MHT CET 2017 Physics Question Paper with Answer and Solution

(C) The acceleration due to gravity at a depth $d$ below the surface of the earth is given by the formula:
$g' = g\left(1 - \frac{d}{R}\right)$
Given that the acceleration due to gravity at depth $d$ is $\frac{1}{n}$ times the value at the surface,we have:
$g' = \frac{g}{n}$
Substituting this into the formula:
$\frac{g}{n} = g\left(1 - \frac{d}{R}\right)$
Dividing both sides by $g$:
$\frac{1}{n} = 1 - \frac{d}{R}$
Rearranging the terms to solve for $d$:
$\frac{d}{R} = 1 - \frac{1}{n}$
$\frac{d}{R} = \frac{n-1}{n}$
$d = R\left(\frac{n-1}{n}\right)$

(C) The binding energy of a satellite at rest on the earth's surface is given by $E_1 = \frac{GMm}{2R}$.
The binding energy of a satellite of mass $m$ revolving in a circular orbit at a height $h$ above the earth's surface is $E_2 = \frac{GMm}{2(R+h)}$.
Taking the ratio of the two energies:
$\frac{E_1}{E_2} = \frac{\frac{GMm}{2R}}{\frac{GMm}{2(R+h)}} = \frac{R+h}{R}$.

(C) For a rigid diatomic molecule,the degrees of freedom $f = 5$.
The molar specific heat at constant volume is $C_V = \frac{f}{2} R = \frac{5}{2} R$.
The molar specific heat at constant pressure is $C_P = C_V + R = \frac{5}{2} R + R = \frac{7}{2} R$.
We are given the relation $R = n C_P$.
Substituting the value of $C_P$,we get $R = n (\frac{7}{2} R)$.
Dividing both sides by $R$,we get $1 = n (\frac{7}{2})$.
Therefore,$n = \frac{2}{7} \approx 0.2857$.

(B) The force equation for the elevator moving upward with acceleration $a$ is given by $T_{actual} = m(g+a)$.
For a safe journey,the maximum tension $T$ that the rope can bear must be at least equal to the actual tension,so $T = m(g+a)$.
The stress in the rope is defined as $\sigma = \frac{T}{A}$,where $A$ is the cross-sectional area of the rope.
Assuming the rope is cylindrical with diameter $d$,the area $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.
Substituting this into the tension equation: $T = \frac{m(g+a)}{\pi d^2 / 4} = \frac{4 m(g+a)}{\pi d^2}$.
Rearranging for $d^2$,we get $d^2 = \frac{4 m(g+a)}{\pi T}$.
Taking the square root of both sides,the minimum diameter is $d = [\frac{4 m(g+a)}{\pi T}]^{1/2}$.

(A) The upward force due to surface tension $(F)$ is given by the formula: $F = T \cdot L$,where $T$ is the surface tension and $L$ is the inner circumference of the capillary tube.
Given: $F = 105 \text{ dyne} = 105 \times 10^{-5} \text{ N} = 1.05 \times 10^{-3} \text{ N}$.
Surface tension $T = 7 \times 10^{-2} \text{ N/m}$.
We need to find the circumference $L$.
Using the formula $L = F / T$:
$L = (1.05 \times 10^{-3} \text{ N}) / (7 \times 10^{-2} \text{ N/m})$
$L = 0.15 \times 10^{-1} \text{ m} = 0.015 \text{ m}$.
Converting to centimeters: $L = 0.015 \times 100 \text{ cm} = 1.5 \text{ cm}$.
Thus,the inner circumference of the capillary tube is $1.5 \text{ cm}$.

(B) Let $R$ be the radius of the bigger drop and $r$ be the radius of a single small water drop.
Since the volume remains conserved,the volume of the big drop equals the sum of the volumes of $n$ small drops:
$\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$
$R^3 = n r^3 \Rightarrow R = n^{1/3} r$
Surface energy of $n$ small drops is $E_n = n \times (4 \pi r^2 T)$,where $T$ is the surface tension.
Surface energy of the big drop is $E = 4 \pi R^2 T$.
The ratio of the surface energy of $n$ droplets to the surface energy of the big drop is:
$\frac{E_n}{E} = \frac{n \times 4 \pi r^2 T}{4 \pi R^2 T} = \frac{n r^2}{R^2}$
Substituting $R = n^{1/3} r$:
$\frac{E_n}{E} = \frac{n r^2}{(n^{1/3} r)^2} = \frac{n r^2}{n^{2/3} r^2} = n^{1 - 2/3} = n^{1/3} = \sqrt[3]{n}$
Thus,the ratio is $\sqrt[3]{n}: 1$.

(B) The thermal expansion of the rod if it were free to expand is given by $\Delta L = \alpha L T$.
To prevent this expansion,a compressive force $F$ must be applied.
The stress produced is $\sigma = \frac{F}{A}$.
The strain produced is $\epsilon = \frac{\Delta L}{L_{new}}$,where $L_{new} = L(1 + \alpha T)$.
Using Young's modulus $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L / L(1 + \alpha T)}$.
Rearranging for $F$,we get $F = \frac{Y A \Delta L}{L(1 + \alpha T)}$.
Substituting $\Delta L = \alpha L T$,we get $F = \frac{Y A (\alpha L T)}{L(1 + \alpha T)}$.
Therefore,$F = \frac{Y A \alpha T}{1 + \alpha T}$.

