MHT CET 2018 Physics Question Paper with Answer and Solution

(C) Let the mass of both spheres be $m$. The initial velocity of the first sphere is $v$ and the second sphere is $0$. Let their final velocities be $V_1$ and $V_2$ respectively.
By the law of conservation of linear momentum:
$m v + m(0) = m V_1 + m V_2$
$v = V_1 + V_2$ --- $(1)$
By the definition of the coefficient of restitution $e$:
$e = \frac{V_2 - V_1}{u_1 - u_2}$
$e = \frac{V_2 - V_1}{v - 0}$
$e v = V_2 - V_1$ --- $(2)$
Adding equations $(1)$ and $(2)$:
$v + e v = (V_1 + V_2) + (V_2 - V_1)$
$v(1 + e) = 2 V_2$
$V_2 = \frac{v(e + 1)}{2}$
The ratio of the final velocity of the second sphere $(V_2)$ to the initial velocity of the first sphere $(v)$ is:
$\frac{V_2}{v} = \frac{e + 1}{2}$

(A) According to the law of conservation of linear momentum,since the bomb is initially at rest,the total initial momentum is zero. Therefore,the total final momentum must also be zero.
Let the momentum of the third part be $\vec{p}_3$.
$\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$
$(-3 p \hat{i}) + (2 p \hat{j}) + \vec{p}_3 = 0$
$\vec{p}_3 = 3 p \hat{i} - 2 p \hat{j}$
The magnitude of the momentum of the third part is:
$|\vec{p}_3| = \sqrt{(3p)^2 + (-2p)^2}$
$|\vec{p}_3| = \sqrt{9p^2 + 4p^2}$
$|\vec{p}_3| = \sqrt{13} p$

(B) The value of $g$ at a height $h$ above the surface of the Earth is given by $g_h = g(1 - \frac{2h}{R})$.
Thus,the change in $g$ is $\Delta g_h = g - g_h = g(\frac{2h}{R})$.
The value of $g$ at a depth $x$ below the surface of the Earth is given by $g_x = g(1 - \frac{x}{R})$.
Thus,the change in $g$ is $\Delta g_x = g - g_x = g(\frac{x}{R})$.
Given that the change in $g$ is the same at height $h$ and depth $x$,we equate the two expressions:
$g(\frac{2h}{R}) = g(\frac{x}{R})$.
Solving for $x$,we get $x = 2h$.

(A) By the law of conservation of energy,the total energy at the surface of the earth is equal to the total energy at the maximum height $h$.
At the surface: $E_i = -\frac{GMm}{R} + \frac{1}{2}mu^2$
At maximum height $h$: $E_f = -\frac{GMm}{R+h} + 0$
Equating $E_i = E_f$: $-\frac{GMm}{R} + \frac{1}{2}mu^2 = -\frac{GMm}{R+h}$
Dividing by $m$ and rearranging: $\frac{GM}{R+h} = \frac{GM}{R} - \frac{u^2}{2}$
Using $GM = gR^2$: $\frac{gR^2}{R+h} = gR - \frac{u^2}{2} = \frac{2gR - u^2}{2}$
$\frac{R+h}{R^2} = \frac{2g}{2gR - u^2}$
$R+h = \frac{2gR^2}{2gR - u^2}$
$h = \frac{2gR^2}{2gR - u^2} - R = \frac{2gR^2 - 2gR^2 + u^2R}{2gR - u^2} = \frac{u^2R}{2gR - u^2}$

(D) According to Mayer's relation for an ideal gas,the difference between molar specific heat at constant pressure $(C_P)$ and constant volume $(C_V)$ is equal to the universal gas constant $(R)$:
$C_P - C_V = R$
We are also given the ratio of molar specific heats as $\gamma = \frac{C_P}{C_V}$,which implies $C_P = \gamma C_V$.
Substituting the expression for $C_P$ into Mayer's relation:
$\gamma C_V - C_V = R$
$C_V(\gamma - 1) = R$
Therefore,$C_V = \frac{R}{\gamma - 1}$.

(C) The tangential acceleration $(a_t)$ is given by the formula $a_t = \alpha r$.
The radial (centripetal) acceleration $(a_r)$ is given by the formula $a_r = \frac{u^2}{r}$.
The ratio of tangential acceleration to radial acceleration is $\frac{a_t}{a_r} = \frac{\alpha r}{u^2 / r}$.
Simplifying this expression,we get $\frac{\alpha r^2}{u^2}$.

