ISC Class 12 Physics Board Exam Question Paper 2020 with Solutions

ISC Class 12 Physics Board Exam Question Paper with Solutions

SECTION A

Question 1

(A) Choose the correct alternative (a), (b), (c) or (d) for each of the questions given below: [5×1]

(i) A point charge 'q' is kept at each of the vertices of an equilateral triangle having each side 'a'. Total electrostatic potential energy of the system is [1 Mark]
(a) \(\left(\frac{1}{4\pi\varepsilon_0}\right)\left(\frac{3q^2}{a^2}\right)\)
(b) \(\left(\frac{1}{4\pi\varepsilon_0}\right)\left(\frac{3q}{a}\right)\)
(c) \(\left(\frac{1}{4\pi\varepsilon_0}\right)\left(\frac{3q^2}{a}\right)\)
(d) \(\left(\frac{1}{4\pi\varepsilon_0}\right)\left(\frac{3q}{a^2}\right)\)

Answer: (c) \(\left(\frac{1}{4\pi\varepsilon_0}\right)\left(\frac{3q^2}{a}\right)\)

Total potential energy \(U = \frac{1}{4\pi\varepsilon_0} \left(\frac{q \cdot q}{a} + \frac{q \cdot q}{a} + \frac{q \cdot q}{a}\right) = \frac{3q^2}{4\pi\varepsilon_0 a}\).

Teacher's Note:
a) The electrostatic potential energy of a system of point charges is the sum of potential energies of all unique pairs.
b) Students often forget that an equilateral triangle has three pairs of charges, leading to a factor of 3.

(ii) Curie temperature is the temperature above which: [1 Mark]
(a) A ferromagnetic substance behaves like a paramagnetic substance
(b) A paramagnetic substance behaves like a diamagnetic substance.
(c) A ferromagnetic substance behaves like a diamagnetic substance.
(d) A paramagnetic substance behaves like a ferromagnetic substance.

Answer: (a) A ferromagnetic substance behaves like a paramagnetic substance

Above the Curie temperature, thermal agitation destroys the alignment of magnetic domains, converting ferromagnetism into paramagnetism.

Teacher's Note:
a) Curie temperature is a critical material-specific threshold property.
b) Do not confuse paramagnetism with diamagnetism when thermal energy disrupts atomic dipoles.

(iii) In an astronomical telescope of refracting type: [1 Mark]
(a) Objective should have small focal length.
(b) Objective should have large focal length.
(c) Eyepiece should have large focal length.
(d) Both objective and eyepieces should have large focal length.

Answer: (b) Objective should have large focal length.

The magnifying power of an astronomical telescope is given by \(m = \frac{f_o}{f_e}\), requiring a large objective focal length \(f_o\) and a small eyepiece focal length \(f_e\).

Teacher's Note:
a) A larger objective lens also increases light-gathering power to view distant celestial objects clearly.
b) Students frequently mix up microscope and telescope focal length requirements.

(iv) In photoelectric effect experiment, the slope of the graph of the stopping potential versus frequency gives the value of: [1 Mark]
(a) \(\frac{h}{e}\)
(b) \(e\)
(c) \(\frac{e}{h}\)
(d) \(\frac{hc}{e}\)

Answer: (a) \(\frac{h}{e}\)

From Einstein equation, \(eV_s = h\nu - \phi\), so \(V_s = \left(\frac{h}{e}\right)\nu - \frac{\phi}{e}\). The slope of \(V_s\) versus \(\nu\) is \(\frac{h}{e}\).

Teacher's Note:
a) The intercept on the frequency axis gives the threshold frequency.
b) Remember to use consistent units for potential and frequency during numerical estimations.

(v) In a nuclear reactor, cadmium rods are used as: [1 Mark]
(a) Control rods
(b) Fuel rods
(c) Coolant
(d) Moderator

Answer: (a) Control rods

Cadmium is a strong neutron absorber and acts as control rods to regulate the fission chain reaction.

Teacher's Note:
a) Control rods can be inserted or withdrawn to control the reactor power level safely.
b) Do not confuse control rods with moderators like heavy water or graphite which slow down neutrons.

(B) Answer the following questions briefly and to the point: [7×1]

(i) State 'Gauss' theorem. [1 Mark]

Answer:
Gauss's theorem states that the total electric flux through any closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the total charge enclosed by that surface, i.e., \(\oint \vec{E} \cdot d\vec{S} = \frac{q}{\varepsilon_0}\).

Teacher's Note:
a) The closed surface is referred to as a Gaussian surface.
b) Always state the mathematical expression along with the definition to secure full credit.

(ii) A metallic wire having a resistance of \(20\ \Omega\) is bent in order to form a complete circle. Calculate the resistance between any two diametrically opposite points on the circle. [1 Mark]

Answer:
When bent into a circle and connected across diametrically opposite points, the wire divides into two equal semicircular halves connected in parallel. Resistance of each half is \(\frac{20}{2} = 10\ \Omega\).
Equivalent resistance \(R_{eq} = \frac{10 \times 10}{10 + 10} = 5\ \Omega\).

