HSC Physics Board Question Paper 2018 - Solutions

Board Question Paper: March 2018

Complete Solutions and Explanations for Physics (Section I & II)

SECTION – I

Q.1. Select and write the most appropriate answer from the given alternatives for each sub-question: [7]

i. In stationary wave, the distance between a node and its adjacent antinode is _______.

Answer: (B) $\frac{\lambda}{4}$
Explanation: The distance between two successive nodes or antinodes is $\lambda/2$. The distance between a node and the nearest antinode is half of that, i.e., $\lambda/4$.
ii. If the source is moving away from the observer, then the apparent frequency _______.

Answer: (D) will decrease
Explanation: According to the Doppler effect, when the source moves away from a stationary observer, the wavelength increases, causing the apparent frequency to decrease.
iii. A particle of mass m performs vertical motion in a circle of radius r. Its potential energy at the highest point is _______.

Answer: (A) 2mgr
Explanation: Taking the lowest point of the circle as the reference level (h=0), the height of the highest point is the diameter, $h = 2r$. Therefore, Potential Energy $PE = mgh = mg(2r) = 2mgr$.
iv. The compressibility of a substance is the reciprocal of _______.

Answer: (B) bulk modulus
Explanation: Compressibility (k) is defined as the reciprocal of Bulk Modulus (K). i.e., $k = \frac{1}{K}$.
v. If the particle starts its motion from mean position, the phase difference between displacement and acceleration is _______.

Answer: (C) $\pi$ rad
Explanation: Displacement $x = A \sin(\omega t)$ and acceleration $a = -\omega^2 A \sin(\omega t) = \omega^2 A \sin(\omega t + \pi)$. The phase difference is $\pi$ radians (180°).
vi. The kinetic energy per molecule of a gas at temperature T is _______.

Answer: (B) $\frac{3}{2} k_B T$
Explanation: According to the kinetic theory of gases, the average kinetic energy per molecule is given by $\frac{3}{2} k_B T$, where $k_B$ is the Boltzmann constant.
vii. A thin ring has mass 0.25 kg and radius 0.5m. Its moment of inertia about an axis passing through its centre and perpendicular to its plane is _______.

Answer: (A) 0.0625 kg m²
Explanation: For a ring, $I = MR^2$ about the central axis perpendicular to the plane.
Given $M = 0.25$ kg, $R = 0.5$ m.
$I = 0.25 \times (0.5)^2 = 0.25 \times 0.25 = 0.0625$ kg m².

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Q.2. Attempt any SIX: [12]

i. State Kepler’s law of orbit and law of equal areas.

Law of Orbit (First Law): All planets move in elliptical orbits around the Sun, with the Sun situated at one of the foci of the ellipse.
Law of Equal Areas (Second Law): The line joining the planet and the Sun sweeps out equal areas in equal intervals of time. (This implies that the areal velocity of the planet is constant).

ii. State any ‘four’ assumptions of kinetic theory of gases.

1. A gas consists of a large number of extremely small particles called molecules.
2. The molecules are perfectly elastic, rigid spheres.
3. The actual volume occupied by the molecules is negligible compared to the total volume of the gas.
4. The molecules are in a state of random motion and collide with each other and the walls of the container.

iii. Define moment of inertia. State its SI unit and dimensions.

Definition: Moment of inertia of a rigid body about an axis of rotation is defined as the sum of the product of the mass of each particle and the square of its perpendicular distance from the axis of rotation. $I = \sum m_i r_i^2$.
SI Unit: kg m²
Dimensions: $[M^1 L^2 T^0]$

iv. Distinguish between centripetal and centrifugal force.

Centripetal Force	Centrifugal Force
It is directed towards the center of the circular path.	It is directed away from the center of the circular path.
It is a real force arising from interaction (e.g., tension, gravity).	It is a pseudo force arising due to the non-inertial frame of reference.
Essential for circular motion.	Effect experienced in a rotating frame.

v. In Melde’s experiment, when tension in the string is 10 g wt then three loops are obtained. Determine the tension in the string required to obtain four loops, if all other conditions are constant.

