ISC Class 12 Physics Board Exam Question Paper 2018 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.

(i) The order of coloured rings in a carbon resistor is red, yellow, blue and silver. The resistance of the carbon resistor is: [1 Mark]
(a) \( 24 \times 10^6 \, \Omega \pm 5\% \)
(b) \( 24 \times 10^6 \, \Omega \pm 10\% \)
(c) \( 34 \times 10^4 \, \Omega \pm 10\% \)
(d) \( 26 \times 10^4 \, \Omega \pm 5\% \)

Answer: (b) \( 24 \times 10^6 \, \Omega \pm 10\% \)

Red = 2, Yellow = 4, Blue = \( 10^6 \), Silver = \( \pm 10\% \).

Teacher's Note:
a) Use the standard colour coding mnemonic (B B ROY Great Britain Very Good Wife) to read resistor values.
b) Remember that the fourth band indicates the tolerance percentage (silver is \( 10\% \), gold is \( 5\% \), and no fourth band is \( 20\% \)).

(ii) A circular coil carrying a current I has radius R and number of turns N. If all the three, i.e. the current I, radius R and number of turns N are doubled, then, magnetic field at its centre becomes: [1 Mark]
(a) Double
(b) Half
(c) Four times
(d) One fourth

Answer: (A) Double

Magnetic field \( B = \frac{\mu_0 N I}{2 R} \). When \( N \), \( I \), and \( R \) are all doubled, \( B^\prime = \frac{\mu_0 (2N) (2I)}{2(2R)} = 2 \left(\frac{\mu_0 N I}{2 R}\right) = 2B \).

Teacher's Note:
a) Write down the formula for the magnetic field at the centre of a circular coil before applying proportionality.
b) Students often make mistakes by incorrectly scaling the radius in the denominator along with numerator terms.

(iii) An object is kept on the principal axis of a concave mirror of focal length 10 cm at a distance of 15 cm from its pole. The image formed by the mirror is: [1 Mark]
(a) Virtual and magnified
(b) Virtual and diminished
(c) Real and magnified
(d) Real and diminished

Answer: (c) Real and magnified

Using mirror formula \( \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \), with \( u = -15 \, \text{cm} \) and \( f = -10 \, \text{cm} \), we get \( v = -30 \, \text{cm} \). Since \( v \) is negative and greater than \( f \), the image is real, inverted, and magnified.

Teacher's Note:
a) Always apply sign conventions strictly for concave mirrors where both \( u \) and \( f \) are negative.
b) When the object lies between \( F \) and \( C \) (between 10 cm and 20 cm for this mirror), the real image is always formed beyond \( C \) and is magnified.

(iv) Einstein's photoelectric equation is: [1 Mark]
(a) \( E_{\text{max}} = h\lambda - \varphi_0 \)
(b) \( E_{\text{max}} = \frac{hc}{\lambda} - \varphi_0 \)
(c) \( E_{\text{max}} = h\nu + \varphi_0 \)
(d) \( E_{\text{max}} = \frac{h\nu}{\lambda} + \varphi_0 \)

Answer: (b) \( E_{\text{max}} = \frac{hc}{\lambda} - \varphi_0 \)

Energy of incident photon is \( h\nu = \frac{hc}{\lambda} \), which equals work function \( \varphi_0 \) plus maximum kinetic energy \( E_{\text{max}} \).

Teacher's Note:
a) Recall that energy of a photon is given by \( E = h\nu = \frac{hc}{\lambda} \).
b) Do not confuse the plus and minus signs in photoelectric equations; the energy of the incident photon splits into work function and maximum kinetic energy.

(v) In Bohr's model of hydrogen atom, radius of the first orbit of an electron is \( r_0 \). Then, radius of the third orbit is: [1 Mark]
(a) \( \frac{r_0}{9} \)
(b) \( r_0 \)
(c) \( 3r_0 \)
(d) \( 9r_0 \)

Answer: (d) \( 9r_0 \)

Radius of the \( n \)th Bohr orbit is proportional to \( n^2 \). Therefore, \( r_3 = 3^2 r_0 = 9r_0 \).

Teacher's Note:
a) Remember the scaling relation for the radius of Bohr orbits: \( r_n \propto n^2 \).
b) Students often mistakenly write linear proportionality instead of quadratic proportionality with principal quantum number \( n \).

(B) Answer the following questions briefly and to the point.