(B) First,calculate the magnitudes squared of the vectors:
$A^2 = |\vec{A}|^2 = 3^2 + (-2)^2 + 1^2 = 9 + 4 + 1 = 14$
$B^2 = |\vec{B}|^2 = 1^2 + (-3)^2 + 5^2 = 1 + 9 + 25 = 35$
$C^2 = |\vec{C}|^2 = 2^2 + 1^2 + (-4)^2 = 4 + 1 + 16 = 21$
We observe that $B^2 = A^2 + C^2$ $(35 = 14 + 21)$,which satisfies the Pythagorean theorem for a right-angled triangle.
Next,check the vector sum relation:
$\vec{A} + \vec{C} = (3+2)\hat{i} + (-2+1)\hat{j} + (1-4)\hat{k} = 5\hat{i} - \hat{j} - 3\hat{k} \neq \vec{B}$
$\vec{B} = \vec{A} + \vec{C}$ is not satisfied. Let us check $\vec{A} = \vec{B} + \vec{C}$:
$\vec{B} + \vec{C} = (1+2)\hat{i} + (-3+1)\hat{j} + (5-4)\hat{k} = 3\hat{i} - 2\hat{j} + \hat{k} = \vec{A}$
Thus,$\vec{A} = \vec{B} + \vec{C}$ and $B^2 = A^2 + C^2$ is the correct condition.

(B) The given physical quantity is $x = \frac{a^2 b^2}{c}$.
Using the formula for propagation of errors,the relative error in $x$ is given by:
$\frac{\Delta x}{x} = 2 \frac{\Delta a}{a} + 2 \frac{\Delta b}{b} + \frac{\Delta c}{c}$.
Given percentage errors are $\frac{\Delta a}{a} \times 100 = 2\%$,$\frac{\Delta b}{b} \times 100 = 3\%$,and $\frac{\Delta c}{c} \times 100 = 4\%$.
Substituting these values into the error equation:
$\frac{\Delta x}{x} \times 100 = 2(2\%) + 2(3\%) + 4\%$.
$\frac{\Delta x}{x} \times 100 = 4\% + 6\% + 4\% = 14\%$.
Therefore,the percentage error in $x$ is $14\%$.

(D) unit vector has a magnitude of $1$. The magnitude of a vector $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ is given by $|\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}$.
Given the vector is $(0.8 \hat{i} + b \hat{j} + 0.4 \hat{k})$,we set its magnitude to $1$:
$\sqrt{(0.8)^2 + b^2 + (0.4)^2} = 1$
Squaring both sides:
$(0.8)^2 + b^2 + (0.4)^2 = 1^2$
$0.64 + b^2 + 0.16 = 1$
$0.80 + b^2 = 1$
$b^2 = 1 - 0.80$
$b^2 = 0.2$
$b = \sqrt{0.2}$

(A) For the wire to float,the downward gravitational force must be balanced by the upward force due to surface tension.
The gravitational force (weight) acting on the wire is $W = mg = (\text{volume} \times \rho) g = (\pi r^2 l) \rho g$,where $r$ is the radius and $l$ is the length of the wire.
The surface tension force acts along the two sides of the wire on the water surface,so the upward force is $F_s = 2Tl$.
Equating the forces for the limiting case of floating: $Mg = 2Tl$.
Substituting the values: $(\pi r^2 l) \rho g = 2Tl$.
Canceling $l$ from both sides: $\pi r^2 \rho g = 2T$.
Solving for $r$: $r^2 = \frac{2T}{\pi \rho g}$,which gives $r = \sqrt{\frac{2T}{\pi \rho g}}$.
Thus,the maximum radius is proportional to $\sqrt{\frac{2T}{\pi \rho g}}$.

(C) The volume flow rate (or discharge) $Q$ from a hole at depth $y$ is given by $Q = A_v \cdot v$,where $A_v$ is the area of the hole and $v = \sqrt{2gy}$ is the velocity of efflux.
For hole $A$ at depth $h$: Area $A_A = L^2$,velocity $v_A = \sqrt{2gh}$. So,$Q_A = L^2 \sqrt{2gh}$.
For hole $B$ at depth $3h$: Area $A_B = \pi r^2$,velocity $v_B = \sqrt{2g(3h)} = \sqrt{6gh}$. So,$Q_B = \pi r^2 \sqrt{6gh}$.
Given that the flow rate is the same,$Q_A = Q_B$:
$L^2 \sqrt{2gh} = \pi r^2 \sqrt{6gh}$
$L^2 = \pi r^2 \frac{\sqrt{6gh}}{\sqrt{2gh}} = \pi r^2 \sqrt{3}$
Taking the square root of both sides:
$L = \sqrt{\pi r^2 \sqrt{3}} = r \sqrt{\pi} (3)^{\frac{1}{4}} = r (\pi)^{\frac{1}{2}} (3)^{\frac{1}{4}}$.