(B) The critical velocity at the topmost point is $v = \sqrt{rg}$.
Using the principle of conservation of energy,the velocity $v'$ when the string becomes horizontal is given by $v'^2 = v^2 + 2g(r)$,where $r$ is the radius of the circle.
Substituting $v^2 = rg$,we get $v'^2 = rg + 2rg = 3rg$.
The centripetal acceleration $a_c$ is given by the formula $a_c = \frac{v'^2}{r}$.
Substituting the value of $v'^2$,we get $a_c = \frac{3rg}{r} = 3g$.

(A) In $S.H.M.$,the displacement of a particle is given by $x(t) = A \cos(\omega t + \phi)$.
Since the particle starts from the extreme position,at $t = 0$,$x = A$,which implies $\phi = 0$. Thus,$x(t) = A \cos(\omega t)$.
The acceleration $a(t)$ is given by the second derivative of displacement: $a(t) = \frac{d^2x}{dt^2} = -\omega^2 A \cos(\omega t)$.
We can rewrite the acceleration as $a(t) = \omega^2 A \cos(\omega t + \pi)$.
Comparing the phase of displacement $(\omega t)$ and acceleration $(\omega t + \pi)$,the phase difference is $\pi \ rad$.

(A) Let the positions of the particle be $x_1$ and $x_2$ at the two instants.
In $S.H.M.$,acceleration $a = -\omega^2 x$. Considering magnitudes,$\alpha = \omega^2 |x_1|$ and $\beta = \omega^2 |x_2|$.
Thus,$|x_1| = \frac{\alpha}{\omega^2}$ and $|x_2| = \frac{\beta}{\omega^2}$.
The velocity in $S.H.M.$ is given by $v^2 = \omega^2(A^2 - x^2)$.
For the first instant: $u^2 = \omega^2 A^2 - \omega^2 x_1^2$ . . . $(i)$
For the second instant: $v^2 = \omega^2 A^2 - \omega^2 x_2^2$ . . . $(ii)$
Subtracting $(ii)$ from $(i)$: $u^2 - v^2 = \omega^2(x_2^2 - x_1^2) = \omega^2(x_2 - x_1)(x_2 + x_1)$.
Since $\alpha = \omega^2 x_1$ and $\beta = \omega^2 x_2$,we have $\alpha + \beta = \omega^2(x_1 + x_2)$.
Substituting this into the equation: $u^2 - v^2 = \omega^2(x_2 - x_1) \cdot \frac{\alpha + \beta}{\omega^2}$.
Therefore,the distance between the two positions is $|x_2 - x_1| = \frac{u^2 - v^2}{\alpha + \beta}$.

(A) The potential energy of a simple harmonic oscillator is given by the formula $U = \frac{1}{2} k A^2$,where $k$ is the force constant and $A$ is the amplitude.
For a simple pendulum,the restoring force is $F = -Mg \sin \theta \approx -Mg \theta$ for small angles.
Since $\theta = \frac{x}{L}$,the force is $F = -\frac{Mg}{L} x$.
Comparing this with $F = -kx$,we get the force constant $k = \frac{Mg}{L}$.
Substituting $k$ into the potential energy formula: $U = \frac{1}{2} (\frac{Mg}{L}) A^2 = \frac{MgA^2}{2L}$.

(D) The displacement of a particle performing $S.H.M.$ starting from the equilibrium position is given by $x = A \sin(\omega t)$.
The velocity of the particle is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.
Given: $v = \pi \ m \ s^{-1}$,$T = 16 \ s$,and $t = 2 \ s$.
The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{16} = \frac{\pi}{8} \ rad \ s^{-1}$.
Substituting the values into the velocity equation:
$\pi = A \times \frac{\pi}{8} \times \cos\left(\frac{\pi}{8} \times 2\right)$
$\pi = A \times \frac{\pi}{8} \times \cos\left(\frac{\pi}{4}\right)$
$1 = \frac{A}{8} \times \frac{1}{\sqrt{2}}$
$A = 8\sqrt{2} \ m$.

(D) The given equation is $y=5(\sin 4 \pi t+\sqrt{3} \cos 4 \pi t)$.
Multiplying by $5$,we get $y=5 \sin 4 \pi t+5 \sqrt{3} \cos 4 \pi t$.
This is in the form $y=A_1 \sin \omega t+A_2 \cos \omega t$,where $A_1=5$ and $A_2=5 \sqrt{3}$.
The resultant amplitude $A$ is given by $A=\sqrt{A_1^2+A_2^2}$.
Substituting the values,$A=\sqrt{(5)^2+(5 \sqrt{3})^2}$.
$A=\sqrt{25+75} = \sqrt{100}$.
Thus,$A=10$ units.

