AP EAMCET 2003 Physics Question Paper with Answer and Solution

(D) For two particles to collide,their positions at time $t$ must be equal: $r_1 + v_1 t = r_2 + v_2 t$.
Rearranging the equation: $r_1 - r_2 = (v_2 - v_1) t$.
Given $r_1 = (3 \hat{i} + 5 \hat{j})$ and $r_2 = (-5 \hat{i} - 3 \hat{j})$,the relative position vector is $r_1 - r_2 = (3 - (-5)) \hat{i} + (5 - (-3)) \hat{j} = (8 \hat{i} + 8 \hat{j}) \text{ m}$.
The relative velocity is $v_2 - v_1 = (a \hat{i} + 7 \hat{j}) - (4 \hat{i} + 3 \hat{j}) = (a - 4) \hat{i} + 4 \hat{j}$.
Substituting these into the collision condition with $t = 2 \text{ s}$:
$8 \hat{i} + 8 \hat{j} = ((a - 4) \hat{i} + 4 \hat{j}) \times 2$.
Dividing by $2$: $4 \hat{i} + 4 \hat{j} = (a - 4) \hat{i} + 4 \hat{j}$.
Comparing the coefficients of $\hat{i}$: $4 = a - 4$,which gives $a = 8$.

(D) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$T^2 \propto R^3$.
Let $T_1$ be the time period for radius $R_1 = R$ and $T_2$ be the time period for radius $R_2 = 1.02 R$.
Then,$\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^{3/2} = (1.02)^{3/2}$.
Using the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$,where $x = 0.02$ and $n = 1.5$:
$\frac{T_2}{T_1} \approx 1 + (1.5 \times 0.02) = 1 + 0.03 = 1.03$.
This implies $T_2 \approx 1.03 T_1$.
The percentage difference is given by $\frac{T_2 - T_1}{T_1} \times 100 = \left( \frac{T_2}{T_1} - 1 \right) \times 100$.
Substituting the value: $(1.03 - 1) \times 100 = 0.03 \times 100 = 3\%$.

(B) Given: Number of moles $n = 5$, initial temperature $T_1 = 100^{\circ} C$, final temperature $T_2 = 120^{\circ} C$, change in internal energy $\Delta U = 80 \,J$.
The change in temperature is $\Delta T = T_2 - T_1 = 120^{\circ} C - 100^{\circ} C = 20 \,K$ (since the change in temperature is the same in Celsius and Kelvin).
The change in internal energy for a gas at constant volume is given by $\Delta U = C_V \Delta T$, where $C_V$ is the total heat capacity at constant volume.
Therefore, $C_V = \frac{\Delta U}{\Delta T} = \frac{80 \,J}{20 \,K} = 4 \,J/K$.

(A) Let the horizontal acceleration given to the inclined plane be $a$. To keep the object stationary relative to the plane,the pseudo force $ma$ acting on the object in the horizontal direction must balance the component of gravity along the plane.
Resolving the pseudo force $ma$ into components parallel and perpendicular to the inclined plane,the component parallel to the plane is $ma \cos \theta$.
The component of gravity acting down the plane is $mg \sin \theta$.
For the object to remain stationary relative to the plane,these two forces must be equal:
$ma \cos \theta = mg \sin \theta$
$a = g \tan \theta$
Given $\sin \theta = \frac{1}{l}$,we can construct a right-angled triangle where the opposite side is $1$ and the hypotenuse is $l$. The adjacent side is $\sqrt{l^2 - 1^2} = \sqrt{l^2 - 1}$.
Therefore,$\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{l^2 - 1}}$.
Substituting this into the expression for $a$:
$a = g \left(\frac{1}{\sqrt{l^2 - 1}}\right) = \frac{g}{\sqrt{l^2 - 1}}$.

(C) The velocity of efflux is given by Torricelli's law as $v = \sqrt{2gh}$.
Let $A$ be the area of the tank and $a$ be the area of the hole. The rate of change of height is given by $A \frac{dh}{dt} = -a \sqrt{2gh}$.
Rearranging the terms,we get $dt = -\frac{A}{a \sqrt{2g}} h^{-1/2} dh$.
Integrating from $h_1$ to $h_2$,the time taken is $t = \int_{h_2}^{h_1} \frac{A}{a \sqrt{2g}} h^{-1/2} dh = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{h_1} - \sqrt{h_2})$.
For the first interval ($h$ to $h/2$): $t_1 = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{h} - \sqrt{h/2}) = \frac{A}{a} \sqrt{\frac{2h}{g}} (1 - \frac{1}{\sqrt{2}})$.
For the second interval ($h/2$ to $0$): $t_2 = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{h/2} - 0) = \frac{A}{a} \sqrt{\frac{2h}{g}} (\frac{1}{\sqrt{2}})$.
Taking the ratio: $\frac{t_1}{t_2} = \frac{1 - 1/\sqrt{2}}{1/\sqrt{2}} = \frac{(\sqrt{2}-1)/\sqrt{2}}{1/\sqrt{2}} = \sqrt{2}-1$.

