarads, so the final answer remains in microfarads.

OR

(ii) Calculate electric potential at a point P which is at a distance of \(9\text{ cm}\) from a point charge of \(50\ \mu\text{C}\). [2 Marks]

Answer:
1. Given data: \(r = 9\text{ cm} = 9 \times 10^{-2}\text{ m}\), \(Q = 50\ \mu\text{C} = 50 \times 10^{-6}\text{ C}\), \(k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\text{ N m}^2\text{ C}^{-2}\).
2. Formula: \(V = \frac{kQ}{r}\).
3. Substitution: \(V = \frac{9 \times 10^9 \times 50 \times 10^{-6}}{9 \times 10^{-2}} = 5 \times 10^6\text{ V}\).

Teacher's Note:
a) Convert all dimensions to SI units (meters) before calculation.
b) The potential is a scalar quantity, so sign of the charge must be included if the charge is negative.

Question 3.

(i) Write balancing condition of a Wheatstone bridge. [1 Mark]

Answer:
When the Wheatstone bridge is balanced, the ratio of resistances of the adjacent arms is equal: \(\frac{P}{Q} = \frac{R}{S}\).

Teacher's Note:
a) Under this condition, no current flows through the galvanometer connected between the opposite junctions.
b) Ensure variable names match standard textbook notations.

(ii) Current 'I' flowing through a metallic wire is related to drift speed \(v_d\) of free electrons as follows: \(I = nAe v_d\). State what symbol 'n' stands for. [1 Mark]

Answer:
The symbol 'n' stands for the number density of free electrons (number of free electrons per unit volume).

Teacher's Note:
a) It is important to specify 'per unit volume' to avoid losing marks.
b) Other terms are \(A\) for cross-sectional area, \(e\) for electronic charge, and \(v_d\) for drift velocity.

Question 4.

When an electric current is passed through a wire or a coil, a magnetic field is produced. Is the reverse phenomenon possible i.e., can a magnetic field produce an electric current? Explain with the help of an appropriate example. [2 Marks]

Answer:
1. Yes, the reverse phenomenon is possible, which is known as electromagnetic induction.
2. When a conductor is moved across a magnetic field or when the magnetic flux linked with a closed circuit changes, an electric current is induced in it. For example, in an AC generator, mechanical rotation of a coil inside a magnetic field continuously changes the magnetic flux linked with it, thereby generating an induced emf and electric current.

Teacher's Note:
a) Mention Faraday's law of electromagnetic induction as the governing principle.
b) Giving a practical example like a transformer or a generator completes the explanation.

Question 5.

(i) A long straight wire \(AB\) carries a current of \(5\text{ A}\). \(P\) is a proton travelling with a velocity of \(2 \times 10^6\text{ m/s}\), parallel to the wire, \(0.2\text{ m}\) from it and in a direction opposite to the current, as shown in Figure 2 below. Calculate the force which magnetic field of the current carrying conductor AB exerts on the proton. [3 Marks]
[Figure: Long straight wire AB with upward current 5A, point P at distance 0.2 m with velocity vector downward of magnitude \(2 \times 10^6\text{ m/s}\)]

Answer:
1. Magnetic field \(B\) at distance \(r = 0.2\text{ m}\) from the wire carrying current \(I = 5\text{ A}\) is given by \(B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi \times 10^{-7})(5)}{2\pi (0.2)} = 1 \times 10^{-5}\text{ T}\).
2. The force on the proton is given by \(F = qvB\sin\theta\). Here \(q = 1.6 \times 10^{-19}\text{ C}\), \(v = 2 \times 10^6\text{ m/s}\), \(B = 1 \times 10^{-5}\text{ T}\), and \(\theta = 90^{\circ}\).
3. Substituting the values: \(F = (1.602 \times 10^{-19})(2 \times 10^6)(1 \times 10^{-5})(1) = 3.2 \times 10^{-18}\text{ N}\) directed downwards.

Teacher's Note:
a) Use right-hand palm rule or Fleming's left-hand rule to determine the direction of force accurately.
b) Ensure proper substitution of the elementary charge of a proton.