(B) soap bubble has two surfaces (inner and outer), so its total surface area is $2 \times 4 \pi R^2 = 8 \pi R^2$.
Initial surface energy $E_i = T \times (8 \pi R^2) = 8 \pi R^2 T$.
When the radius is doubled, the new radius becomes $R' = 2R$.
The new surface area is $2 \times 4 \pi (2R)^2 = 2 \times 4 \pi (4R^2) = 32 \pi R^2$.
Final surface energy $E_f = T \times (32 \pi R^2) = 32 \pi R^2 T$.
Work done $W = E_f - E_i$.
$W = 32 \pi R^2 T - 8 \pi R^2 T = 24 \pi R^2 T$.

(B) The height of water rise in a capillary tube is given by the formula $h = \frac{2T \cos \theta}{r \rho g}$,where $T$ is surface tension,$\theta$ is the contact angle,$r$ is the radius of the tube,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
Since $T, \theta, \rho,$ and $g$ are constant,we have $h \propto \frac{1}{r}$,which means $rh = \text{constant}$.
The area of cross-section is $A = \pi r^2$,so $r = \sqrt{\frac{A}{\pi}}$,which implies $r \propto \sqrt{A}$.
Given the initial area $A_1 = A$ and final area $A_2 = \frac{A}{9}$.
The ratio of radii is $\frac{r_2}{r_1} = \sqrt{\frac{A_2}{A_1}} = \sqrt{\frac{A/9}{A}} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.
Using the relation $r_1 h_1 = r_2 h_2$,we get $h_2 = h_1 \left( \frac{r_1}{r_2} \right)$.
Substituting the values,$h_2 = h \times 3 = 3h$.

(C) Young's modulus $Y$ is defined as the ratio of longitudinal stress to longitudinal strain.
$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$
Rearranging the formula to solve for the change in length $\Delta L$:
$Y = \frac{F L}{A \Delta L}$
$\Delta L = \frac{F L}{A Y}$
Thus,the decrease in length is $\frac{F L}{A Y}$.

(B) Given: Diameter $d = 50 \ cm = 0.5 \ m$.
Radius $r = d/2 = 0.25 \ m = 25 \times 10^{-2} \ m$.
Frequency $f = 2 \ Hz$.
Angular velocity $\omega = 2 \pi f = 2 \pi \times 2 = 4 \pi \ rad/s$.
The centripetal acceleration $a$ in $U.C.M.$ is given by $a = r \omega^2$.
Substituting the values: $a = (25 \times 10^{-2}) \times (4 \pi)^2$.
$a = 0.25 \times 16 \pi^2$.
$a = 4 \pi^2 \ m/s^2$.

(B) The path length of oscillation is the total distance between the two extreme positions, which is equal to $2a$, where $a$ is the amplitude.
Given, $2a = 16 \,cm$, so the amplitude $a = 8 \,cm$.
The length of the pendulum is $l = 1 \,m$.
The angular frequency of a simple pendulum is given by $\omega = \sqrt{\frac{g}{l}}$.
Substituting $g = \pi^2 \,m/s^2$ and $l = 1 \,m$, we get $\omega = \sqrt{\frac{\pi^2}{1}} = \pi \,rad/s$.
The maximum velocity $v_{max}$ in simple harmonic motion is given by $v_{max} = a\omega$.
Substituting the values, $v_{max} = 8 \,cm \times \pi \,rad/s = 8\pi \,cm/s$.

(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: $F_{max} = 1 \times 4 \times \pi^2 \times (50)^2 \times 0.005$.
$F_{max} = 4 \times \pi^2 \times 2500 \times 0.005$.
$F_{max} = 10000 \times \pi^2 \times 0.005 = 50 \pi^2 \ N$.

(D) The frequency of $S.H.M.$ is given by $n = 5 Hz$. The angular frequency is $\omega = 2 \pi n = 2 \pi \times 5 = 10 \pi rad s^{-1}$.
At the highest point of oscillation,the spring is unstretched,meaning the extension $x = 0$. In $S.H.M.$,the equilibrium position is where the spring force balances gravity,$k x_0 = mg$,where $x_0$ is the static extension.
The amplitude $A$ of the oscillation is equal to this static extension $x_0$,because the particle reaches the unstretched position $(x=0)$ at the top of its path. Thus,$A = x_0 = \frac{mg}{k}$.
We know that $\omega^2 = \frac{k}{m}$,so $k = m \omega^2 = m(10 \pi)^2 = 100 \pi^2 m$.
Substituting $k$ into the amplitude equation: $A = \frac{mg}{100 \pi^2 m} = \frac{g}{100 \pi^2} = \frac{10}{100 \pi^2} = \frac{1}{10 \pi} m$.
The maximum speed is $V_{max} = \omega A = (10 \pi) \times (\frac{1}{10 \pi}) = 1 m s^{-1}$.
Wait,re-evaluating the calculation: $V_{max} = \omega A = (10 \pi) \times \frac{g}{\omega^2} = \frac{g}{\omega} = \frac{10}{10 \pi} = \frac{1}{\pi} m s^{-1}$.