Teacher's Note:
a) Resistance is directly proportional to the length of the conductor.
b) Parallel combination formula must be applied correctly to the divided halves.

(iii) How can a moving coil galvanometer be converted into a voltmeter? [1 Mark]

Answer:
A moving coil galvanometer can be converted into a voltmeter by connecting a very high resistance (multiplier) in series with the galvanometer coil.

Teacher's Note:
a) An ideal voltmeter has infinite resistance.
b) Specify clearly that the high resistance is connected in series, whereas for an ammeter a low shunt resistance is connected in parallel.

(iv) Write Biot-Savart's law in vector form. [1 Mark]

Answer:
In vector form, Biot-Savart's law is given by:
\(d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \vec{r})}{r^3} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \hat{r})}{r^2}\)

Teacher's Note:
a) Mention the significance of each term in the formula if required, especially unit vectors.
b) Vector cross-product notation must be precise to indicate the direction of magnetic field.

(v) What is the phase difference between any two points lying on the same wave front? [1 Mark]

Answer:
The phase difference between any two points lying on the same wave front is zero (or an integral multiple of \(2\pi\)).

Teacher's Note:
a) A wavefront is defined as the locus of all points vibrating in the same phase.
b) Keep the answer concise and direct.

(vi) Name the physical principle on the basis of which optical fibers work. [1 Mark]

Answer:
Total Internal Reflection (TIR).

Teacher's Note:
a) Light must travel from a denser medium (core) to a rarer medium (cladding) at an angle greater than the critical angle.
b) Spelling of 'Total Internal Reflection' must be accurate.

SECTION B

Question 2 [2 Marks]

(a) A uniform copper wire having a cross sectional area of \(1\text{ mm}^2\) carries a current of \(5\text{ A}\). Calculate the drift speed of free electrons in it.
(Free electron number density of copper \(= 2 \times 10^{28}\text{ m}^{-3}\).)

Answer:
Given: \(A = 1\text{ mm}^2 = 1 \times 10^{-6}\text{ m}^2\), \(I = 5\text{ A}\), \(n = 2 \times 10^{28}\text{ m}^{-3}\), \(e = 1.6 \times 10^{-19}\text{ C}\).
Formula: \(I = nAeAv_d \implies v_d = \frac{I}{nAe}\)
\(v_d = \frac{5}{(2 \times 10^{28}) \times (1 \times 10^{-6}) \times (1.6 \times 10^{-19})}
= \frac{5}{3.2 \times 10^{3}} = 1.56 \times 10^{-3}\text{ m/s}\).

Teacher's Note:
a) Always convert cross-sectional area from square millimetres to square metres properly.
b) State units clearly in the final calculated answer.

OR

(b) An electric bulb is rated as \(250\text{ V}, 750\text{ W}\). Calculate the:
(i) Electric current flowing through it when it is operated on a \(250\text{ V}\) supply.
(ii) Resistance of its filament. [2 Marks]

Answer:
Given: \(V = 250\text{ V}\), \(P = 750\text{ W}\).
(i) Current \(I = \frac{P}{V} = \frac{750}{250} = 3\text{ A}\).
(ii) Resistance \(R = \frac{V}{I} = \frac{250}{3} = 83.33\ \Omega\) (or \(R = \frac{V^2}{P} = \frac{250^2}{750} = 83.33\ \Omega\)).

Teacher's Note:
a) Use standard power formulas \(P = VI\) and \(P = \frac{V^2}{R}\).
b) Express resistance to appropriate decimal places with proper units.

Question 3 [2 Marks]
Write an expression for force per unit length between two long current carrying wires, kept parallel to each other, in vacuum and hence define an ampere, the SI unit of current.

Answer:
Force per unit length between two parallel current-carrying wires is given by:
\(\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi r}\)
Definition of Ampere: One ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to \(2 \times 10^{-7}\text{ newton per metre}\) of length.

Teacher's Note:
a) Mention all conditions like 'infinite length', 'negligible cross-section', and '1 metre apart' while defining the ampere.
b) Check subscripts and constants in the formula carefully.

Question 4 [2 Marks]
(i) Define angle of dip.
(ii) State the relation between magnetic susceptibility (\(\chi\)) and relative permeability (\(\mu_r\)) of a magnetic substance.

Answer:
(i) Angle of dip is defined as the angle made by the total Earth's magnetic field vector with the horizontal direction in the magnetic meridian.
(ii) The relation is: \(\mu_r = 1 + \chi\)

Teacher's Note:
a) Ensure the phrase 'magnetic meridian' is included in the definition of angle of dip.
b) Define symbols used in the relation if not explicitly stated.