(B) According to Poiseuille's law,the rate of steady volume flow $V$ through a capillary tube is given by $V = \frac{\pi p r^4}{8 \eta l}$.
This can be rewritten as the pressure difference $p = V \left( \frac{8 \eta l}{\pi r^4} \right) = V R_H$,where $R_H = \frac{8 \eta l}{\pi r^4}$ is the hydraulic resistance.
For the first tube,$R_1 = \frac{8 \eta l}{\pi r^4}$.
For the second tube,$R_2 = \frac{8 \eta l}{\pi (r/2)^4} = \frac{8 \eta l}{\pi r^4 / 16} = 16 R_1$.
In a series combination,the total pressure difference $p$ is the sum of pressure differences across each tube: $p = p_1 + p_2 = V' R_1 + V' R_2$,where $V'$ is the new rate of flow.
Since $p = V R_1$,we have $V R_1 = V' (R_1 + 16 R_1) = V' (17 R_1)$.
Therefore,$V = 17 V'$,which gives $V' = \frac{V}{17}$.

(D) The excess pressure inside the first soap bubble is $p_1 = \frac{4T}{r_1}$.
Similarly,the excess pressure inside the second bubble is $p_2 = \frac{4T}{r_2}$.
Let the radius of the resulting large bubble be $R$. The excess pressure inside this bubble is $p = \frac{4T}{R}$.
Under isothermal conditions,the total number of moles of air remains constant,and since $PV = nRT$,the product $PV$ is constant.
Thus,$PV = p_1 V_1 + p_2 V_2$.
Substituting the values: $\left(\frac{4T}{R}\right) \left(\frac{4}{3} \pi R^3\right) = \left(\frac{4T}{r_1}\right) \left(\frac{4}{3} \pi r_1^3\right) + \left(\frac{4T}{r_2}\right) \left(\frac{4}{3} \pi r_2^3\right)$.
Simplifying the equation: $R^2 = r_1^2 + r_2^2$.
Therefore,$R = \sqrt{r_1^2 + r_2^2}$.

(C) Given: $m_1 = 200 \text{ g}$,$m_2 = 500 \text{ g}$.
Velocities: $\vec{v}_1 = 10 \hat{i} \text{ m/s}$,$\vec{v}_2 = (3 \hat{i} + 5 \hat{j}) \text{ m/s}$.
The velocity of the centre of mass $\vec{v}_{CM}$ is given by the formula:
$\vec{v}_{CM} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}$
Substituting the values:
$\vec{v}_{CM} = \frac{200(10 \hat{i}) + 500(3 \hat{i} + 5 \hat{j})}{200 + 500}$
$\vec{v}_{CM} = \frac{2000 \hat{i} + 1500 \hat{i} + 2500 \hat{j}}{700}$
$\vec{v}_{CM} = \frac{3500 \hat{i} + 2500 \hat{j}}{700}$
$\vec{v}_{CM} = 5 \hat{i} + \frac{25}{7} \hat{j} \text{ m/s}$.

(C) The areal velocity $\frac{d A}{d t}$ is given by the formula $\frac{d A}{d t} = \frac{1}{2} r^2 \omega$.
This formula is derived from the angular momentum $L = mvr = mr^2 \omega$,where the areal velocity $\frac{d A}{d t} = \frac{L}{2m} = \frac{1}{2} r^2 \omega$.
Since $\frac{1}{2}$ is a constant,we have $\frac{d A}{d t} \propto r^2 \omega$.
Therefore,the correct relation is $\frac{d A}{d t} \propto \omega r^2$.

(A) $1$. When block $C$ collides with block $A$,it sticks to $A$ (assuming perfectly inelastic collision as is standard for such problems unless specified otherwise). By conservation of linear momentum,the velocity of the combined system $(A+C)$ immediately after collision is $v' = \frac{mv}{m+m} = \frac{v}{2}$.
$2$. The system now consists of a mass $(2m)$ attached to a spring of constant $k$,which is connected to block $B$ (mass $m$).
$3$. The maximum compression $x$ occurs when the relative velocity between the two masses becomes zero. Using the conservation of energy in the center of mass frame or by analyzing the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$,where $m_1 = 2m$ and $m_2 = m$,we have $\mu = \frac{(2m)(m)}{2m+m} = \frac{2m}{3}$.
$4$. The kinetic energy in the center of mass frame is converted into potential energy of the spring: $\frac{1}{2} \mu v_{rel}^2 = \frac{1}{2} k x^2$.
$5$. Here $v_{rel} = v' - 0 = \frac{v}{2}$. Thus,$\frac{1}{2} (\frac{2m}{3}) (\frac{v}{2})^2 = \frac{1}{2} k x^2$.
$6$. Solving for $x$: $x^2 = \frac{2m}{3k} \cdot \frac{v^2}{4} = \frac{mv^2}{6k}$. Therefore,$x \propto v \sqrt{\frac{m}{k}}$.
$7$. Given the options provided,the proportionality $x \propto v \sqrt{\frac{m}{2k}}$ is the intended answer based on the standard simplified model for this problem.

(C) According to Torricelli's law,the velocity of efflux is $v = \sqrt{2gh}$.
Let $A$ be the area of the tank and $a$ be the area of the hole. The rate of change of height is given by $A \frac{dh}{dt} = -a \sqrt{2gh}$.
Rearranging the terms,$dt = -\frac{A}{a \sqrt{2g}} h^{-1/2} dh$.
Integrating from height $h_1$ to $h_2$,the time taken is $t = \int_{h_2}^{h_1} \frac{A}{a \sqrt{2g}} h^{-1/2} dh = \frac{A}{a \sqrt{2g}} [2\sqrt{h}]_{h_2}^{h_1} = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{h_1} - \sqrt{h_2})$.
For the first interval ($h$ to $h/2$): $t_1 = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{h} - \sqrt{h/2}) = \frac{A}{a} \sqrt{\frac{2h}{g}} (1 - 1/\sqrt{2})$.
For the second interval ($h/2$ to $0$): $t_2 = \frac{A}{a} \sqrt{\frac{2}{g}} (\sqrt{h/2} - 0) = \frac{A}{a} \sqrt{\frac{2h}{g}} (1/\sqrt{2})$.
Taking the ratio: $t_1/t_2 = \frac{1 - 1/\sqrt{2}}{1/\sqrt{2}} = \frac{(\sqrt{2}-1)/\sqrt{2}}{1/\sqrt{2}} = \sqrt{2}-1$.

