AIEEE 2006 Physics Question Paper with Answer and Solution

(B) The initial moment of inertia of the ring about its axis is $I = Mr^2$.
The initial angular momentum is $L = I\omega = Mr^2\omega$.
When two objects of mass $m$ are attached to the opposite ends of a diameter,the new moment of inertia $I'$ becomes the sum of the ring's moment of inertia and the moment of inertia of the two point masses: $I' = Mr^2 + m(r)^2 + m(r)^2 = (M + 2m)r^2$.
According to the principle of conservation of angular momentum,the external torque is zero,so $L_{initial} = L_{final}$.
$Mr^2\omega = (M + 2m)r^2\omega'$
Solving for the new angular velocity $\omega'$:
$\omega' = \frac{Mr^2\omega}{(M + 2m)r^2} = \frac{M\omega}{M + 2m}$.

(B) Given the velocity $v = \frac{dx}{dt} = \alpha \sqrt{x}$.
Rearranging the terms to separate the variables,we get:
$\frac{dx}{\sqrt{x}} = \alpha dt$.
Integrating both sides with the initial condition $x=0$ at $t=0$:
$\int_{0}^{x} x^{-1/2} dx = \int_{0}^{t} \alpha dt$.
Solving the integral:
$[2x^{1/2}]_{0}^{x} = \alpha [t]_{0}^{t}$.
This simplifies to:
$2\sqrt{x} = \alpha t$.
Squaring both sides:
$4x = \alpha^2 t^2$.
Therefore,the displacement $x$ is proportional to $t^2$:
$x \propto t^2$.

(D) Let the velocity of the ball just when it leaves the hand be $v$. For the motion of the ball after leaving the hand,using $v_f^2 - v_i^2 = 2as$:
$0^2 - v^2 = 2(-10)(2)$
$v^2 = 40 \ m^2/s^2$
Now,consider the motion of the ball while in the hand. Let the acceleration be $a'$. Using $v^2 - u^2 = 2a's'$ where $u=0$ and $s'=0.2 \ m$:
$40 - 0 = 2(a')(0.2)$
$40 = 0.4a'$
$a' = 100 \ m/s^2$
Applying Newton's Second Law for the ball while in the hand:
$F - mg = ma'$
$F - (0.2)(10) = (0.2)(100)$
$F - 2 = 20$
$F = 22 \ N$

(D) Let the length of the string be $\ell$. When the mass is displaced by an angle $\theta = 45^\circ$,the vertical height gained by the mass is $h = \ell - \ell \cos 45^\circ = \ell(1 - \frac{1}{\sqrt{2}})$.
The horizontal displacement of the mass is $x = \ell \sin 45^\circ = \frac{\ell}{\sqrt{2}}$.
Using the work-energy theorem,the work done by the external horizontal force $F$ must equal the change in potential energy of the mass (assuming it starts and ends at rest,so $\Delta K = 0$):
$W_F + W_g = 0$
$F \cdot x - Mg \cdot h = 0$
$F \cdot (\frac{\ell}{\sqrt{2}}) = Mg \cdot \ell(1 - \frac{1}{\sqrt{2}})$
$F = Mg \cdot \sqrt{2}(1 - \frac{1}{\sqrt{2}})$
$F = Mg(\sqrt{2} - 1)$

(C) Let the mass and velocity of the first piece be $m_1 = 4 \ kg$ and $v_1,$ and the mass and velocity of the second piece be $m_2 = 12 \ kg$ and $v_2 = 4 \ m s^{-1}.$
Since the bomb is initially at rest,the initial momentum is $0.$
By the law of conservation of linear momentum,the final momentum must also be $0.$
$m_1 v_1 + m_2 v_2 = 0$
$4 v_1 + 12 \times 4 = 0$
$4 v_1 = -48$
$v_1 = -12 \ m s^{-1}$
The magnitude of the velocity is $12 \ m s^{-1}.$
The kinetic energy of the $4 \ kg$ mass is:
$K.E. = \frac{1}{2} m_1 v_1^2$
$K.E. = \frac{1}{2} \times 4 \times (12)^2$
$K.E. = 2 \times 144 = 288 \ J.$

(A) The mass of the particle is $m = 100 \ g = 0.1 \ kg$.
The initial velocity is $u = 5 \ m/s$.
At the highest point,the final velocity is $v = 0 \ m/s$.
According to the work-energy theorem,the work done by all forces is equal to the change in kinetic energy: $W_{total} = \Delta K.E$.
Here,the only force acting during the upward motion is gravity $(W_g)$.
$W_g = K.E_{final} - K.E_{initial} = 0 - \frac{1}{2} m u^2$.
$W_g = -\frac{1}{2} \times 0.1 \times (5)^2 = -0.5 \times 0.1 \times 25 = -1.25 \ J$.