Solution:
In Melde's experiment (transverse arrangement), the law of tension is $T p^2 = \text{constant}$, where $p$ is the number of loops.
Given: $T_1 = 10$ g wt, $p_1 = 3$, $p_2 = 4$.
Formula: $T_1 p_1^2 = T_2 p_2^2$
Calculation:
$10 \times (3)^2 = T_2 \times (4)^2$
$10 \times 9 = 16 T_2$
$90 = 16 T_2$
$T_2 = \frac{90}{16} = 5.625$ g wt.
Answer: The required tension is 5.625 g wt.

vi. Calculate the work done in increasing the radius of a soap bubble in air from 1 cm to 2 cm. The surface tension of soap solution is 30 dyne/cm. ($\pi = 3.142$)

Solution:
Initial radius $r_1 = 1$ cm, Final radius $r_2 = 2$ cm.
Surface Tension $T = 30$ dyne/cm.
A soap bubble has two free surfaces.
Increase in surface area $\Delta A = 2 \times 4\pi (r_2^2 - r_1^2)$.
$\Delta A = 8\pi (2^2 - 1^2) = 8\pi (4 - 1) = 24\pi$ cm².
Work done $W = T \times \Delta A = 30 \times 24\pi$.
$W = 720 \times 3.142$
$W = 2262.24$ ergs.
Answer: Work done is 2262.24 ergs (or $2.26 \times 10^{-4}$ Joules).

vii. A flat curve on a highway has a radius of curvature 400 m. A car goes around a curve at a speed of 32 m/s. What is the minimum value of coefficient of friction that will prevent the car from sliding? (g = 9.8 m/s²)

Solution:
Given: $r = 400$ m, $v = 32$ m/s, $g = 9.8$ m/s².
For a car not to slide on a level road, the centripetal force is provided by friction.
$\frac{mv^2}{r} \leq \mu mg \implies \mu \geq \frac{v^2}{rg}$
Minimum $\mu = \frac{v^2}{rg}$
$\mu = \frac{32 \times 32}{400 \times 9.8} = \frac{1024}{3920}$
$\mu \approx 0.2612$
Answer: The minimum coefficient of friction is approximately 0.26.

viii. A particle performing linear S. H. M. has maximum velocity of 25 cm/s and maximum acceleration of 100 cm/s². Find the amplitude and period of oscillation. ($\pi = 3.142$)

Solution:
Given: $v_{max} = A\omega = 25$ cm/s
$a_{max} = A\omega^2 = 100$ cm/s²
Dividing $a_{max}$ by $v_{max}$:
$\frac{A\omega^2}{A\omega} = \frac{100}{25} \implies \omega = 4$ rad/s.
Substituting $\omega$ in $v_{max}$:
$A(4) = 25 \implies A = \frac{25}{4} = 6.25$ cm.
Period $T = \frac{2\pi}{\omega} = \frac{2 \times 3.142}{4} = \frac{6.284}{4} = 1.571$ s.
Answer: Amplitude = 6.25 cm, Period = 1.571 s.

Q.3. Attempt any THREE: [9]

i. Derive Laplace’s law for a spherical membrane.

Consider a spherical membrane (like a drop) of radius $R$. Let $P_i$ be the internal pressure and $P_o$ be the external pressure. The excess pressure is $P = P_i - P_o$.
Let the radius increase by a small amount $dR$.
Work done by excess pressure = Force $\times$ Distance = $(P \times Area) \times dR = P (4\pi R^2) dR$.
Increase in surface area (for a drop/one surface): $dA = 4\pi(R+dR)^2 - 4\pi R^2 \approx 8\pi R dR$.
Work done against surface tension = $T \times dA = T(8\pi R dR)$.
Equating both works: $P(4\pi R^2) dR = T(8\pi R dR)$.
$P = \frac{2T}{R}$.
For a soap bubble (2 surfaces), $dA = 2 \times 8\pi R dR$, so $P = \frac{4T}{R}$. (Note: The question says "spherical membrane", typically implying a bubble-like structure or just the general relation. The formula $P = 2T/R$ is for a drop, $4T/R$ for a bubble).

ii. State and prove principle of conservation of angular momentum.