Question 5 [2 Marks]
(a) Figure 1 below shows a metallic rod MN of length \(l = 80\text{ cm}\), kept in a uniform magnetic field of flux density \(B = 0.5\text{ T}\), on two parallel metallic rails P and Q. Calculate the emf that will be induced between its two ends, when it is moved towards right with a constant velocity \(v\) of \(36\text{ km/h}\).
[Figure: A uniform magnetic field directed into the page marked with crosses (B), with parallel rails P and Q horizontal, a vertical rod MN of length l = 80 cm moving right with velocity v, and an arrow indicating B pointing upwards/downwards beside the rail structure.]

Answer:
Given: \(l = 80\text{ cm} = 0.8\text{ m}\), \(B = 0.5\text{ T}\), \(v = 36\text{ km/h} = 36 \times \frac{5}{18} = 10\text{ m/s}\).
Motional emf \(\varepsilon = Blv = 0.5 \times 0.8 \times 10 = 4\text{ V}\).

Teacher's Note:
a) Convert speed from km/h to m/s by multiplying by \(\frac{5}{18}\).
b) Formula \(\varepsilon = Blv\) applies when velocity, magnetic field, and length are mutually perpendicular.

OR

(b) When current flowing through one coil changes from \(0\text{ A}\) to \(15\text{ A}\) in \(0.2\text{ s}\), an emf of \(750\text{ V}\) is induced in an adjacent coil. Calculate the coefficient of mutual inductance of the two coils. [2 Marks]

Answer:
Given: \(\Delta I = 15 - 0 = 15\text{ A}\), \(\Delta t = 0.2\text{ s}\), \(\varepsilon = 750\text{ V}\).
Formula: \(\varepsilon = M \frac{\Delta I}{\Delta t} \implies M = \frac{\varepsilon}{(\Delta I / \Delta t)} = \frac{750}{15 / 0.2} = \frac{750}{75} = 10\text{ H}\).

Teacher's Note:
a) Mutual inductance unit is Henry (H) or \(\text{V}\cdot\text{s/A}\).
b) Pay close attention to rate of change of current calculations.

Question 6 [2 Marks]
(i) State any one use of infrared radiations.
(ii) State any one source of ultraviolet radiations.

Answer:
(i) Infrared radiations are used in remote controls of TV/electronic devices (or physical therapy/night vision goggles).

(ii) Ultraviolet radiations are produced by the Sun (or high-voltage mercury lamps/electric arcs).

Teacher's Note:
a) Accept any one valid application and source as per standard textbook listings.
b) Ensure clarity and precision in naming sources.

Question 7 [2 Marks]
Where will you keep an object in front of a:
(i) Convex lens in order to get a virtual and magnified image?
(ii) Concave mirror to get a real and diminished image?

Answer:
(i) Between the optical centre and the principal focus of the convex lens.
(ii) Beyond the centre of curvature (or between C and infinity) in front of the concave mirror.

Teacher's Note:
a) Precise position descriptions relative to the focus or centre of curvature are essential.
b) Distinguish clearly between mirror and lens sign conventions and terms.

Question 8 [2 Marks]
Draw a labelled graph of angle of deviation (\(\delta\)) versus angle of incidence (\(\mathcal{i}\)) for a prism.

Answer:
[Figure: A curved graph showing \(\delta\) on the y-axis and \(i\) on the x-axis, starting high, decreasing to a minimum deviation \(\delta_m\) where \(i = e\), and then increasing again.]
The graph is an asymmetric curve showing that for a given deviation (except minimum deviation), there are two values of angle of incidence, \(i\) and \(e\).

Teacher's Note:
a) Label both axes clearly along with the minimum deviation point \(\delta_m\).
b) The curve should not be symmetrical; the slope is steeper on the side of smaller angles of incidence.

Question 9 [2 Marks]
(i) State de Broglie hypothesis.
(ii) What conclusion can be drawn from Davisson and Germer's experiment?

Answer:
(i) De Broglie hypothesis states that all material particles in motion possess a wave-like character, and the wavelength associated with a particle of momentum \(p\) is given by \(\lambda = \frac{h}{p} = \frac{h}{mv}\).
(ii) Davisson and Germer's experiment conclusively proved the wave nature of electrons (experimental verification of de Broglie hypothesis).

Teacher's Note:
a) Mention both wave-particle duality and the formula for de Broglie wavelength.
b) State clearly that electron diffraction observed in the experiment confirms de Broglie's wave theory.

Question 10 [2 Marks]
Calculate binding energy of oxygen nucleus (\(^{16}_{8}\text{O}\)) from the data given below:
Mass of a proton = \(1.007825\text{ u}\)
Mass of a neutron = \(1.008665\text{ u}\)
Mass of (\(^{16}_{8}\text{O}\)) = \(15.994915\text{ u}\)

Answer:
Number of protons \(Z = 8\), Number of neutrons \(N = 16 - 8 = 8\).
Total mass of nucleons = \((8 \times 1.007825\text{ u}) + (8 \times 1.008665\text{ u})\)
\(= 8.06260\text{ u} + 8.06932\text{ u} = 16.13192\text{ u}\).
Mass defect \(\Delta m = 16.13192\text{ u} - 15.994915\text{ u} = 0.137005\text{ u}\).
Binding Energy = \(\Delta m \times 931.5\text{ MeV} = 0.137005 \times 931.5\text{ MeV} \approx 127.62\text{ MeV}\).

