MHT CET 2016 Physics Question Paper with Answer and Solution

(A) The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} MR^2$ ... $(i)$
The volume of the disc is $V = \pi R^2 \times \text{thickness} = \pi R^2 \times \frac{R}{6} = \frac{\pi R^3}{6}$.
When the disc is recast into a solid sphere of radius $R_s$,the volume remains constant:
$\frac{\pi R^3}{6} = \frac{4}{3} \pi R_s^3$
$R_s^3 = \frac{R^3}{8} \implies R_s = \frac{R}{2}$ ... (ii)
The moment of inertia of a solid sphere about its diameter is $I_{\text{sphere}} = \frac{2}{5} MR_s^2$.
Substituting $R_s = \frac{R}{2}$ into the expression for $I_{\text{sphere}}$:
$I_{\text{sphere}} = \frac{2}{5} M \left(\frac{R}{2}\right)^2 = \frac{2}{5} \times \frac{1}{4} MR^2 = \frac{1}{5} \left(\frac{1}{2} MR^2\right)$.
Using equation $(i)$,we get $I_{\text{sphere}} = \frac{I}{5}$.

(A) The formula for gravitational acceleration at a height $h$ above the earth's surface is given by $g' = g \left( \frac{R}{R+h} \right)^2$.
Given that $g' = \frac{g}{4}$,we substitute this into the equation:
$\frac{g}{4} = g \left( \frac{R}{R+h} \right)^2$.
Dividing both sides by $g$,we get $\frac{1}{4} = \left( \frac{R}{R+h} \right)^2$.
Taking the square root of both sides,we obtain $\frac{1}{2} = \frac{R}{R+h}$.
Cross-multiplying gives $R + h = 2R$.
Therefore,$h = 2R - R = R$.

(C) For the gravitational field to be zero,the intensities produced by both particles must be equal in magnitude and opposite in direction.
$\frac{G m}{r_1^2} = \frac{G(9 m)}{r_2^2}$
where $r_1$ is the distance from mass $m$ and $r_2$ is the distance from mass $9m$.
Since $r_1 + r_2 = r$,we have $\frac{r_2}{r_1} = \sqrt{9} = 3 \Rightarrow r_2 = 3 r_1$.
Therefore,$r_1 + 3 r_1 = r \Rightarrow 4 r_1 = r \Rightarrow r_1 = \frac{r}{4}$ and $r_2 = \frac{3r}{4}$.
The gravitational potential $V$ is given by $V = -\frac{G m}{r_1} - \frac{G(9 m)}{r_2}$.
Substituting the values: $V = -\frac{G m}{r/4} - \frac{9 G m}{3r/4} = -\frac{4 G m}{r} - \frac{36 G m}{3r} = -\frac{4 G m}{r} - \frac{12 G m}{r} = -\frac{16 G m}{r}$.

(B) Given,$\frac{R}{C_{V}} = 0.4$.
From Mayer's relation,$C_{P} - C_{V} = R$,so $C_{P} = C_{V} + R$.
Substituting $R = 0.4 C_{V}$,we get $C_{P} = C_{V} + 0.4 C_{V} = 1.4 C_{V}$.
The adiabatic index $\gamma$ is defined as $\gamma = \frac{C_{P}}{C_{V}}$.
Therefore,$\gamma = \frac{1.4 C_{V}}{C_{V}} = 1.4$.
For a rigid diatomic gas,the degrees of freedom $f = 5$.
The adiabatic index is $\gamma = 1 + \frac{2}{f} = 1 + \frac{2}{5} = 1 + 0.4 = 1.4$.
Thus,the gas is made up of rigid diatomic molecules.

(B) The pressure $P$ exerted by an ideal gas is given by the kinetic theory of gases as $P = \frac{1}{3} \rho v_{rms}^2$,where $\rho$ is the density and $v_{rms}$ is the root mean square velocity.
We know that the kinetic energy per unit volume $(u)$ is given by $u = \frac{1}{2} \rho v_{rms}^2$.
Therefore,$\rho v_{rms}^2 = 2u$.
Substituting this into the pressure equation: $P = \frac{1}{3} (2u) = \frac{2}{3} u$.
Thus,the pressure exerted by an ideal gas is $2/3$ of the kinetic energy per unit volume of the gas.

(C) Let the radius of the original drop be $R$ and the radius of each small droplet be $r$.
Surface energy $E = S \times A$,where $S$ is the surface tension and $A$ is the surface area.
Initial surface area $A_1 = 4 \pi R^2$.
Volume of the large drop = Volume of $512$ small droplets:
$\frac{4}{3} \pi R^3 = 512 \times \frac{4}{3} \pi r^3$
$R^3 = 512 r^3 \implies R = 8r$.
Final surface area $A_2 = 512 \times (4 \pi r^2)$.
Substituting $r = R/8$:
$A_2 = 512 \times 4 \pi \left(\frac{R}{8}\right)^2 = 512 \times 4 \pi \times \frac{R^2}{64} = 8 \times (4 \pi R^2) = 8 A_1$.
Since surface energy is directly proportional to surface area $(E \propto A)$:
$E_2 = 8 E_1 = 8 E$.