OR

(ii) A moving coil galvanometer of resistance \(55\,\Omega\) produces a full scale deflection for a current of \(250\text{ mA}\). How will you convert it into an ammeter having a range of \(0 - 3\text{ A}\)? [3 Marks]

Answer:
1. To convert a galvanometer into an ammeter, a small shunt resistance \(R_s\) must be connected in parallel with it.
2. Formula for shunt resistance: \(R_s = \frac{I_g R_g}{I - I_g}\).
3. Given values: \(I_g = 250\text{ mA} = 0.25\text{ A}\), \(R_g = 55\,\Omega\), \(I = 3\text{ A}\).
4. Calculation: \(R_s = \frac{0.25 \times 55}{3 - 0.25} = \frac{13.75}{2.75} = 5\,\Omega\).
Therefore, a shunt resistance of \(5\,\Omega\) must be connected in parallel.

Teacher's Note:
a) Emphasize that the shunt must always be connected in parallel, not in series.
b) Verify units of current (convert milliamperes to amperes) before substitution.

Question 6.

(i) State how vectors \(\vec{E}\), \(\vec{B}\) and \(\vec{c}\) are oriented in an electromagnetic wave [1 Mark]

Answer:
In an electromagnetic wave, the electric field vector (\(\vec{E}\)), magnetic field vector (\(\vec{B}\)), and the wave propagation vector (\(\vec{c}\)) are mutually perpendicular to each other.

Teacher's Note:
a) This mutual perpendicularity can be verified using the right-hand screw rule or cross product relation \(\vec{E} \times \vec{B}\) along direction of propagation.
b) The oscillations of both fields are in the same phase.

(ii) Name the electromagnetic wave / radiation which is used to study crystal structure. [1 Mark]

Answer:
X-rays are used to study the atomic crystal structure of solids.

Teacher's Note:
a) This application relies on the phenomenon of X-ray diffraction by crystal lattices (Bragg's law).
b) X-rays have wavelengths comparable to interatomic spacing in crystals.

Question 7.

Name any two phenomenon which takes place in the formation of a rainbow. [2 Marks]

Answer:
1. Dispersion of white sunlight into its constituent colors.
2. Total internal reflection of light inside water droplets.

Teacher's Note:
a) Refraction also occurs at the entry and exit boundaries of the water droplet.
b) Listing dispersion and total internal reflection covers the essential physical processes.

Question 8.

With reference to semiconductor physics, answer the following questions.

(i) What is meant by "Forbidden band" of energy levels? [1 Mark]

Answer:
The forbidden band is the energy gap between the valence band and the conduction band where no allowed electron energy states exist.

Teacher's Note:
a) Electrons cannot occupy states within this forbidden energy gap under normal conditions.
b) The width of this band determines whether the material is an insulator, semiconductor, or conductor.

(ii) In which material "Forbidden band" is absent? [1 Mark]

Answer:
The forbidden band is absent in conductors (metals), where the valence and conduction bands overlap.

Teacher's Note:
a) Due to overlapping bands, electrons can move freely into the conduction band without requiring external thermal energy.
b) Insulators have the largest forbidden band gap.

Question 9.

Show that intensity of electric field at a point in broadside position of an electric dipole is given by: \(E = \left(\frac{1}{4\pi\varepsilon_0}\right) \frac{p}{(r^2 + l^2)^{3/2}}\) where the terms have their usual meaning. [3 Marks]

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) line at a distance \(r\) from the center of the dipole.
2. The electric field due to individual charges at point \(P\) are equal in magnitude: \(E_+ = E_- = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + l^2}$.
3. The vertical components of the electric fields cancel out, and the horizontal components add up: \(E = 2 E_+ \cos\theta\).
4. Since \(\cos\theta = \frac{l}{\sqrt{r^2 + l^2}}\) and \(p = q(2l)\), substituting these gives \(E = \frac{1}{4\pi\varepsilon_0} \frac{p}{(r^2 + l^2)^{3/2}}\).

Teacher's Note:
a) Draw a neat diagram showing vector resolutions for full credit.
b) State clearly that vertical components cancel due to symmetry.

Question 10.

(i) Eight identical cells, each of emf \(2\text{ V}\) and internal resistance \(3\,\Omega\), are connected in series to form a row. Six such rows are connected in parallel to form a battery. This battery is now connected to an external resistor R of resistance \(6\,\Omega\). Calculate: [3 Marks]
(a) emf of the battery.
(b) internal resistance of the battery.
(c) current flowing through R.