(A) The total mechanical energy $E$ is the sum of kinetic energy $K$ and potential energy $U$: $E = K + U = 2 \ J$.
To find the maximum speed,we need to maximize the kinetic energy $K = E - U$.
This occurs when the potential energy $U(x)$ is at its minimum.
To find the minimum of $U(x)$,we set the derivative $\frac{dU}{dx} = 0$:
$\frac{dU}{dx} = \frac{4x^3}{4} - \frac{2x}{2} = x^3 - x = 0$.
$x(x^2 - 1) = 0$,which gives $x = 0, 1, -1$.
Evaluating $U(x)$ at these points:
$U(0) = 0 \ J$.
$U(1) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \ J$.
$U(-1) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \ J$.
The minimum potential energy is $U_{\min} = -\frac{1}{4} \ J$.
Thus,$K_{\max} = E - U_{\min} = 2 - (-0.25) = 2.25 \ J = \frac{9}{4} \ J$.
Using $K_{\max} = \frac{1}{2}mv_{\max}^2$ with $m = 1 \ kg$:
$\frac{1}{2} \times 1 \times v_{\max}^2 = \frac{9}{4}$.
$v_{\max}^2 = \frac{9}{2}$.
$v_{\max} = \frac{3}{\sqrt{2}} \ m/s$.

(C) The total power radiated by the Sun is given by the Stefan-Boltzmann law: $P_{sun} = \sigma T^4 \times (4\pi R^2)$.
The intensity $I$ of the radiation at a distance $r$ from the Sun is the power per unit area: $I = \frac{P_{sun}}{4\pi r^2} = \frac{\sigma T^4 \times 4\pi R^2}{4\pi r^2} = \frac{\sigma T^4 R^2}{r^2}$.
The Earth intercepts this radiation over its cross-sectional area,which is $\pi r_0^2$.
Therefore,the total radiant power incident on Earth is $P_{earth} = I \times \pi r_0^2 = \frac{\sigma T^4 R^2}{r^2} \times \pi r_0^2 = \frac{\pi r_0^2 R^2 \sigma T^4}{r^2}$.

(D) Let the center of mass be at the origin. Initially,the positions of the particles are $-x_1$ and $x_2$ such that $m_1(-x_1) + m_2(x_2) = 0$,which implies $m_1 x_1 = m_2 x_2$ (Equation $1$).
When the first particle is moved by distance $d$ towards the center of mass,its new position becomes $-(x_1 - d)$. Let the second particle be moved by distance $d'$ towards the center of mass,so its new position becomes $(x_2 - d')$.
To keep the center of mass at the origin,the new condition is:
$m_1(-(x_1 - d)) + m_2(x_2 - d') = 0$
$-m_1 x_1 + m_1 d + m_2 x_2 - m_2 d' = 0$
Since $m_1 x_1 = m_2 x_2$ from Equation $1$,the terms $m_1 x_1$ and $m_2 x_2$ cancel out:
$m_1 d - m_2 d' = 0$
$m_2 d' = m_1 d$
$d' = \frac{m_1}{m_2} d$

(D) The position vector $\vec{r}$ of the origin $O(0, 0)$ with respect to the point $P(1, -1)$ is given by $\vec{r} = (0 - 1)\hat{i} + (0 - (-1))\hat{j} = -\hat{i} + \hat{j}$.
The force acting at the origin is $\vec{F} = -F\hat{k}$.
The torque $\vec{\tau}$ about the point $P$ is defined as $\vec{\tau} = \vec{r} \times \vec{F}$.
Substituting the values,we get $\vec{\tau} = (-\hat{i} + \hat{j}) \times (-F\hat{k})$.
Using the cross product rules $\hat{i} \times \hat{k} = -\hat{j}$ and $\hat{j} \times \hat{k} = \hat{i}$,we have:
$\vec{\tau} = F(\hat{i} \times \hat{k}) - F(\hat{j} \times \hat{k})$
$\vec{\tau} = F(-\hat{j}) - F(\hat{i})$
$\vec{\tau} = -F(\hat{i} + \hat{j})$.