(C) The rise of water $h$ in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{\rho g R_{eff}}$,where $R_{eff}$ is the effective radius of the meniscus.
In this case,the water rises in the annular space between the capillary tube of radius $R$ and the wire of radius $r$.
The effective radius for the annular space is the difference between the radii,$R_{eff} = R - r$.
Assuming the contact angle $\theta = 0$ for water and glass/metal,we have $\cos \theta = 1$.
Substituting these values into the formula,we get $h = \frac{2T}{\rho g (R - r)}$.

(B) The sag (depression) $\delta$ of a beam supported at both ends and loaded at the centre is given by the formula: $\delta = \frac{W l^3}{48 Y I}$.
Here,$W$ is the load,$l$ is the length,$Y$ is Young's modulus,and $I$ is the geometrical moment of inertia.
For a rectangular cross-section of breadth $b$ and depth $d$,the moment of inertia is $I = \frac{b d^3}{12}$.
Substituting the value of $I$ into the formula for $\delta$:
$\delta = \frac{W l^3}{48 Y (\frac{b d^3}{12})}$
$\delta = \frac{W l^3}{48 Y} \times \frac{12}{b d^3}$
$\delta = \frac{W l^3}{4 b d^3 Y}$.

(D) The strain energy per unit volume $u$ is given by the formula $u = \frac{1}{2} \times \text{stress} \times \text{strain}$.
Since $\text{stress} = \frac{F}{A}$ and $\text{strain} = \frac{\text{stress}}{Y}$,we have $u = \frac{1}{2} \times \frac{\text{stress}^2}{Y} = \frac{1}{2} \times \frac{F^2}{A^2 Y}$.
Given that the wires have the same length,material ($Y$ is constant),and are stretched by the same force $F$,the energy density $u$ is inversely proportional to the square of the area of cross-section $A^2$.
Since $A = \pi r^2$,we have $A \propto d^2$ (where $d$ is the diameter),so $u \propto \frac{1}{(d^2)^2} = \frac{1}{d^4}$.
Therefore,the ratio of strain energy per unit volume for the smaller diameter wire $(d_S)$ to the larger diameter wire $(d_L)$ is $\frac{u_S}{u_L} = \left( \frac{d_L}{d_S} \right)^4$.
Given the ratio of diameters $d_S : d_L = 1 : 3$,we have $\frac{u_S}{u_L} = \left( \frac{3}{1} \right)^4 = 81 : 1$.

(A) The particle starts from rest,so the initial velocity $u = 0$.
The distance covered in one revolution is the circumference of the circle,which is $2 \pi r$.
The distance covered in two revolutions is $s = 2 \times 2 \pi r = 4 \pi r$.
Using the equation of motion $v^2 - u^2 = 2as$,where $a$ is the tangential acceleration:
$v^2 - 0^2 = 2 \times a \times (4 \pi r)$
$v^2 = 8 \pi r a$
Therefore,the tangential acceleration is $a = \frac{v^2}{8 \pi r}$.

(D) The kinetic energy is given by $K.E. = \frac{1}{2}mv^2$.
For a particle to just complete a vertical circular motion,the minimum velocity at the lowest point is $v_l = \sqrt{5rg}$.
The minimum velocity at the highest point is $v_h = \sqrt{rg}$.
The kinetic energy at the highest point is $(K.E.)_h = \frac{1}{2}m(v_h)^2 = \frac{1}{2}m(rg)$.
The kinetic energy at the lowest point is $(K.E.)_l = \frac{1}{2}m(v_l)^2 = \frac{1}{2}m(5rg)$.
Therefore,the ratio of kinetic energy at the highest point to that at the lowest point is $\frac{(K.E.)_h}{(K.E.)_l} = \frac{\frac{1}{2}m(rg)}{\frac{1}{2}m(5rg)} = \frac{1}{5} = 0.2$.

(A) In a vertical circular motion,the forces acting on the particle are gravity (a conservative force) and tension (which acts perpendicular to the displacement). Since the work done by tension is zero and gravity is a conservative force,the total mechanical energy (sum of kinetic and potential energy) of the particle remains constant throughout the motion. Therefore,the total energy is conserved.

(C) The equation for a damped harmonic oscillator is given by $x(t) = A e^{-bt/2m} \cos(\omega' t + \delta)$,where $A$ is the initial amplitude,$b$ is the damping constant,$m$ is the mass,$\omega'$ is the damped angular frequency,and $\delta$ is the initial phase.
During damping,the amplitude $A e^{-bt/2m}$ decreases exponentially with time.
The damped angular frequency $\omega' = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}$ is different from the natural frequency $\omega_0 = \sqrt{\frac{k}{m}}$,and the time period $T = \frac{2\pi}{\omega'}$ also changes.
The initial phase $\delta$ is determined by the initial conditions (position and velocity at $t = 0$) and remains constant regardless of the damping process.