(B) In one complete oscillation,a particle in $SHM$ travels from the mean position to one extreme $(A)$,back to the mean position $(A)$,to the other extreme $(A)$,and back to the mean position $(A)$.
Total distance travelled in $1$ oscillation $= A + A + A + A = 4 \ A$.
The time taken for one complete oscillation is the time period $T$.
The frequency of oscillation is $n = \frac{1}{T}$,which implies $T = \frac{1}{n}$.
Average speed is defined as the total distance divided by the total time.
$\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{4 \ A}{T}$.
Substituting $T = \frac{1}{n}$,we get $\text{Average speed} = 4 \ A \ n$.

(B) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} M R^2$.
Using the rotational analog of Newton's second law,$\tau = I \alpha$,where $\tau$ is the torque and $\alpha$ is the angular acceleration.
The tangential force $F$ applied at the rim provides a torque $\tau = F \times R$.
The angular acceleration is given by $\alpha = \frac{\omega}{t}$.
Substituting these into the torque equation: $F \times R = (\frac{1}{2} M R^2) \times (\frac{\omega}{t})$.
Solving for $F$: $F = \frac{M R \omega}{2 t}$.

(B) Initial state: Moment of inertia $I_1 = I$,angular velocity $\omega_1 = \omega$. Initial kinetic energy $KE_i = \frac{1}{2} I \omega^2$.
Since no external torque acts on the system,the angular momentum is conserved: $L_i = L_f$.
$I \omega = (I + I) \omega_2$,where $\omega_2$ is the final angular velocity.
$I \omega = 2I \omega_2 \implies \omega_2 = \frac{\omega}{2}$.
Final kinetic energy $KE_f = \frac{1}{2} (2I) \omega_2^2 = I \left(\frac{\omega}{2}\right)^2 = \frac{I \omega^2}{4}$.
Loss in kinetic energy $\Delta KE = KE_i - KE_f = \frac{1}{2} I \omega^2 - \frac{1}{4} I \omega^2 = \frac{1}{4} I \omega^2$.

(B) The moment of inertia of a thin uniform rod of length $L$ and mass $M$ about an axis passing through its center of mass $(CM)$ and perpendicular to the rod is $I_{CM} = \frac{M L^2}{12}$.
The distance of the given axis from the center of mass is $x = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}$.
Using the parallel axis theorem,$I = I_{CM} + M x^2$.
Substituting the values,we get $I = \frac{M L^2}{12} + M \left( \frac{L}{6} \right)^2$.
$I = \frac{M L^2}{12} + \frac{M L^2}{36}$.
Taking the least common multiple,$I = \frac{3 M L^2 + M L^2}{36} = \frac{4 M L^2}{36} = \frac{M L^2}{9}$.

(B) The total incident heat energy rate is $P_i = 1000 \ J \ min^{-1}$.
According to the law of conservation of energy for heat radiation,the sum of the coefficients of reflection $(r)$,absorption $(a)$,and transmission $(t)$ is equal to $1$,i.e.,$r + a + t = 1$.
Given $r = 0.1$ and $a = 0.8$,we can find the coefficient of transmission $(t)$:
$t = 1 - (r + a) = 1 - (0.1 + 0.8) = 1 - 0.9 = 0.1$.
The rate of transmitted heat energy is $P_t = t \times P_i = 0.1 \times 1000 \ J \ min^{-1} = 100 \ J \ min^{-1}$.
For a time duration of $5 \ minutes$,the total transmitted heat energy $(Q_t)$ is:
$Q_t = P_t \times \text{time} = 100 \ J \ min^{-1} \times 5 \ min = 500 \ J$.

(B) For a string fixed at both ends,the $n^{th}$ overtone corresponds to the $(n+1)^{th}$ harmonic.
Here,the fifth overtone is the sixth harmonic $(n=6)$.
The length of the string $L = 2.4 \ m$.
The condition for the $n^{th}$ harmonic is $L = n \frac{\lambda}{2}$.
Substituting the values: $2.4 = 6 \times \frac{\lambda}{2}$.
This gives $\frac{\lambda}{2} = \frac{2.4}{6} = 0.4 \ m$.
Thus,$\lambda = 0.8 \ m$.
The distance between a node and the successive antinode is always $\frac{\lambda}{4}$.
Therefore,distance $= \frac{0.8 \ m}{4} = 0.2 \ m$.