(D) According to the principle of conservation of angular momentum,since no external torque acts on the system,the initial angular momentum equals the final angular momentum:
$I_1 \omega_1 = (I_1 + I_2) \omega_2$
$\omega_2 = \frac{I_1 \omega_1}{I_1 + I_2}$
The initial rotational kinetic energy is $E_1 = \frac{1}{2} I_1 \omega_1^2$.
The final rotational kinetic energy is $E_2 = \frac{1}{2} (I_1 + I_2) \omega_2^2$.
Substituting $\omega_2$ into the expression for $E_2$:
$E_2 = \frac{1}{2} (I_1 + I_2) \left( \frac{I_1 \omega_1}{I_1 + I_2} \right)^2 = \frac{1}{2} \frac{I_1^2 \omega_1^2}{I_1 + I_2}$.
The energy lost is $\Delta E = E_1 - E_2 = \frac{1}{2} I_1 \omega_1^2 - \frac{1}{2} \frac{I_1^2 \omega_1^2}{I_1 + I_2}$.
$\Delta E = \frac{1}{2} I_1 \omega_1^2 \left( 1 - \frac{I_1}{I_1 + I_2} \right) = \frac{1}{2} I_1 \omega_1^2 \left( \frac{I_1 + I_2 - I_1}{I_1 + I_2} \right)$.
$\Delta E = \frac{1}{2} \left[ \frac{I_1 I_2}{I_1 + I_2} \right] \omega_1^2$.

(C) Given: Moment of inertia $I = 2 \ kg \ m^2$, initial angular velocity $\omega_0 = 60 \ rad \ s^{-1}$, and time to stop $t_{total} = 5 \ min = 300 \ s$.
Since the wheel comes to rest, final angular velocity $\omega_f = 0$.
The angular acceleration $\alpha$ is given by $\alpha = \frac{\omega_f - \omega_0}{t_{total}} = \frac{0 - 60}{300} = -0.2 \ rad \ s^{-2}$.
We need the angular momentum $3 \ minutes$ before it stops. This corresponds to the time $t = 5 - 3 = 2 \ minutes$ from the start.
$t = 2 \ min = 120 \ s$.
The angular velocity at $t = 120 \ s$ is $\omega = \omega_0 + \alpha t = 60 + (-0.2)(120) = 60 - 24 = 36 \ rad \ s^{-1}$.
The angular momentum $L$ is $L = I\omega = 2 \times 36 = 72 \ kg \ m^2 \ s^{-1}$.

(C) Let the initial angular velocity be $\omega_0$ and uniform angular retardation be $\alpha$.
Using the equation of rotational motion $\omega^2 = \omega_0^2 - 2\alpha\theta$,where $\theta = 2\pi n$.
At time $t$,$\omega = \frac{\omega_0}{4}$.
So,$(\frac{\omega_0}{4})^2 = \omega_0^2 - 2\alpha(2\pi n) \implies \frac{\omega_0^2}{16} = \omega_0^2 - 4\pi n\alpha$.
$4\pi n\alpha = \omega_0^2(1 - \frac{1}{16}) = \frac{15\omega_0^2}{16}$.
Thus,$2\alpha = \frac{15\omega_0^2}{32\pi n}$.
Now,for the fan to come to rest,final angular velocity $\omega_f = 0$.
Let $n'$ be the total revolutions from switch off to rest.
$0^2 = \omega_0^2 - 2\alpha(2\pi n')$.
$2\alpha(2\pi n') = \omega_0^2$.
Substituting $2\alpha$: $(\frac{15\omega_0^2}{32\pi n})(2\pi n') = \omega_0^2$.
$\frac{15n'}{16n} = 1 \implies n' = \frac{16n}{15}$.

(D) Given: Initial angular velocity $\omega_0 = 0 \ rad \ s^{-1}$,final angular velocity $\omega = 24 \ rad \ s^{-1}$,and time $t = 8 \ s$.
First,calculate the constant angular acceleration $\alpha$ using the formula $\alpha = \frac{\omega - \omega_0}{t}$.
$\alpha = \frac{24 - 0}{8} = 3 \ rad \ s^{-2}$.
Now,calculate the total angle $\theta$ turned through using the kinematic equation $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$.
$\theta = 0 \times 8 + \frac{1}{2} \times 3 \times (8)^2$.
$\theta = \frac{1}{2} \times 3 \times 64 = 3 \times 32 = 96 \ rad$.

(C) The total kinetic energy of a rolling solid sphere is the sum of its translational and rotational kinetic energies.
$KE_{total} = \frac{1}{2} m V^2 + \frac{1}{2} I \omega^2$
For a solid sphere,the moment of inertia $I = \frac{2}{5} m r^2$ and the rolling condition is $V = r \omega$.
$KE_{total} = \frac{1}{2} m V^2 + \frac{1}{2} (\frac{2}{5} m r^2) (\frac{V}{r})^2 = \frac{1}{2} m V^2 + \frac{1}{5} m V^2 = \frac{7}{10} m V^2$.
When the sphere compresses the spring by a maximum distance $x$,all its kinetic energy is converted into the potential energy of the spring $(U = \frac{1}{2} k x^2)$.
$\frac{1}{2} k x^2 = \frac{7}{10} m V^2$
$x^2 = \frac{14}{10} \frac{m V^2}{k} = \frac{1.4 \times 2 \times 6^2}{36} = \frac{1.4 \times 2 \times 36}{36} = 2.8$.
$x = \sqrt{2.8} \approx 1.67 \ m$.
Note: Given the options provided,there appears to be a discrepancy in the calculation or the provided options. Based on the standard physics approach,the result is $\sqrt{2.8} \ m$.

