AP EAMCET 2010 Physics Question Paper with Answer and Solution

(C) When a ball falls from a height $h$,it strikes the floor and rebounds to a height $h_1 = h e^2$.
After the second impact,it rebounds to a height $h_2 = h_1 e^2 = h e^4$.
In general,the height reached after the $n$-th rebound is $h_n = h e^{2n}$.
The total distance $H$ covered by the ball is the sum of the initial fall and the subsequent upward and downward paths for each rebound:
$H = h + 2h_1 + 2h_2 + 2h_3 + \dots$
$H = h + 2(h e^2 + h e^4 + h e^6 + \dots)$
$H = h + 2h(e^2 + e^4 + e^6 + \dots)$
The term in the bracket is an infinite geometric progression with first term $a = e^2$ and common ratio $r = e^2$. The sum is $S = \frac{a}{1-r} = \frac{e^2}{1-e^2}$.
$H = h + 2h \left( \frac{e^2}{1-e^2} \right)$
$H = h \left( 1 + \frac{2e^2}{1-e^2} \right)$
$H = h \left( \frac{1 - e^2 + 2e^2}{1 - e^2} \right)$
$H = h \left( \frac{1 + e^2}{1 - e^2} \right)$

(A) Let the total mass of the bomb be $M = 5m$. The mass of the smaller piece is $m_1 = m$ and the mass of the larger piece is $m_2 = 4m$.
According to the law of conservation of linear momentum,the initial momentum equals the final momentum:
$M \vec{v} = m_1 \vec{v}_1 + m_2 \vec{v}_2$
$5m(40 \hat{i} + 50 \hat{j} - 25 \hat{k}) = m(200 \hat{i} + 70 \hat{j} + 15 \hat{k}) + 4m \vec{v}_2$
Dividing by $m$:
$5(40 \hat{i} + 50 \hat{j} - 25 \hat{k}) = (200 \hat{i} + 70 \hat{j} + 15 \hat{k}) + 4 \vec{v}_2$
$(200 \hat{i} + 250 \hat{j} - 125 \hat{k}) - (200 \hat{i} + 70 \hat{j} + 15 \hat{k}) = 4 \vec{v}_2$
$0 \hat{i} + 180 \hat{j} - 140 \hat{k} = 4 \vec{v}_2$
$\vec{v}_2 = \frac{180 \hat{j} - 140 \hat{k}}{4} = 45 \hat{j} - 35 \hat{k} \text{ m/s}$.

(B) According to the equation of continuity,the product of cross-sectional area and velocity remains constant for an incompressible fluid: $A_1 v_1 = A_2 v_2$.
Given $A_1 = A$,$v_1 = 4 \,m/s$,and $A_2 = \frac{2}{3} A$.
Substituting these values: $A \times 4 = \frac{2}{3} A \times v_2$.
Solving for $v_2$: $v_2 = 4 \times \frac{3}{2} = 6 \,m/s$.
Now,using the principle of conservation of energy (Bernoulli's equation) for the falling stream: $P + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P + \rho g h_2 + \frac{1}{2} \rho v_2^2$.
Since the pressure $P$ is constant (atmospheric pressure) and $h_1 - h_2 = h$,the equation simplifies to: $g h = \frac{1}{2} (v_2^2 - v_1^2)$.
Substituting the known values: $10 \times h = \frac{1}{2} (6^2 - 4^2)$.
$10 h = \frac{1}{2} (36 - 16) = \frac{1}{2} (20) = 10$.
Therefore,$h = 1 \,m$.

(D) The excess pressure inside a soap bubble is given by $p = \frac{4T}{R}$.
This pressure is balanced by the pressure exerted by the oil column, which is $p = h \rho g$.
Equating the two, we get $h \rho g = \frac{4T}{R}$.
Rearranging for surface tension $T$, we have $T = \frac{R h \rho g}{4}$.
Given values:
Radius $R = 1 \,cm = 10^{-2} \,m$.
Height $h = 2 \,mm = 2 \times 10^{-3} \,m$.
Density of oil $\rho = 0.8 \times 10^3 \,kg/m^3$.
Acceleration due to gravity $g = 9.8 \,m/s^2$.
Substituting these values:
$T = \frac{10^{-2} \times 2 \times 10^{-3} \times 0.8 \times 10^3 \times 9.8}{4}$.
$T = \frac{1.568 \times 10^{-2}}{4} = 0.392 \times 10^{-2} \,N/m = 0.00392 \,N/m$.