Statement: If the resultant external torque acting on a rotating body is zero, its total angular momentum is conserved (remains constant).
Proof:
Angular momentum $L = I\omega$ or vector form $\vec{L} = \vec{r} \times \vec{p}$.
Differentiating with respect to time: $\frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p})$.
$= \vec{r} \times \frac{d\vec{p}}{dt} + \frac{d\vec{r}}{dt} \times \vec{p}$.
Since $\frac{d\vec{r}}{dt} = \vec{v}$ and $\vec{p} = m\vec{v}$, their cross product is zero.
So, $\frac{d\vec{L}}{dt} = \vec{r} \times \vec{F} = \vec{\tau}$ (Torque).
If external torque $\vec{\tau} = 0$, then $\frac{d\vec{L}}{dt} = 0$, which implies $\vec{L} = \text{constant}$.

iii. Calculate the strain energy per unit volume in a brass wire of length 3 m and area of cross-section 0.6 mm² when it is stretched by 3 mm and a force of 6 kgwt is applied to its free end.

Solution:
Length $L = 3$ m.
Area $A = 0.6 \text{ mm}^2 = 0.6 \times 10^{-6} \text{ m}^2$.
Extension $l = 3 \text{ mm} = 3 \times 10^{-3} \text{ m}$.
Force $F = 6 \text{ kgwt} = 6 \times 9.8 = 58.8$ N.
Strain Energy per unit volume $U = \frac{1}{2} \times \text{Stress} \times \text{Strain}$.
Stress = $\frac{F}{A} = \frac{58.8}{0.6 \times 10^{-6}} = 98 \times 10^6$ N/m².
Strain = $\frac{l}{L} = \frac{3 \times 10^{-3}}{3} = 10^{-3}$.
$U = \frac{1}{2} \times (98 \times 10^6) \times (10^{-3})$
$U = \frac{1}{2} \times 98 \times 10^3 = 49 \times 10^3$ J/m³.
Answer: Strain energy per unit volume is $4.9 \times 10^4$ J/m³.

iv. What is the decrease in weight of a body of mass 500 kg when it is taken into a mine of depth 1000 km? (Radius of earth R = 6400 km, g = 9.8 m/s²)

Solution:
Mass $m = 500$ kg, Depth $d = 1000$ km, $R = 6400$ km.
Acceleration due to gravity at depth d: $g_d = g(1 - \frac{d}{R})$.
Original Weight $W = mg$.
Weight at depth $W_d = mg_d = mg(1 - \frac{d}{R}) = W - W(\frac{d}{R})$.
Decrease in weight $\Delta W = W - W_d = mg(\frac{d}{R})$.
$\Delta W = 500 \times 9.8 \times \frac{1000}{6400}$
$\Delta W = \frac{4900 \times 1000}{6400} = \frac{4900}{6.4} = 765.625$ N.
Answer: The decrease in weight is 765.625 N.

Q.4. [7 Marks]

A. State the differential equation of linear simple harmonic motion.

The differential equation of linear SHM is: $$\frac{d^2x}{dt^2} + \omega^2 x = 0$$ where $x$ is displacement and $\omega$ is the angular frequency.

B. Hence obtain the expression for acceleration, velocity and displacement of a particle performing linear S. H. M.

Acceleration: From the diff. eq., $\frac{d^2x}{dt^2} = -\omega^2 x$. Thus, $a = -\omega^2 x$.
Velocity: $a = \frac{dv}{dt} = v \frac{dv}{dx} = -\omega^2 x$. Integrating $\int v dv = -\omega^2 \int x dx$, we get $v = \omega \sqrt{A^2 - x^2}$.
Displacement: $v = \frac{dx}{dt} = \omega \sqrt{A^2 - x^2}$. Integrating $\int \frac{dx}{\sqrt{A^2-x^2}} = \int \omega dt$, we get $\sin^{-1}(x/A) = \omega t + \phi$. Thus, $x = A \sin(\omega t + \phi)$.