(B) Case $(i)$: When a load $W$ is hung from a wire of length $L$ and cross-sectional area $A$,the tension in the wire is $T = W$. The elongation $l$ is given by Young's modulus formula: $Y = \frac{W/A}{l/L} \Rightarrow l = \frac{WL}{AY}$.
Case $(ii)$: When the wire passes over a pulley and two weights $W$ each are hung at the two ends,the tension in the wire remains $T = W$ throughout. The total length of the wire is still $L$ and the cross-sectional area is $A$. Therefore,the elongation $l'$ is given by $l' = \frac{TL}{AY} = \frac{WL}{AY}$.
Comparing the two cases,the elongation remains the same,which is $l$.

(D) The terminal velocity $v_T$ of a sphere of radius $r$ and density $\rho$ falling through a liquid of density $\sigma$ and viscosity $\eta$ is given by the formula: $v_T = \frac{2r^2(\rho - \sigma)g}{9\eta}$.
Since the radius $r$, the liquid density $\sigma$, the viscosity $\eta$, and the acceleration due to gravity $g$ are constant for both spheres, we have $v_T \propto (\rho - \sigma)$.
Therefore, $\frac{v_{T, \text{silver}}}{v_{T, \text{gold}}} = \frac{\rho_{\text{silver}} - \sigma}{\rho_{\text{gold}} - \sigma}$.
Given $\rho_{\text{gold}} = 19.5 \times 10^3 \ kg/m^3$, $\rho_{\text{silver}} = 10.5 \times 10^3 \ kg/m^3$, $\sigma = 1.5 \times 10^3 \ kg/m^3$, and $v_{T, \text{gold}} = 0.2 \ m/s$.
Substituting the values: $\frac{v_{T, \text{silver}}}{0.2} = \frac{10.5 - 1.5}{19.5 - 1.5} = \frac{9}{18} = 0.5$.
Thus, $v_{T, \text{silver}} = 0.2 \times 0.5 = 0.1 \ m/s$.

(B) For an adiabatic process,the work done $W$ on the gas is given by $W = \frac{nR\Delta T}{1 - \gamma}$.
Here,$n = 1 \text{ kilo mole} = 1000 \text{ moles}$,$R = 8.3 \ J \ mol^{-1} K^{-1}$,$\Delta T = 7 \ K$,and $W = -146 \ kJ = -146000 \ J$ (work is done on the gas).
Substituting the values: $-146000 = \frac{1000 \times 8.3 \times 7}{1 - \gamma}$.
$1 - \gamma = -\frac{58100}{146000} \approx -0.3979 \approx -0.4$.
$1 - \gamma = -0.4 \Rightarrow \gamma = 1.4$.
Since $\gamma = 1.4$ for a diatomic gas,the gas is diatomic.

(D) Since the boxes are in thermal contact and isolated from the surroundings,the heat lost by the helium gas equals the heat gained by the nitrogen gas.
For helium (monatomic gas),$C_v = \frac{3}{2}R$. For nitrogen (diatomic gas),$C_v = \frac{5}{2}R$.
Let $n_1 = 1$ mole of He and $n_2 = 1$ mole of $N_2$.
Heat lost by He = $n_1 C_{v,He} (T_{initial,He} - T_f) = 1 \cdot \frac{3}{2}R \cdot (\frac{7}{3}T_0 - T_f)$.
Heat gained by $N_2$ = $n_2 C_{v,N_2} (T_f - T_{initial,N_2}) = 1 \cdot \frac{5}{2}R \cdot (T_f - T_0)$.
Equating the two: $\frac{3}{2}R (\frac{7}{3}T_0 - T_f) = \frac{5}{2}R (T_f - T_0)$.
Dividing by $R/2$: $3(\frac{7}{3}T_0 - T_f) = 5(T_f - T_0)$.
$7T_0 - 3T_f = 5T_f - 5T_0$.
$12T_0 = 8T_f$.
$T_f = \frac{12}{8}T_0 = \frac{3}{2}T_0$.