(D) According to the Stefan-Boltzmann law,the power $P$ radiated by a black body of surface area $A$ and temperature $T$ is given by $P = \sigma A T^4$.
For a spherical body,the surface area $A = 4 \pi r^2$.
Thus,the power radiated is $P = \sigma (4 \pi r^2) T^4$.
Given that both bodies radiate the same power,we have $P_1 = P_2$.
Therefore,$\sigma (4 \pi r_1^2) T_1^4 = \sigma (4 \pi r_2^2) T_2^4$.
Simplifying the equation,we get $r_1^2 T_1^4 = r_2^2 T_2^4$.
Rearranging to find the ratio $\frac{r_2}{r_1}$,we get $\frac{r_2^2}{r_1^2} = \frac{T_1^4}{T_2^4}$.
Taking the square root of both sides,we get $\frac{r_2}{r_1} = \frac{T_1^2}{T_2^2} = \left(\frac{T_1}{T_2}\right)^2$.

(A) The ideal gas equation is given by $P V = n R T$.
Since $n = \frac{m_{total}}{M}$,where $m_{total}$ is the total mass and $M$ is the molar mass,we have $P V = \frac{m_{total}}{M} R T$.
Rearranging for density $\rho = \frac{m_{total}}{V}$,we get $P = \frac{\rho R T}{M}$.
We know that $R = N_A K$,where $N_A$ is Avogadro's number and $K$ is the Boltzmann constant.
Substituting $R$ and noting that the molar mass $M = N_A m$ (where $m$ is the mass of one molecule),we get:
$P = \frac{\rho (N_A K) T}{N_A m} = \frac{\rho K T}{m}$.
Solving for density $\rho$,we get $\rho = \frac{P m}{K T}$.

(A) According to the Doppler effect,when the observer moves towards a stationary source with velocity $v_0$,the apparent frequency $n^{\prime}$ is given by $n^{\prime} = n \left( \frac{v + v_0}{v} \right)$.
When the observer moves away from the stationary source with velocity $v_0$,the apparent frequency $n^{\prime \prime}$ is given by $n^{\prime \prime} = n \left( \frac{v - v_0}{v} \right)$.
The difference between the apparent frequencies is $\Delta n = n^{\prime} - n^{\prime \prime}$.
Substituting the values: $\Delta n = n \left( \frac{v + v_0}{v} \right) - n \left( \frac{v - v_0}{v} \right)$.
$\Delta n = \frac{n}{v} (v + v_0 - v + v_0) = \frac{n}{v} (2 v_0) = \frac{2 n v_0}{v}$.

(D) Let the equations of the two waves be $y_1 = a \sin(\omega t - kx)$ and $y_2 = a \sin(\omega t - kx + \phi)$.
When they superpose,the resultant wave is $y = y_1 + y_2$.
Using the trigonometric identity $\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$,we get:
$y = 2a \cos(\frac{\phi}{2}) \sin(\omega t - kx + \frac{\phi}{2})$.
The resultant amplitude is $A_R = |2a \cos(\frac{\phi}{2})|$.
Given that the resultant amplitude is equal to the individual amplitude $a$,we have:
$a = |2a \cos(\frac{\phi}{2})| \implies \cos(\frac{\phi}{2}) = \pm \frac{1}{2}$.
Considering the magnitude,$\cos(\frac{\phi}{2}) = \frac{1}{2}$.
Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we have $\frac{\phi}{2} = \frac{\pi}{3}$.
Therefore,the phase difference is $\phi = \frac{2\pi}{3}$.

(D) For a pipe closed at one end,the fundamental frequency is given by $n_1 = \frac{V}{4L} = 100 \ Hz$.
When the same pipe is open at both ends,the new fundamental frequency $n'_1$ is given by $n'_1 = \frac{V}{2L}$.
Comparing the two expressions,we get $n'_1 = 2 \times \frac{V}{4L} = 2 \times 100 \ Hz = 200 \ Hz$.
In an open pipe,all harmonics (integer multiples of the fundamental frequency) are produced.
Therefore,the frequencies produced are $200 \ Hz, 400 \ Hz, 600 \ Hz, 800 \ Hz, .....$.