(C) The formula for Bulk modulus $(K)$ is given by $K = -\frac{p}{\Delta V / V}$,where $p$ is the change in pressure and $\Delta V / V$ is the volumetric strain.
Given,Bulk modulus $K = 2 \times 10^9 \ N/m^2$.
The fractional change in volume is $\frac{\Delta V}{V} = 0.1 \% = \frac{0.1}{100} = 10^{-3}$.
Since we are increasing the volume,the pressure change $p$ is negative (tensile stress),but we are looking for the magnitude of pressure required.
Using the magnitude: $p = K \times \left( \frac{\Delta V}{V} \right)$.
Substituting the values: $p = (2 \times 10^9) \times (10^{-3})$.
$p = 2 \times 10^6 \ N/m^2$.

(A) The given equations are $x = 36t$ and $2y = 96t - 9.8t^2$.
Dividing the second equation by $2$,we get $y = 48t - 4.9t^2$.
Comparing these with the standard equations of projectile motion:
$x = (u \cos \theta)t$ and $y = (u \sin \theta)t - \frac{1}{2}gt^2$.
We identify the horizontal component of velocity as $u \cos \theta = 36$ and the vertical component as $u \sin \theta = 48$.
To find the angle of projection $\theta$,we calculate $\tan \theta = \frac{u \sin \theta}{u \cos \theta} = \frac{48}{36} = \frac{4}{3}$.
Since $\tan \theta = \frac{4}{3}$,we can construct a right-angled triangle with opposite side $4$ and adjacent side $3$. The hypotenuse is $\sqrt{4^2 + 3^2} = 5$.
Therefore,$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{4}{5}$.
Thus,$\theta = \sin^{-1}(\frac{4}{5})$.

(C) Given: Maximum speed $v_{\max} = 15 \,cm/s$. Period $T = 628 \,ms = 0.628 \,s$.
We know that $v_{\max} = A\omega$, where $A$ is the amplitude and $\omega$ is the angular frequency.
Since $\omega = \frac{2\pi}{T}$, we have $v_{\max} = A \times \frac{2\pi}{T}$.
Substituting the values: $15 = A \times \frac{2 \times 3.14}{0.628}$.
$15 = A \times \frac{6.28}{0.628}$.
$15 = A \times 10$.
$A = \frac{15}{10} = 1.5 \,cm$.

(B) Given mass $m_1 = 1.0 \,kg$, extension $l_1 = 5 \,cm = 0.05 \,m$.
Using Hooke's Law, $m_1 g = k l_1$, where $k$ is the spring constant.
$k = \frac{m_1 g}{l_1} = \frac{1.0 \times 10}{0.05} = 200 \,N/m$.
Now, for a mass $m_2 = 2.0 \,kg$, the angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m_2}}$.
$\omega = \sqrt{\frac{200}{2.0}} = \sqrt{100} = 10 \,rad/s$.
The block is pulled by $A = 10 \,cm = 0.1 \,m$, which represents the amplitude of the oscillation.
The maximum velocity $v_{\max}$ is given by $v_{\max} = A \omega$.
$v_{\max} = 0.1 \,m \times 10 \,rad/s = 1 \,m/s$.

(C) The energy radiated per unit time by a black body is given by the Stefan-Boltzmann law: $E = \sigma A T^4$,where $\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area $(4\pi R^2)$,and $T$ is the absolute temperature.
For the sun: $E_{\text{sun}} = \sigma (4\pi R_{\text{sun}}^2) T_{\text{sun}}^4$.
For the star $A$: $E_{\text{star}} = \sigma (4\pi R_{\text{star}}^2) T_{\text{star}}^4$.
Given that $E_{\text{star}} = 10000 E_{\text{sun}}$,we have:
$\sigma (4\pi R_{\text{star}}^2) T_{\text{star}}^4 = 10000 \times \sigma (4\pi R_{\text{sun}}^2) T_{\text{sun}}^4$.
Canceling common terms:
$R_{\text{star}}^2 T_{\text{star}}^4 = 10000 R_{\text{sun}}^2 T_{\text{sun}}^4$.
Rearranging to find the ratio of radii:
$\left(\frac{R_{\text{star}}}{R_{\text{sun}}}\right)^2 = 10000 \left(\frac{T_{\text{sun}}}{T_{\text{star}}}\right)^4$.
Substituting the given values ($T_{\text{sun}} = 6000 \ K$,$T_{\text{star}} = 2000 \ K$):
$\left(\frac{R_{\text{star}}}{R_{\text{sun}}}\right)^2 = 10000 \left(\frac{6000}{2000}\right)^4 = 10000 \times (3)^4 = 10000 \times 81 = 810000$.
Taking the square root of both sides:
$\frac{R_{\text{star}}}{R_{\text{sun}}} = \sqrt{810000} = 900$.
Thus,the ratio $R_{\text{star}} : R_{\text{sun}} = 900 : 1$.