Teacher's Note:
a) Calculate mass defect precisely by subtracting the actual nuclear mass from the sum of individual nucleon masses.
b) Multiply by \(931.5\text{ MeV}\) (or \(931\text{ MeV}\)) to convert atomic mass units into energy units.

Question 11 [2 Marks]
For a radioactive substance, write the relation between:
(i) Half life (\(T\)) and disintegration constant (\(\lambda\))
(ii) Mean life (\(\tau\)) and disintegration constant (\(\lambda\))

Answer:
(i) \(T = \frac{\ln 2}{\lambda} = \frac{0.693}{\lambda}\)
(ii) \(\tau = \frac{1}{\lambda}\)

Teacher's Note:
a) Both relations are standard derivations from radioactive decay law.
b) Ensure correct symbols are used for half-life and mean life.

Question 12 [2 Marks]
With reference to communication systems, what is meant by:
(i) Modulation?
(ii) Demodulation?

Answer:
(i) Modulation is the process of superimposing a low-frequency baseband (information) signal onto a high-frequency carrier wave for efficient transmission.
(ii) Demodulation is the reverse process of recovering the original baseband signal from the modulated carrier wave at the receiver end.

Teacher's Note:
a) Emphasise that low-frequency signals cannot travel long distances directly, necessitating modulation.
b) Keep definitions concise and technically accurate.

SECTION C

Question 13 [3 Marks]
Show that intensity of electric field \(E\) at a point in broadside-on position is given by:
\(E = \left(\frac{1}{4\pi\varepsilon_0}\right) \frac{p}{(r^2 + l^2)^{\frac{3}{2}}}\)
Where the terms have their usual meaning.

Answer:
Consider an electric dipole consisting of charges \(-q\) and \(+q\) separated by distance \(2l\). Let \(P\) be a point on the equatorial (broadside-on) line at a perpendicular distance \(r\) from the center of the dipole.
The distance from each charge to point \(P\) is \(\sqrt{r^2 + l^2}\).
The magnitude of electric field due to each charge is \(E_1 = E_2 = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + l^2}\).
Resolving components, the vertical components cancel out and horizontal components add up along the direction parallel to the dipole axis (opposite to dipole moment direction).
\(E = 2E_1 \cos\theta\)
From geometry, \(\cos\theta = \frac{l}{\sqrt{r^2 + l^2}}\).
Substituting values:
\(E = 2 \left(\frac{1}{4\pi\varepsilon_0} \frac{q}{r^2 + l^2}\right) \left(\frac{l}{\sqrt{r^2 + l^2}}\right) = \frac{1}{4\pi\varepsilon_0} \frac{2ql}{(r^2 + l^2)^{\frac{3}{2}}}\)
Since dipole moment \(p = 2ql\), we get:
\(E = \left(\frac{1}{4\pi\varepsilon_0}\right) \frac{p}{(r^2 + l^2)^{\frac{3}{2}}}\)

Teacher's Note:
a) Draw a clear diagram showing charge positions, distances, and electric field vector resolutions.
b) Clearly substitute \(p = 2ql\) in the final step.

Question 14 [3 Marks]
A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric constant (relative permittivity) \(K\) is now introduced between its two plates in order to occupy the space completely.
State, in terms of \(K\), its effect on the following:
(i) The capacitance of the capacitor.
(ii) The potential difference between its plates.
(iii) The energy stored in the capacitor.

Answer:
(i) Capacitance increases by a factor of \(K\). (\(C' = KC\))
(ii) Potential difference decreases by a factor of \(K\). (\(V' = \frac{V}{K}\))
(iii) Energy stored decreases by a factor of \(K\). (\(U' = \frac{U}{K}\))

Teacher's Note:
a) Since the battery is disconnected, charge \(q\) remains constant.
b) Use \(C' = KC\), \(V' = \frac{q}{C'} = \frac{V}{K}\), and \(U' = \frac{q^2}{2C'} = \frac{U}{K}\) for justification.

Question 15 [3 Marks]
(a) \(E_1\) and \(E_2\) are two batteries having emfs of \(3\text{ V}\) and \(4\text{ V}\) and internal resistances of \(2\ \Omega\) and \(1\ \Omega\) respectively. They are connected as shown in Figure 2 below. Using Kirchhoff's laws of electrical circuits, calculate the currents \(I_1\) and \(I_2\).
[Figure: A multi-loop circuit diagram showing branches with batteries \(E_1 (3\text{ V}, 2\ \Omega)\), \(E_2 (4\text{ V}, 1\ \Omega)\), resistors \(R_1 = 4\ \Omega\), \(R_2 = 7\ \Omega\), \(R_3 = 8\ \Omega\), nodes A, B, C and bottom junctions F, E, D, with current directions \(I_1\) and \(I_2\).]