(A) In the second overtone of a string fixed at both ends,the string vibrates in $3$ loops (segments).
For a string of length $L$ vibrating in $n$ loops,the positions of the antinodes (where amplitude is maximum) are given by $x = \frac{(2k-1)L}{2n}$,where $k = 1, 2, ..., n$.
Here,$n = 3$ (second overtone corresponds to the $3^{rd}$ harmonic).
For $k = 1$: $x_1 = \frac{(2(1)-1)L}{2(3)} = \frac{L}{6}$.
For $k = 2$: $x_2 = \frac{(2(2)-1)L}{2(3)} = \frac{3L}{6} = \frac{L}{2}$.
For $k = 3$: $x_3 = \frac{(2(3)-1)L}{2(3)} = \frac{5L}{6}$.
Thus,the amplitude is maximum at $\frac{L}{6}, \frac{L}{2}, \text{ and } \frac{5L}{6}$.

(A) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$.
Thus,$f = \frac{1}{2Lr} \sqrt{\frac{T}{\pi \rho}}$.
The first overtone of the first wire is $f_1 = 2f_1 = \frac{2}{2L_1 r_1} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{L_1 r_1} \sqrt{\frac{T}{\pi \rho}}$.
The second overtone of the second wire is $f_2 = 3f_2 = \frac{3}{2L_2 r_2} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_1 = f_2$,we have $\frac{1}{L_1 r_1} = \frac{3}{2L_2 r_2}$.
Rearranging for the ratio of lengths: $\frac{L_1}{L_2} = \frac{2r_2}{3r_1}$.
Since $r_1 = 2r_2$,we substitute: $\frac{L_1}{L_2} = \frac{2r_2}{3(2r_2)} = \frac{2}{6} = \frac{1}{3}$.

(D) The true power consumed in an $AC$ circuit is given by the formula: $P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \phi$.
Since $I_{\text{rms}} = \frac{V_{\text{rms}}}{Z}$ and the power factor $\cos \phi = \frac{R}{Z}$,we can write:
$P = V_{\text{rms}} \cdot \left( \frac{V_{\text{rms}}}{Z} \right) \cdot \left( \frac{R}{Z} \right) = \frac{V_{\text{rms}}^2 \cdot R}{Z^2}$.
Given $V_{\text{rms}} = 220 \ V$,$R = 18 \ \Omega$,and $Z = 33 \ \Omega$:
$P = \frac{220 \times 220 \times 18}{33 \times 33}$.
Simplifying the expression:
$P = \left( \frac{220}{33} \right) \times \left( \frac{220}{33} \right) \times 18 = \left( \frac{20}{3} \right) \times \left( \frac{20}{3} \right) \times 18$.
$P = \frac{400}{9} \times 18 = 400 \times 2 = 800 \ W$.

(D) In an $LC$ parallel resonant circuit,the impedance is maximum at the resonant frequency $f_r$,and the current is minimum at the resonant frequency $f_r$.
Graph $(4)$ shows that the current is minimum at the resonant frequency $f_r$,which is the correct characteristic for an $LC$ parallel resonant circuit.

(A) The frequency of a spectral line is given by $f = Rc \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the series limit of the Balmer series $(n_1=2, n_2=\infty)$: $f_1 = Rc \left( \frac{1}{2^2} - 0 \right) = \frac{Rc}{4}$.
For the series limit of the Paschen series $(n_1=3, n_2=\infty)$: $f_3 = Rc \left( \frac{1}{3^2} - 0 \right) = \frac{Rc}{9}$.
For the first line of the Balmer series $(n_1=2, n_2=3)$: $f_2 = Rc \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = Rc \left( \frac{1}{4} - \frac{1}{9} \right)$.
Substituting the expressions for $f_1$ and $f_3$ into the equation for $f_2$:
$f_2 = \frac{Rc}{4} - \frac{Rc}{9} = f_1 - f_3$.
Rearranging the terms,we get $f_1 - f_2 = f_3$.

(D) The magnetic moment $M_0$ of an electron due to its orbital motion is given by the formula: $M_0 = \frac{e}{2m_e} L_0$,where $e$ is the charge of the electron,$m_e$ is the mass of the electron,and $L_0$ is the orbital angular momentum.
According to Bohr's quantization condition,the orbital angular momentum $L_0$ is given by $L_0 = \frac{nh}{2\pi}$,where $n$ is the principal quantum number and $h$ is Planck's constant.
Substituting this into the magnetic moment expression,we get $M_0 = \frac{e}{2m_e} \times \frac{nh}{2\pi}$.
Since $e$,$m_e$,$h$,and $\pi$ are constants,it follows that $M_0 \propto n$.

(B) When a dielectric is inserted into a capacitor while it remains connected to a battery,the potential difference $V$ remains constant. However,if the capacitor is disconnected from the battery,the charge $Q$ remains constant.
Assuming the capacitor is disconnected from the battery:
$1$. The new capacitance becomes $C' = KC$,where $K > 1$.
$2$. Since the charge $Q$ is constant,the new potential $V' = Q/C' = Q/(KC) = V/K$. Since $K > 1$,$V' < V$,so the potential decreases.
$3$. The new energy $E' = Q^2 / (2C') = Q^2 / (2KC) = E/K$. Since $K > 1$,$E' < E$,so the energy decreases.
Therefore,both $V$ and $E$ decrease.