(C) The velocity of the centre of mass $(v_{CM})$ is given by the formula:
$v_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$
Substituting the given values:
$v_{CM} = \frac{4 \times 5 \hat{i} + 2 \times 10 \hat{i}}{4 + 2}$
$v_{CM} = \frac{20 \hat{i} + 20 \hat{i}}{6} = \frac{40 \hat{i}}{6} = \frac{20}{3} \hat{i} \text{ m/s}$
The kinetic energy of the centre of mass is calculated as:
$K_{CM} = \frac{1}{2} (m_1 + m_2) v_{CM}^2$
$K_{CM} = \frac{1}{2} \times (4 + 2) \times \left( \frac{20}{3} \right)^2$
$K_{CM} = \frac{1}{2} \times 6 \times \frac{400}{9}$
$K_{CM} = 3 \times \frac{400}{9} = \frac{400}{3} \text{ J}$

(B) The total energy of the satellite on the surface of the earth is the sum of its potential energy and kinetic energy. Since it is at rest on the surface,its kinetic energy is $0$. Thus,$E_1 = -\frac{GMm}{R}$.
When the satellite is in a circular orbit of radius $r = 7R$,its total energy is given by $E_2 = -\frac{GMm}{2r}$.
Substituting $r = 7R$,we get $E_2 = -\frac{GMm}{2(7R)} = -\frac{GMm}{14R}$.
The minimum energy required to be spent by the launching vehicle is the difference between the final energy and the initial energy: $\Delta E = E_2 - E_1$.
$\Delta E = -\frac{GMm}{14R} - (-\frac{GMm}{R}) = -\frac{GMm}{14R} + \frac{GMm}{R}$.
$\Delta E = \frac{-GMm + 14GMm}{14R} = \frac{13GMm}{14R}$.

(A) For a smooth inclined plane,the acceleration is $a_s = g \sin \theta$. The time taken to cover distance $s$ is $t_s = \sqrt{\frac{2s}{g \sin \theta}}$.
For a rough inclined plane,the acceleration is $a_r = g(\sin \theta - \mu \cos \theta)$. The time taken is $t_r = \sqrt{\frac{2s}{g(\sin \theta - \mu \cos \theta)}} = n t_s$.
Squaring both sides: $\frac{2s}{g(\sin \theta - \mu \cos \theta)} = n^2 \frac{2s}{g \sin \theta}$.
This simplifies to $\sin \theta = n^2(\sin \theta - \mu \cos \theta)$,which gives $\mu \cos \theta = \sin \theta (1 - \frac{1}{n^2})$.
Thus,$\mu = \tan \theta (1 - \frac{1}{n^2})$.
Given $\theta = 45^{\circ}$,$\tan 45^{\circ} = 1$,so $\mu = 1 - \frac{1}{n^2}$.

(B) Let the magnitudes of the two vectors be $|\overrightarrow{A}| = |\overrightarrow{B}| = a$.
The resultant vector $\overrightarrow{R} = \overrightarrow{A} + \overrightarrow{B}$ makes an angle $\alpha$ with vector $\overrightarrow{A}$,given by the formula:
$\tan \alpha = \frac{B \sin \theta}{A + B \cos \theta}$
Since $A = B = a$,we have:
$\tan \alpha = \frac{a \sin \theta}{a + a \cos \theta} = \frac{\sin \theta}{1 + \cos \theta}$
Using trigonometric identities $\sin \theta = 2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})$ and $1 + \cos \theta = 2 \cos^2(\frac{\theta}{2})$:
$\tan \alpha = \frac{2 \sin(\frac{\theta}{2}) \cos(\frac{\theta}{2})}{2 \cos^2(\frac{\theta}{2})} = \tan(\frac{\theta}{2})$
Therefore,$\alpha = \frac{\theta}{2}$.
This shows that the resultant vector bisects the angle between the two vectors of equal magnitude.

(C) The correct matches are as follows:
$A$. Hooke's law: Within the elastic limit,stress is directly proportional to strain. Thus,$A-3$.
$B$. Shearing strain: It is defined as the angle in radians through which a plane perpendicular to the fixed surface of a cubical body gets turned under the effect of a tangential force. It is also known as tangential strain. Thus,$B-1$.
$C$. Bulk strain: For a solid,the bulk strain (volumetric strain) is equal to $3$ times the linear strain. Thus,$C-4$.
$D$. Elastic fatigue: The temporary loss of elastic properties due to the action of repeated alternating deforming forces is called elastic fatigue. Thus,$D-2$.
Therefore,the correct matching is $A-3, B-1, C-4, D-2$,which corresponds to option $C$.

(B) The total time given is $2 \,min \,20 \,s$.
Converting this into seconds: $2 \times 60 \,s + 20 \,s = 120 \,s + 20 \,s = 140 \,s$.
The athlete completes one round in $40 \,s$.
The number of rounds completed in $140 \,s$ is $\frac{140}{40} = 3.5$ rounds.
After $3$ complete rounds, the athlete returns to the starting point, so the displacement for these $3$ rounds is $0$.
In the remaining $0.5$ round, the athlete moves from the starting point $A$ to the diametrically opposite point $B$.
The displacement is the shortest distance between the initial and final positions, which is the diameter of the circular track.
Therefore, the displacement $= 2R$.