Problem: A body cools from 80°C to 70°C in 5 minutes and to 62°C in the next 5 minutes. Calculate the temperature of the surroundings.

Solution:
Newton's Law of Cooling: $\frac{d\theta}{dt} = K(\theta_{avg} - \theta_0)$.
Case 1: $80 \to 70$ in 5 min.
Rate of cooling = $\frac{80-70}{5} = 2$ °C/min.
Avg Temp = $\frac{80+70}{2} = 75$ °C.
Eq 1: $2 = K(75 - \theta_0)$.
Case 2: $70 \to 62$ in 5 min.
Rate of cooling = $\frac{70-62}{5} = \frac{8}{5} = 1.6$ °C/min.
Avg Temp = $\frac{70+62}{2} = 66$ °C.
Eq 2: $1.6 = K(66 - \theta_0)$.
Dividing Eq 1 by Eq 2:
$\frac{2}{1.6} = \frac{75 - \theta_0}{66 - \theta_0}$
$1.25 = \frac{75 - \theta_0}{66 - \theta_0}$
$1.25(66 - \theta_0) = 75 - \theta_0$
$82.5 - 1.25\theta_0 = 75 - \theta_0$
$82.5 - 75 = 1.25\theta_0 - \theta_0$
$7.5 = 0.25\theta_0$
$\theta_0 = \frac{7.5}{0.25} = 30$ °C.
Answer: The temperature of the surroundings is 30°C.

OR

A. What is meant by harmonics? Show that only odd harmonics are present as overtones in the case of an air column vibrating in a pipe closed at one end.

Harmonics: The fundamental frequency and all integral multiples of the fundamental frequency are called harmonics.
Pipe closed at one end:
Fundamental mode: $L = \lambda/4 \implies n = v/4L$. (1st Harmonic)
First Overtone: $L = 3\lambda_1/4 \implies n_1 = 3v/4L = 3n$. (3rd Harmonic)
Second Overtone: $L = 5\lambda_2/4 \implies n_2 = 5v/4L = 5n$. (5th Harmonic)
Since frequencies are $n, 3n, 5n...$, only odd harmonics are present.

B. The wavelengths of two sound waves in air are $\frac{81}{173}$ m and $\frac{81}{170}$ m. They produce 10 beats per second. Calculate the velocity of sound in air.

Solution:
$\lambda_1 = \frac{81}{173}$ m, $\lambda_2 = \frac{81}{170}$ m.
Since $\lambda_2 > \lambda_1$, frequency $n_1 > n_2$.
$n_1 = \frac{v}{\lambda_1} = \frac{173v}{81}$
$n_2 = \frac{v}{\lambda_2} = \frac{170v}{81}$
Beats $n_1 - n_2 = 10$.
$\frac{173v}{81} - \frac{170v}{81} = 10$
$\frac{3v}{81} = 10$
$\frac{v}{27} = 10 \implies v = 270$ m/s.
Answer: The velocity of sound is 270 m/s.

SECTION – II

Q.5. Select and write the most appropriate answer from the given alternatives for each sub-question: [7]

i. The reflected waves from an ionosphere are _______.

Answer: (B) sky waves
Explanation: Sky waves are radio waves that are reflected back to Earth from the ionosphere.
ii. In interference pattern, using two coherent sources of light; the fringe width is _______.

Answer: (A) directly proportional to wavelength.
Explanation: Fringe width $X = \frac{\lambda D}{d}$. Thus $X \propto \lambda$.
iii. Electric intensity outside a charged cylinder having the charge per unit length ‘$\lambda$’ at a distance r from its axis is _______.