(i) In a potentiometer experiment, balancing length is found to be 120 cm for a cell \( E_1 \) of emf 2V. What will be the balancing length for another cell \( E_2 \) of emf 1.5V? (No other changes are made in the experiment.) [1 Mark]

Answer:
Using the relation \( \frac{E_1}{E_2} = \frac{\ell_1}{\ell_2} \), we have \( \frac{2}{1.5} = \frac{120}{\ell_2} \), which gives \( \ell_2 = 90 \, \text{cm} \).

Teacher's Note:
a) Emf of a cell in a potentiometer is directly proportional to its balancing length (\( E \propto \ell \)).
b) Ensure that the potential gradient remains constant throughout the comparison.

(ii) How will you convert a moving coil galvanometer into a voltmeter? [1 Mark]

Answer:
By connecting a high resistance in series with the galvanometer.

Teacher's Note:
a) A voltmeter must have a very high resistance to draw negligible current from the circuit.
b) Contrast this with an ammeter, which requires a low shunt resistance connected in parallel.

(iii) A moving charged particle \( q \) travelling along the positive x-axis enters a uniform magnetic field B. When will the force acting on \( q \) be maximum? [1 Mark]

Answer:
When the charged particle enters perpendicular to the magnetic field direction.

Teacher's Note:
a) Magnetic force is given by \( F = qvB\sin\theta \), where \( \theta \) is the angle between velocity and magnetic field vectors.
b) Force is maximum when \( \theta = 90^{\circ} \) so that \( \sin 90^{\circ} = 1 \).

(iv) Why is the core of a transformer laminated? [1 Mark]

Answer:
To reduce energy losses due to eddy currents.

Teacher's Note:
a) Lamination breaks the path of eddy currents, increasing resistance and minimizing heating loss.
b) This is a standard conceptual question frequently asked in electromagnetic induction.

(v) Ordinary (i.e. unpolarised) light is incident on the surface of a transparent material at the polarising angle. If it is partly reflected and partly refracted, what is the angle between the reflected and the refracted rays? [1 Mark]

Answer:
\( 90^{\circ} \) (Brewster's law).

Teacher's Note:
a) At the polarising angle, the reflected ray and refracted ray are mutually perpendicular.
b) State Brewster's law relation \( \tan i_p = n \) clearly if asked in longer formats.

(vi) Define coherent sources of light. [1 Mark]

Answer:
Coherent sources of light are those sources which emit light waves of the same frequency and wavelength having zero or a constant phase difference.

Teacher's Note:
a) Constant phase relationship is mandatory to observe sustained interference patterns.
b) Ordinary independent light sources cannot be coherent.

(vii) Name a material which is used in making control rods in a nuclear reactor. [1 Mark]

Answer:
Boron (or Cadmium).

Teacher's Note:
a) Control rods absorb neutrons to regulate or halt the nuclear fission chain reaction.
b) Boron and cadmium are preferred due to their high neutron absorption cross-section.

SECTION B

Question 2 [2 Marks]
Define current density. Write an expression which connects current density with drift speed.

Answer:
Current density at any point in a conductor is defined as the amount of current flowing per unit cross-sectional area normal to the direction of current.
Expression connecting current density (\( J \)) with drift speed (\( v_d \)):
\( J = ne v_d \)
where \( n \) is electron density and \( e \) is electronic charge.

Teacher's Note:
a) Current density is a vector quantity represented as \( \vec{J} = \sigma \vec{E} \).
b) Ensure proper definition mentioning unit normal cross-sectional area.

Question 3

(a) A long horizontal wire P carries a current of 50A. It is rigidly fixed. Another wire Q is placed directly above and parallel to P, as shown in Figure 1 below. The weight per unit length of the wire Q is 0.025 Nm^-1 and it carries a current of 25A. Find the distance 'r' of the wire Q from the wire P so that the wire Q remains at rest. [3 Marks]
[Figure: Two parallel horizontal wires P and Q separated by distance r in vacuum. Wire P carries current 50A to the right, wire Q carries current 25A to the left.]

Answer:
Force per unit length between two parallel current-carrying wires is given by:
\( \frac{F}{\ell} = \frac{\mu_0}{4\pi} \frac{2 I_1 I_2}{r} \)
Given \( \frac{F}{\ell} = 0.025 \, \text{N/m} \), \( I_1 = 50 \, \text{A} \), \( I_2 = 25 \, \text{A} \), and \( \frac{\mu_0}{4\pi} = 10^{-7} \, \text{T m A}^{-1} \).
\( 0.025 = 10^{-7} \times \frac{2 \times 50 \times 25}{r} \)
\( r = \frac{2500 \times 10^{-7}}{0.025} = 0.01 \, \text{m} \)

Teacher's Note:
a) Equate the magnetic upward or downward force per unit length to the gravitational weight per unit length of wire Q.
b) Check all power-of-ten conversions carefully during substitution.