(B) The displacement of a body starting from the origin in $SHM$ is given by $x = A \sin(\omega t)$.
Velocity is $v = \frac{dx}{dt} = A \omega \cos(\omega t)$.
Kinetic Energy $(K.E.)$ is given by $K.E. = \frac{1}{2} m v^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t)$.
Total Energy $(T.E.)$ is given by $T.E. = \frac{1}{2} m A^2 \omega^2$.
Given that $K.E. = 75\% \text{ of } T.E. = 0.75 \ T.E.$
Substituting the expressions: $\frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) = 0.75 \times \frac{1}{2} m A^2 \omega^2$.
$\cos^2(\omega t) = 0.75 = \frac{3}{4}$.
Taking the square root,$\cos(\omega t) = \frac{\sqrt{3}}{2}$.
This implies $\omega t = \frac{\pi}{6}$.
Since $\omega = \frac{2\pi}{T}$ and $T = 2 \ s$,we have $\omega = \frac{2\pi}{2} = \pi \ rad/s$.
Substituting $\omega$: $\pi t = \frac{\pi}{6}$.
Therefore,$t = \frac{1}{6} \ s$.

(B) The maximum velocity of a particle in simple harmonic motion is given by $v_{\max} = a \omega$,where $a$ is the amplitude and $\omega$ is the angular frequency.
Since $\omega = \frac{2 \pi}{T}$,we have $v_{\max} = a \times \frac{2 \pi}{T}$.
Rearranging for the period $T$,we get $T = \frac{2 \pi a}{v_{\max}}$.
Given $a = 7 \ mm = 7 \times 10^{-3} \ m$ and $v_{\max} = 4.4 \ m/s$.
Substituting the values: $T = \frac{2 \times 3.14 \times 7 \times 10^{-3}}{4.4}$.
$T = \frac{43.96 \times 10^{-3}}{4.4} \approx 9.99 \times 10^{-3} \ s \approx 0.01 \ s$.

(D) The apparent frequency $f'$ heard by a stationary observer when the source approaches with velocity $v_s$ is given by the Doppler effect formula: $f' = f \left( \frac{v}{v - v_s} \right)$.
Given: source frequency $f = 9500 \ Hz$,speed of sound $v = 300 \ ms^{-1}$,and maximum audible frequency $f' = 10000 \ Hz$.
Substituting the values: $10000 = 9500 \left( \frac{300}{300 - v} \right)$.
Simplifying the equation: $\frac{10000}{9500} = \frac{300}{300 - v} \Rightarrow \frac{20}{19} = \frac{300}{300 - v}$.
Cross-multiplying: $20(300 - v) = 19 \times 300 \Rightarrow 6000 - 20v = 5700$.
$20v = 300 \Rightarrow v = 15 \ ms^{-1}$.
Thus,the maximum value of $v$ is $15 \ ms^{-1}$.

(C) Let the square be $ABCD$ with side length $l$. The masses are at $A, B, C, D$.
The axis passes through $A$ and is parallel to the diagonal $BD$.
The distance of point $A$ from the axis is $0$.
The distance of point $B$ from the axis is $d_B = \frac{l}{\sqrt{2}}$.
The distance of point $D$ from the axis is $d_D = \frac{l}{\sqrt{2}}$.
The distance of point $C$ from the axis is $d_C = \sqrt{l^2 + (l/\sqrt{2})^2} = \sqrt{l^2 + l^2/2} = \sqrt{3l^2/2} = l\sqrt{3/2}$.
Alternatively,the distance of $C$ from the diagonal $BD$ is $l\sqrt{2}$.
The moment of inertia $I$ is given by $\sum m_i r_i^2$.
$I = m(0)^2 + m(l/\sqrt{2})^2 + m(l\sqrt{2})^2 + m(l/\sqrt{2})^2$
$I = 0 + m(l^2/2) + 2ml^2 + m(l^2/2)$
$I = ml^2 + 2ml^2 = 3ml^2$.