Answer:
Applying Kirchhoff's Current Law at node B, current through middle branch \(R_3\) is \((I_1 + I_2)\).
Applying Kirchhoff's Voltage Law to loop ABFE:
\(-I_1(4 + 2) - (I_1 + I_2)(8) + 3 = 0 \implies -6I_1 - 8I_1 - 8I_2 = -3 \implies 14I_1 + 8I_2 = 3\) --(1)
Applying Kirchhoff's Voltage Law to loop BCDE:
\(-I_2(7 + 1) - (I_1 + I_2)(8) + 4 = 0 \implies -8I_2 - 8I_1 - 8I_2 = -4 \implies 8I_1 + 16I_2 = 4\) --(2)
Multiplying equation (1) by 2: \(28I_1 + 16I_2 = 6\) --(3)
Subtracting (2) from (3): \((28I_1 - 8I_1) = 6 - 4 \implies 20I_1 = 2 \implies I_1 = 0.1\text{ A}\).
Substitute \(I_1 = 0.1\) in equation (2):
\(8(0.1) + 16I_2 = 4 \implies 0.8 + 16I_2 = 4 \implies 16I_2 = 3.2 \implies I_2 = 0.2\text{ A}\).

Teacher's Note:
a) Carefully assign loop directions and sign conventions for emfs and potential drops.
b) Double-check simultaneous linear equation solutions to avoid calculation errors.

OR

(b) A potentiometer circuit is shown in Figure 3 below. AB is a uniform metallic wire having length of \(2\text{ m}\) and resistance of \(8\ \Omega\). The batteries \(E_1\) and \(E_2\) have emfs of \(4\text{ V}\) and \(1.5\text{ V}\) and their internal resistances are \(1\ \Omega\) and \(2\ \Omega\) respectively.
[Figure: Potentiometer circuit with primary circuit having battery \(E_1 (4\text{ V}, 1\ \Omega)\), rheostat \(R (7\ \Omega)\), wire AB (\(2\text{ m}, 8\ \Omega\)), and secondary circuit with cell \(E_2 (1.5\text{ V}, 2\ \Omega)\), galvanometer G, and jockey J.]
(i) When the jockey J does not touch the wire AB, calculate:
(a) the current flowing through the potentiometer wire AB.
(b) the potential gradient across the wire AB.
(c) Now the jockey J is made to touch the wire AB at a point C such that the galvanometer (G) shows no deflection. Calculate the length AC. [3 Marks]

Answer:
(a) Total resistance of primary circuit \(R_{total} = r_1 + R + R_{AB} = 1 + 7 + 8 = 16\ \Omega\).
Current \(I = \frac{E_1}{R_{total}} = \frac{4}{16} = 0.25\text{ A}\).
(b) Potential drop across wire AB: \(V_{AB} = I \times R_{AB} = 0.25 \times 8 = 2\text{ V}\).
Potential gradient \(k = \frac{V_{AB}}{L} = \frac{2\text{ V}}{2\text{ m}} = 1\text{ V/m}\).
(c) At null point, emf of secondary cell \(E_2 = k \times l_{AC}\).
\(1.5 = 1 \times l_{AC} \implies l_{AC} = 1.5\text{ m}\).

Teacher's Note:
a) Internal resistance of the secondary cell does not affect the balancing length at null deflection.
b) Keep track of unit conversions for length and resistance.

Question 16 [3 Marks]
For two thin lenses kept in contact with each other, show that:
\(\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}\)
Where the terms have their usual meaning.

Answer:
Consider two thin lenses of focal lengths \(f_1\) and \(f_2\) placed in contact coaxially in a medium. Let an object be placed at point \(O\) at distance \(u\) from the lens combination.
Lens 1 alone would form an image at distance \(v_1\):
\(\frac{1}{v_1} - \frac{1}{u} = \frac{1}{f_1}\) --(1)
This image acts as a virtual object for Lens 2, forming the final image at distance \(v\):
\(\frac{1}{v} - \frac{1}{v_1} = \frac{1}{f_2}\) --(2)
Adding equations (1) and (2):
\(\left(\frac{1}{v_1} - \frac{1}{u}\right) + \left(\frac{1}{v} - \frac{1}{v_1}\right) = \frac{1}{f_1} + \frac{1}{f_2}\)
\(\frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2}\)
If the combination acts as a single equivalent lens of focal length \(F\) for an object at \(u\) and image at \(v\):
\(\frac{1}{v} - \frac{1}{u} = \frac{1}{F}\)
Therefore, \(\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}\).

Teacher's Note:
a) Clearly state the intermediate image formation steps for lens combinations.
b) This derivation can also be extended to power as \(P = P_1 + P_2\).