OR

(b) Calculate force per unit length acting on the wire B due to the current flowing in the wire A. (See Figure 2 below) [3 Marks]
[Figure: Two parallel horizontal wires A and B separated by distance \( r = 1 \, \text{cm} \) in vacuum. Wire A carries current 75A to the right, wire B carries current 20A to the right.]

Answer:
\( \frac{F}{\ell} = \frac{\mu_0}{4\pi} \frac{2 I_1 I_2}{r} \)
Given \( I_1 = 75 \, \text{A} \), \( I_2 = 20 \, \text{A} \), \( r = 1 \, \text{cm} = 0.01 \, \text{m} \).
\( \frac{F}{\ell} = 10^{-7} \times \frac{2 \times 75 \times 20}{0.01} \)
\( \frac{F}{\ell} = \frac{3000 \times 10^{-7}}{10^{-2}} = 0.03 \, \text{N/m} \)

Teacher's Note:
a) Convert distance from centimeters to SI meters before substitution.
b) State the final unit clearly as Newton per meter.

Question 4 [2 Marks]
(i) Explain Curie's law for a paramagnetic substance.
(ii) A rectangular coil having 60 turns and area of \( 0.4 \, \text{m}^2 \) is held at right angles to a uniform magnetic field of flux density \( 5 \times 10^{-5} \, \text{T} \). Calculate the magnetic flux passing through it.

Answer:
(i) Curie's law states that the magnetic susceptibility (\( \chi \)) of a paramagnetic substance is inversely proportional to its absolute temperature (\( T \)), i.e., \( \chi \propto \frac{1}{T} \) or \( \chi = \frac{C}{T} \).
(ii) Magnetic flux \( \phi = N B A \cos\theta \). Since the coil is held at right angles to the magnetic field, the angle between normal to the coil and field is \( \theta = 0^{\circ} \).
\( \phi = 60 \times (5 \times 10^{-5}) \times 0.4 \times \cos(0^{\circ}) = 120 \times 10^{-5} \, \text{Wb} = 1.2 \times 10^{-3} \, \text{Wb} \).

Teacher's Note:
a) Note that "at right angles to the magnetic field" means the area vector is parallel to the field, making \( \theta = 0^{\circ} \).
b) Do not miss the number of turns \( N \) in magnetic flux calculations for coils.

Question 5 [2 Marks]
What is motional emf? State any two factors on which it depends.

Answer:
Motional emf is the electromotive force induced across a conductor when it moves in a magnetic field, cutting magnetic flux lines.
Expression: \( e = Bv\ell \).
It depends on:
1. Magnetic field strength (\( B \)).
2. Velocity of the conductor (\( v \)).
3. Length of the conductor (\( \ell \)). (Any two)

Teacher's Note:
a) Motional emf arises due to the magnetic Lorentz force acting on free electrons inside the moving conductor.
b) Mentioning any two valid factors from magnetic field, velocity, or length is sufficient for full marks.

Question 6 [2 Marks]
(i) What is the ratio of the speed of gamma rays to that of radio waves in vacuum?
(ii) Name an electromagnetic wave which is used in the radar system used in aircraft navigation.

Answer:
(i) 1 : 1 (All electromagnetic waves travel at the same speed \( c \) in vacuum).
(ii) Microwaves (or Radio waves).

Teacher's Note:
a) In vacuum, all electromagnetic radiation travels at speed \( c = 3 \times 10^8 \, \text{m/s} \).
b) Microwaves have short wavelengths suitable for radar detection and tracking systems.

Question 7 [2 Marks]
A biconvex lens made of glass (refractive index 1.5) has two spherical surfaces having radii 20 cm and 30 cm. Calculate its focal length.