Answer:
(a) Since cells in each row are in series, the emf of one row of 8 cells is \(E_{\text{row}} = 8 \times 2\text{ V} = 16\text{ V}\). For parallel connection of rows, the total emf of the battery remains equal to the emf of a single row, so \(E_{\text{battery}} = 16\text{ V}\).
(b) Internal resistance of one row of 8 cells in series is \(r_{\text{row}} = 8 \times 3\,\Omega = 24\,\Omega\). For 6 such rows in parallel, equivalent internal resistance \(r_{\text{int}}\) is given by \(\frac{1}{r_{\text{int}}} = \frac{1}{24} \times 6 = \frac{6}{24} = \frac{1}{4}\), so \(r_{\text{int}} = 4\,\Omega\).
(c) Total resistance of the circuit is \(R_{\text{total}} = R + r_{\text{int}} = 6\,\Omega + 4\,\Omega = 10\,\Omega\). Using Ohm's law, current \(I = \frac{E}{R_{\text{total}}} = \frac{16\text{ V}}{10\,\Omega} = 1.6\text{ A}\).

Teacher's Note:
a) The official key contains a calculation error in part (b) taking internal resistance as 32 instead of 3, yielding an incorrect internal resistance of \(5.33\,\Omega\) and current \(0.178\text{ A}\). The correct value based on data is \(4\,\Omega\) and current \(1.6\text{ A}\).
b) Always combine series resistors within rows first, then apply parallel formulas.

OR

(ii) In the circuit shown in Figure 3 below, \(E_1\) and \(E_2\) are batteries having emfs of \(25\text{ V}\) and \(26\text{ V}\). They have an internal resistance of \(1\,\Omega\) and \(5\,\Omega\) respectively. Applying Kirchhoff's laws of electrical networks, calculate the currents \(I_1\) and \(I_2\). [3 Marks]
[Figure: Multi-loop circuit with batteries \(E_1 = 25\text{ V}, r_1 = 1\,\Omega\) and \(E_2 = 26\text{ V}, r_2 = 5\,\Omega\) with resistors \(4\,\Omega\), \(3\,\Omega\), \(2\,\Omega\)]

Answer:
1. Applying Kirchhoff's Voltage Law to loop ABCDEF: \(3I_2 + 2(I_1 + I_2) - (26 - 5I_2) = 0 \implies 2I_1 + 10I_2 = 26 \implies I_1 + 5I_2 = 13\).
2. Applying Kirchhoff's Voltage Law to loop HICDGH: \(4I_1 + 2(I_1 + I_2) - (25 - I_1) = 0 \implies 7I_1 + 2I_2 = 25\).
3. Solving the simultaneous equations: multiplying first by 2 gives \(2I_1 + 10I_2 = 26\) and multiplying second gives \(7I_1 + 2I_2 = 25\). Solving yields \(I_1 = 3\text{ A}\) and \(I_2 = 2\text{ A}\).

Teacher's Note:
a) Clearly define loop directions and current junctions before setting up Kirchhoff equations.
b) Double-check algebraic steps when solving simultaneous linear equations.

Question 11.

Using Ampere's circuital law, obtain an expression for magnetic flux density 'B' at a point near an infinitely long and straight conductor, carrying a current I. [3 Marks]
[Figure: Long straight wire with current element dl and a circular Amperian loop of radius r containing point B]

Answer:
1. According to Ampere's circuital law, the line integral of magnetic field \(\vec{B}\) around a closed loop is equal to \(\mu_0\) times the total current \(I\) enclosed by the loop: \(\oint \vec{B} \cdot d\vec{l} = \mu_0 I\).
2. For a circular Amperian loop of radius \(a\) centered on the wire, the magnetic field has constant magnitude \(B\) and is parallel to the length element \(dl\) everywhere along the loop.
3. Therefore, \(\oint B \, dl = B \oint dl = B(2\pi a) = \mu_0 I\).
4. Solving for \(B\), we get \(B = \frac{\mu_0 I}{2\pi a}\) or \(B = \left(\frac{\mu_0}{4\pi}\right) \frac{2I}{a}\).

Teacher's Note:
a) State Ampere's circuital law clearly as the starting step.
b) Mention symmetry arguments to take \(B\) outside the integral.

Question 12.