(A) According to the Doppler effect,the apparent frequency $n_a$ heard by an observer is given by $n_a = n \left[ \frac{v \pm v_0}{v \mp v_s} \right]$.
Since the observer is stationary,$v_0 = 0$.
As the source moves towards the observer,the denominator becomes $(v - v_s)$,so $n_a = n \left[ \frac{v}{v - v_s} \right]$.
Since $(v - v_s) < v$,the apparent frequency $n_a$ increases.
Because the speed of sound $v$ in the medium remains constant and $v = n_a \lambda_a$,an increase in frequency $n_a$ results in a decrease in the observed wavelength $\lambda_a$.

(C) For a pipe closed at one end,the natural frequencies are given by $f_k = (2k - 1) \frac{V}{4L}$,where $k = 1, 2, 3, \dots$ is the mode number.
Given $V = 332 \ m/s$ and $L = 83 \ cm = 0.83 \ m$.
The fundamental frequency $(k=1)$ is $f_1 = \frac{V}{4L} = \frac{332}{4 \times 0.83} = \frac{332}{3.32} = 100 \ Hz$.
The possible frequencies are odd multiples of the fundamental frequency: $f_k = (2k - 1) \times 100 \ Hz$.
We need to find the number of frequencies such that $f_k < 1000 \ Hz$.
$(2k - 1) \times 100 < 1000 \implies 2k - 1 < 10 \implies 2k < 11 \implies k < 5.5$.
Since $k$ must be a positive integer,$k$ can be $1, 2, 3, 4, 5$.
The frequencies are $100 \ Hz, 300 \ Hz, 500 \ Hz, 700 \ Hz, 900 \ Hz$.
There are $5$ such frequencies.

(D) The given alternating voltage is $e = e_0 \sin(\omega t)$, where $e_0 = 200 \sqrt{2} \text{ V}$ and $\omega = 100 \text{ rad/s}$.
The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 1 \times 10^{-6}} = \frac{1}{10^{-4}} = 10^4 \Omega$.
The $RMS$ voltage is $V_{rms} = \frac{e_0}{\sqrt{2}} = \frac{200 \sqrt{2}}{\sqrt{2}} = 200 \text{ V}$.
The reading of the $AC$ ammeter gives the $RMS$ current $I_{rms} = \frac{V_{rms}}{X_C}$.
Substituting the values, $I_{rms} = \frac{200}{10^4} = 2 \times 10^{-2} \text{ A} = 20 \text{ mA}$.

(B) For an ideal transformer,the input power is equal to the output power.
Given: $P = 4.4 kW = 4400 W$,$V_p = 220 V$,$V_s = 3.3 kV = 3300 V$.
Since $P = V_s \times I_s$ for the secondary coil:
$I_s = \frac{P}{V_s} = \frac{4400}{3300} A$.
$I_s = \frac{44}{33} A = \frac{4}{3} A$.
Thus,the alternating current in the secondary coil is $\frac{4}{3} A$.

(A) For the series combination of $N_1$ capacitors of capacity $C_1$,the equivalent capacitance is $C_{eq1} = \frac{C_1}{N_1}$.
The energy stored in this combination is $E_1 = \frac{1}{2} C_{eq1} (3V)^2 = \frac{1}{2} \left( \frac{C_1}{N_1} \right) 9V^2 = \frac{9 C_1 V^2}{2 N_1}$.
For the parallel combination of $N_2$ capacitors of capacity $C_2$,the equivalent capacitance is $C_{eq2} = N_2 C_2$.
The energy stored in this combination is $E_2 = \frac{1}{2} C_{eq2} V^2 = \frac{1}{2} N_2 C_2 V^2$.
Given that the total energy stored in both combinations is the same $(E_1 = E_2)$:
$\frac{9 C_1 V^2}{2 N_1} = \frac{N_2 C_2 V^2}{2}$.
Canceling $\frac{V^2}{2}$ from both sides,we get $\frac{9 C_1}{N_1} = N_2 C_2$.
Solving for $C_1$,we get $C_1 = \frac{C_2 N_1 N_2}{9}$.

(D) In amplitude modulation,the amplitude of the high-frequency carrier wave is varied in accordance with the instantaneous amplitude of the modulating (information) signal,while the frequency and phase of the carrier wave remain constant.