Answer: (C) $\frac{\lambda}{2\pi \epsilon_0 r}$
Explanation: Using Gauss's Law for a line charge or cylinder, $E \cdot 2\pi r L = \frac{Q}{\epsilon_0} = \frac{\lambda L}{\epsilon_0} \implies E = \frac{\lambda}{2\pi \epsilon_0 r}$. Note: The option uses $K$ for dielectric constant or simply $1$ for air. The closest correct form is C (rewritten in standard notation). The paper likely implies $K$ is dielectric constant, so $E = \frac{\lambda}{2\pi \epsilon_0 K r}$. Option C in the image is $E = \frac{\lambda}{2\pi \epsilon_0 K r}$.
iv. SI unit of potential gradient is _______.

Answer: (D) $\frac{V}{m}$
Explanation: Potential gradient is change in potential per unit length $dV/dx$. Unit is Volt/meter.
v. The momentum associated with photon is given by _______.

Answer: (B) $\frac{h\nu}{c}$
Explanation: Energy $E = h\nu$. Momentum $p = E/c = h\nu/c$. Also $p = h/\lambda$.
vi. A pure semiconductor is _______.

Answer: (B) an intrinsic semiconductor
Explanation: Pure semiconductors without any doping are called intrinsic semiconductors.
vii. Glass plate of refractive index 1.732 is to be used as a polariser, its polarizing angle is _______.

Answer: (C) 60°
Explanation: According to Brewster's Law, $\tan \theta_p = \mu$.
$\tan \theta_p = 1.732 = \sqrt{3}$.
$\theta_p = 60^\circ$.

Q.6. Attempt any SIX: [12]

i. State the conditions to get constructive and destructive interference of light.

Constructive Interference: The path difference between the two waves must be an integral multiple of wavelength ($n\lambda$) or phase difference must be an even multiple of $\pi$ ($2n\pi$).
Destructive Interference: The path difference must be an odd integral multiple of half-wavelength ($(2n-1)\lambda/2$) or phase difference must be an odd multiple of $\pi$ ($(2n-1)\pi$).

ii. State and explain Ampere’s circuital law.

Statement: The line integral of magnetic field induction $\vec{B}$ around any closed path in free space is equal to $\mu_0$ times the total current enclosed by the path.
$\oint \vec{B} \cdot \vec{dl} = \mu_0 I$.
Explanation: This relates the magnetic field to the electric current flowing through the loop. $\mu_0$ is the permeability of free space.

iii. Draw a neat and labelled block diagram of a receiver.

(Imagine a diagram with the following blocks in order):
Receiving Antenna $\rightarrow$ Amplifier $\rightarrow$ IF Stage (Intermediate Frequency) $\rightarrow$ Detector/Demodulator $\rightarrow$ Audio Amplifier $\rightarrow$ Loudspeaker.

iv. Define magnetization. Write its SI unit and dimensions.

Definition: Magnetization (M) is defined as the net magnetic dipole moment per unit volume of the material. $M = m_{net}/V$.
SI Unit: Ampere per meter (A/m).
Dimensions: $[L^{-1} M^0 T^0 A^1]$.

v. The electron in the hydrogen atom is moving with a speed of $2.3 \times 10^6$ m/s in an orbit of radius 0.53 Å. Calculate the period of revolution of electron. ($\pi = 3.142$)

Solution:
$v = 2.3 \times 10^6$ m/s.
$r = 0.53$ Å $= 0.53 \times 10^{-10}$ m.
Period $T = \frac{2\pi r}{v}$.
$T = \frac{2 \times 3.142 \times 0.53 \times 10^{-10}}{2.3 \times 10^6}$
$T = \frac{3.3305 \times 10^{-10}}{2.3 \times 10^6}$
$T \approx 1.45 \times 10^{-16}$ s.
Answer: Period is $1.45 \times 10^{-16}$ s.

vi. A capacitor of capacitance 0.5 µF is connected to a source of alternating e.m.f. of frequency 100 Hz. What is the capacitive reactance? ($\pi = 3.142$)