Answer:
Using Lens Maker's Formula:
\( \frac{1}{f} = (\mu - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \)
Given \( \mu = 1.5 \), \( R_1 = +20 \, \text{cm} \), \( R_2 = -30 \, \text{cm} \).
\( \frac{1}{f} = (1.5 - 1) \left(\frac{1}{20} - \frac{1}{-30}\right) \)
\( \frac{1}{f} = 0.5 \left(\frac{1}{20} + \frac{1}{30}\right) = 0.5 \left(\frac{3 + 2}{60}\right) = 0.5 \times \frac{5}{60} = \frac{2.5}{60} = \frac{1}{24} \)
\( f = 24 \, \text{cm} \)

Teacher's Note:
a) Apply sign conventions correctly: for a biconvex lens, \( R_1 \) is positive and \( R_2 \) is negative.
b) Double-check fraction additions to avoid calculation mistakes.

Question 8 [2 Marks]
State any two difference between primary rainbow and secondary rainbow.

Answer:
1. A primary rainbow is formed by two refractions and one total internal reflection, whereas a secondary rainbow is formed by two refractions and two total internal reflections.
2. In a primary rainbow, the intensity of colours is brighter with red on the outer edge and violet on the inner edge; in a secondary rainbow, the colours are fainter with the order reversed (violet on the outer edge and red on the inner edge).

Teacher's Note:
a) Secondary rainbows are fainter because additional reflection causes loss of light energy.
b) Mentioning the reflection count or colour order distinction guarantees full credit.

Question 9 [2 Marks]
(i) State de Broglie hypothesis.
(ii) With reference to photoelectric effect, define threshold wavelength.

Answer:
(i) de Broglie hypothesis states that every moving material particle is associated with a wave-like character, and the wavelength associated with the particle is given by \( \lambda = \frac{h}{p} \).
(ii) Threshold wavelength is the maximum wavelength of incident radiation required to eject photoelectrons from a photosensitive metal surface.

Teacher's Note:
a) Clearly mention wave-particle duality in de Broglie's hypothesis.
b) Threshold wavelength corresponds to threshold frequency, beyond which photoelectric emission stops.

Question 10 [2 Marks]
Calculate the minimum wavelength of the spectral line present in Balmer series of hydrogen.

Answer:
For the Balmer series, the wavelength formula is:
\( \frac{1}{\lambda} = R \left(\frac{1}{2^2} - \frac{1}{n^2}\right) \) where \( n = 3, 4, 5, \dots \)
For minimum wavelength (series limit), \( n = \infty \):
\( \frac{1}{\lambda} = R \left(\frac{1}{4} - \frac{1}{\infty}\right) = \frac{R}{4} \)
\( \lambda = \frac{4}{R} = \frac{4}{1.097 \times 10^7 \, \text{m}^{-1}} \approx 3.646 \times 10^{-7} \, \text{m} = 3646 \, \text{\AA} \approx 3700 \, \text{\AA} \)

Teacher's Note:
a) Minimum wavelength corresponds to maximum energy transition where \( n_2 = \infty \).
b) Remember to convert meters to Angstroms (\( 1 \, \text{\AA} = 10^{-10} \, \text{m} \)) if required.

Question 11

(a) What is meant by pair annihilation? Write a balanced equation for the same. [2 Marks]

Answer:
Pair annihilation is the process in which an elementary particle and its antiparticle collide and destroy each other, converting their mass entirely into electromagnetic radiation (gamma ray photons).
Balanced equation:
\( e^{-1} + e^{+1} \rightarrow 2\gamma \)

Teacher's Note:
a) Pair annihilation is the exact reverse of pair production.
b) Conservation of charge and total energy must be maintained in the equation.

OR

(b) What is meant by the terms half-life of a radioactive substance and binding energy of a nucleus? [2 Marks]

Answer:
Half-life: The time interval in which half of the radioactive nuclei initially present decay or disintegrate.
Binding energy: The energy required to separate all the nucleons of a nucleus completely to infinite distance from one another.

Teacher's Note:
a) Half-life is denoted by \( T_{1/2} = \frac{0.693}{\lambda} \).
b) Binding energy per nucleon determines nuclear stability.

Question 12 [2 Marks]
In a communication system, what is meant by modulation? State any two types of modulation.

Answer:
Modulation is the process of superimposing a low-frequency audio/message signal on a high-frequency carrier wave so that information can be transmitted efficiently over long distances.
Two types of modulation:
1. Amplitude Modulation (AM)
2. Frequency Modulation (FM)

Teacher's Note:
a) Modulation is necessary to reduce antenna size and prevent overlapping of signals.
b) Mentioning any two standard modulation types earns full credit.

SECTION C

Question 13 [5 Marks]
Obtain an expression for intensity of electric field at a point in end on position, i.e., axial position of an electric dipole.