(B) Let $m$ be the mass of the coin and $N$ be the normal reaction force exerted by the platform on the coin.
The equation of motion for the coin in the vertical direction is given by:
$mg - N = m a$
where $a$ is the acceleration of the platform. For simple harmonic motion,the acceleration is $a = -\omega^2 x$,where $x$ is the displacement from the mean position.
Substituting this into the equation of motion:
$mg - N = m(-\omega^2 x)$
$N = m(g + \omega^2 x)$
However,when the platform moves upwards,the acceleration is directed downwards,so $a = -\omega^2 x$ (where $x$ is positive upwards). When the platform is at the highest point,$x = A$ (amplitude),and the acceleration is $a = -\omega^2 A$.
The equation becomes $mg - N = m(-\omega^2 A)$,which implies $N = m(g + \omega^2 A)$. This does not lead to $N=0$.
When the platform moves downwards,the acceleration is directed upwards. At the highest point of the oscillation,the platform's acceleration is directed downwards. The coin loses contact when the normal force $N$ becomes zero.
This happens when the downward acceleration of the platform exceeds the acceleration due to gravity $g$.
The maximum downward acceleration is $\omega^2 A$. Thus,the coin leaves contact when $\omega^2 A \ge g$,or $A \ge \frac{g}{\omega^2}$.
Therefore,the coin will leave contact for the first time when the amplitude reaches $\frac{g}{\omega^2}$ at the highest position of the platform.

(A) The resonant frequencies of a string fixed at both ends are given by $f_n = n \times f_1$,where $f_1 = \frac{v}{2L}$ is the fundamental frequency and $n = 1, 2, 3, \dots$ is an integer.
Given two consecutive resonant frequencies $f_n = 315\, Hz$ and $f_{n+1} = 420\, Hz$.
We know that the difference between two consecutive resonant frequencies is equal to the fundamental frequency $f_1$.
$f_1 = f_{n+1} - f_n = 420\, Hz - 315\, Hz = 105\, Hz$.
Alternatively,$\frac{f_{n+1}}{f_n} = \frac{n+1}{n} = \frac{420}{315} = \frac{4}{3}$.
This implies $n = 3$,so $f_3 = 315\, Hz$ and $f_4 = 420\, Hz$.
Since $f_3 = 3 \times f_1 = 315\, Hz$,we get $f_1 = \frac{315}{3} = 105\, Hz$.
The lowest resonant frequency is the fundamental frequency $f_1 = 105\, Hz$.

(C) Given: Mass $m = 150 \ g = 0.15 \ kg$,Initial velocity $u = 20 \ m/s$,Final velocity $v = 0 \ m/s$,Time interval $\Delta t = 0.1 \ s$.
According to Newton's second law of motion,the force $F$ exerted is equal to the rate of change of momentum:
$F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}$
$F = \frac{0.15 \times (0 - 20)}{0.1}$
$F = \frac{0.15 \times (-20)}{0.1} = \frac{-3}{0.1} = -30 \ N$
The magnitude of the force is $|F| = 30 \ N$.

(D) The resistance $R$ of the bulb is constant and is calculated using the rated values: $R = \frac{V_{rated}^2}{P_{rated}} = \frac{220^2}{100} = \frac{48400}{100} = 484\,\Omega$.
When the bulb is operated at a new voltage $V' = 110\, V$,the power consumed $P'$ is given by: $P' = \frac{(V')^2}{R} = \frac{110^2}{484} = \frac{12100}{484} = 25\, W$.

(B) According to the experimental observations of the photoelectric effect,the emission of photoelectrons from a metal surface is an instantaneous process. The time lag between the incidence of a photon and the emission of a photoelectron is approximately $10^{-10} \ s$.

(A) Kirchhoff's first law,also known as the Junction Rule,states that the algebraic sum of currents meeting at a junction is zero. This is a direct consequence of the law of conservation of charge,as charge cannot accumulate at a junction.
Kirchhoff's second law,also known as the Loop Rule,states that the algebraic sum of potential differences in any closed loop is zero. This is a direct consequence of the law of conservation of energy,as the work done in moving a unit charge around a closed loop in an electrostatic field is zero.