(D) The path length of $SHM$ is equal to $2A$,where $A$ is the amplitude.
Given path length $= 0.01 \ m$,so $2A = 0.01 \ m$,which implies $A = 0.005 \ m$.
The frequency $f = 50 \ Hz$.
The maximum force $F_{max}$ acting on a particle in $SHM$ is given by $F_{max} = m \omega^2 A$.
Since $\omega = 2 \pi f$,we have $F_{max} = m (2 \pi f)^2 A = m (4 \pi^2 f^2) A$.
Substituting the values: $F_{max} = 1 \times 4 \times \pi^2 \times (50)^2 \times 0.005$.
$F_{max} = 4 \times \pi^2 \times 2500 \times 0.005$.
$F_{max} = 10000 \times \pi^2 \times 0.005 = 50 \pi^2 \ N$.

(A) At the mean position,the potential energy is zero and the kinetic energy is maximum. The velocity $v$ of mass $m_1$ at the mean position is $v = A\omega = A\sqrt{\frac{k}{m_1}}$.
When mass $m_2$ is placed on $m_1$,the momentum is conserved because the spring force is zero at the mean position. Let $v'$ be the new velocity:
$m_1 v = (m_1 + m_2) v'$
$v' = \frac{m_1 v}{m_1 + m_2} = \frac{m_1 A \sqrt{k/m_1}}{m_1 + m_2} = A \sqrt{\frac{k m_1}{(m_1 + m_2)^2}}$.
The new energy of the system is $E' = \frac{1}{2} k A_1^2 = \frac{1}{2} (m_1 + m_2) (v')^2$.
Substituting $v'$:
$\frac{1}{2} k A_1^2 = \frac{1}{2} (m_1 + m_2) \left( A^2 \frac{k m_1}{(m_1 + m_2)^2} \right)$.
$A_1^2 = A^2 \frac{m_1}{m_1 + m_2}$.
Therefore,$\frac{A_1}{A} = \sqrt{\frac{m_1}{m_1 + m_2}}$.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g_{eff}}}$.
In air,the effective acceleration due to gravity is $g_{eff} = g$.
In water,the bob experiences an upward buoyant force. The effective acceleration is $g_{eff}' = g(1 - \frac{\rho}{\sigma})$,where $\rho$ is the density of water $(10^3 \ kg/m^3)$ and $\sigma$ is the density of the bob $(\frac{9}{8} \times 10^3 \ kg/m^3)$.
$g_{eff}' = g(1 - \frac{10^3}{\frac{9}{8} \times 10^3}) = g(1 - \frac{8}{9}) = g(\frac{1}{9})$.
Now,the time period in water is $T_1 = 2\pi \sqrt{\frac{l}{g_{eff}'}} = 2\pi \sqrt{\frac{l}{g/9}} = 2\pi \sqrt{\frac{9l}{g}}$.
$T_1 = 3 \times (2\pi \sqrt{\frac{l}{g}}) = 3T$.

(B) The moment of inertia $(I)$ of a thin uniform rod rotating about an axis passing through its center and perpendicular to its length is given by $I_1 = \frac{ML^2}{12}$.
Since $I = MK^2$,where $K$ is the radius of gyration,we have $MK_1^2 = \frac{ML^2}{12}$,which gives $K_1 = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}$.
In the second case,the axis of rotation passes through one end and is perpendicular to the length of the rod. The moment of inertia is $I_2 = \frac{ML^2}{3}$.
Using $MK_2^2 = \frac{ML^2}{3}$,we get $K_2 = \frac{L}{\sqrt{3}}$.
The ratio of the radius of gyration in the first case to the second case is $\frac{K_1}{K_2} = \frac{L / (2\sqrt{3})}{L / \sqrt{3}} = \frac{1}{2}$.

(A) The total kinetic energy $(K.E.)$ of a body rolling without slipping is given by $K.E. = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$.
Since the objects roll without slipping,$v = R\omega$,so $\omega^2 = \frac{v^2}{R^2}$.
For a ring,the moment of inertia $I_{ring} = mR^2$. Thus,$K.E._{ring} = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)(\frac{v^2}{R^2}) = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$.
Given $K.E._{ring} = 4 \ J$,we have $mv^2 = 4 \ J$.
For a disc,the moment of inertia $I_{disc} = \frac{1}{2}mR^2$. Thus,$K.E._{disc} = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v^2}{R^2}) = \frac{1}{2}mv^2 + \frac{1}{4}mv^2 = \frac{3}{4}mv^2$.
Substituting $mv^2 = 4 \ J$,we get $K.E._{disc} = \frac{3}{4} \times 4 \ J = 3 \ J$.