(A) Initially,two capacitors of capacity $C$ are in series. The equivalent capacitance $C_1$ is given by $\frac{1}{C_1} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$,so $C_1 = \frac{C}{2}$.
After filling one capacitor with a dielectric of constant $K$,its new capacity becomes $KC$. The new equivalent capacitance $C_2$ is given by $\frac{1}{C_2} = \frac{1}{C} + \frac{1}{KC} = \frac{1}{C} \left(1 + \frac{1}{K}\right) = \frac{K+1}{KC}$.
Thus,$C_2 = \frac{KC}{K+1}$.
The change in effective capacitance $\Delta C = C_2 - C_1 = \frac{KC}{K+1} - \frac{C}{2}$.
$\Delta C = C \left[ \frac{K}{K+1} - \frac{1}{2} \right] = C \left[ \frac{2K - (K+1)}{2(K+1)} \right] = \frac{C}{2} \left[ \frac{K-1}{K+1} \right]$.

(A) Let the capacity of each capacitor be $C$.
When three capacitors of capacity $C$ are connected in parallel,their equivalent capacity is $C_p = C + C + C = 3C$.
Now,this combination is connected in series with another capacitor of capacity $C$.
The resultant equivalent capacity $C_{eq}$ is given by the series formula:
$\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C} = \frac{1}{3C} + \frac{1}{C} = \frac{1 + 3}{3C} = \frac{4}{3C}$.
Therefore,$C_{eq} = \frac{3C}{4}$.
Given $C_{eq} = 3.75 \mu F$,we have:
$3.75 = \frac{3C}{4}$
$C = \frac{3.75 \times 4}{3} = 1.25 \times 4 = 5.00 \mu F$.
Thus,the capacity of each capacitor is $5 \mu F$.

(C) The process of superimposing a low-frequency message signal (baseband signal) onto a high-frequency carrier wave is called $Modulation$. This is necessary because low-frequency signals cannot travel long distances efficiently,and high-frequency waves act as carriers to transmit information over large distances.

(B) The resistance $R$ of a wire is given by $R = \frac{\rho l}{A}$, where $\rho$ is resistivity, $l$ is length, and $A$ is the area of cross-section.
Potential gradient is defined as the potential drop per unit length, given by $x = \frac{V}{l}$.
Using Ohm's law, $V = IR$, so $x = \frac{IR}{l} = I \left( \frac{R}{l} \right)$.
From the resistance formula, $\frac{R}{l} = \frac{\rho}{A}$.
Substituting the given values: $\frac{R}{l} = \frac{40 \times 10^{-8} \Omega \text{ m}}{8 \times 10^{-6} \text{ m}^2} = 5 \times 10^{-2} \Omega \text{ m}^{-1}$.
Now, calculating the potential gradient: $x = I \times \left( \frac{R}{l} \right) = 0.2 \text{ A} \times 5 \times 10^{-2} \Omega \text{ m}^{-1} = 10^{-2} \text{ V m}^{-1}$.

(C) Let the resistance of the galvanometer be $G$ and the shunt resistance be $S$. Given $S = \frac{G}{8}$.
The current sensitivity of a galvanometer is defined as the deflection per unit current. When a shunt $S$ is connected in parallel,the total current $I$ is divided such that the current through the galvanometer $I_g$ is given by $I_g = I \left( \frac{S}{S+G} \right)$.
The new sensitivity $s^{\prime}$ is the ratio of deflection $\theta$ to the total current $I$. Since $\theta = k I_g$ (where $k$ is a constant),we have $s^{\prime} = \frac{\theta}{I} = k \frac{I_g}{I} = k \left( \frac{S}{S+G} \right)$.
The original sensitivity $s = k$. Therefore,$s^{\prime} = s \left( \frac{S}{S+G} \right)$.
Substituting $S = \frac{G}{8}$:
$s^{\prime} = s \left( \frac{G/8}{G/8 + G} \right) = s \left( \frac{G/8}{9G/8} \right) = \frac{s}{9}$.

(B) For a photon,the energy is given by $E = \frac{hc}{\lambda_p}$.
Therefore,$\lambda_p = \frac{hc}{E} \dots (i)$.
For a non-relativistic electron,the de-Broglie wavelength is $\lambda_e = \frac{h}{p}$.
Since $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Thus,$\lambda_e = \frac{h}{\sqrt{2mE}}$,which implies $E = \frac{h^2}{2m\lambda_e^2}$.
Substituting $E$ into equation $(i)$:
$\lambda_p = \frac{hc}{(h^2 / 2m\lambda_e^2)} = \frac{2mc}{h} \lambda_e^2$.
Therefore,$\lambda_p \propto \lambda_e^2$.

(B) According to Einstein's photoelectric equation: $K_{max} = h\nu - W_0$,where $W_0$ is the work function.
Initially: $0.4 = h\nu - W_0 \implies h\nu = 0.4 + W_0$ ... $(i)$
When frequency is increased by $30 \%$,the new frequency $\nu' = 1.3\nu$. The new kinetic energy is $0.9 \ eV$.
So,$0.9 = 1.3h\nu - W_0$ ... (ii)
Substitute $(i)$ into (ii): $0.9 = 1.3(0.4 + W_0) - W_0$
$0.9 = 0.52 + 1.3W_0 - W_0$
$0.9 - 0.52 = 0.3W_0$
$0.38 = 0.3W_0$
$W_0 = \frac{0.38}{0.3} = 1.267 \ eV$.