Answer:
Consider an electric dipole consisting of charges \( -q \) and \( +q \) separated by a distance \( 2r \).
Let P be a point on the axial line at a distance \( x \) from the centre O of the dipole.
Electric field at P due to \( +q \) charge:
\( E_1 = \frac{1}{4\pi\varepsilon_0} \frac{q}{(x - r)^2} \) (directed away from dipole)
Electric field at P due to \( -q \) charge:
\( E_2 = \frac{1}{4\pi\varepsilon_0} \frac{q}{(x + r)^2} \) (directed towards dipole)
Resultant electric field \( E = E_1 - E_2 \):
\( E = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{(x - r)^2} - \frac{1}{(x + r)^2} \right] \)
\( E = \frac{q}{4\pi\varepsilon_0} \left[ \frac{(x + r)^2 - (x - r)^2}{(x^2 - r^2)^2} \right] = \frac{q}{4\pi\varepsilon_0} \frac{4rx}{(x^2 - r^2)^2} = \frac{1}{4\pi\varepsilon_0} \frac{2px}{(x^2 - r^2)^2} \) (since dipole moment \( p = q \times 2r \))
For a short dipole where \( r \ll x \), \( r^2 \) can be neglected compared to \( x^2 \):
\( E = \frac{1}{4\pi\varepsilon_0} \frac{2p}{x^3} \).

Teacher's Note:
a) Clearly state the direction of individual electric fields and resultant field vector.
b) Mention the short dipole approximation clearly at the final step.

Question 14 [5 Marks]
Deduce an expression for equivalent capacitance C when three capacitors \( C_1 \), \( C_2 \) and \( C_3 \) connected in parallel.

Answer:
[Figure: Three capacitors \( C_1 \), \( C_2 \), \( C_3 \) connected in parallel across a DC voltage source V with charges \( Q_1, Q_2, Q_3 \).]
Consider three capacitors of capacitance \( C_1, C_2, C_3 \) connected in parallel across a voltage source \( V \).
The potential difference across each capacitor is the same equal to \( V \), but charges stored on them are different:.
\( Q_1 = C_1 V \)
\( Q_2 = C_2 V \)
\( Q_3 = C_3 V \)
Total charge \( Q \) supplied by the source is:
\( Q = Q_1 + Q_2 + Q_3 \)
\( Q = C_1 V + C_2 V + C_3 V = V(C_1 + C_2 + C_3) \)
If the combination is replaced by a single equivalent capacitor \( C_p \) such that \( Q = C_p V \), then:
\( C_p V = V(C_1 + C_2 + C_3) \)
\( C_p = C_1 + C_2 + C_3 \).

Teacher's Note:
a) Emphasize that in parallel combination, potential difference across each capacitor is identical.
b) Total charge is the algebraic sum of individual charges.

Question 15

(a) \( \varepsilon_1 \) and \( \varepsilon_2 \) are two batteries having emf of 34V and 10V respectively and internal resistance of \( 1 \, \Omega \) and \( 2 \, \Omega \) respectively. They are connected as shown in figure below. Using Kirchhoff's Laws of electrical networks, calculate the currents \( I_1 \) and \( I_2 \). [3 Marks]
[Figure: Multi-loop circuit diagram with batteries 34V (\( 1\,\Omega \)), 10V (\( 2\,\Omega \)), and resistors of \( 4\,\Omega \), \( 5\,\Omega \), \( 7\,\Omega \).]

(b) An electrical bulb is marked 200V, 100W. Calculate electrical resistance of its filament. If five such bulbs are connected in series to a 200V supply, how much current will flow through them? [2 Marks]

Answer:
(a) Applying Kirchhoff's Voltage Law to loop ABEFA:
\( -4I_1 - 5(I_1 + I_2) - 7I_1 - 1I_1 + 34 = 0 \)
\( -17I_1 - 5I_2 = -34 \implies 17I_1 + 5I_2 = 34 \) ...(1)
Applying KVL to loop BCDEB:
\( 4I_2 - 10 + 2I_2 + 7I_2 + 5(I_1 + I_2) = 0 \)
\( 18I_2 + 5I_1 = 10 \) ...(2)
Solving equations (1) and (2) simultaneously gives \( I_1 = 2 \, \text{A} \) and \( I_2 = 0 \, \text{A} \).
(b) Given \( V = 200 \, \text{V} \), \( P = 100 \, \text{W} \).
Resistance of one bulb \( R = \frac{V^2}{P} = \frac{200^2}{100} = 400 \, \Omega \).
When five such bulbs are connected in series, total resistance \( R_T = 5 \times 400 = 2000 \, \Omega \).
Current flowing through them \( I = \frac{V}{R_T} = \frac{200}{2000} = 0.1 \, \text{A} \).