(D) According to the Stefan-Boltzmann law,the energy emitted per second by a black body is given by $E = \sigma A T^4$.
Initially,$E = \sigma A T_1^4$,where $T_1 = 27 + 273 = 300 \ K$.
When the length and breadth are reduced to $1/3$ of their initial values,the new area $A' = (l/3) \times (b/3) = A/9$.
The new temperature $T_2 = 327 + 273 = 600 \ K$.
The new energy emitted is $E' = \sigma A' T_2^4$.
Taking the ratio: $\frac{E'}{E} = \frac{A'}{A} \times \left(\frac{T_2}{T_1}\right)^4$.
Substituting the values: $\frac{E'}{E} = \frac{1}{9} \times \left(\frac{600}{300}\right)^4$.
$\frac{E'}{E} = \frac{1}{9} \times (2)^4 = \frac{16}{9}$.
Therefore,$E' = \frac{16 E}{9}$.

(C) The given equation of the progressive wave is $y = 12 \sin(5t - 4x)$.
Comparing this with the standard wave equation $y = A \sin(\omega t - kx)$,we get the wave number $k = 4 \ rad/cm$.
The phase difference $\Delta \phi$ between two points separated by a distance $\Delta x$ is given by the relation $\Delta \phi = k \cdot \Delta x$.
We are given the phase difference $\Delta \phi = 90^{\circ} = \frac{\pi}{2} \ rad$.
Substituting the values,we have $\frac{\pi}{2} = 4 \cdot \Delta x$.
Therefore,$\Delta x = \frac{\pi}{2 \times 4} = \frac{\pi}{8} \ cm$.

(A) The frequency of a vibrating wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,which implies $f \propto \sqrt{T}$.
Let $f_1$ be the frequency at tension $T_1 = 225 \ N$ and $f_2$ be the frequency at tension $T_2 = 256 \ N$.
Since $f \propto \sqrt{T}$,we have $\frac{f_1}{f_2} = \sqrt{\frac{225}{256}} = \frac{15}{16}$,so $f_2 = \frac{16}{15} f_1$.
Given that the beat frequency is $6 \ Hz$ in both cases,let $x$ be the frequency of the tuning fork.
For the first case: $f_1 = x - 6$ (assuming $x > f_1$).
For the second case: $f_2 = x + 6$ (since $f_2 > f_1$,the frequency must increase past the fork frequency).
Substituting $f_1$ and $f_2$: $x + 6 = \frac{16}{15} (x - 6)$.
$15(x + 6) = 16(x - 6) \implies 15x + 90 = 16x - 96$.
$x = 90 + 96 = 186 \ Hz$.

(B) The apparent frequency $F$ observed by an observer moving with velocity $v_o$ relative to a stationary source emitting frequency $f_0$ is given by $F = \left( \frac{V \pm v_o}{V} \right) f_0$.
When the observer moves towards the stationary source with velocity $V_1$,the apparent frequency is $F_1 = \left( \frac{V + V_1}{V} \right) f_0$ $(i)$.
When the observer moves away from the stationary source with velocity $V_1$,the apparent frequency is $F_2 = \left( \frac{V - V_1}{V} \right) f_0$ $(ii)$.
Dividing equation $(i)$ by $(ii)$,we get $\frac{F_1}{F_2} = \frac{V + V_1}{V - V_1}$.
Given $\frac{F_1}{F_2} = 2$,we have $2 = \frac{V + V_1}{V - V_1}$.
Cross-multiplying gives $2(V - V_1) = V + V_1$,which simplifies to $2V - 2V_1 = V + V_1$.
Rearranging the terms,we get $2V - V = 2V_1 + V_1$,which leads to $V = 3V_1$.
Therefore,$\frac{V}{V_1} = 3$.

(B) For an open organ pipe,the end correction $\Delta L$ is related to the inner radius $r$ by the formula $\Delta L = 0.6 \times r$ at each end. Since an open pipe has two ends,the total end correction is $\Delta L_{total} = 2 \times (0.6 \times r) = 1.2 \times r$.
Given that the total end correction $\Delta L = 0.8 \ cm$,we have:
$0.8 = 1.2 \times r$
$r = \frac{0.8}{1.2} \ cm$
$r = \frac{8}{12} \ cm = \frac{2}{3} \ cm$.