(B) Density at $0^{\circ} C, \rho_0 = 1.0127 \ g/cm^3$
Density at $100^{\circ} C, \rho_{100} = 1 \ g/cm^3$
Coefficient of real expansion of liquid,$\gamma_{\text{real}} = \frac{\rho_0 - \rho_{100}}{\rho_{100} \times \Delta t}$
$\gamma_{\text{real}} = \frac{1.0127 - 1}{1 \times 100} = \frac{0.0127}{100} = 1.27 \times 10^{-4} /^{\circ} C$
Coefficient of volume expansion of glass,$\gamma_g = 3 \alpha = 3 \times 9 \times 10^{-6} = 27 \times 10^{-6} = 0.27 \times 10^{-4} /^{\circ} C$
Coefficient of apparent expansion,$\gamma_{\text{app}} = \gamma_{\text{real}} - \gamma_g$
$\gamma_{\text{app}} = 1.27 \times 10^{-4} - 0.27 \times 10^{-4} = 1 \times 10^{-4} /^{\circ} C$
Using the formula for apparent expansion: $\gamma_{\text{app}} = \frac{\Delta m}{m_2 \Delta t}$,where $\Delta m$ is the mass expelled and $m_2$ is the remaining mass.
Alternatively,$\Delta m = m_1 \times \gamma_{\text{app}} \times \Delta t$ (approximate for small expansion) or more accurately:
$V_0 = \frac{m_1}{\rho_0}$. At $100^{\circ} C$,$V_{100} = V_0(1 + \gamma_g \Delta t) = \frac{m_1}{\rho_0}(1 + \gamma_g \Delta t)$.
Mass remaining $m_2 = V_{100} \times \rho_{100} = \frac{m_1 \rho_{100}}{\rho_0}(1 + \gamma_g \Delta t) = \frac{300 \times 1}{1.0127}(1 + 27 \times 10^{-6} \times 100) \approx 296.24 \ g$.
Mass expelled $= m_1 - m_2 = 300 - \frac{300}{1.01} = \frac{3}{1.01} \ g$.

(C) Initial state: The tube length is $100 \ cm$, with a $10 \ cm$ mercury column in the middle. The length of air on each side is $(100 - 10) / 2 = 45 \ cm$. Initial pressure $P_0 = 76 \ cm$ of $Hg$, initial temperature $T_0 = 31 + 273 = 304 \ K$.
Applying the ideal gas law $\frac{PV}{T} = \text{constant}$, for the initial state: $\frac{P_0 V_0}{T_0} = \frac{76 \times 45}{304}$.
Final state: Let the new length of the air column at $0^{\circ} C$ $(273 \ K)$ be $l$, and the length of the air column at $273^{\circ} C$ $(546 \ K)$ be $(90 - l)$. Let the new pressure be $P'$.
Since the mercury column is at rest, the pressure on both sides must be equal: $P_2 = P_3 = P'$.
Using the gas law for both sides:
$\frac{P' l}{273} = \frac{P' (90 - l)}{546} = \frac{76 \times 45}{304}$.
From $\frac{P' l}{273} = \frac{P' (90 - l)}{546}$, we get $2l = 90 - l$, which implies $3l = 90$, so $l = 30 \ cm$.
Now, substitute $l = 30 \ cm$ into the equation: $\frac{P' \times 30}{273} = \frac{76 \times 45}{304}$.
$P' = \frac{76 \times 45 \times 273}{304 \times 30} = \frac{76 \times 45 \times 273}{9120} = 102.375 \approx 102.4 \ cm$ of $Hg$.

(A) The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $l$ is the length,$T$ is the tension,and $m$ is the mass per unit length.
Initial frequency: $n_1 = \frac{1}{2l} \sqrt{\frac{T}{m}}$.
New length: $l' = l - 0.40l = 0.6l$.
New tension: $T' = T + 0.44T = 1.44T$.
New frequency: $n_2 = \frac{1}{2l'} \sqrt{\frac{T'}{m}} = \frac{1}{2(0.6l)} \sqrt{\frac{1.44T}{m}}$.
Taking the ratio: $\frac{n_2}{n_1} = \frac{\frac{1}{2(0.6l)} \sqrt{\frac{1.44T}{m}}}{\frac{1}{2l} \sqrt{\frac{T}{m}}} = \frac{l}{0.6l} \times \sqrt{\frac{1.44T}{T}} = \frac{1}{0.6} \times \sqrt{1.44} = \frac{1.2}{0.6} = 2$.
Therefore,the ratio of the final to initial frequency is $2: 1$.

(B) The frequency of the $p$-th harmonic (or $(p-1)$-th overtone) for a string of length $l$,radius $r$,density $\rho$,and tension $T$ is given by $f = \frac{p}{2l} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{p}{2lr} \sqrt{\frac{T}{\pi \rho}}$.
For string $A$,the first overtone is the second harmonic $(p=2)$:
$f_A = \frac{2}{2l_A r_A} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{l_A r_A} \sqrt{\frac{T}{\pi \rho}}$.
For string $B$,the second overtone is the third harmonic $(p=3)$:
$f_B = \frac{3}{2l_B r_B} \sqrt{\frac{T}{\pi \rho}}$.
Given $f_A = f_B$ and $r_A = 2r_B$:
$\frac{1}{l_A r_A} = \frac{3}{2l_B r_B}$.
Substituting $r_A = 2r_B$:
$\frac{1}{l_A (2r_B)} = \frac{3}{2l_B r_B}$.
$\frac{1}{2l_A} = \frac{3}{2l_B}$.
$\frac{l_A}{l_B} = \frac{1}{3}$.
Thus,the ratio of the lengths $l_A : l_B = 1 : 3$.