Question 17 [3 Marks]
(a) A compound microscope consists of two convex lenses having focal length of \(1.5\text{ cm}\) and \(5\text{ cm}\). When an object is kept at a distance of \(1.6\text{ cm}\) from the objective, the final image is virtual and lies at a distance of \(25\text{ cm}\) from the eyepiece. Calculate magnifying power of the compound microscope in this set-up.

Answer:
Given: \(f_o = 1.5\text{ cm}\), \(f_e = 5\text{ cm}\), \(u_o = -1.6\text{ cm}\), \(v_e = -25\text{ cm}\).
For objective lens: \(\frac{1}{v_o} - \frac{1}{u_o} = \frac{1}{f_o} \implies \frac{1}{v_o} - \frac{1}{-1.6} = \frac{1}{1.5}\)
\(\frac{1}{v_o} + \frac{1}{1.6} = \frac{1}{1.5} \implies \frac{1}{v_o} = \frac{1}{1.5} - \frac{1}{1.6} = \frac{1.6 - 1.5}{2.4} = \frac{0.1}{2.4} = \frac{1}{24}\)
\(v_o = 24\text{ cm}\).
Magnification by objective \(m_o = \frac{v_o}{u_o} = \frac{24}{-1.6} = -15\).
For eyepiece: \(\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e} \implies \frac{1}{-25} - \frac{1}{u_e} = \frac{1}{5} \implies -\frac{1}{u_e} = \frac{1}{5} + \frac{1}{25} = \frac{6}{25} \implies u_e = -\frac{25}{6}\text{ cm}\).
Magnifying power of eyepiece \(m_e = 1 + \frac{D}{f_e} = 1 + \frac{25}{5} = 6\).
Total magnifying power \(M = m_o \times m_e = -15 \times 6 = -90\).

Teacher's Note:
a) Sign convention must be strictly followed for all distances.
b) The negative sign in magnification indicates an inverted final image relative to the object.

(b) In Young's double slit experiment, the screen is kept at a distance of \(1.2\text{ m}\) from the plane of the slits. The two slits are separated by \(5\text{ mm}\) and illuminated with monochromatic light having wavelength \(600\text{ nm}\). Calculate:
(i) Fringe width i.e. fringe separation of the interference pattern.
(ii) Distance of \(10^{\text{th}}\) bright fringe from the center of the pattern. [3 Marks]

Answer:
Given: \(D = 1.2\text{ m}\), \(d = 5\text{ mm} = 5 \times 10^{-3}\text{ m}\), \(\lambda = 600\text{ nm} = 600 \times 10^{-9}\text{ m}\).
(i) Fringe width \(\beta = \frac{\lambda D}{d} = \frac{600 \times 10^{-9} \times 1.2}{5 \times 10^{-3}} = \frac{7.2 \times 10^{-7}}{5 \times 10^{-3}} = 1.44 \times 10^{-4}\text{ m} = 0.144\text{ mm}\).
(ii) Distance of \(10^{\text{th}}\) bright fringe \(x_{10} = 10 \beta = 10 \times 1.44 \times 10^{-4}\text{ m} = 1.44 \times 10^{-3}\text{ m} = 1.44\text{ mm}\).

Teacher's Note:
a) Convert all parameters into SI units (metres) before calculation.
b) Formula for \(n^{\text{th}}\) bright fringe is \(x_n = n\beta\).

Question 18 [3 Marks]
Draw the energy level diagram of hydrogen atom and show the transitions responsible for:
(I) absorption lines of Lyman series.
(II) emission lines of Balmer series.

Answer:
[Figure: Energy level diagram of hydrogen atom showing principal quantum numbers \(n = 1, 2, 3, 4, 5, \infty\) with energy values \(-13.6\text{ eV}\), \(-3.4\text{ eV}\), \(-1.51\text{ eV}\), etc., with upward arrows from \(n = 1\) to higher levels for Lyman absorption, and downward arrows terminating at \(n = 2\) from higher levels for Balmer emission.]
(I) Absorption lines of Lyman series: Transitions starting from the ground state (\(n = 1\)) to excited states (\(n = 2, 3, 4, \dots\)).
(II) Emission lines of Balmer series: Transitions originating from higher excited states (\(n = 3, 4, 5, \dots\)) and terminating at the first excited state (\(n = 2\)).

Teacher's Note:
a) Clearly label energy levels, arrows, and series names in the diagram.
b) Absorption involves upward transitions from the lowest state, while emission involves downward transitions.

Question 19 [3 Marks]
(i) State any one difference between energy band diagram of conductors and that of insulators.
(ii) Give a relation between \(\alpha\) and \(\beta\) for a transistor. (Derivation is not required.)
(iii) What is the advantage of an LED bulb over the filament electric bulb?

Answer:
(i) In conductors, the valence band and conduction band overlap (or there is no forbidden energy gap), whereas in insulators, there is a large forbidden energy gap (typically \(> 3\text{ eV}\)) between the valence band and conduction band.
(ii) \(\beta = \frac{\alpha}{1 - \alpha}\) (or \(\alpha = \frac{\beta}{1 + \beta}\)).
(iii) LED bulbs consume much less electrical power, have higher energy efficiency, and have a much longer operating lifespan compared to traditional filament bulbs.