Teacher's Note:
a) Ensure sign conventions are followed correctly when traversing loops in KVL.
b) For the bulb numerical, use \( R = \frac{V^2}{P} \) as the standard formula.

Question 16

(a) For any prism, prove that: \( n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \) where the terms have their usual meaning. [3 Marks]

(b) When two thin lenses are kept in contact, prove that their combined or effective focal length F is given by: \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \) where the terms have their usual meaning. [2 Marks]

Answer:
(a) In a prism, deviation \( \delta = (i + i^\prime) - (r + r^\prime) \).
Also, angle of prism \( A = r + r^\prime \).
At minimum deviation condition: \( i = i^\prime \) and \( r = r^\prime = \frac{A}{2} \), and \( \delta = \delta_m \), so \( i = \frac{A + \delta_m}{2} \).
Applying Snell's law, refractive index \( n = \frac{\sin i}{\sin r} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \).
(b) Consider two thin lenses \( L_1 \) and \( L_2 \) of focal lengths \( f_1 \) and \( f_2 \) in contact.
Lens formula for first lens: \( \frac{1}{v^\prime} - \frac{1}{u} = \frac{1}{f_1} \) ...(1)
For second lens acting on real/virtual image \( v^\prime \): \( \frac{1}{v} - \frac{1}{v^\prime} = \frac{1}{f_2} \) ...(2)
Adding (1) and (2): \( \frac{1}{v} - \frac{1}{u} = \frac{1}{f_1} + \frac{1}{f_2} \) ...(3)
For equivalent lens of focal length \( F \): \( \frac{1}{v} - \frac{1}{u} = \frac{1}{F} \) ...(4)
Comparing (3) and (4): \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \).

Teacher's Note:
a) Prisms and lens combinations are frequent 5-mark derivation questions in board exams.
b) State all geometrical and optical approximations clearly during derivation steps.

Question 17 [2 Marks]
(i) In Young's double slit experiment, show graphically how intensity of light varies with distance.
(ii) In Fraunhofer diffraction, how is the angular width of the central bright fringe affected when slit separation is increased?

Answer:
(i) [Figure: Graph showing uniform periodic alternating peaks of equal intensity with distance in interference pattern.] The intensity remains uniform across all bright fringes in ideal Young's double slit interference.
(ii) Angular width of the central bright fringe in diffraction is given by \( \theta = \frac{2\lambda}{a} \). When slit separation/width \( a \) is increased, the angular width of the central bright fringe decreases.

Teacher's Note:
a) Interference fringes have uniform intensity, whereas diffraction fringes have decreasing intensity away from the centre.
b) Note the inverse proportionality between slit width and angular fringe width in diffraction.

Question 18 [3 Marks]
Write one balanced equation each to show:
(i) Nuclear fission
(ii) Nuclear fusion
(iii) Emission of \( \beta^{-} \) (i.e. a negative beta particle)

Answer:
(i) Nuclear fission:
\( _{92}\text{U}^{235} + _{0}\text{n}^{1} \rightarrow _{92}\text{U}^{236} \rightarrow _{56}\text{Ba}^{144} + _{36}\text{Kr}^{89} + 3_{0}\text{n}^{1} + \text{Energy} \)
(ii) Nuclear fusion:
\( _{1}\text{H}^{2} + _{1}\text{H}^{2} \rightarrow _{1}\text{H}^{3} + _{1}\text{H}^{1} + 4.0 \, \text{MeV} \)
(iii) Emission of \( \beta^{-} \):
\( _{15}\text{P}^{32} \rightarrow _{16}\text{S}^{32} + _{-1}\text{e}^{0} + \overline{\nu} \) (or neutron decay: \( \text{n} \rightarrow \text{p} + e^{-} + \overline{\nu} \)).

Teacher's Note:
a) Ensure mass numbers and atomic numbers balance on both sides of every nuclear equation.
b) Include antineutrino for beta-minus decay for absolute correctness.

Question 19 [2 Marks]
With reference to semiconductor devices, define a p-type semiconductor and a Zener diode. What is the use of Zener diode?