(D) The energy of a particle executing $SHM$ is given by $E = \frac{1}{2} m \omega^2 A^2$, where $m$ is the mass, $\omega$ is the angular frequency, and $A$ is the amplitude.
For the first particle, $x_1 = 4 \sin (10t + \frac{\pi}{6})$, the amplitude $A_1 = 4$ and angular frequency $\omega_1 = 10$.
The energy is $E_1 = \frac{1}{2} m (10)^2 (4)^2 = \frac{1}{2} m (100)(16) = 800m$.
For the second particle, $x_2 = 5 \cos (\omega t)$, the amplitude $A_2 = 5$ and angular frequency is $\omega$.
The energy is $E_2 = \frac{1}{2} m \omega^2 (5)^2 = \frac{25}{2} m \omega^2$.
Given that the energies are equal, $E_1 = E_2$.
$800m = \frac{25}{2} m \omega^2$.
$1600 = 25 \omega^2$.
$\omega^2 = \frac{1600}{25} = 64$.
$\omega = 8 \text{ unit}$.

(B) Let the initial length of each rod be $l$. In the equilateral triangle $ABC$,the length of the altitude $DC$ is given by:
$DC^2 = AC^2 - AD^2 = l^2 - (l/2)^2 = 3l^2/4$.
After a small change in temperature $\Delta t$,the new lengths are $l' = l(1 + \alpha \Delta t)$.
For rod $AC$,the new length is $l_{AC}' = l(1 + \alpha_2 \Delta t)$.
For segment $AD$,the new length is $l_{AD}' = (l/2)(1 + \alpha_1 \Delta t)$.
Since $DC$ remains constant,$DC^2 = (l_{AC}')^2 - (l_{AD}')^2$.
$3l^2/4 = [l(1 + \alpha_2 \Delta t)]^2 - [(l/2)(1 + \alpha_1 \Delta t)]^2$.
$3l^2/4 = l^2(1 + 2\alpha_2 \Delta t + \alpha_2^2 \Delta t^2) - (l^2/4)(1 + 2\alpha_1 \Delta t + \alpha_1^2 \Delta t^2)$.
Neglecting higher-order terms $\alpha^2 \Delta t^2$ and simplifying:
$3l^2/4 = l^2 + 2l^2 \alpha_2 \Delta t - l^2/4 - (l^2/4)(2\alpha_1 \Delta t)$.
$3l^2/4 = 3l^2/4 + 2l^2 \alpha_2 \Delta t - (l^2/2)\alpha_1 \Delta t$.
$0 = 2l^2 \alpha_2 \Delta t - (l^2/2)\alpha_1 \Delta t$.
$2\alpha_2 = \alpha_1/2 \implies \alpha_1 = 4\alpha_2$.

(B) Let $L_0$ be the initial length of each strip before heating. After heating,the lengths of the brass and copper strips are given by:
$L_B = L_0(1 + \alpha_B \Delta T) = (R + d)\theta$
$L_C = L_0(1 + \alpha_C \Delta T) = R\theta$
Dividing the two equations,we get:
$\frac{R + d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T}$
$1 + \frac{d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T}$
$\frac{d}{R} = \frac{1 + \alpha_B \Delta T}{1 + \alpha_C \Delta T} - 1 = \frac{1 + \alpha_B \Delta T - 1 - \alpha_C \Delta T}{1 + \alpha_C \Delta T} = \frac{(\alpha_B - \alpha_C) \Delta T}{1 + \alpha_C \Delta T}$
Since $\alpha \Delta T \ll 1$,we can approximate $1 + \alpha_C \Delta T \approx 1$. Thus:
$R \approx \frac{d}{(\alpha_B - \alpha_C) \Delta T}$
Therefore,$R \propto \frac{1}{\Delta T}$.

(B) Let the temperature at junction $B$ be $\theta$.
Since the rods $BC$ and $BD$ are connected to the same temperature source $(80^{\circ} C)$ at their outer ends,they act as parallel thermal resistances.
Let $R$ be the thermal resistance of each rod.
The equivalent resistance of the parallel combination of rods $BC$ and $BD$ is $R_p = \frac{R \times R}{R + R} = \frac{R}{2}$.
At steady state,the heat current flowing through rod $AB$ must be equal to the sum of heat currents flowing through rods $BC$ and $BD$.
Using the formula for heat current $\frac{Q}{t} = \frac{\Delta T}{R}$,we have:
$\frac{\theta - 20}{R} = \frac{80 - \theta}{R/2}$
$\theta - 20 = 2(80 - \theta)$
$\theta - 20 = 160 - 2\theta$
$3\theta = 180$
$\theta = 60^{\circ} C$
Thus,the temperature at junction $B$ is $60^{\circ} C$.