(D) Let $G$ be the resistance of the galvanometer and $I$ be the total current. When shunted by $S_1 = 3 \Omega$,the current through the galvanometer is $I_g = \frac{I}{4}$.
Using the current divider rule: $I_g = I \left( \frac{S_1}{G + S_1} \right) \Rightarrow \frac{I}{4} = I \left( \frac{3}{G + 3} \right)$.
Solving for $G$: $G + 3 = 12 \Rightarrow G = 9 \Omega$.
Now,an additional shunt $S_2 = 2 \Omega$ is connected in parallel to $S_1 = 3 \Omega$. The equivalent shunt resistance $S_{eq}$ is:
$S_{eq} = \frac{S_1 \times S_2}{S_1 + S_2} = \frac{3 \times 2}{3 + 2} = \frac{6}{5} = 1.2 \Omega$.
The new current through the galvanometer $I_g'$ is:
$I_g' = I \left( \frac{S_{eq}}{G + S_{eq}} \right) = I \left( \frac{1.2}{9 + 1.2} \right) = I \left( \frac{1.2}{10.2} \right) = I \left( \frac{12}{102} \right) = I \left( \frac{1}{8.5} \right)$.
Thus,the deflection falls to $\left(\frac{1}{8.5}\right)^{th}$ of the initial value.

(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $A = \frac{\pi d^2}{4}$.
Since the volume of the wire remains constant during stretching,$V = A_1 L_1 = A_2 L_2$,which implies $L_1 d_1^2 = L_2 d_2^2$,or $\frac{L_1}{L_2} = \frac{d_2^2}{d_1^2}$.
The ratio of resistances is $\frac{R_1}{R_2} = \frac{L_1}{L_2} \times \frac{A_2}{A_1} = \frac{L_1}{L_2} \times \frac{d_2^2}{d_1^2}$.
Substituting $\frac{L_1}{L_2} = \frac{d_2^2}{d_1^2}$ into the equation,we get $\frac{R_1}{R_2} = \left(\frac{d_2^2}{d_1^2}\right) \times \left(\frac{d_2^2}{d_1^2}\right) = \frac{d_2^4}{d_1^4}$.

(C) The given circuit is a Wheatstone bridge. Let the nodes be $A, B, C, D$. The resistances are $R_{AB} = 15\Omega$,$R_{BC} = 3\Omega$,$R_{AD} = 20\Omega$,$R_{CD} = 4\Omega$,and the galvanometer resistance $R_G = 6\Omega$.
First,check for the balanced condition: $\frac{R_{AB}}{R_{AD}} = \frac{15}{20} = 0.75$ and $\frac{R_{BC}}{R_{CD}} = \frac{3}{4} = 0.75$.
Since $\frac{R_{AB}}{R_{AD}} = \frac{R_{BC}}{R_{CD}}$,the bridge is balanced. Therefore,no current flows through the galvanometer $(G)$.
This simplifies the circuit to two parallel branches: one with $(15+3) = 18\Omega$ and the other with $(20+4) = 24\Omega$.
The total current $I = 2.1 A$ divides into $I_1$ (through the $18\Omega$ branch) and $I_2$ (through the $24\Omega$ branch).
Using the current divider rule: $I_1 = I \times \frac{R_{parallel2}}{R_{parallel1} + R_{parallel2}} = 2.1 \times \frac{24}{18+24} = 2.1 \times \frac{24}{42} = 2.1 \times \frac{4}{7} = 1.2 A$.

(B) The de-Broglie wavelength $\lambda$ is related to the momentum $p$ by the equation $\lambda = \frac{h}{p}$.
Since kinetic energy $K.E. = \frac{p^2}{2m}$,we can express momentum as $p = \sqrt{2m(K.E.)}$.
Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2m(K.E.)}}$.
Squaring both sides,we obtain $\lambda^2 = \frac{h^2}{2m(K.E.)}$.
Rearranging the terms to solve for kinetic energy,we get $K.E. = \frac{h^2}{2m\lambda^2}$.

(C) The de-Broglie wavelength $\lambda$ is given by the relation $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mKE}}$.
From this,the kinetic energy $KE$ can be expressed as $KE = \frac{h^2}{2m\lambda^2}$.
Since the de-Broglie wavelength $\lambda$ is the same for both the electron and the proton,we have $KE \propto \frac{1}{m}$.
Because the mass of an electron $(m_e \approx 9.11 \times 10^{-31} \ kg)$ is much smaller than the mass of a proton $(m_p \approx 1.67 \times 10^{-27} \ kg)$,the kinetic energy of the electron will be greater than that of the proton.