Solution:
$C = 0.5 \mu F = 0.5 \times 10^{-6}$ F.
$f = 100$ Hz.
Reactance $X_c = \frac{1}{2\pi f C}$.
$X_c = \frac{1}{2 \times 3.142 \times 100 \times 0.5 \times 10^{-6}}$
$X_c = \frac{1}{314.2 \times 10^{-6}}$
$X_c = \frac{10^6}{314.2} \approx 3182.68$ $\Omega$.
Answer: Capacitive reactance is approximately 3182.7 $\Omega$.

vii. Calculate the de-Broglie wavelength of an electron moving with one fifth of the speed of light. Neglect relativistic effects. (h = $6.63 \times 10^{-34}$ J.s., c = $3 \times 10^8$ m/s, mass of electron = $9 \times 10^{-31}$ kg)

Solution:
Velocity $v = \frac{c}{5} = \frac{3 \times 10^8}{5} = 0.6 \times 10^8 = 6 \times 10^7$ m/s.
$\lambda = \frac{h}{mv}$
$\lambda = \frac{6.63 \times 10^{-34}}{9 \times 10^{-31} \times 6 \times 10^7}$
$\lambda = \frac{6.63}{54 \times 10^{-24}} \times 10^{-34} = \frac{6.63}{54} \times 10^{-10}$
$\lambda \approx 0.1227 \times 10^{-10}$ m.
Answer: Wavelength is 0.123 Å.

viii. In a cyclotron, magnetic field of 1.4 Wb/m² is used. To accelerate protons, how rapidly should the electric field between the Dees be reversed? ($\pi = 3.142$, $M_p = 1.67 \times 10^{-27}$ kg, $e = 1.6 \times 10^{-19}$ C)

Solution:
The electric field must reverse every half time period ($T/2$) of the cyclotron frequency.
Cyclotron period $T = \frac{2\pi m}{qB}$.
Reversal time $t = \frac{T}{2} = \frac{\pi m}{qB}$.
$t = \frac{3.142 \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 1.4}$
$t = \frac{5.247 \times 10^{-27}}{2.24 \times 10^{-19}}$
$t \approx 2.34 \times 10^{-8}$ s.
Answer: The field should be reversed every $2.34 \times 10^{-8}$ s.

Q.7. Attempt any THREE: [9]

i. Explain with a neat circuit diagram how will you determine unknown resistance ‘X’ by using meter bridge.

Connect the unknown resistance X in the left gap and a known resistance box R in the right gap of the meter bridge. A cell, key, and rheostat are connected in series with the 1m wire. A galvanometer is connected between the central point B and the jockey.
Move the jockey to find the null point (D). Measure length $l_x$ (left side) and $l_R$ (right side).
By Wheatstone's principle: $\frac{X}{R} = \frac{l_x}{l_R}$.
$X = R \left( \frac{l_x}{100 - l_x} \right)$.

ii. What is Zener diode? How is it used as a voltage regulator?

Zener Diode: A specially designed highly doped p-n junction diode which operates in the reverse breakdown region without damage.
Voltage Regulator: The Zener diode is connected in reverse bias parallel to the load. When the input voltage increases, the current through the Zener diode increases sharply, but the voltage drop across it remains constant (at $V_z$). This keeps the voltage across the load constant.

iii. In a biprism experiment, light of wavelength 5200 Å is used to get an interference pattern on the screen. The fringe width changes by 1.3 mm when the screen is moved towards biprism by 50 cm. Find the distance between two virtual images of the slit.

Solution:
$\lambda = 5200 \text{ \AA} = 5.2 \times 10^{-7}$ m.
Change in fringe width $\Delta X = 1.3 \text{ mm} = 1.3 \times 10^{-3}$ m.
Change in distance $\Delta D = 50 \text{ cm} = 0.5$ m.
Formula: $X = \frac{\lambda D}{d} \implies \Delta X = \frac{\lambda \Delta D}{d}$.
$d = \frac{\lambda \Delta D}{\Delta X}$
$d = \frac{5.2 \times 10^{-7} \times 0.5}{1.3 \times 10^{-3}}$
$d = \frac{2.6 \times 10^{-7}}{1.3 \times 10^{-3}} = 2 \times 10^{-4}$ m = 0.2 mm.
Answer: The distance between the two virtual images (d) is 0.2 mm.