Teacher's Note:
a) Contrast band gaps clearly for band diagram differences.
b) State transistor current gain relations accurately without error.

SECTION D

Question 20 [5 Marks]
(a) (i) A \(400\ \Omega\) resistor, a \(3\text{ H}\) inductor and a \(5\ \mu\text{F}\) capacitor are connected in series to a \(220\text{ V}\), \(50\text{ Hz}\) AC source. Calculate the:
(1) Impedance of the circuit.
(2) Current flowing through the circuit.
(ii) Draw a labelled graph showing the variation of impedance (\(Z\)) of a series LCR circuit versus frequency (\(f\)) of the AC supply.

Answer:
Given: \(R = 400\ \Omega\), \(L = 3\text{ H}\), \(C = 5\ \mu\text{F} = 5 \times 10^{-6}\text{ F}\), \(V_{rms} = 220\text{ V}\), \(f = 50\text{ Hz}\).
Inductive reactance \(X_L = 2\pi f L = 2 \times \frac{222}{7} \times 50 \times 3 \approx 314.28\ \Omega\) (or \(2\pi \times 50 \times 3 = 300\pi \approx 942.48\ \Omega\)).
Capacitive reactance \(X_C = \frac{1}{2\pi f C} = \frac{1}{2 \times \pi \times 50 \times 5 \times 10^{-6}} = \frac{10^6}{500\pi} = \frac{2000}{\pi} \approx 636.62\ \Omega\).
(1) Impedance \(Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{400^2 + (636.62 - 942.48)^2}\) (using \(\pi \approx 3.1416\), \(X_L = 942.5\ \Omega\), \(X_C = 636.6\ \Omega\)):
\(X_L - X_C = 942.5 - 636.6 = 305.9\ \Omega\).
\(Z = \sqrt{400^2 + (305.9)^2} = \sqrt{160000 + 93574.81} = \sqrt{253574.81} \approx 503.56\ \Omega\).
(2) Current \(I_{rms} = \frac{V_{rms}}{Z} = \frac{220}{503.56} \approx 0.437\text{ A}\).
(ii) [Figure: A resonance curve graph with \(Z\) on the y-axis and frequency \(f\) on the x-axis, showing a sharp minimum at resonant frequency \(f_r\), where \(Z = R\).]

Teacher's Note:
a) Calculate reactances \(X_L\) and \(X_C\) carefully before finding impedance \(Z\).
b) Label resonance frequency clearly on the impedance versus frequency graph.

(b) (i) When an alternating emf \(e = 310 \sin(100\pi t)\text{ V}\) is applied to a series LCR circuit, current flowing through it is \(i = 5 \sin\left(100\pi t + \frac{\pi}{3}\right)\text{ A}\).
(1) What is the phase difference between the current and the emf?
(2) Calculate the average power consumed by the circuit.
(ii) Obtain an expression for the resonant frequency (\(f_r\)) of a series LCR circuit. [5 Marks]

Answer:
(i) (1) Phase difference \(\phi = \frac{\pi}{3}\) radians (or \(60^{\circ}\)). Current leads emf.
(2) Peak emf \(E_0 = 310\text{ V}\), Peak current \(I_0 = 5\text{ A}\).
\(V_{rms} = \frac{310}{\sqrt{2}}\text{ V}\), \(I_{rms} = \frac{5}{\sqrt{2}}\text{ A}\).
Power factor \(\cos\phi = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\).
Average power \(P_{avg} = V_{rms} I_{rms} \cos\phi = \left(\frac{310}{\sqrt{2}}\right) \left(\frac{5}{\sqrt{2}}\right) \left(\frac{1}{2}\right) = \frac{1550}{4} = 387.5\text{ W}\).
(ii) Resonance condition in a series LCR circuit occurs when inductive reactance equals capacitive reactance:
\(X_L = X_C \implies 2\pi f_r L = \frac{1}{2\pi f_r C}\)
\(f_r^2 = \frac{1}{4\pi^2 L C} \implies f_r = \frac{1}{2\pi \sqrt{L C}}\).

Teacher's Note:
a) Use RMS values of voltage and current when calculating average AC power.
b) State the resonance condition clearly for deriving the resonant frequency expression.

Question 21 [5 Marks]
(a) (i) Derive an expression for refraction at a single (convex) spherical surface, i.e., a relation between \(u\), \(v\), \(R\), \(n_1\) (rarer medium) and \(n_2\) (denser medium).
(ii) Name the phenomenon due to which the sun appears reddish at sunset.