(A) According to Einstein's photoelectric equation,the maximum kinetic energy $(K_{max})$ of emitted photoelectrons is given by:
$K_{max} = hf - \phi$
Since the stopping potential $(V_s)$ is related to the maximum kinetic energy by $K_{max} = eV_s$,we can write:
$eV_s = hf - \phi$
$V_s = (h/e)f - (\phi/e)$
This equation is of the form $y = mx + c$,which represents a straight line.
Here,the slope is $(h/e)$ (a positive constant) and the intercept on the $V_s$-axis is $(-\phi/e)$.
When $V_s = 0$,we have $hf = \phi$,which gives $f = \phi/h = f_0$ (the threshold frequency).
Thus,the graph of $V_s$ versus $f$ is a straight line starting from $f = f_0$ on the frequency axis.
Graph $(1)$ correctly represents this linear relationship.

(B) According to Einstein's photoelectric equation: $e V_0 = h f - \Phi$, where $\Phi = h f_0$ is the work function.
Rearranging the equation for stopping potential $V_0$: $V_0 = \frac{h}{e} f - \frac{h f_0}{e}$.
Since $h$, $e$, and $f_0$ are constants, $V_0$ is a linear function of the incident frequency $f$.
As the frequency $f$ of the incident radiation increases, the term $\frac{h}{e} f$ increases.
Therefore, the stopping potential $V_0$ increases.

(A) According to Gauss's Law,the electric field intensity $E$ at a point outside a uniformly charged thin infinite plane sheet with surface charge density $\sigma$ is given by the formula:
$E = \frac{\sigma}{2 \varepsilon_0}$
Here,$\sigma$ is the surface charge density and $\varepsilon_0$ is the permittivity of free space.
Since the expression for the electric field does not contain the distance $d$,the electric field intensity is independent of the distance $d$ from the plane sheet.

(B) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 I}{2R}$.
The magnetic field at a distance $x$ along the axis of the coil is given by $B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$.
We are given that $B_x = \frac{B}{8}$. Substituting the expression for $B$:
$\frac{B}{8} = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$
$\frac{1}{8} \left( \frac{\mu_0 I}{2R} \right) = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$
$\frac{1}{8R} = \frac{R^2}{(R^2 + x^2)^{3/2}}$
$\frac{1}{8R^3} = \frac{1}{(R^2 + x^2)^{3/2}}$
Taking the cube root of both sides:
$\frac{1}{2R} = \frac{1}{(R^2 + x^2)^{1/2}}$
Squaring both sides:
$\frac{1}{4R^2} = \frac{1}{R^2 + x^2}$
$R^2 + x^2 = 4R^2$
$x^2 = 3R^2$
$x = R\sqrt{3}$.

(A) In a cyclotron,the frequency of the alternating electric field $f$ is equal to the cyclotron frequency $f_c = \frac{eB}{2\pi m}$.
From this,the magnetic field $B$ is given by $B = \frac{2\pi mf}{e}$.
The maximum velocity $v$ of the proton at the exit radius $R$ is given by $v = \omega R = (2\pi f)R$.
The kinetic energy $K.E.$ is given by $K.E. = \frac{1}{2}mv^2$.
Substituting $v = 2\pi f R$,we get $K.E. = \frac{1}{2}m(2\pi f R)^2 = \frac{1}{2}m(4\pi^2 f^2 R^2) = 2\pi^2 mf^2 R^2$.
Thus,the magnetic field is $\frac{2\pi mf}{e}$ and the kinetic energy is $2\pi^2 mf^2 R^2$.

(B) According to Curie's law for paramagnetic materials,the magnetization $M_z$ is directly proportional to the external magnetic field $B$ and inversely proportional to the absolute temperature $T$.
Mathematically,this is expressed as $M_z \propto \frac{B}{T}$.
Introducing the Curie constant $C$,we get the relation $M_z = C \frac{B}{T}$ or $M_z = \frac{C B}{T}$.
Therefore,the correct relation is $M_z = \frac{C B}{T}$.

(B) The magnetic field on the axis of a bar magnet is given by $B = \frac{\mu_0}{4\pi} \frac{2Md}{(d^2-l^2)^2}$,where $d$ is the distance from the center and $2l$ is the magnetic length.
Given $d_1 = 10 \ cm$ and $d_2 = 10 + 10 = 20 \ cm$.
The ratio is $\frac{B_1}{B_2} = \frac{d_1}{d_2} \left( \frac{d_2^2 - l^2}{d_1^2 - l^2} \right)^2 = \frac{25}{2}$.
Substituting the values: $\frac{10}{20} \left( \frac{20^2 - l^2}{10^2 - l^2} \right)^2 = \frac{25}{2}$.
$\frac{1}{2} \left( \frac{400 - l^2}{100 - l^2} \right)^2 = \frac{25}{2} \Rightarrow \left( \frac{400 - l^2}{100 - l^2} \right)^2 = 25$.
Taking the square root: $\frac{400 - l^2}{100 - l^2} = 5$.
$400 - l^2 = 500 - 5l^2 \Rightarrow 4l^2 = 100 \Rightarrow l^2 = 25$.
So,$l = 5 \ cm$.
The magnetic length is $2l = 2 \times 5 \ cm = 10 \ cm$.