(C) The energy of an electron in the $n^{\text{th}}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$.
For the ground state $(n=1)$,$E_1 = -13.6 \ eV$.
For the second orbit $(n=2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \ eV$.
The energy of the emitted photon is the difference between these two states: $E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$.
According to Einstein's photoelectric equation,$h\nu = \phi_0 + eV_s$,where $h\nu$ is the photon energy,$\phi_0$ is the work function,and $V_s$ is the stopping potential.
Given $\phi_0 = 4.2 \ eV$ and $h\nu = 10.2 \ eV$,we have $10.2 \ eV = 4.2 \ eV + eV_s$.
$eV_s = 10.2 \ eV - 4.2 \ eV = 6 \ eV$.
Therefore,the stopping potential $V_s = 6 \ V$.

(D) The magnetic field inside a long solenoid is given by $B = \frac{\mu_0 NI}{L}$.
Since magnetic flux $\phi = B \cdot A$, we have $B = \frac{\phi}{A}$.
Equating the two expressions for $B$: $\frac{\phi}{A} = \frac{\mu_0 NI}{L}$.
Rearranging to find the magnetic moment $M = NIA$: $NIA = \frac{\phi L}{\mu_0}$.
Given $\phi = 1.57 \times 10^{-6} \, Wb$, $L = 0.6 \, m$, and $\mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A$.
Substituting the values: $M = \frac{1.57 \times 10^{-6} \times 0.6}{4 \pi \times 10^{-7}}$.
Using $\pi \approx 3.14$, $4 \pi \approx 12.56$.
$M = \frac{1.57 \times 10^{-6} \times 0.6}{12.56 \times 10^{-7}} = \frac{0.942 \times 10^{-6}}{12.56 \times 10^{-7}} = \frac{9.42}{12.56} \approx 0.75 \, Am^2$.

(D) The motional electromotive force (emf) induced in a straight conductor of length $l$ moving with velocity $v$ in a uniform magnetic field $B$ is given by the formula:
$e = Bvl \sin \theta$
where $\theta$ is the angle between the velocity vector and the length of the conductor.
Given:
Length of the conductor,$l = 0.4 \ m$
Speed of the conductor,$v = 7 \ ms^{-1}$
Magnetic field intensity,$B = 0.9 \ Wb \ m^{-2}$
Since the conductor is moving perpendicular to the magnetic field,the effective length is perpendicular to the velocity,so $\theta = 90^\circ$ and $\sin 90^\circ = 1$.
Therefore,the induced emf is:
$e = Bvl = 0.9 \times 7 \times 0.4$
$e = 2.52 \ V$

(C) The potential energy of a magnetic dipole in a magnetic field is given by $U = -M B \cos \theta$.
Initially,the dipole is oriented along the field,so $\theta_1 = 0^\circ$. The initial potential energy is $U_i = -M B \cos(0^\circ) = -M B$.
To perform maximum work,the dipole must be rotated to the position of maximum potential energy,which is $\theta_2 = 180^\circ$. The final potential energy is $U_f = -M B \cos(180^\circ) = M B$.
The work done by an external agent is $W = U_f - U_i = M B - (-M B) = 2 M B$.

(B) The rate of disintegration $R$ is given by $R = R_0 e^{-\lambda t}$.
Given $R_0 = 10,000$ disintegrations per minute,$R = 2500$ disintegrations per minute,and $t = 4$ minutes.
Substituting these values into the equation:
$\frac{2500}{10000} = e^{-\lambda \times 4}$
$\frac{1}{4} = e^{-4 \lambda}$
Taking the natural logarithm on both sides:
$\ln(\frac{1}{4}) = -4 \lambda$
$-\ln(4) = -4 \lambda$
$\ln(2^2) = 4 \lambda$
$2 \ln(2) = 4 \lambda$
$\lambda = \frac{2}{4} \ln(2)$
$\lambda = 0.5 \log _e 2$ per minute.

(C) The number of waves $N$ in a medium of thickness $t$ is given by $N = \frac{t}{\lambda_m}$,where $\lambda_m = \frac{\lambda_0}{\mu}$ is the wavelength in the medium and $\lambda_0$ is the wavelength in vacuum.
Thus,$N = \frac{t \cdot \mu}{\lambda_0}$.
Given that the number of waves in the glass slab is equal to the number of waves in the water column:
$\frac{t_g \cdot \mu_g}{\lambda_0} = \frac{t_w \cdot \mu_w}{\lambda_0}$
$\therefore \mu_g \cdot t_g = \mu_w \cdot t_w$
Substituting the given values: $\mu_g = 1.5$,$t_g = 6 \ cm$,and $t_w = 7 \ cm$:
$1.5 \times 6 = \mu_w \times 7$
$9 = 7 \cdot \mu_w$
$\mu_w = \frac{9}{7} \approx 1.286$.