Answer:
(i) Consider a point object \(O\) placed in a rarer medium of refractive index \(n_1\) in front of a convex spherical refracting surface of radius of curvature \(R\) separating it from a denser medium of refractive index \(n_2\).
Let a ray starting from \(O\) be refracted at point \(N\) on the surface and meet the principal axis at \(I\) to form a real image.
Let angle of incidence be \(i\), angle of refraction be \(r\), and angle of rays with principal axis be \(\alpha, \beta, \gamma\).
From Snell's law: \(n_1 \sin i = n_2 \sin r\). For small angles, \(n_1 i = n_2 r\).
From geometry of triangle ONC: \(i = \alpha + \gamma\).
From triangle INC: \(\gamma = r + \beta \implies r = \gamma - \beta\).
Substituting in Snell's law: \(n_1 (\alpha + \gamma) = n_2 (\gamma - \beta) \implies n_1 \alpha + n_2 \beta = (n_2 - n_1) \gamma\).
Using small angle approximations for tangents: \(\alpha \approx \tan\alpha \approx \frac{AM}{-u}\), \(\beta \approx \tan\beta \approx \frac{AM}{v}\), \(\gamma \approx \tan\gamma \approx \frac{AM}{R}\).
Substituting these into the equation:
\(n_1 \left(-\frac{1}{u}\right) + n_2 \left(\frac{1}{v}\right) = (n_2 - n_1) \left(\frac{1}{R}\right)\)
\(\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}\).
(ii) Scattering of light (Rayleigh scattering).

Teacher's Note:
a) Draw a neat ray diagram showing angles \(\alpha, \beta, \gamma, i, r\) and point of incidence.
b) Sign convention must be applied correctly while substituting distances \(u\), \(v\), and \(R\).

OR

(b) (i) Draw a labelled graph of intensity of diffracted light (\(I\)) versus angle (\(\theta\)) in the Fraunhofer diffraction experiment for a single slit diffraction.
(ii) State the law of Malus.
(iii) How will you distinguish experimentally between ordinary light and plane polarized light? [5 Marks]

Answer:
(i) [Figure: Intensity distribution graph for single slit diffraction showing a central maximum of high intensity at \(\theta = 0\), flanked by secondary minima at \(\sin\theta = \pm \frac{\lambda}{a}, \pm \frac{2\lambda}{a}\) and smaller secondary maxima in between.]
(ii) Law of Malus states that when completely plane-polarized light is incident on an analyzer, the intensity (\(I\)) of the transmitted light varies directly as the square of the cosine of the angle (\(\theta\)) between the transmission axis of the analyzer and the polarizer. (\(I = I_0 \cos^2\theta\)).
(iii) To distinguish experimentally, pass the light through a Nicol prism or Polaroid sheet acting as an analyzer. If the transmitted light intensity does not change when the analyzer is rotated through \(360^{\circ}\), the light is ordinary (unpolarized). If the intensity varies between maximum and zero (or minimum), the light is plane polarized.

Teacher's Note:
a) Ensure the central maximum is drawn twice as wide as secondary maxima in the diffraction graph.
b) Clearly explain the rotation of the analyzer in the experimental distinction.

Question 22 [5 Marks]
(a) (i) In a semiconductor diode, what is meant by potential barrier?
(ii) Draw a labelled circuit diagram of a Zener diode as a voltage regulator.
(iii) Show with the help of a diagram, how you will obtain an AND gate using only NAND gates. (Truth table is not required.)

Answer:
(i) Potential barrier is the potential difference built across the p-n junction due to the diffusion of majority charge carriers and formation of a depletion region, which opposes further movement of charge carriers across the junction.
(ii) [Figure: Circuit diagram of Zener diode voltage regulator showing unregulated DC input voltage, series resistor \(R_s\), Zener diode connected in reverse bias across output terminals, and load resistor \(R_L\).]\
(iii) [Figure: Logic diagram showing two inputs A and B connected to a NAND gate, followed by another single-input NAND gate (or NAND used as NOT) to invert the output, yielding an AND gate equivalent.]

Teacher's Note:
a) Explain depletion region and barrier potential clearly.
b) Label all components in the Zener diode voltage regulator circuit diagram correctly.

OR

(b) (i) Draw a labelled circuit diagram of a transistor acting as a common emitter amplifier. What is meant by phase reversal?
(ii) Draw the symbol of a NAND gate and write its truth table. [5 Marks]

Answer:
(i) [Figure: Circuit diagram of common emitter npn transistor amplifier showing input signal \(v_i\), base bias battery \(V_{BB}\), collector bias battery \(V_{CC}\), load resistance \(R_L\), and output voltage \(v_o\).]\
Phase reversal means that in a common emitter amplifier circuit, an alternating input signal voltage and the corresponding output amplified voltage are out of phase by \(180^{\circ}\) (or \(\pi\) radians with respect to each other).
(ii) [Figure: Logic symbol of a NAND gate showing two inputs A and B with an AND gate shape followed by a bubble inversion circle leading to output Y.]
Truth table for NAND gate:

Teacher's Note:
a) Clearly indicate biasing batteries and signal directions in the common emitter amplifier diagram.
b) Verify truth table logic values for NAND gate carefully.