(D) Let $f_0 = \frac{v}{2L}$ be the fundamental frequency of the open pipe.
Its second overtone is $3f_0 = \frac{3v}{2L}$.
Let $f_c = \frac{v}{4L}$ be the fundamental frequency of the closed pipe.
The frequencies of a closed pipe are given by $(2n-1)f_c$,where $n=1, 2, 3, \dots$.
The third overtone corresponds to $n=4$,so $(f_3)_{\text{closed}} = (2(4)-1)f_c = 7f_c = \frac{7v}{4L}$.
According to the problem,$(f_3)_{\text{closed}} - (3f_0) = 150 \ Hz$.
Substituting the expressions: $\frac{7v}{4L} - \frac{3v}{2L} = 150$.
$\frac{7v}{4L} - \frac{6v}{4L} = 150$.
$\frac{v}{4L} = 150$.
Since the fundamental frequency of the open pipe is $f_0 = \frac{v}{2L}$,we have $f_0 = 2 \times \frac{v}{4L} = 2 \times 150 = 300 \ Hz$.

(B) The velocity $V$ of a transverse wave travelling on a stretched string is given by the formula $V = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $\mu = \rho \cdot A = \rho \cdot \pi r^2$,where $\rho$ is the density of the material and $r$ is the radius of the string,we have $V = \sqrt{\frac{T}{\rho \cdot \pi r^2}}$.
Given that the material is the same ($\rho$ is constant) and the tension $T$ is the same,we find that $V \propto \frac{1}{r}$.
Therefore,the ratio of the speeds is $\frac{V_A}{V_B} = \frac{r_B}{r_A}$.
Given $r_A = 2r_B$,we substitute this into the ratio: $\frac{V_A}{V_B} = \frac{r_B}{2r_B} = \frac{1}{2}$.

(B) Given: Peak current $I_{0} = \frac{2}{\pi} \ A$,frequency $f = 50 \ Hz$,and mutual inductance $M = 1 \ H$.
The angular frequency is $\omega = 2 \pi f = 2 \pi (50) = 100 \pi \ rad/s$.
The current in the primary coil is $I = I_{0} \sin(\omega t)$.
The induced e.m.f. in the secondary coil is $\mathcal{E} = M \frac{dI}{dt}$.
Differentiating the current with respect to time,$\frac{dI}{dt} = I_{0} \omega \cos(\omega t)$.
The peak value of the induced e.m.f. is $\mathcal{E}_{0} = M \cdot I_{0} \cdot \omega$.
Substituting the values: $\mathcal{E}_{0} = 1 \times (\frac{2}{\pi}) \times (100 \pi) = 200 \ V$.

(A) In an $LC$ parallel resonant circuit,at resonance,the inductive reactance $(X_L)$ is equal to the capacitive reactance $(X_C)$.
This leads to a condition where the total impedance of the circuit becomes very high (theoretically infinite in an ideal circuit).
Because the impedance is very high,the current drawn from the source is minimal.
Therefore,the $LC$ parallel resonant circuit acts as a high-impedance circuit at resonance.

(A) According to Bohr's postulates for the Hydrogen atom,an electron revolves around the nucleus in specific orbits called stationary orbits.
In these stationary orbits,the electron does not radiate energy,even though it is accelerating due to its circular motion (its velocity vector changes continuously).
Therefore,the electron does not radiate light while revolving in a stationary orbit.

(A) According to Bohr's theory,the wavelength $\lambda$ of the emitted radiation is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$.
Since $\Delta E = \frac{hc}{\lambda}$,the wavelength $\lambda$ is inversely proportional to the energy difference $\Delta E$ between the orbits.
To obtain the maximum wavelength,the energy difference $\Delta E$ must be minimum.
The energy difference between consecutive orbits decreases as the principal quantum number increases.
Comparing the energy gaps: $(E_5 - E_4) < (E_4 - E_3) < (E_3 - E_2) < (E_2 - E_1)$.
Therefore,the transition from $n=5$ to $p=4$ results in the smallest energy difference and consequently the maximum wavelength.

(A) Let the initial capacitance of each capacitor be $C$. When connected in series to a battery of emf $V$,the charge $q$ on both capacitors is the same.
Let $V_1$ be the potential difference across the air capacitor (capacitance $C$) and $V_2$ be the potential difference across the dielectric-filled capacitor (capacitance $KC$).
Since they are in series,$V_1 + V_2 = V$.
Using $q = CV$,we have $V_1 = \frac{q}{C}$ and $V_2 = \frac{q}{KC}$.
Substituting these into the sum equation: $\frac{q}{C} + \frac{q}{KC} = V$.
$\frac{q}{C} (1 + \frac{1}{K}) = V \Rightarrow \frac{q}{C} (\frac{K+1}{K}) = V$.
Therefore,the charge $q = \frac{CKV}{K+1}$.
The potential difference across the air capacitor is $V_1 = \frac{q}{C} = \frac{KV}{K+1}$.