Answer:
p-type semiconductor: An intrinsic semiconductor doped with trivalent impurities (like boron, aluminium) creating an excess of holes as majority charge carriers.
Zener diode: A heavily doped p-n junction diode designed to operate in the reverse breakdown region.
Use: It is used as a voltage stabilizer/regulator in electronic circuits.

Teacher's Note:
a) Trivalent impurities create acceptor energy levels just above the valence band.
b) Zener diodes maintain a constant voltage across loads despite input supply fluctuations.

SECTION D

Question 20

(a) An alternating emf of 220 V is applied to a circuit containing a resistor R having resistance of \( 160 \, \Omega \) and a capacitor 'C' in series. The current is found to lead the supply voltage by an angle \( \theta = \tan^{-1}(3/4) \).
(i) Calculate: (1) The capacitive reactance
(2) Impedance of the circuit
(3) Current flowing in the circuit
(ii) If the frequency of the applied emf is 50 Hz, what is the value of the capacitance of the capacitor 'C'? [5 Marks]
[Figure: Series RC circuit with resistor R = 160 ohms, capacitor C connected across 220V AC supply.]

Answer:
(i) (1) Given \( \tan\theta = \frac{3}{4} \). In an RC circuit, \( \tan\theta = \frac{X_c}{R} \).
\( \frac{X_c}{160} = \frac{3}{4} \implies X_c = \frac{3}{4} \times 160 = 120 \, \Omega \).
(2) Impedance \( Z = \sqrt{R^2 + X_c^2} = \sqrt{160^2 + 120^2} = \sqrt{25600 + 14400} = \sqrt{40000} = 200 \, \Omega \).
(3) Current \( I = \frac{V}{Z} = \frac{220}{200} = 1.1 \, \text{A} \).
(ii) Given \( f = 50 \, \text{Hz} \).
\( X_c = \frac{1}{2\pi f C} \implies C = \frac{1}{2\pi f X_c} = \frac{1}{2 \times \pi \times 50 \times 120} = \frac{1}{12000\pi} \approx 2.65 \times 10^{-5} \, \text{F} = 26.5 \, \mu\text{F} \).

Teacher's Note:
a) Phase angle in an RC circuit shows current leading voltage.
b) Verify unit conversions between Farads and microfarads carefully.

OR

(b) An A.C. generator generating an emf of \( \varepsilon = 300\sin(100\pi t) \, \text{V} \) is connected to a series combination of \( 16 \, \mu\text{F} \) capacitor, 1 H inductor and \( 100 \, \Omega \) resistor.
Calculate:
(i) Impedance of the circuit at the given frequency.
(ii) Resonant frequency \( f_0 \).
(iii) Power factor at resonant frequency \( f_0 \). [5 Marks]
[Figure: LCR series circuit connected to sinusoidal AC source.]

Answer:
From \( \varepsilon = 300\sin(100\pi t) \), angular frequency \( \omega = 100\pi \, \text{rad/s} \), so frequency \( f = 50 \, \text{Hz} \).
Inductive reactance \( X_L = \omega L = 100\pi \times 1 = 314.1 \, \Omega \).
Capacitive reactance \( X_c = \frac{1}{\omega C} = \frac{1}{100\pi \times 16 \times 10^{-6}} \approx 198.94 \, \Omega \).
(i) Impedance \( Z = \sqrt{R^2 + (X_L - X_c)^2} = \sqrt{100^2 + (314.1 - 198.94)^2} = \sqrt{10000 + 13262} \approx 153.18 \, \Omega \).
(ii) Resonant frequency \( f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{1 \times 16 \times 10^{-6}}} = \frac{1}{2\pi \times 4 \times 10^{-3}} = \frac{125}{\pi} \approx 39.8 \, \text{Hz} \).
(iii) At resonance, \( X_L = X_c \), so \( Z = R \).
Power factor \( \cos\theta = \frac{R}{Z} = \frac{R}{R} = 1 \).

Teacher's Note:
a) At resonance, impedance is purely resistive, making power factor unity.
b) Compare operating frequency with resonant frequency to determine circuit nature.

Question 21

(a) Draw a labelled ray diagram of an image formed by a refracting telescope with final image formed at infinity. Derive an expression for its magnifying power with the final image at infinity. [5 Marks]
[Figure: Ray diagram of astronomical refracting telescope showing objective and eye lenses with parallel rays from infinity forming final image at infinity.]