(A) We know the relationship between the current gain parameters $\alpha_{dc}$ and $\beta_{dc}$ is given by $\beta_{dc} = \frac{\alpha_{dc}}{1 - \alpha_{dc}}$.
Rearranging this equation,we get $1 - \alpha_{dc} = \frac{\alpha_{dc}}{\beta_{dc}}$.
Now,we need to evaluate the expression $\frac{\beta_{dc} - \alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}}$.
This can be written as $\frac{\beta_{dc}}{\alpha_{dc} \cdot \beta_{dc}} - \frac{\alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}} = \frac{1}{\alpha_{dc}} - \frac{1}{\beta_{dc}}$.
Substituting $\frac{1}{\alpha_{dc}} = \frac{1 + \beta_{dc}}{\beta_{dc}}$ is not necessary here; instead,use $\frac{1}{\alpha_{dc}} - \frac{1}{\beta_{dc}} = \frac{\beta_{dc} - \alpha_{dc}}{\alpha_{dc} \cdot \beta_{dc}}$.
From the relation $\beta_{dc} = \frac{\alpha_{dc}}{1 - \alpha_{dc}}$,we have $\frac{1}{\beta_{dc}} = \frac{1 - \alpha_{dc}}{\alpha_{dc}} = \frac{1}{\alpha_{dc}} - 1$.
Therefore,$\frac{1}{\alpha_{dc}} - \frac{1}{\beta_{dc}} = 1$.
Thus,the value of the expression is $1$.

(A) photodiode is a semiconductor $p-n$ junction device that is specifically designed to operate under reverse bias conditions. When light of energy greater than the bandgap energy falls on the junction,it generates electron-hole pairs. The reverse bias electric field sweeps these charge carriers across the junction,creating a photocurrent that is proportional to the intensity of the incident light.

(B) According to Brewster's Law,the refractive index $\mu$ of a medium is related to the polarising angle $\theta$ by the equation $\mu = \tan \theta$.
By definition,the refractive index $\mu$ is the ratio of the speed of light in air $(c)$ to the speed of light in the medium $(v)$,so $\mu = \frac{c}{v}$.
Equating the two expressions for $\mu$,we get $\tan \theta = \frac{c}{v}$.
Taking the reciprocal of both sides,we have $\cot \theta = \frac{1}{\tan \theta} = \frac{v}{c}$.
Therefore,the relation is $\theta = \cot ^{-1}\left(\frac{v}{c}\right)$.

(C) The position of the slits is at $y = \pm d/2$. The second minimum occurs at $y = d/2$.
For a minimum in Young's double-slit experiment,the path difference is given by $\Delta x = (n - 1/2)\lambda$,where $n = 2$ for the second minimum.
Thus,$\Delta x = (2 - 1/2)\lambda = (3/2)\lambda$.
Also,the path difference is given by $\Delta x = d \sin \theta \approx d \tan \theta = d(y/D)$.
Substituting $y = d/2$,we get $\Delta x = d(d/2D) = d^2 / 2D$.
Equating the two expressions for path difference: $(3/2)\lambda = d^2 / 2D$.
Solving for $\lambda$,we get $\lambda = (d^2 / 2D) \times (2/3) = d^2 / 3D$.

(A) The position of the $n^{th}$ bright fringe (maxima) is given by the formula:
$y_n = \frac{n D \lambda}{d}$
Rearranging the formula to solve for wavelength $\lambda$:
$\lambda = \frac{y_n d}{n D}$
Given values:
$n = 3$
$y_n = 1.2 \ mm = 1.2 \times 10^{-3} \ m$
$D = 1 \ m$
$d = 1 \ mm = 1 \times 10^{-3} \ m$
Substituting these values into the formula:
$\lambda = \frac{(1.2 \times 10^{-3} \ m) \times (1 \times 10^{-3} \ m)}{3 \times 1 \ m}$
$\lambda = \frac{1.2 \times 10^{-6}}{3} \ m$
$\lambda = 0.4 \times 10^{-6} \ m = 4 \times 10^{-7} \ m$
Converting to $\mathring{A}$s $(\mathring{A})$, where $1 \ \mathring{A} = 10^{-10} \ m$:
$\lambda = 4000 \times 10^{-10} \ m = 4000 \ \mathring{A}$

(B) Let the intensity of each individual wave be $I$.
When two identical waves with phase difference $\phi$ superpose,the resultant intensity $I_R$ is given by the formula:
$I_R = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$
Since the waves are identical,$I_1 = I_2 = I$.
Substituting these values:
$I_R = I + I + 2\sqrt{I \cdot I} \cos \phi$
$I_R = 2I + 2I \cos \phi$
$I_R = 2I(1 + \cos \phi)$
Using the trigonometric identity $1 + \cos \phi = 2 \cos^2 \frac{\phi}{2}$:
$I_R = 2I(2 \cos^2 \frac{\phi}{2}) = 4I \cos^2 \frac{\phi}{2}$
Therefore,the resultant intensity is proportional to $\cos^2 \frac{\phi}{2}$.