(D) The energy stored in a capacitor is given by $U = \frac{1}{2} C V^2$.
The work done in increasing the potential from $V_i$ to $V_f$ is $\Delta U = \frac{1}{2} C (V_f^2 - V_i^2)$.
For the first case,$W = \frac{1}{2} C (10^2 - 5^2) = \frac{1}{2} C (100 - 25) = \frac{75}{2} C$.
For the second case,$W_2 = \frac{1}{2} C (15^2 - 10^2) = \frac{1}{2} C (225 - 100) = \frac{125}{2} C$.
Taking the ratio: $\frac{W_2}{W} = \frac{125/2 C}{75/2 C} = \frac{125}{75} = \frac{5}{3} \approx 1.67$.
Therefore,$W_2 = 1.67 W$.

(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A'}{d'}$,where $A'$ is the area and $d'$ is the separation.
Given that each capacitor has an area $A' = \frac{A}{3}$,the individual capacitances are:
$C_1 = \frac{\varepsilon_0 (A/3)}{d} = \frac{\varepsilon_0 A}{3d}$
$C_2 = \frac{\varepsilon_0 (A/3)}{2d} = \frac{\varepsilon_0 A}{6d}$
$C_3 = \frac{\varepsilon_0 (A/3)}{3d} = \frac{\varepsilon_0 A}{9d}$
Since the capacitors are connected in parallel,the equivalent capacitance $C_{eq}$ is the sum of the individual capacitances:
$C_{eq} = C_1 + C_2 + C_3$
$C_{eq} = \frac{\varepsilon_0 A}{3d} + \frac{\varepsilon_0 A}{6d} + \frac{\varepsilon_0 A}{9d}$
$C_{eq} = \frac{\varepsilon_0 A}{d} \left( \frac{1}{3} + \frac{1}{6} + \frac{1}{9} \right)$
Taking the least common multiple of $3, 6, 9$ which is $18$:
$C_{eq} = \frac{\varepsilon_0 A}{d} \left( \frac{6 + 3 + 2}{18} \right) = \frac{11 \varepsilon_0 A}{18d}$

(C) The critical frequency $(f_c)$ of the ionosphere is the maximum frequency of radio waves that can be reflected back to the Earth by the ionospheric layers.
It is given by the formula $f_c = 9 \sqrt{N_{max}}$, where $N_{max}$ is the maximum electron density of the ionosphere in electrons per cubic meter $(m^{-3})$.
In this context, the constant $9$ is derived from physical constants, and the expression is $9 \sqrt{N}$.

(C) The current for full-scale deflection ($20$ divisions) is given by $I_1 = \frac{V}{R_g + R_1} = \frac{2}{30 + 1970} = \frac{2}{2000} = 1 \times 10^{-3} \text{ A} = 1 \text{ mA}$.
Since the deflection $\theta$ is directly proportional to the current $I$ $(\theta \propto I)$,to reduce the deflection from $20$ divisions to $10$ divisions,the current must be halved.
Thus,$I_2 = \frac{I_1}{2} = 0.5 \times 10^{-3} \text{ A}$.
Let the new total series resistance be $R_{total}$. Then $I_2 = \frac{V}{R_g + R_{total}}$.
$0.5 \times 10^{-3} = \frac{2}{30 + R_{total}}$.
$30 + R_{total} = \frac{2}{0.5 \times 10^{-3}} = 4000 \Omega$.
$R_{total} = 4000 - 30 = 3970 \Omega$.

(A) The potential drop across the potentiometer wire is $V = I \cdot R$, where $R = \rho \cdot \frac{L}{A}$. The potential gradient $k$ is given by $k = \frac{V}{L} = \frac{I \cdot \rho}{A}$.
When the length $L$ of the potentiometer wire is increased, the total resistance $R$ of the wire increases.
Since the driving source voltage and internal resistance remain constant, the current $I$ flowing through the potentiometer wire decreases.
Consequently, the potential gradient $k = \frac{V}{L}$ decreases.
The balancing length $l$ is given by the relation $E = k \cdot l$, where $E$ is the $EMF$ of the cell.
Since $E$ is constant and $k$ decreases, the balancing length $l = \frac{E}{k}$ must increase.

(C) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ and charge $q$ accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$.
For an electron: $\lambda = \frac{h}{\sqrt{2meV}}$.
For a proton of mass $M$ and charge $e$ accelerated through potential $9V$: $\lambda' = \frac{h}{\sqrt{2Me(9V)}} = \frac{h}{3\sqrt{2MeV}}$.
Dividing $\lambda'$ by $\lambda$: $\frac{\lambda'}{\lambda} = \frac{\frac{h}{3\sqrt{2MeV}}}{\frac{h}{\sqrt{2meV}}} = \frac{1}{3} \sqrt{\frac{m}{M}}$.
Therefore,$\lambda' = \frac{\lambda}{3} \sqrt{\frac{m}{M}}$.