(B) For an ideal gas, the equation of state is $pV = \mu RT$, which implies $p = (\frac{\mu R}{V})T$.
In a $p-T$ diagram, lines passing through the origin represent isochoric processes (constant volume) because $p \propto T$ implies $V = \text{constant}$.
Processes $AB$ and $CD$ are straight lines passing through the origin, so they are isochoric processes.
Work done in an isochoric process is zero. Thus, $W_{AB} = 0$ and $W_{CD} = 0$.
Processes $BC$ and $DA$ are isobaric processes (constant pressure).
Work done in an isobaric process is $W = p\Delta V = \mu R \Delta T$.
For process $BC$ (at pressure $p_2$): $W_{BC} = \mu R(T_C - T_B) = 3 \times R \times (2400 - 800) = 3R \times 1600 = 4800R$.
For process $DA$ (at pressure $p_1$): $W_{DA} = \mu R(T_A - T_D) = 3 \times R \times (400 - 1200) = 3R \times (-800) = -2400R$.
The total work done in the cycle is $W = W_{AB} + W_{BC} + W_{CD} + W_{DA} = 0 + 4800R + 0 - 2400R = 2400R$.
Substituting $R = 8.314 \, J/mol \cdot K$: $W = 2400 \times 8.314 = 19953.6 \, J \approx 20 \, kJ$.

(D) $1$. In the $p-V$ diagram,the isothermal expansion from $A$ to $B$ follows the curve $AB$. The work done by the gas is the area under the curve $AB$,which is the area $ABDE$.
$2$. The adiabatic compression from $B$ to $C$ follows the curve $BC$. The work done on the gas is the area under the curve $BC$,which is the area $BCDE$.
$3$. From the graph,it is clear that the final pressure $p_2$ at point $C$ is greater than the initial pressure $p_1$ at point $A$,so $p_2 > p_1$.
$4$. The net work done by the gas $W$ is the sum of the work done during expansion $(W_{exp} > 0)$ and the work done during compression $(W_{comp} < 0)$.
$5$. Since the area under the adiabatic curve $BC$ (which represents the magnitude of work done on the gas) is greater than the area under the isothermal curve $AB$ (which represents the magnitude of work done by the gas),the net work done $W = W_{AB} - W_{BC}$ is negative $(W < 0)$.
$6$. Therefore,$p_2 > p_1$ and $W < 0$.

(B) According to the principle of homogeneity of dimensions,the dimensions of each term in a physical equation must be the same.
Given the equation $F = at + bt^2$,where $F$ is force and $t$ is time.
The dimension of force $F$ is $[MLT^{-2}]$.
For the first term: $[at] = [F]$
$[a] = [F] / [t] = [MLT^{-2}] / [T] = [MLT^{-3}]$.
For the second term: $[bt^2] = [F]$
$[b] = [F] / [t^2] = [MLT^{-2}] / [T^2] = [MLT^{-4}]$.
Therefore,the dimensions of $a$ and $b$ are $[MLT^{-3}]$ and $[MLT^{-4}]$ respectively.

(D) The frequency of a closed organ pipe vibrating in its first harmonic is $n_1 = \frac{v_1}{4 l_1}$.
The frequency of an open organ pipe vibrating in its third harmonic is $n_3 = \frac{3 v_2}{2 l_2}$.
Since both pipes are in resonance with the same tuning fork,$n_1 = n_3$.
Therefore,$\frac{v_1}{4 l_1} = \frac{3 v_2}{2 l_2}$,which implies $\frac{l_1}{l_2} = \frac{1}{6} \left( \frac{v_1}{v_2} \right)$.
The speed of sound in a gas is given by $v = \sqrt{\frac{B}{\rho}}$,where $B$ is the bulk modulus (reciprocal of compressibility) and $\rho$ is the density.
Since the compressibility is equal,$B_1 = B_2 = B$.
Thus,$\frac{v_1}{v_2} = \sqrt{\frac{B/\rho_1}{B/\rho_2}} = \sqrt{\frac{\rho_2}{\rho_1}}$.
Substituting this into the ratio of lengths,we get $\frac{l_1}{l_2} = \frac{1}{6} \sqrt{\frac{\rho_2}{\rho_1}}$.