Using Huygen's wave theory of light, show that the angle of incidence is equal to the angle of reflection. Draw a neat and labelled diagram. [3 Marks]
[Figure: Reflection of a plane wavefront from a reflecting surface showing incident and reflected wavefronts, angles i and r]

Answer:
1. Consider a plane wavefront \(PA\) incident obliquely on a reflecting surface \(XY\) at an angle of incidence \(i\).
2. Let \(c\) be the speed of light. The time taken by the disturbance to travel from point \(Q\) to \(Q'\) via point \(K\) is analyzed. From geometry, right-angled triangles show that distance \(QK = AK \sin i\) and \(KQ' = KP' \sin r\).
3. The total time \(t\) for rays to travel from incident wavefront to reflected wavefront is independent of point \(K\) only if \(\sin i - \sin r = 0\).
4. Therefore, \(\sin i = \sin r \implies i = r\), proving that the angle of incidence equals the angle of reflection.

Teacher's Note:
a) Drawing congruent triangles (\(\Delta APB\) and \(\Delta A'PB\)) is an alternative, simpler geometrical proof.
b) Label the incident wavefront, reflected wavefront, normal, and angles clearly in the diagram.

Question 13.

(i) For any prism, obtain a relation between angle of the prism (A), angle of minimum deviation (\(\delta_m\)) and refractive index of its material (\(\mu\) or \(n\)). [5 Marks]
[Figure: Ray diagram showing refraction through a triangular prism with prism angle A, angle of deviation \(\delta\), incidence angle \(i_1\), and emergence angle \(i_2\)]

Answer:
1. For any prism, the relation between prism angle \(A\), angle of deviation \(\delta\), angle of incidence \(i_1\), and angle of emergence \(i_2\) is given by \(A + \delta = i_1 + i_2\).
2. Also, the sum of refracting angles inside the prism is \(r_1 + r_2 = A\).
3. In the condition of minimum deviation: \(\delta = \delta_m\), \(i_1 = i_2 = i\), and \(r_1 = r_2 = r\).
4. Substituting these into the equations gives \(2r = A \implies r = \frac{A}{2}\) and \(2i = A + \delta_m \implies i = \frac{A + \delta_m}{2}\).
5. According to Snell's law, refractive index \(\mu = \frac{\sin i}{\sin r} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}\).

Teacher's Note:
a) This is a standard and important derivation frequently asked in board exams.
b) Clearly state the conditions for minimum deviation (\(i_1 = i_2\) and \(r_1 = r_2\)) before substituting.

OR

(ii) Obtain an expression for refraction at a single convex spherical surface i.e., the relation between \(\mu_1\) (rarer medium), \(\mu_2\) (denser medium), object distance u, image distance v and the radius of curvature [5 Marks]
[Figure: Refraction at a convex spherical surface separating media of refractive indices \(\mu_1\) and \(\mu_2\) with object O, image I, center of curvature C]

Answer:
1. Consider a convex spherical surface separating a rarer medium of refractive index \(\mu_1\) from a denser medium of refractive index \(\mu_2\).
2. From Snell's law in small angle approximation: \(\mu_1 i = \mu_2 r\).
3. Using exterior angle property in triangles involving angles \(\alpha\), \(\beta\), and \(\gamma\): \(i = \alpha + \gamma\) and \(\gamma = r + \beta \implies r = \gamma - \beta\).
4. Substituting into Snell's law: \(\mu_1(\alpha + \gamma) = \mu_2(\gamma - \beta) \implies \mu_1 \alpha + \mu_2 \beta = (\mu_2 - \mu_1)\gamma\).
5. Substituting paraxial approximations for angles (\(\alpha \approx \frac{h}{-u}\), \(\beta \approx \frac{h}{v}\), \(\gamma \approx \frac{h}{R}\)), we obtain the refraction formula: \(\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}\).

Teacher's Note:
a) State sign conventions clearly during substitution of distances \(u\), \(v\), and \(R\).
b) Assume paraxial rays (small aperture) for valid linear approximations.

Question 14.

(i) What is the essential condition for obtaining a sustained interference? [2 Marks]

Answer:
1. The two sources of light must be coherent (i.e., they must emit light waves of the same frequency/wavelength with a constant phase difference or zero phase difference).
2. The sources must be monochromatic and preferably narrow and close to each other.

Teacher's Note:
a) Non-coherent sources produce random phase changes, resulting in general illumination instead of stable interference fringes.
b) Mentioning coherence is the most critical key point for this question.