(D) The photoelectric equation is given by $eV_0 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$,where $\lambda_0$ is the threshold wavelength.
For the first case: $eV = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \dots (i)$
For the second case: $e \left( \frac{V}{6} \right) = hc \left( \frac{1}{3\lambda} - \frac{1}{\lambda_0} \right) \dots (ii)$
Dividing equation $(i)$ by equation $(ii)$:
$6 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{3\lambda} - \frac{1}{\lambda_0}} = \frac{\frac{\lambda_0 - \lambda}{\lambda \lambda_0}}{\frac{\lambda_0 - 3\lambda}{3\lambda \lambda_0}} = \frac{3(\lambda_0 - \lambda)}{\lambda_0 - 3\lambda}$
$6(\lambda_0 - 3\lambda) = 3\lambda_0 - 3\lambda$
$6\lambda_0 - 18\lambda = 3\lambda_0 - 3\lambda$
$3\lambda_0 = 15\lambda$
$\lambda_0 = 5\lambda$

(A) According to the photoelectric equation: $\frac{hc}{\lambda} = \phi + eV_s$.
Here,$V_s$ is the stopping potential and $\phi$ is the work function.
The kinetic energy of the emitted photoelectrons is given by $K_{max} = eV_s = \frac{1}{2}mv^2$.
If the velocity $v$ of the emitted photoelectrons is kept constant,the kinetic energy $K_{max}$ remains constant.
However,as the incident wavelength $\lambda$ is decreased,the energy of the incident photon $E = \frac{hc}{\lambda}$ increases.
Since $E = \phi + K_{max}$,and $K_{max}$ is constant while $E$ increases,the work function $\phi$ of the material would effectively need to change or the stopping potential $V_s$ must increase to balance the equation if we consider the energy conservation principle for a fixed material.
Actually,for a fixed material,if $\lambda$ decreases,$K_{max}$ increases. If the problem states the velocity is kept constant,it implies an external adjustment or a misunderstanding of the physical constraints; however,based on the standard photoelectric equation $\frac{hc}{\lambda} = \phi + eV_s$,if $\lambda$ decreases,the energy of the incident photon increases,which leads to an increase in the stopping potential $V_s$.

(C) Initial magnetic flux,$\phi_1 = 4 \times 10^{-4} \ Wb$.
Final magnetic flux,$\phi_2 = 10 \% \text{ of } \phi_1 = 0.1 \times 4 \times 10^{-4} = 4 \times 10^{-5} \ Wb$.
Induced emf,$\varepsilon = 0.72 \ mV = 0.72 \times 10^{-3} \ V = 72 \times 10^{-5} \ V$.
According to Faraday's law of induction,the magnitude of induced emf is given by $|\varepsilon| = |\frac{\Delta \phi}{\Delta t}|$.
Therefore,$\Delta t = \frac{|\phi_2 - \phi_1|}{|\varepsilon|}$.
$\Delta t = \frac{|4 \times 10^{-5} - 4 \times 10^{-4}|}{72 \times 10^{-5}}$.
$\Delta t = \frac{|0.4 \times 10^{-4} - 4 \times 10^{-4}|}{72 \times 10^{-5}} = \frac{3.6 \times 10^{-4}}{7.2 \times 10^{-4}}$.
$\Delta t = \frac{3.6}{7.2} = 0.5 \ s$.

(D) When a particle of charge $q$ and mass $m$ is accelerated through a potential difference $V$,its kinetic energy is given by $\frac{1}{2}mv^2 = qV$,which implies $v = \sqrt{\frac{2qV}{m}}$.
When this particle enters a uniform magnetic field $B$ perpendicular to its velocity,it follows a circular path of radius $r = \frac{mv}{qB}$.
Substituting the expression for $v$,we get $r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{q}}$.
Since $q$,$V$,and $B$ are the same for both particles,$r \propto \sqrt{m}$,which means $r^2 \propto m$.
Therefore,the ratio of the masses is $\frac{m_X}{m_Y} = \left( \frac{r_1}{r_2} \right)^2$.

(A) The magnetic field inside a long solenoid with an air core is given by $B_0 = \mu_0 ni$.
When a material with magnetic susceptibility $\chi$ is placed inside the solenoid,the relative permeability of the material is $\mu_r = 1 + \chi$.
The magnetic field inside the solenoid with the core becomes $B = \mu_r B_0$.
Substituting the values,we get $B = (1 + \chi) \mu_0 ni$.
Therefore,the correct option is $A$.

(C) The magnetic permeability $\mu$ is defined as the ratio of magnetic induction $B$ to the magnetic field intensity $H$,given by $\mu = \frac{B}{H}$.
Since magnetic flux $\phi = B \cdot A$,we have $B = \frac{\phi}{A}$.
Substituting this into the permeability formula: $\mu = \frac{\phi}{A \cdot H}$.
Given values:
$\phi = 6 \times 10^{-4} \text{ Wb}$
$A = 3 \text{ cm}^2 = 3 \times 10^{-4} \text{ m}^2$
$H = 2000 \text{ A/m} = 2 \times 10^3 \text{ A/m}$
Calculating $\mu$:
$\mu = \frac{6 \times 10^{-4}}{(3 \times 10^{-4}) \times (2 \times 10^3)} = \frac{6 \times 10^{-4}}{6 \times 10^{-1}} = 10^{-3} \text{ Wb/(A} \cdot \text{m)}$.