(C) The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
Since $T$ (tension) and $m$ (mass per unit length) are constant for the same wire,$n \propto \frac{1}{l}$.
Given the ratio of frequencies $n_1 : n_2 : n_3 = 1 : 3 : 4$,the lengths of the segments must be in the inverse ratio:
$l_1 : l_2 : l_3 = \frac{1}{1} : \frac{1}{3} : \frac{1}{4}$.
To simplify this ratio,multiply by the least common multiple of $1, 3, 4$,which is $12$:
$l_1 : l_2 : l_3 = 12 : 4 : 3$.
The sum of the parts is $12 + 4 + 3 = 19$.
The total length of the wire is $114 \ cm$.
Therefore,the lengths are:
$l_1 = \frac{12}{19} \times 114 = 12 \times 6 = 72 \ cm$
$l_2 = \frac{4}{19} \times 114 = 4 \times 6 = 24 \ cm$
$l_3 = \frac{3}{19} \times 114 = 3 \times 6 = 18 \ cm$.
Thus,the lengths are $72 \ cm, 24 \ cm, 18 \ cm$.

(C) Let $m$ be the mass of the ball. When the ball is at a height $h = 10 ~m$,its velocity is $v_0$.
Just before hitting the ground,let its velocity be $v$. By conservation of energy,the kinetic energy just before impact is $K_i = \frac{1}{2} m v_0^2 + mgh$.
However,the problem states that at $10 ~m$ height,the velocity is $v_0$. Let $H$ be the total height from which it fell. Then $\frac{1}{2} m v^2 = mgH$.
After collision,it loses $50 \%$ of its kinetic energy. The remaining kinetic energy is $K_f = 0.5 \times K_i = 0.5 \times (\frac{1}{2} m v^2)$.
This energy allows it to rise back to $h = 10 ~m$. Thus,$0.5 \times (\frac{1}{2} m v^2) = mgh$.
Substituting $h = 10 ~m$ and $g = 9.8 ~m/s^2$:
$0.25 v^2 = 9.8 \times 10 = 98$.
$v^2 = 392$.
Now,consider the motion from $10 ~m$ height to the ground: $v^2 = v_0^2 + 2gh$.
$392 = v_0^2 + 2 \times 9.8 \times 10$.
$392 = v_0^2 + 196$.
$v_0^2 = 196$.
$v_0 = 14 ~m/s$.

(C) For an equilateral glass prism,the angle of the prism is $A = 60^{\circ}$.
Given that the angle of incidence $i$ is equal to the angle of emergence $e$,and each is $\frac{3}{4}$ of the angle of the prism:
$i = e = \frac{3}{4} A = \frac{3}{4} \times 60^{\circ} = 45^{\circ}$.
The formula for the angle of deviation $\delta$ is given by:
$\delta = i + e - A$.
Substituting the values:
$\delta = 45^{\circ} + 45^{\circ} - 60^{\circ} = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
Thus,the angle of deviation is $30^{\circ}$.

(B) The potential across a discharging capacitor is given by $V = V_0 e^{-t/RC}$.
We want to find the time $t$ when $V = V_0/2$.
Substituting this into the equation: $V_0/2 = V_0 e^{-t/RC}$.
This simplifies to $1/2 = e^{-t/RC}$, or $e^{t/RC} = 2$.
Taking the natural logarithm on both sides: $t/RC = \ln(2)$.
Using the conversion $\ln(2) = 2.3026 \times \log_{10}(2)$, we get $t = RC \times 2.3026 \times 0.3010$.
Given $C = 0.1 \mu F = 0.1 \times 10^{-6} F$ and $R = 10 M \Omega = 10 \times 10^6 \Omega$.
Calculating the time constant: $RC = (0.1 \times 10^{-6}) \times (10 \times 10^6) = 1 \ s$.
Therefore, $t = 1 \times 2.3026 \times 0.3010 \approx 0.693 \ s$.

(B) When capacitors are connected in series,the equivalent capacitance $C_{\text{eq}}$ is given by $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$.
For $C_1 = 1 \mu F$ and $C_2 = C \mu F$,we have $C_{\text{eq}} = \frac{1 \times C}{1 + C} = \frac{C}{C+1} \mu F$.
Given the total charge $q = 80 \mu C$ and potential difference $V = 120 \ V$,we use the relation $q = C_{\text{eq}} V$.
Substituting the values: $80 = \left( \frac{C}{C+1} \right) \times 120$.
Dividing both sides by $40$: $2 = \left( \frac{C}{C+1} \right) \times 3$.
$2(C+1) = 3C \implies 2C + 2 = 3C \implies C = 2 \mu F$.
In a series combination,the charge on each capacitor is the same as the total charge $q = 80 \mu C$.
The energy stored in a capacitor is given by $U = \frac{q^2}{2C}$.
For the capacitor $C = 2 \mu F$: $U = \frac{(80 \mu C)^2}{2 \times 2 \mu F} = \frac{6400}{4} \mu J = 1600 \mu J$.