(C) The magnetic susceptibility $\chi$ is a measure of how easily a material can be magnetized in an external magnetic field.
For paramagnetic materials,the magnetic susceptibility $\chi$ is small and positive,meaning they are weakly attracted to an external magnetic field.
For diamagnetic materials,the magnetic susceptibility $\chi$ is small and negative,meaning they are weakly repelled by an external magnetic field.
Therefore,the correct sequence is small,positive for paramagnetic and small,negative for diamagnetic materials.

(D) In a compound microscope,the objective lens is placed very close to the object to form a real,inverted,and magnified image. To achieve high magnification and better resolution,the objective lens must have a short focal length and a small aperture.

(C) The limit of resolution $(d)$ of a microscope is given by the formula: $d = \frac{1.22 \lambda}{2 NA}$,where $\lambda$ is the wavelength of light and $NA$ is the numerical aperture.
From this relation,it is clear that the limit of resolution $(d)$ is inversely proportional to the numerical aperture $(NA)$: $d \propto \frac{1}{NA}$.
Therefore,if the numerical aperture $(NA)$ of a microscope is increased,the limit of resolution $(d)$ decreases,which improves the resolving power of the microscope.

(A) The angle of incidence is defined as the angle between the incident ray and the normal to the surface at the point of incidence.
However,the angle between the incident ray and the reflecting surface itself is known as the glancing angle.
In the provided diagram,the angle $\theta$ represents the angle between the incident ray $I$ and the surface $XY$,which is the glancing angle.
Therefore,the correct option is $A$.

(B) Let the distance of the source from the surface be $x$. The thickness of the glass slab is $d = 5 \ cm$,and its refractive index is $\mu = 1.6$.
The time taken by light to travel a distance $x$ in air (where speed is $c$) is $t_1 = \frac{x}{c}$.
The time taken by light to travel through the glass slab of thickness $d$ (where speed is $v = \frac{c}{\mu}$) is $t_2 = \frac{d}{v} = \frac{d}{c/\mu} = \frac{\mu d}{c}$.
According to the problem,$t_1 = t_2$,so:
$\frac{x}{c} = \frac{\mu d}{c}$
$x = \mu d$
$x = 1.6 \times 5 \ cm = 8 \ cm$.
Therefore,the distance of the source from the surface is $8 \ cm$.

(C) When a $p-n$ junction is in forward bias,the positive terminal of the battery is connected to the $p$-side and the negative terminal to the $n$-side.
This reduces the width of the depletion layer and lowers the potential barrier.
As a result,the diode offers very low resistance and allows current to flow through it.
Therefore,in forward bias,the $p-n$ junction diode acts as a closed switch.

(B) The correct option is $B$.
Concept: The current gain $\beta$ is given by $\beta = \frac{\Delta I_C}{\Delta I_B}$.
Given: $\Delta I_C = 1.5 \ mA = 1.5 \times 10^{-3} \ A$ and $\Delta I_B = 20 \ \mu A = 20 \times 10^{-6} \ A$.
Calculating $\beta$: $\beta = \frac{1.5 \times 10^{-3}}{20 \times 10^{-6}} = \frac{1500}{20} = 75$.
The voltage gain $A_v$ is given by $A_v = \beta \times \frac{R_L}{R_i}$.
Given: Load resistance $R_L = 2 \ k\Omega = 2000 \ \Omega$ and input resistance $R_i = 150 \ \Omega$.
Substituting the values: $A_v = 75 \times \frac{2000}{150} = 75 \times \frac{40}{3} = 25 \times 40 = 1000$.

(B) The intensity of a wave is proportional to the square of its amplitude,$I \propto a^2$,so $I = ka^2$. Let $I_1 = ka_1^2$ and $I_2 = ka_2^2$.
The maximum intensity during interference is given by $I_{\max} = k(a_1 + a_2)^2 = k(a_1^2 + a_2^2 + 2a_1a_2)$.
The minimum intensity during interference is given by $I_{\min} = k(a_1 - a_2)^2 = k(a_1^2 + a_2^2 - 2a_1a_2)$.
The sum of the maximum and minimum intensities is $I_{\max} + I_{\min} = k(a_1^2 + a_2^2 + 2a_1a_2) + k(a_1^2 + a_2^2 - 2a_1a_2)$.
$I_{\max} + I_{\min} = k(2a_1^2 + 2a_2^2) = 2(ka_1^2 + ka_2^2)$.
Substituting the intensities back,we get $I_{\max} + I_{\min} = 2(I_1 + I_2)$.