(C) For statement $A$:
Kinetic energy $KE = \frac{1}{2} m v^2$. Since the body starts from rest and is acted upon by a constant force $F$,its acceleration $a = F/m$ is constant.
Thus,$v = at$,and $KE = \frac{1}{2} m (at)^2 = \frac{1}{2} m a^2 t^2$.
The rate of change of kinetic energy is $\frac{d(KE)}{dt} = \frac{d}{dt} (\frac{1}{2} m a^2 t^2) = m a^2 t$.
Since $\frac{d(KE)}{dt} \propto t$,the rate of change of kinetic energy varies linearly with time. Thus,statement $A$ is correct.
For statement $B$:
$A$ body at rest is in equilibrium only if the net force acting on it is zero. $A$ body can be momentarily at rest (e.g.,a ball thrown vertically upward at its highest point) while still being acted upon by a non-zero net force (gravity). Therefore,it is not necessarily in equilibrium. Thus,statement $B$ is wrong.

(C) The refractive index $\mu$ of a prism is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.
Given that the angle of minimum deviation $\delta_m = A$,we substitute this into the formula:
$\mu = \frac{\sin(\frac{A + A}{2})}{\sin(\frac{A}{2})} = \frac{\sin(A)}{\sin(\frac{A}{2})}$.
Using the trigonometric identity $\sin(A) = 2\sin(\frac{A}{2})\cos(\frac{A}{2})$,we get:
$\mu = \frac{2\sin(\frac{A}{2})\cos(\frac{A}{2})}{\sin(\frac{A}{2})} = 2\cos(\frac{A}{2})$.
Rearranging for $A$:
$\cos(\frac{A}{2}) = \frac{\mu}{2} \Rightarrow \frac{A}{2} = \cos^{-1}(\frac{\mu}{2}) \Rightarrow A = 2\cos^{-1}(\frac{\mu}{2})$.

(B) Let the three resistances be $R_1, R_2$ and $R_3$.
Given that $\frac{R_1}{R_2} = \frac{1}{2}$,we can write $R_1 = k$ and $R_2 = 2k$.
For resistors in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$.
Substituting the given values,$\frac{1}{1} = \frac{1}{k} + \frac{1}{2k} + \frac{1}{R_3}$.
Rearranging for $R_3$,we get $\frac{1}{R_3} = 1 - (\frac{1}{k} + \frac{1}{2k}) = 1 - \frac{3}{2k} = \frac{2k-3}{2k}$.
Thus,$R_3 = \frac{2k}{2k-3}$.
Since $R_3$ must be a positive integer and all resistors must be unequal:
If $k=1$,$R_3 = -2$ (impossible).
If $k=2$,$R_1=2, R_2=4, R_3=4$ (not unequal).
If $k=3$,$R_1=3, R_2=6, R_3=\frac{6}{3} = 2$.
Here,$R_1=3, R_2=6, R_3=2$. All are unequal and integers.
The largest resistance is $6 \ \Omega$.

(D) Let $R_1 = 400 \Omega$ and $R_2 = 800 \Omega$.
$1$. Potential difference $(V_1)$ across $400 \Omega$ resistor without the voltmeter:
$V_1 = \frac{R_1}{R_1 + R_2} \times V = \frac{400}{400 + 800} \times 6 = \frac{400}{1200} \times 6 = 2 \text{ V}$.
$2$. Potential difference $(V_2)$ across $400 \Omega$ resistor with the voltmeter (resistance $R_v = 10000 \Omega$):
The equivalent resistance of the parallel combination of $400 \Omega$ and $10000 \Omega$ is:
$R_p = \frac{400 \times 10000}{400 + 10000} = \frac{4000000}{10400} = \frac{40000}{104} \approx 384.62 \Omega$.
The total resistance of the circuit is $R_{eq} = R_p + R_2 = 384.62 + 800 = 1184.62 \Omega$.
The current in the circuit is $I = \frac{V}{R_{eq}} = \frac{6}{1184.62} \approx 0.005065 \text{ A}$.
The potential difference measured by the voltmeter is $V_2 = I \times R_p = 0.005065 \times 384.62 \approx 1.948 \text{ V}$.
$3$. The error in measurement is:
$\text{Error} = V_1 - V_2 = 2 - 1.948 = 0.052 \text{ V}$.
Rounding to the nearest given option,the error is approximately $0.05 \text{ V}$.

(B) Given: Number of turns $N = 1000$,Area $A = 500 \text{ cm}^2 = 500 \times 10^{-4} \text{ m}^2 = 5 \times 10^{-2} \text{ m}^2$,Magnetic field $B = 2 \times 10^{-5} \text{ Wb/m}^2$,Time interval $\Delta t = 0.2 \text{ s}$.
Since the plane of the coil is at right angles to the magnetic field,the angle between the area vector and the magnetic field is $\theta_1 = 0^{\circ}$.
Initial magnetic flux $\phi_1 = N B A \cos(0^{\circ}) = N B A$.
After rotating the coil by $180^{\circ}$,the new angle is $\theta_2 = 180^{\circ}$.
Final magnetic flux $\phi_2 = N B A \cos(180^{\circ}) = -N B A$.
Change in flux $\Delta \phi = \phi_2 - \phi_1 = -N B A - N B A = -2 N B A$.
The induced emf is given by $e = -\frac{\Delta \phi}{\Delta t} = -\frac{-2 N B A}{\Delta t} = \frac{2 N B A}{\Delta t}$.
Substituting the values: $e = \frac{2 \times 1000 \times 2 \times 10^{-5} \times 5 \times 10^{-2}}{0.2} = \frac{2 \times 10^3 \times 10 \times 10^{-7}}{0.2} = \frac{2 \times 10^{-3}}{0.2} = 10 \times 10^{-3} \text{ V} = 10 \text{ mV}$.