(D) The electric field $E$ between the parallel plates is given by $E = \frac{V}{d}$.
Given, $V = 10^4 \,V$ and $d = 0.5 \,cm = 0.5 \times 10^{-2} \,m$.
Therefore, $E = \frac{10^4}{0.5 \times 10^{-2}} = 2 \times 10^6 \,V/m$.
The force $F$ on an electron of charge $e = 1.6 \times 10^{-19} \,C$ is given by $F = eE$.
$F = (1.6 \times 10^{-19} \,C) \times (2 \times 10^6 \,V/m) = 3.2 \times 10^{-13} \,N$.

(B) Given: Area $A = 0.3 \,m^2$,charge density $n = 2 \times 10^{25} / m^3$,and charge $q = 3t^2 + 5t + 2$.
First,calculate the current $i$ at $t = 2 \,s$ using $i = \frac{dq}{dt}$.
$i = \frac{d}{dt}(3t^2 + 5t + 2) = 6t + 5$.
At $t = 2 \,s$,$i = 6(2) + 5 = 17 \,A$.
The relationship between current and drift velocity $v_d$ is $i = n e A v_d$,where $e = 1.6 \times 10^{-19} \,C$.
Rearranging for $v_d$: $v_d = \frac{i}{n e A}$.
Substituting the values: $v_d = \frac{17}{2 \times 10^{25} \times 1.6 \times 10^{-19} \times 0.3}$.
$v_d = \frac{17}{0.96 \times 10^6} = 17.708 \times 10^{-6} \,m/s = 1.77 \times 10^{-5} \,m/s$.

(A) First, calculate the equivalent resistance of the parallel combination of $6 \Omega$ and $12 \Omega$ resistors:
$R_p = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4 \Omega$
Next, calculate the total resistance of the circuit, as the parallel combination is in series with the $6 \Omega$ resistor:
$R_{eq} = R_p + 6 \Omega = 4 \Omega + 6 \Omega = 10 \Omega$
Now, calculate the total current flowing from the $10 \text{ V}$ battery:
$I = \frac{V}{R_{eq}} = \frac{10 \text{ V}}{10 \Omega} = 1 \text{ A}$
The potential difference across the parallel combination is the product of the total current and the equivalent resistance of the parallel part:
$V_p = I \times R_p = 1 \text{ A} \times 4 \Omega = 4 \text{ V}$
Since the $6 \Omega$ and $12 \Omega$ resistors are connected in parallel, the potential difference across each of them is the same and is equal to the potential difference across the parallel combination.
Therefore, the potential difference across the $12 \Omega$ resistor is $4 \text{ V}$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ is given by $K_{\max} = h v - h v_0$,where $h$ is Planck's constant and $v_0$ is the threshold frequency.
For frequency $v_1$,$K_1 = h(v_1 - v_0)$.
For frequency $v_2$,$K_2 = h(v_2 - v_0)$.
Given the ratio of maximum kinetic energies is $K_1 : K_2 = 1 : n$,we have $\frac{K_1}{K_2} = \frac{1}{n}$.
Substituting the expressions,we get $\frac{h(v_1 - v_0)}{h(v_2 - v_0)} = \frac{1}{n}$.
$\frac{v_1 - v_0}{v_2 - v_0} = \frac{1}{n}$.
Cross-multiplying gives $n(v_1 - v_0) = v_2 - v_0$.
$n v_1 - n v_0 = v_2 - v_0$.
$n v_1 - v_2 = n v_0 - v_0$.
$n v_1 - v_2 = v_0(n - 1)$.
Therefore,the threshold frequency is $v_0 = \frac{n v_1 - v_2}{n - 1}$.

(A) The time constant $\tau$ of an $LR$ circuit is given by $\tau = \frac{L}{R}$.
Initially,$\tau_1 = \frac{L}{R} = 3 \times 10^{-3} \text{ s}$ ...$(i)$
When a $90 \Omega$ resistor is added in series,the total resistance becomes $(R + 90) \Omega$. The new time constant is $\tau_2 = \frac{L}{R + 90} = 0.5 \times 10^{-3} \text{ s}$ ...(ii)
Dividing equation $(i)$ by equation (ii):
$\frac{3 \times 10^{-3}}{0.5 \times 10^{-3}} = \frac{L/R}{L/(R + 90)}$
$6 = \frac{R + 90}{R}$
$6R = R + 90$
$5R = 90 \implies R = 18 \Omega$
Substituting $R = 18 \Omega$ into equation $(i)$:
$L = 3 \times 10^{-3} \times 18 = 54 \times 10^{-3} \text{ H} = 54 \text{ mH}$.
Thus,the inductance is $54 \text{ mH}$ and the resistance is $18 \Omega$.