(ii) In Young's double slit experiment, the distance of the 4th bright fringe from the centre of the interference pattern is \(1.5\text{ mm}\). The distance between the slits and the screen is \(1.5\text{ m}\) and the wavelength of light used is \(500\text{ nm}\). Calculate the distance between the two slits. [3 Marks]

Answer:
1. Formula for the position of the \(n\text{th}\) bright fringe: \(y_n = \frac{n\lambda D}{d}\).
2. Given data: \(n = 4\), \(y_4 = 1.5\text{ mm} = 1.5 \times 10^{-3}\text{ m}\), \(D = 1.5\text{ m}\), \(\lambda = 500\text{ nm} = 500 \times 10^{-9}\text{ m}\).
3. Rearranging for slit separation \(d\): \(d = \frac{n\lambda D}{y_4} = \frac{4 \times (500 \times 10^{-9}\text{ m}) \times 1.5\text{ m}}{1.5 \times 10^{-3}\text{ m}}\).
4. Calculation: \(d = \frac{3 \times 10^{-6}}{1.5 \times 10^{-3}} = 2 \times 10^{-3}\text{ m} = 2\text{ mm}\).

Teacher's Note:
a) The official key has a minor calculation slip yielding \(0.5\text{ mm}\), but substituting \(n=4\) correctly gives \(2\text{ mm}\).
b) Always convert all parameters into SI units (meters) before final evaluation.

Question 15.
Monochromatic light of wavelength \(396\text{ nm}\) is incident on the surface of a metal whose work function is \(1.125\text{ eV}\). Calculate: (i) the energy of an incident photon in eV. (ii) the maximum kinetic energy of photoelectrons in eV. [3 Marks]

Answer:
1. Given data: \(\lambda = 396\text{ nm} = 396 \times 10^{-9}\text{ m}\), Work function \(\phi_0 = 1.125\text{ eV}\), Planck's constant \(h = 6.63 \times 10^{-34}\text{ J s}\), speed of light \(c = 3 \times 10^8\text{ m/s}\).\br />2. (i) Energy of incident photon in Joules: \(E = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{396 \times 10^{-9}} = 5.02 \times 10^{-19}\text{ J}\).
Converting to eV by dividing by \(1.6 \times 10^{-19}\text{ C}\): \(E = \frac{5.02 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.14\text{ eV}\) (or \(3.11\text{ eV}\) using \(hc = 1242\text{ eV nm}\)).
3. (ii) Maximum kinetic energy of photoelectrons: \(K_{\max} = E - \phi_0 = 3.14\text{ eV} - 1.125\text{ eV} = 2.015\text{ eV}\) (or \(1.985\text{ eV}\) with key values).

Teacher's Note:
a) Remember the convenient conversion \(hc \approx 1242\text{ eV nm}\) or \(1240\text{ eV nm}\) for quick photon energy calculations.
b) Ensure proper subtraction of work function from photon energy.

Question 16.
Name any two essential parts of a nuclear reactor. State the function of any one of them. [2 Marks]

Answer:
1. Two essential parts of a nuclear reactor are: (i) Fuel (Uranium-235) and (ii) Control rods (Cadmium or boron).
2. Function of Control rods: They absorb neutrons to regulate and control the rate of the nuclear fission chain reaction in the reactor core.

Teacher's Note:
a) Other parts include moderator (heavy water or graphite) and coolant.
b) Clearly state both the components and the specified function to secure full marks.

Question 17.
Draw a labelled circuit diagram of a full wave rectifier. Show graphically how the output voltage varies with time [3 Marks]

Answer:
[Figure: Centre-tap transformer connected to two diodes \(D_1\) and \(D_2\), load resistor \(R_L\), with input AC waveform and pulsating DC output waveform showing continuous positive half-cycles]
1. The circuit consists of a centre-tap transformer, two junction diodes (\(D_1\) and \(D_2\)), and a load resistor \(R_L\).
2. During the positive half-cycle of input AC, diode \(D_1\) is forward biased and conducts current. During the negative half-cycle, diode \(D_2\) is forward biased and conducts current in the same direction through \(R_L\).
3. The output voltage across \(R_L\) consists of continuous unidirectional pulses for both half-cycles.

Teacher's Note:
a) Label input AC wave, diodes, transformer, and output rectified wave clearly.
b) Explain diode conduction alternation for both half cycles.

Question 18.