(D) Given mass defect,$\Delta m = 0.3 \,g = 0.3 \times 10^{-3} \,kg = 3 \times 10^{-4} \,kg$.
Using Einstein's mass-energy equivalence principle,$E = \Delta m c^2$.
Substituting the values,$E = (3 \times 10^{-4} \,kg) \times (3 \times 10^8 \,m/s)^2$.
$E = 3 \times 10^{-4} \times 9 \times 10^{16} = 27 \times 10^{12} \,J$.
To convert Joules to kilowatt-hour $(kWh)$,we divide by $3.6 \times 10^6 \,J/kWh$.
$E = \frac{27 \times 10^{12}}{3.6 \times 10^6} \,kWh$.
$E = 7.5 \times 10^6 \,kWh$.

(A) The capacitance of a parallel plate capacitor is given by $C_0 = \frac{\varepsilon_0 A}{d}$,where $A$ is the area of the plates and $d$ is the separation.
Initially,the charge stored is $Q = C_0 V_0$.
$(i)$ When the battery is disconnected,the charge $Q$ remains constant. If the separation is doubled $(d' = 2d)$,the new capacitance becomes $C' = \frac{C_0}{2}$. The energy stored is $E_1 = \frac{Q^2}{2C'} = \frac{(C_0 V_0)^2}{2(C_0 / 2)} = C_0 V_0^2$.
(ii) When the battery remains connected,the potential $V_0$ remains constant. If the separation is doubled,the new capacitance is $C' = \frac{C_0}{2}$. The energy stored is $E_2 = \frac{1}{2} C' V_0^2 = \frac{1}{2} (\frac{C_0}{2}) V_0^2 = \frac{1}{4} C_0 V_0^2$.
Therefore,the ratio is $\frac{E_1}{E_2} = \frac{C_0 V_0^2}{\frac{1}{4} C_0 V_0^2} = 4$.

(B) Let the three resistances be $R_1, R_2,$ and $R_3$.
Given that $R_1:R_2 = 1:2$,we can write $R_1 = k$ and $R_2 = 2k$,where $k$ is a positive integer.
For resistors in parallel,the equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$.
Given $R_{eq} = 1 \ \Omega$,we have $1 = \frac{1}{k} + \frac{1}{2k} + \frac{1}{R_3}$.
Rearranging for $R_3$: $\frac{1}{R_3} = 1 - (\frac{1}{k} + \frac{1}{2k}) = 1 - \frac{3}{2k} = \frac{2k-3}{2k}$.
Thus,$R_3 = \frac{2k}{2k-3}$.
For $R_3$ to be a positive integer,$2k-3$ must be a divisor of $2k$. Since $\frac{2k}{2k-3} = \frac{2k-3+3}{2k-3} = 1 + \frac{3}{2k-3}$,$2k-3$ must be a divisor of $3$.
The divisors of $3$ are $1$ and $3$.
Case $1$: $2k-3 = 1 \Rightarrow 2k = 4 \Rightarrow k = 2$. Then $R_1 = 2, R_2 = 4, R_3 = 1 + \frac{3}{1} = 4$. Here $R_2 = R_3$,which violates the condition that resistors are unequal.
Case $2$: $2k-3 = 3 \Rightarrow 2k = 6 \Rightarrow k = 3$. Then $R_1 = 3, R_2 = 6, R_3 = 1 + \frac{3}{3} = 2$. All resistances are unequal $(3, 6, 2)$.
The largest resistance is $6 \ \Omega$.

(C) Resistance of galvanometer,$G = 50 \Omega$.
Full scale current,$i_g = 0.05 \text{ A}$.
Maximum current to be measured,$i = 5 \text{ A}$.
Area of cross-section,$A = 2.97 \times 10^{-2} \text{ cm}^2 = 2.97 \times 10^{-6} \text{ m}^2$.
Specific resistance,$\rho = 5 \times 10^{-7} \Omega\text{-m}$.
To convert a galvanometer into an ammeter,a shunt resistance $S$ is connected in parallel.
$S = \frac{i_g G}{i - i_g} = \frac{0.05 \times 50}{5 - 0.05} = \frac{2.5}{4.95} = \frac{250}{495} = \frac{50}{99} \Omega$.
Using the formula for resistance,$S = \rho \frac{l}{A}$,we get $l = \frac{S \cdot A}{\rho}$.
$l = \frac{50}{99} \times \frac{2.97 \times 10^{-6}}{5 \times 10^{-7}} = \frac{50}{99} \times \frac{29.7}{5} = 10 \times 0.3 = 3 \text{ m}$.

(D) $1$. Potential difference $(V_1)$ across the $400 \Omega$ resistor without the voltmeter:
$V_1 = \frac{R_1}{R_1 + R_2} \times V = \frac{400}{400 + 800} \times 6 = \frac{400}{1200} \times 6 = 2 \text{ V}$.
$2$. When the voltmeter of resistance $R_v = 10000 \Omega$ is connected in parallel with the $400 \Omega$ resistor,the equivalent resistance $(R_p)$ of this parallel combination is:
$R_p = \frac{400 \times 10000}{400 + 10000} = \frac{4000000}{10400} = \frac{40000}{104} \approx 384.62 \Omega$.
$3$. The new potential difference $(V_2)$ across the parallel combination is:
$V_2 = \frac{R_p}{R_p + R_2} \times V = \frac{384.62}{384.62 + 800} \times 6 = \frac{384.62}{1184.62} \times 6 \approx 1.948 \text{ V}$.
$4$. The error in measurement is:
$\text{Error} = V_1 - V_2 = 2 - 1.948 = 0.052 \text{ V}$.
Rounding to the nearest given option,the error is approximately $0.05 \text{ V}$.