iv. The refractive indices of water and diamond are $\frac{4}{3}$ and 2.42 respectively. Find the speed of light in water and diamond. (c = $3 \times 10^8$ m/s)

Solution:
$v = c / \mu$.
For Water: $v_w = \frac{3 \times 10^8}{4/3} = \frac{9 \times 10^8}{4} = 2.25 \times 10^8$ m/s.
For Diamond: $v_d = \frac{3 \times 10^8}{2.42} \approx 1.24 \times 10^8$ m/s.
Answer: Speed in water = $2.25 \times 10^8$ m/s, Speed in diamond $\approx$ $1.24 \times 10^8$ m/s.

Q.8. [7 Marks]

A. Prove theoretically the relation between e.m.f. induced in a coil and rate of change of magnetic flux in electromagnetic induction.

Consider a rectangular loop moving into a uniform magnetic field.
Magnetic flux $\phi = B \times A = B l x$.
Rate of change of flux $\frac{d\phi}{dt} = B l \frac{dx}{dt} = B l v$.
The Lorentz force on the charge carriers creates an induced emf $e = B l v$.
Comparing, $e = \frac{d\phi}{dt}$. By Lenz's law, $e = -\frac{d\phi}{dt}$.

B. A parallel plate air condenser has a capacity of 20 µF. What will be the new capacity if:
i. the distance between the two plates is doubled?
ii. a marble slab of dielectric constant 8 is introduced between the two plates?

Solution:
Initial $C = \frac{A\epsilon_0}{d} = 20 \mu F$.
i. Distance doubled ($d' = 2d$):
$C' = \frac{A\epsilon_0}{2d} = \frac{1}{2} C = \frac{20}{2} = 10 \mu F$.
ii. Dielectric introduced ($k=8$):
$C'' = k C = 8 \times 20 = 160 \mu F$.

OR

A. Draw a neat and labelled energy level diagram and explain Balmer series and Brackett series of spectral lines for hydrogen atom.

Balmer Series: Electrons transition from higher outer orbits ($n_2 = 3, 4, 5...$) to the second inner orbit ($n_1 = 2$). This falls in the Visible region.
Brackett Series: Electrons transition from higher outer orbits ($n_2 = 5, 6, 7...$) to the fourth inner orbit ($n_1 = 4$). This falls in the Near Infrared region.

B. The work function for a metal surface is 2.2 eV. If light of wavelength 5000Å is incident on the surface of the metal, find the threshold frequency and incident frequency. Will there be an emission of photoelectrons or not? (c = $3 \times 10^8$ m/s, 1 eV = $1.6 \times 10^{-19}$ J, h = $6.63 \times 10^{-34}$ J.s.)

Solution:
Work Function $\Phi_0 = 2.2$ eV = $2.2 \times 1.6 \times 10^{-19} = 3.52 \times 10^{-19}$ J.
Wavelength $\lambda = 5000$ Å = $5 \times 10^{-7}$ m.
1. Threshold Frequency ($\nu_0$):
$\nu_0 = \frac{\Phi_0}{h} = \frac{3.52 \times 10^{-19}}{6.63 \times 10^{-34}}$
$\nu_0 \approx 0.531 \times 10^{15} = 5.31 \times 10^{14}$ Hz.
2. Incident Frequency ($\nu$):
$\nu = \frac{c}{\lambda} = \frac{3 \times 10^8}{5 \times 10^{-7}}$
$\nu = 0.6 \times 10^{15} = 6.0 \times 10^{14}$ Hz.
3. Emission Check:
Since Incident Frequency ($6.0 \times 10^{14}$ Hz) > Threshold Frequency ($5.31 \times 10^{14}$ Hz), emission of photoelectrons will take place.

OMTEX AD 2

HSC Physics 2018 Board Question Paper - Full Solutions & Answers