Answer:
Magnifying power \( m \) is defined as the ratio of angle subtended by the image at the eye (\( \beta \)) to the angle subtended by the object at the unaided eye (\( \alpha \)):
\( m = \frac{\beta}{\alpha} \)
For small angles, \( \alpha \approx \tan\alpha \) and \( \beta \approx \tan\beta \).
From objective lens triangle: \( \tan\alpha = \frac{h}{f_o} \) (where \( h \) is height of intermediate image).
From eyepiece triangle: \( \tan\beta = \frac{h}{f_e} \).
Therefore, \( m = \frac{h/f_e}{h/f_o} = -\frac{f_o}{f_e} \).
The negative sign indicates the final image is inverted with respect to the object.

Teacher's Note:
a) Ensure all principal rays, focal lengths \( f_o \) and \( f_e \), and angles are clearly marked in the ray diagram.
b) Telescope magnification is the ratio of objective focal length to eyepiece focal length.

OR

(b) (i) Using Huygen's wave theory, derive Snell's law of refraction.
(ii) With the help of an experiment, state how will you identify whether a given beam of light is polarised or unpolarised. [5 Marks]

Answer:
(i) Consider a plane wavefront AB incident obliquely on a refracting surface separating two media with speeds \( c_1 \) and \( c_2 \) and refractive indices \( n_1 \) and \( n_2 \).
Using secondary wavelets, time taken by wavefront to cover distance \( BD = c_1 t \) equals time taken by secondary wavelet from A to cover distance \( AC = c_2 t \).
From triangle BAD, \( \sin i = \frac{BD}{AD} = \frac{c_1 t}{AD} \).
From triangle ACD, \( \sin r = \frac{AC}{AD} = \frac{c_2 t}{AD} \).
Dividing both: \( \frac{\sin i}{\sin r} = \frac{c_1}{c_2} = \frac{n_2}{n_1} \implies n_1 \sin i = n_2 \sin r \) (Snell's law).
(ii) Pass the given light beam through an analyser (tourmaline crystal or polaroid) and rotate it. If the transmitted intensity remains completely unchanged, the light is unpolarised. If the intensity varies between maximum and zero (or minimum), the light is polarised.

Teacher's Note:
a) Huygen's construction diagrams must show wavefronts and secondary wavelets clearly.
b) Polarisation detection relies on Malus law variation during analyser rotation.

Question 22

(a) (i) The forward characteristic curve of a junction diode is shown in Figure 4 below:
Calculate the resistance of the diode at:
(1) V = 0.5 V
(2) I = 60 mA
(ii) Draw separate energy band diagram for conductors, semi-conductors and insulators and label each of them. [5 Marks]
[Figure: Forward characteristic curve of a junction diode showing voltage along x-axis and current in mA along y-axis.]

Answer:
(i) (1) At \( V = 0.5 \, \text{V} \), from graph current \( I = 40 \, \text{mA} = 40 \times 10^{-3} \, \text{A} \).
Resistance \( R = \frac{V}{I} = \frac{0.5}{40 \times 10^{-3}} = 12.5 \, \Omega \).
(2) At \( I = 60 \, \text{mA} \), from graph voltage \( V = 0.6 \, \text{V} \).
Resistance \( R = \frac{V}{I} = \frac{0.6}{60 \times 10^{-3}} = 10 \, \Omega \).
(ii) [Figure: Energy band diagrams for Conductor (valence and conduction bands overlapping), Semiconductor (small forbidden gap \( < 1 \, \text{eV} \)), and Insulator (large forbidden gap).]

Teacher's Note:
a) Always extract correct corresponding values of voltage and current from the given diode characteristic graph.
b) Label valence band, conduction band, and forbidden energy gap clearly in band diagrams.

OR

(b) (i) The arrangement given below represents a logic gate:
Copy the following truth table in your answer booklet and complete it showing outputs at C and D.
(ii) Draw a labelled diagram of a common emitter amplifier, showing waveforms of signal voltage and output voltage. [5 Marks]
[Figure: Logic gate combination consisting of an AND gate followed by a NOT gate.]

Answer:
(i) The given combination is an AND gate followed by a NOT gate, which forms a NAND gate.
(ii) [Figure: Circuit diagram of common emitter n-p-n transistor amplifier showing input signal wave and phase-inverted amplified output waveform.]

Teacher's Note:
a) Recognize gate combinations quickly; an AND gate followed by a NOT gate is equivalent to a NAND gate.
b) In a common emitter amplifier, the output voltage is out of phase by \( 180^{\circ} \) relative to the input signal.