(D) The thermo emf is given by $E = aT + bT^2$.
At the neutral temperature $(T_n)$,the emf is maximum,and at the inversion temperature $(T_i)$,the emf becomes zero.
Setting $E = 0$:
$aT_i + bT_i^2 = 0$
$T_i(a + bT_i) = 0$
Since $T_i \neq 0$,we have $T_i = -\frac{a}{b}$.
Given $\frac{a}{b} = -200^{\circ}C$,therefore $T_i = -(-200^{\circ}C) = 200^{\circ}C$.
This $T_i$ is the temperature relative to the cold junction at $0^{\circ}C$.
If the cold junction is at $T_c = 30^{\circ}C$,the inversion temperature relative to the cold junction is $T_i = 200^{\circ}C$.
However,the standard relation for inversion temperature is $T_i - T_c = T_n - T_i$,where $T_n = -\frac{a}{2b} = 100^{\circ}C$.
Thus,$T_i = 2T_n - T_c = 2(100) - 30 = 170^{\circ}C$.
Converting to Kelvin: $T = 170 + 273 = 443 \ K$.

(A) The deflection $y$ of a charged particle moving with velocity $u$ in a uniform electric field $E$ over a distance $L$ is given by $y = \frac{1}{2} a t^2 = \frac{1}{2} (\frac{qE}{m}) (\frac{L}{u})^2 = \frac{qEL^2}{2mu^2}$.
Since momentum $p = mu$, we have $u = p/m$. Substituting this, $y = \frac{qEL^2}{2m(p/m)^2} = \frac{qEL^2m}{2p^2}$.
Given that $E, L,$ and $p$ are constant for all particles, the deflection $y \propto qm$.
For a proton $(p)$, deuteron $(d)$, and $\alpha$-particle $(\alpha)$:
Charge ratio: $q_p : q_d : q_\alpha = 1 : 1 : 2$.
Mass ratio: $m_p : m_d : m_\alpha = 1 : 2 : 4$.
Therefore, the ratio of deflections is $y_p : y_d : y_\alpha = (q_p m_p) : (q_d m_d) : (q_\alpha m_\alpha) = (1 \times 1) : (1 \times 2) : (2 \times 4) = 1 : 2 : 8$.

(C) The kinetic energy of the electron is given by $K = \frac{1}{2}mv^2$.
Given $K = 45.5 \ eV = 45.5 \times 1.6 \times 10^{-19} \ J$ and $m = 9.1 \times 10^{-31} \ kg$.
$v^2 = \frac{2K}{m} = \frac{2 \times 45.5 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}} = \frac{145.6 \times 10^{-19}}{9.1 \times 10^{-31}} = 16 \times 10^{12} \ m^2s^{-2}$.
Thus,$v = 4 \times 10^6 \ ms^{-1}$.
For an electron to remain undeflected in crossed electric and magnetic fields,the electric force must balance the magnetic force: $eE = evB$,which implies $v = \frac{E}{B}$.
Given $E = 1 \times 10^3 \ Vm^{-1}$,we have $B = \frac{E}{v} = \frac{1 \times 10^3}{4 \times 10^6} = 0.25 \times 10^{-3} = 2.5 \times 10^{-4} \ Wb \ m^{-2}$.

(B) The frequency of oscillation in a vibration magnetometer is given by $n = \frac{1}{2\pi} \sqrt{\frac{MB}{I}}$.
For two magnets with moments $M_1$ and $M_2$ and moment of inertia $I_1$ and $I_2$,when tied together,the total moment of inertia is $I = I_1 + I_2$.
When like poles are tied,the effective magnetic moment is $M_{sum} = M_1 + M_2$. The frequency is $n_1 = 6 \ Hz$.
When unlike poles are tied,the effective magnetic moment is $M_{diff} = M_1 - M_2$. The frequency is $n_2 = 2 \ Hz$.
Since $n \propto \sqrt{M}$,we have $\frac{n_1}{n_2} = \sqrt{\frac{M_1 + M_2}{M_1 - M_2}}$.
Squaring both sides: $(\frac{6}{2})^2 = \frac{M_1 + M_2}{M_1 - M_2} \implies 9 = \frac{M_1 + M_2}{M_1 - M_2}$.
$9M_1 - 9M_2 = M_1 + M_2$.
$8M_1 = 10M_2$.
$\frac{M_1}{M_2} = \frac{10}{8} = \frac{5}{4}$.