(i) A \(60\,\Omega\) resistor, a \(1.0\text{ H}\) inductor and \(4\ \mu\text{F}\) capacitor are connected in series to an ac supply generating an emf \(e = 300\sin(500t)\text{V}\). calculate: [3 Marks]
A. Impedance of the circuit.
B. Peak value of the current flowing through the circuit.
C. Phase difference between the current and the supply voltage.

Answer:
1. Given data: \(R = 60\,\Omega\) (Note: solution text uses \(602\,\Omega\) or \(60\,\Omega\); taking \(60\,\Omega\)), \(L = 1.0\text{ H}\), \(C = 4\ \mu\text{F} = 4 \times 10^{-6}\text{ F}\), angular frequency \(\omega = 500\text{ rad/s}\), peak voltage \(E_0 = 300\text{ V}\).
2. Inductive reactance: \(X_L = \omega L = 500 \times 1.0 = 500\,\Omega\).
3. Capacitive reactance: \(X_C = \frac{1}{\omega C} = \frac{1}{500 \times 4 \times 10^{-6}} = \frac{10^6}{2000} = 500\,\Omega\).
4. A. Impedance \(Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{60^2 + (500 - 500)^2} = 60\,\Omega\).
5. B. Peak current \(I_0 = \frac{E_0}{Z} = \frac{300\text{ V}}{60\,\Omega} = 5\text{ A}\).
6. C. Phase difference \(\tan\phi = \frac{X_L - X_C}{R} = 0 \implies \phi = 0^{\circ}\) (circuit is in resonance).

Teacher's Note:
a) The official key contains arithmetic discrepancies taking \(R = 602\,\Omega\) and incorrect reactances; the correct resonance condition gives \(X_L = X_C = 500\,\Omega\), \(Z = 60\,\Omega\), and \(I_0 = 5\text{ A}\).
b) Recognize that when \(X_L = X_C\), the circuit is in electrical resonance.

OR

(ii)
a) An ac generator generates an emf which is given by \(e = 311\sin(240\pi t)\text{V}\). Calculate: [3 Marks]
1. Frequency of the emf.
2. r.m.s value of the emf.
b) The primary coil of a transformer has 60 turns whereas its secondary coil has 3000 turns. [3 Marks]
1. If a \(220\text{ V}\) ac voltage is applied to the primary coil, how much emf is induced in the secondary coil?
2. If a current of \(5\text{ A}\) flows in the primary coil, how much current will flow in a load in the secondary coil? State the assumption you have made regarding the transformer, in this calculation.

Answer:
(a) 1. Comparing \(e = 311\sin(240\pi t)\) with standard form \(e = E_0\sin(\omega t)\), angular frequency \(\omega = 240\pi\). Frequency \(f = \frac{\omega}{2\pi} = \frac{240\pi}{2\pi} = 120\text{ Hz}\).
2. RMS emf \(E_{\text{rms}} = \frac{E_0}{\sqrt{2}} = \frac{311}{\sqrt{2}} \approx 220\text{ V}\).
(b) 1. Transformer ratio: \(\frac{V_s}{V_p} = \frac{N_s}{N_p} \implies V_s = 220 \times \frac{3000}{60} = 220 \times 50 = 11000\text{ V}\) (Note: official key solved with \(V_p = 220\text{ V}\) giving \(11\text{ V}\) assuming step-down ratio error in key, but standard formula gives \(11000\text{ V}\)).
2. Using ideal transformer power conservation \(I_p V_p = I_s V_s\), with key's values \(5\text{ A} \times 220\text{ V} = I_s \times 11\text{ V} \implies I_s = 100\text{ A}\).
Assumption: The transformer is assumed to be 100% efficient (ideal transformer with no energy losses).

Teacher's Note:
a) Always state the assumption of an ideal transformer when equating input power to output power.
b) Pay careful attention to step-up and step-down turn ratios.

Question 19.

(i) (a) Name the series of lines of hydrogen spectrum which lies in the (1) ultraviolet region. (2) visible region. [2 Marks]
(b) How much is the angular momentum of an electron when it is orbiting in the second Bohr orbit of a hydrogen atom? [1 Mark]
(c) With reference to Nuclear Physics, answer the following questions. [2 Marks]
(1) What is meant by "Isotopes"?
(2) Define 1u (where u stands for unified atomic mass unit).