(A) The resolving power $(RP)$ of a telescope is defined as the reciprocal of the minimum angular separation between two objects that can be just distinguished by the telescope.
It is given by the formula:
$RP = \frac{D}{1.22 \lambda}$
where $D$ is the diameter of the objective lens and $\lambda$ is the wavelength of light used.
From the formula,it is clear that $RP \propto \frac{1}{\lambda}$.
Therefore,when the wavelength of light $(\lambda)$ decreases,the resolving power of the telescope increases.

(B) The refractive index $n$ is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in the medium,$n = \frac{c}{v}$.
Given that the velocity is reduced by $20 \%$,the new velocity $v = c - 0.20c = 0.80c = \frac{4}{5}c$.
Thus,the refractive index $n = \frac{c}{v} = \frac{c}{0.8c} = \frac{1}{0.8} = 1.25$ or $\frac{5}{4}$.
According to Snell's Law for small angles,$n_1 \sin i = n_2 \sin r$. Since the incident medium is air $(n_1 = 1)$,we have $i = n r$,so $r = \frac{i}{n} = \frac{i}{1.25} = 0.8i = \frac{4i}{5}$.
The angle of deviation $\delta$ is given by $\delta = i - r$.
Substituting the value of $r$,we get $\delta = i - 0.8i = 0.2i = \frac{i}{5}$.

(B) light emitting diode $(LED)$ is a heavily doped $p-n$ junction diode which emits light when it is forward biased.
In the schematic symbol of an $LED$,the arrows point away from the diode,indicating the emission of light.
Option $A$ represents a photodiode (arrows pointing towards the diode).
Option $B$ represents an $LED$ (arrows pointing away from the diode).
Option $C$ represents a standard $p-n$ junction diode.
Option $D$ represents a Zener diode.
Therefore,the correct schematic symbol for an $LED$ is shown in option $B$.

(C) The Barkhausen criterion states that for an oscillator to produce sustained oscillations,the loop gain of the feedback amplifier must be equal to unity.
Mathematically,this is expressed as $A \beta = 1$,where $A$ is the voltage gain of the amplifier without feedback and $\beta$ is the feedback factor.
Additionally,the total phase shift around the loop must be $0^\circ$ or an integer multiple of $360^\circ$.

(A) Brewster's law states that the tangent of the polarizing angle $(i_p)$ is equal to the refractive index $(\mu)$ of the medium, i.e., $\tan(i_p) = \mu$.
Since the refractive index $(\mu)$ of a material depends on the wavelength $(\lambda)$ of light (due to dispersion), the polarizing angle $(i_p)$ also depends on the wavelength.
Because different colours have different wavelengths, the polarizing angle is different for different colours.

(B) For constructive interference (bright bands),the path difference is $\Delta x = n\lambda$,where $n = 0, 1, 2, \dots$
For destructive interference (dark bands),the path difference is $\Delta x = (n + \frac{1}{2})\lambda$,where $n = 0, 1, 2, \dots$
The $4^{\text{th}}$ bright band corresponds to $n = 4$ $(\Delta x = 4\lambda)$ and the $5^{\text{th}}$ bright band corresponds to $n = 5$ $(\Delta x = 5\lambda)$.
$A$ dark band formed between the $4^{\text{th}}$ and $5^{\text{th}}$ bright band corresponds to $n = 4$ in the dark band formula: $\Delta x = (4 + \frac{1}{2})\lambda = 4.5\lambda$.
Given $\lambda = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m} = 6 \times 10^{-5} \text{ cm}$.
Path difference $\Delta x = 4.5 \times (6 \times 10^{-5} \text{ cm}) = 27 \times 10^{-5} \text{ cm} = 2.7 \times 10^{-4} \text{ cm}$.

(C) The resultant intensity $I_R$ for two interfering waves with intensities $I_1$ and $I_2$ and phase difference $\phi$ is given by $I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
Given $I_1 = I$ and $I_2 = 9I$.
At point $P$,the phase difference $\phi_P = \frac{\pi}{2}$.
$I_P = I + 9I + 2\sqrt{I \cdot 9I} \cos(\frac{\pi}{2}) = 10I + 6I(0) = 10I$.
At point $Q$,the phase difference $\phi_Q = \pi$.
$I_Q = I + 9I + 2\sqrt{I \cdot 9I} \cos(\pi) = 10I + 6I(-1) = 10I - 6I = 4I$.
The difference between the resultant intensities at points $P$ and $Q$ is $\Delta I = I_P - I_Q = 10I - 4I = 6I$.