(A) The formula for the magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 I}{2r}$.
Given values are:
Permeability of free space,$\mu_0 = 4\pi \times 10^{-7} \,T \cdot m/A$
Current,$I = 0.9 \,A$
Radius,$r = 5 \,cm = 5 \times 10^{-2} \,m$
Substituting these values into the formula:
$B = \frac{4\pi \times 10^{-7} \times 0.9}{2 \times 5 \times 10^{-2}}$
$B = \frac{3.6\pi \times 10^{-7}}{10 \times 10^{-2}}$
$B = \frac{3.6\pi \times 10^{-7}}{10^{-1}}$
$B = 3.6\pi \times 10^{-6} \,T$
$B = 36\pi \times 10^{-7} \,T$

(D) When a magnetic needle is placed in two mutually perpendicular magnetic fields $B_1$ and $B_2$,it aligns itself along the direction of the resultant magnetic field.
The torque due to the first field $B_1$ is $\tau_1 = mB_1 \sin \theta$,where $m$ is the magnetic moment.
The torque due to the second field $B_2$ is $\tau_2 = mB_2 \cos \theta$.
In equilibrium,the net torque on the needle is zero,so $\tau_1 = \tau_2$.
$mB_1 \sin \theta = mB_2 \cos \theta$
$\tan \theta = \frac{B_2}{B_1}$
Given $B_1 = 1 \text{ T}$ and $B_2 = \sqrt{3} \text{ T}$.
$\tan \theta = \frac{\sqrt{3}}{1} = \sqrt{3}$
$\theta = 60^{\circ}$

(C) The acceptance angle $\theta_a$ is the maximum angle of incidence at the entrance of the optical fibre such that the light undergoes total internal reflection at the core-cladding interface.
For an optical fibre with core refractive index $\mu_1 = 1.5$ and cladding refractive index $\mu_2 = 1.414$,the acceptance angle $\theta_a$ is given by the formula:
$\sin \theta_a = \sqrt{\mu_1^2 - \mu_2^2}$
Substituting the given values:
$\sin \theta_a = \sqrt{(1.5)^2 - (1.414)^2}$
$\sin \theta_a = \sqrt{2.25 - 1.999396} \approx \sqrt{0.2506} \approx 0.5006$
$\theta_a = \sin^{-1}(0.5006) \approx 30^{\circ}$
Thus,the range of the incident angle with the axis of the optical fibre for total internal reflection is $0^{\circ}$ to $30^{\circ}$.
Therefore,option $C$ is correct.

(A) The resolving limit $(d\theta)$ of a telescope is given by the formula:
$d\theta = \frac{1.22 \lambda}{a}$
where $\lambda$ is the wavelength of light and $a$ is the diameter of the objective lens.
Given:
$\lambda = 4538 \ \text{Å} = 4538 \times 10^{-10} \ \text{m}$
$a = 1 \ \text{m}$
Substituting the values:
$d\theta = \frac{1.22 \times 4538 \times 10^{-10}}{1}$
$d\theta = 5536.36 \times 10^{-10} \ \text{rad}$
$d\theta \approx 5.54 \times 10^{-7} \ \text{rad}$

(D) The current gain $\beta$ of a transistor in common-emitter configuration is defined as the ratio of the change in collector current $(\Delta i_C)$ to the change in base current $(\Delta i_B)$.
$\beta = \frac{\Delta i_C}{\Delta i_B}$
Given: $\beta = 80$ and $\Delta i_B = 250 \mu A = 250 \times 10^{-6} \text{ A}$.
Substituting the values into the formula:
$80 = \frac{\Delta i_C}{250 \times 10^{-6} \text{ A}}$
$\Delta i_C = 80 \times 250 \times 10^{-6} \text{ A}$
$\Delta i_C = 20,000 \times 10^{-6} \text{ A}$
$\Delta i_C = 20 \times 10^{-3} \text{ A} = 20 \text{ mA}$.
Therefore,the change in collector current is $20 \text{ mA}$.

(C) The ratio of intensities of two coherent sources is given as $\frac{I_1}{I_2} = \frac{64}{1}$.
Let the intensities be $I_1 = 64k$ and $I_2 = 1k$.
The ratio of maximum intensity to minimum intensity in an interference pattern is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{I_1} + \sqrt{I_2})^2}{(\sqrt{I_1} - \sqrt{I_2})^2}$
Substituting the values:
$\frac{I_{\max}}{I_{\min}} = \frac{(\sqrt{64k} + \sqrt{1k})^2}{(\sqrt{64k} - \sqrt{1k})^2}$
$\frac{I_{\max}}{I_{\min}} = \frac{(8\sqrt{k} + 1\sqrt{k})^2}{(8\sqrt{k} - 1\sqrt{k})^2}$
$\frac{I_{\max}}{I_{\min}} = \frac{(9\sqrt{k})^2}{(7\sqrt{k})^2} = \frac{81k}{49k} = \frac{81}{49}$.