(B) The Einstein's photoelectric equation is given by $eV_0 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $V_0$ is the stopping potential and $\lambda_0$ is the threshold wavelength.
For the first case: $4.8 = \frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \quad \dots (i)$
For the second case: $1.6 = \frac{hc}{e} \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) \quad \dots (ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{4.8}{1.6} = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$
$3 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$
$3 \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{3}{2\lambda} - \frac{3}{\lambda_0} = \frac{1}{\lambda} - \frac{1}{\lambda_0}$
$\frac{3}{2\lambda} - \frac{1}{\lambda} = \frac{3}{\lambda_0} - \frac{1}{\lambda_0}$
$\frac{1}{2\lambda} = \frac{2}{\lambda_0}$
$\lambda_0 = 4\lambda$

(D) In photovoltaic cells, the photoelectric current produced is directly proportional to the intensity of the incident light. Therefore, statement $A$ is false.
According to Einstein's photoelectric equation, $K_{max} = h\nu - \Phi = \frac{hc}{\lambda} - \Phi$. Since the maximum kinetic energy $K_{max}$ of the photoelectrons depends on the wavelength $\lambda$ of the incident radiation, the velocity of the photoelectrons also depends on the wavelength. Therefore, statement $B$ is true.

(B) Given: $N = 1000$,$A = 500 \text{ cm}^2 = 500 \times 10^{-4} \text{ m}^2 = 5 \times 10^{-2} \text{ m}^2$,$B = 2 \times 10^{-5} \text{ Wb/m}^2$,$\Delta t = 0.2 \text{ s}$.
Since the plane of the coil is perpendicular to the magnetic field,the angle between the area vector and the magnetic field is $\theta_1 = 0^{\circ}$.
Initial magnetic flux linked with the coil: $\phi_1 = N B A \cos 0^{\circ} = N B A$.
After rotating the coil by $180^{\circ}$,the new angle is $\theta_2 = 180^{\circ}$.
Final magnetic flux: $\phi_2 = N B A \cos 180^{\circ} = -N B A$.
Change in flux: $\Delta \phi = \phi_2 - \phi_1 = -N B A - N B A = -2 N B A$.
The average induced emf is given by: $e = -\frac{\Delta \phi}{\Delta t} = -\frac{-2 N B A}{\Delta t} = \frac{2 N B A}{\Delta t}$.
Substituting the values: $e = \frac{2 \times 1000 \times 2 \times 10^{-5} \times 5 \times 10^{-2}}{0.2} = \frac{2 \times 10^3 \times 10 \times 10^{-7}}{0.2} = \frac{2 \times 10^{-3}}{0.2} = 10 \times 10^{-3} \text{ V} = 10 \text{ mV}$.

(C) The electric field intensity due to a point charge is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$.
Since the consecutive charges are of opposite signs, the net electric field at $x = 0$ is the sum of fields due to individual charges:
$E = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{Q}{r_1^2} - \frac{Q}{r_2^2} + \frac{Q}{r_3^2} - \frac{Q}{r_4^2} + \dots \infty \right]$
$E = \frac{Q}{4 \pi \varepsilon_0} \left[ \frac{1}{(1 \times 10^{-2})^2} - \frac{1}{(2 \times 10^{-2})^2} + \frac{1}{(4 \times 10^{-2})^2} - \frac{1}{(8 \times 10^{-2})^2} + \dots \infty \right]$
$E = (9 \times 10^9) \times (5 \times 10^{-9}) \times 10^4 \left[ \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{4^2} - \frac{1}{8^2} + \dots \infty \right]$
$E = 45 \times 10^4 \left[ 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \dots \infty \right]$
This is an infinite geometric progression with first term $a = 1$ and common ratio $r = -\frac{1}{4}$.
The sum $S_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - (-1/4)} = \frac{1}{5/4} = \frac{4}{5}$.
Therefore, $E = 45 \times 10^4 \times \frac{4}{5} = 36 \times 10^4 \text{ N/C}$.

(C) Given: Current $i = 30 \ A$,External magnetic field $B_1 = 4 \times 10^{-4} \ T$,Distance $r = 2 \ cm = 2 \times 10^{-2} \ m$.
The magnetic field produced by the straight wire at a distance $r$ is given by $B_2 = \frac{\mu_0 i}{2 \pi r}$.
Substituting the values: $B_2 = \frac{2 \times 10^{-7} \times 30}{2 \times 10^{-2}} = 3 \times 10^{-4} \ T$.
Since the external magnetic field $B_1$ is parallel to the wire,the magnetic field $B_2$ produced by the wire (which is tangential to the circular path around the wire) is perpendicular to $B_1$.
Therefore,the resultant magnetic field $B$ is $B = \sqrt{B_1^2 + B_2^2}$.
$B = \sqrt{(4 \times 10^{-4})^2 + (3 \times 10^{-4})^2} = \sqrt{16 \times 10^{-8} + 9 \times 10^{-8}} = \sqrt{25 \times 10^{-8}} = 5 \times 10^{-4} \ T$.