Answer:
(a) (1) Ultraviolet region: Lyman series.
(2) Visible region: Balmer series.
(b) Angular momentum is given by \(L = \frac{nh}{2\pi}\). For the second Bohr orbit (\(n = 2\)), \(L = \frac{2h}{2\pi} = \frac{h}{\pi}\).
(c) (1) Isotopes are atoms of the same element having the same atomic number (number of protons) but different mass numbers (number of neutrons).
(2) One unified atomic mass unit (\(1\text{ u}\)) is defined as \(\frac{1}{12}\)th of the mass of an unbound neutral carbon-12 atom at rest.

Teacher's Note:
a) Memorize all spectral series of hydrogen: Lyman (UV), Balmer (Visible), Paschen, Brackett, and Pfund (Infrared).
b) Ensure precise phrasing when defining atomic mass units and isotopes.

OR

(ii)
a) Using Bohr’s theory of hydrogen atom, obtain an expression for the velocity of an electron in nth orbit of an atom. [3 Marks]
b) What is meant by Binding Energy per nucleon of a nucleus? State its physical significance. [2 Marks]

Answer:
(a) 1. Equating electrostatic force to centripetal force: \(\frac{1}{4\pi\varepsilon_0} \frac{Ze^2}{r^2} = \frac{mv^2}{r}\).
2. From Bohr's quantization condition for angular momentum: \(mvr = \frac{nh}{2\pi}\).
3. Eliminating radius \(r\) from these equations yields velocity \(v = \frac{Ze^2}{2\varepsilon_0 nh}\).\br />(b) Binding energy per nucleon is the total binding energy of a nucleus divided by its total number of nucleons (mass number \(A\)).
Physical significance: It is a direct measure of the nuclear stability; higher binding energy per nucleon indicates a more stable nucleus.

Teacher's Note:
a) Show both centripetal force balance and quantization condition clearly in derivations.
b) Emphasize that binding energy per nucleon peaks around iron (\(A \approx 56\)), indicating maximum stability.

Question 20.
Read the passage given below and answer the questions that follow.
There are two types of lenses: Converging lenses and Diverging lenses, depending on whether they converge or diverge an incident beam of light. They are also called convex or concave lenses. Lenses are usually made of glass. Convex lenses are more popular as they form a real image of an object. They are widely used in our daily life, for instance, in microscopes, telescopes, projectors, cameras, spectacles etc. Microscopes are used to view small and nearby objects whereas telescopes are used to see distant objects.

i. State any one factor on which focal length of a lens depends. [1 Mark]

Answer:
The focal length of a lens depends on the refractive index of the lens material and the surrounding medium (or the radii of curvature of its surfaces).

Teacher's Note:
a) Lens Maker's formula explicitly shows dependence on refractive indices and radii of curvature.
b) Stating refractive index is sufficient for a single-mark question.

ii. Give an example where a convex lens behaves like a diverging lens. [1 Mark]

Answer:
A convex lens behaves like a diverging lens when it is immersed in a transparent liquid whose refractive index is greater than the refractive index of the lens material.

Teacher's Note:
a) According to Lens Maker's formula, the focal length changes sign when the surrounding medium's refractive index exceeds that of the lens.
b) An example is a glass lens immersed in carbon disulfide.

iii. What type of lens is used in a camera? [1 Mark]

Answer:
A converging (convex) lens system is used in a camera to focus light onto the film or image sensor.

Teacher's Note:
a) Cameras use compound convex lens configurations to minimize chromatic and spherical aberrations.
b) A simple convex lens forms real and inverted images on the sensor plane.

iv. Write an expression for magnifying power of a compound microscope when its final image lies at the least distance of distinct vision (D). [1 Mark]

Answer:
The magnifying power is given by \(M = -\frac{L}{f_o} \left(1 + \frac{D}{f_e}\right)\) or \(M = \left(\frac{D}{f_1}\right) \left(1 + \frac{d}{f_2}\right)\).

Teacher's Note:
a) Ensure correct identification of objective and eyepiece focal lengths (\(f_o\) and \(f_e\)).
b) The negative sign indicates that the final image is inverted with respect to the object.

v. State any one difference between a reflecting telescope and a refracting telescope. [1 Mark]

Answer:
A refracting telescope uses lenses to refract and focus light, whereas a reflecting telescope uses a concave mirror to reflect and focus light.

Teacher's Note:
a) Reflecting telescopes are free from chromatic aberration, which is a major drawback in refracting telescopes.
b) Large reflecting telescopes are easier to support mechanically from behind compared to large lenses.