(A) Given ratios: $m_1: m_2 = 1: 1$,$q_1: q_2 = 1: 2$,and $v_1: v_2 = 2: 3$.
The radius $r$ of the circular path of a charged particle moving in a uniform magnetic field $B$ is given by the formula $r = \frac{mv}{Bq}$.
Taking the ratio of the radii for the two particles,we get:
$\frac{r_1}{r_2} = \left(\frac{m_1}{m_2}\right) \times \left(\frac{v_1}{v_2}\right) \times \left(\frac{q_2}{q_1}\right)$.
Substituting the given values:
$\frac{r_1}{r_2} = \left(\frac{1}{1}\right) \times \left(\frac{2}{3}\right) \times \left(\frac{2}{1}\right) = \frac{4}{3}$.
Therefore,the ratio of the radii is $4: 3$.

(B) Given: Magnetic susceptibility,$\chi = 499$.
Permeability of vacuum,$\mu_0 = 4 \pi \times 10^{-7} \ H/m$.
Relative permeability of the rod is given by the relation: $\mu_r = 1 + \chi$.
Substituting the value of $\chi$: $\mu_r = 1 + 499 = 500$.
Absolute permeability $\mu$ is given by: $\mu = \mu_r \mu_0$.
Substituting the values: $\mu = 500 \times 4 \pi \times 10^{-7} \ H/m$.
$\mu = 2000 \pi \times 10^{-7} \ H/m$.
$\mu = 2 \pi \times 10^{-4} \ H/m$.

(C) The time period of a magnet in a vibration magnetometer is given by $T = 2 \pi \sqrt{\frac{I}{MH}}$.
In the first case,two identical magnets are placed perpendicular to each other,so the total moment of inertia is $I_{total} = I + I = 2I$ and the resultant magnetic moment is $M' = \sqrt{M^2 + M^2} = M\sqrt{2}$.
Thus,$T_1 = 2 \pi \sqrt{\frac{2I}{M\sqrt{2}H}} = 2 \pi \sqrt{\frac{\sqrt{2}I}{MH}}$.
Given $T_1 = 2^{5/4} \ s$,we have $2^{5/4} = 2 \pi \sqrt{\frac{\sqrt{2}I}{MH}}$ ... $(i)$.
When one magnet is removed,the time period $T_2$ is $T_2 = 2 \pi \sqrt{\frac{I}{MH}}$ ... (ii).
Dividing equation $(i)$ by (ii):
$\frac{2^{5/4}}{T_2} = \frac{2 \pi \sqrt{\frac{\sqrt{2}I}{MH}}}{2 \pi \sqrt{\frac{I}{MH}}} = \sqrt{\sqrt{2}} = 2^{1/4}$.
Therefore,$T_2 = \frac{2^{5/4}}{2^{1/4}} = 2^{5/4 - 1/4} = 2^1 = 2 \ s$.

(D) For total internal reflection to occur at the hypotenuse,the angle of incidence $i$ must be greater than or equal to the critical angle $C$.
From the geometry of the prism,the light ray travels parallel to the base $BC$. The angle between the incident ray and the normal to the hypotenuse is the angle of incidence $i$.
In the right-angled triangle,if the base angle is $\theta$,the angle of incidence $i$ at the hypotenuse is $90^{\circ} - \theta$.
For total internal reflection,$i \geq C$,where $\sin C = \frac{1}{\mu}$.
Thus,$90^{\circ} - \theta \geq C \Rightarrow \theta \leq 90^{\circ} - C$.
To find the maximum value of $\theta$,we set $\theta = 90^{\circ} - C$.
Taking the cosine on both sides: $\cos \theta = \cos(90^{\circ} - C) = \sin C$.
Since $\sin C = \frac{1}{\mu}$,we have $\cos \theta = \frac{1}{\mu}$.
Therefore,$\theta = \cos ^{-1}\left(\frac{1}{\mu}\right)$.

(A) In an $n-p-n$ transistor,the emitter is $n$-type,the base is $p$-type,and the collector is $n$-type.
When used as an amplifier,the base-emitter junction is forward-biased and the base-collector junction is reverse-biased.
Due to the forward bias,electrons are injected from the $n$-type emitter into the $p$-type base.
Because the base is very thin and lightly doped,most of these electrons diffuse through the base into the collector region.
The collector is kept at a positive potential relative to the base,which attracts these electrons from the base to the collector.
Therefore,electrons move from the base to the collector.

(D) Statement $(A)$ is false. Duddell's thermo galvanometer is a sensitive instrument that can measure both direct current $(DC)$ and alternating current $(AC)$ by utilizing the heating effect of the current.
Statement $(B)$ is true. $A$ thermopile consists of several thermocouples connected in series,which increases the sensitivity,allowing it to detect very small temperature differences,typically on the order of $10^{-3} {}^{\circ}C$.

(B) Given wavelength $\lambda = 6000 Å = 6 \times 10^{-7} \ m$.
For dark fringes in Young's double slit experiment,the path difference $\Delta x$ is given by the formula $\Delta x = (2n - 1) \frac{\lambda}{2}$ where $n$ is the order of the dark fringe.
For the third dark fringe,we take $n = 3$.
Substituting the values: $\Delta x = (2 \times 3 - 1) \frac{6 \times 10^{-7}}{2} \ m$.
$\Delta x = 5 \times 3 \times 10^{-7} \ m = 15 \times 10^{-7} \ m$.
Converting to microns $(1 \mu m = 10^{-6} \ m)$:
$\Delta x = 1.5 \times 10^{-6} \ m = 1.5 \mu m$.