(C) The work function is given by $\Phi = 6.2\, eV$.
The stopping potential is $V_s = 5\, V$,so the maximum kinetic energy is $K_{\max} = e V_s = 5\, eV$.
According to Einstein's photoelectric equation,the energy of the incident photon is $E = \Phi + K_{\max} = 6.2\, eV + 5\, eV = 11.2\, eV$.
The wavelength $\lambda$ is given by $\lambda = \frac{hc}{E}$. Using $hc \approx 12400\, eV\cdot\mathring{A}$,we get $\lambda = \frac{12400}{11.2} \approx 1107\, \mathring{A}$.
Since the wavelength $1107\, \mathring{A}$ falls in the range of $100\, \mathring{A}$ to $4000\, \mathring{A}$,it lies in the ultraviolet region.

(C) At the distance of closest approach $(r_0)$,the entire initial kinetic energy of the alpha particle is converted into electrostatic potential energy.
The kinetic energy is given by $K = \frac{1}{2}mv^2$.
The potential energy at distance $r_0$ is $U = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$,where $2e$ is the charge of the alpha particle.
Equating kinetic energy to potential energy:
$\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2Ze^2}{r_0}$
Solving for $r_0$:
$r_0 = \frac{4Ze^2}{4\pi\varepsilon_0 mv^2} = \frac{Ze^2}{\pi\varepsilon_0 m v^2}$
From this expression,we can see that $r_0 \propto \frac{1}{m}$.
Therefore,the distance of closest approach is proportional to $\frac{1}{m}$.

(D) In a non-uniform electric field,the electric field intensity at the positions of the two charges $+q$ and $-q$ of the dipole is different.
Let the electric field at the position of $+q$ be $E_1$ and at the position of $-q$ be $E_2$.
The force on the positive charge is $F_1 = qE_1$ and the force on the negative charge is $F_2 = -qE_2$.
Since $E_1 \neq E_2$,the magnitudes of the forces are unequal $(|F_1| \neq |F_2|)$.
Because the forces are unequal and act at different points,they do not cancel each other out,resulting in a net translational force.
Additionally,since the forces are not collinear,they exert a torque on the dipole.
Therefore,the dipole experiences both a torque and a translational force.

(B) The work done by the electric field on the electron is equal to its change in kinetic energy.
$W = \Delta K$
$e V = \frac{1}{2} m_e v^2 - 0$
Here,$V = 20\ V$,$e = 1.6 \times 10^{-19}\ C$,and $m_e = 9.11 \times 10^{-31}\ kg$.
$v = \sqrt{\frac{2 e V}{m_e}}$
$v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 20}{9.11 \times 10^{-31}}}$
$v = \sqrt{\frac{64 \times 10^{-19}}{9.11 \times 10^{-31}}}$
$v = \sqrt{7.025 \times 10^{12}}$
$v \approx 2.65 \times 10^6 \ m/s$

(C) When two conductors are connected by a conducting wire,charge flows until their potentials become equal. Thus,$V_A = V_B$.
Using the formula for potential $V = \frac{KQ}{r}$,we have:
$\frac{KQ_A}{r_A} = \frac{KQ_B}{r_B}$
$\Rightarrow \frac{Q_A}{r_A} = \frac{Q_B}{r_B} \Rightarrow \frac{Q_A}{Q_B} = \frac{r_A}{r_B}$
The electric field at the surface of a spherical conductor is given by $E = \frac{KQ}{r^2}$.
Therefore,the ratio of the electric fields $E_A$ and $E_B$ is:
$\frac{E_A}{E_B} = \frac{K Q_A / r_A^2}{K Q_B / r_B^2} = \frac{Q_A}{Q_B} \times \left( \frac{r_B}{r_A} \right)^2$
Substituting $\frac{Q_A}{Q_B} = \frac{r_A}{r_B}$ into the equation:
$\frac{E_A}{E_B} = \left( \frac{r_A}{r_B} \right) \times \left( \frac{r_B}{r_A} \right)^2 = \frac{r_B}{r_A}$
Given $r_A = 1 \ mm$ and $r_B = 2 \ mm$:
$\frac{E_A}{E_B} = \frac{2 \ mm}{1 \ mm} = \frac{2}{1} = 2 : 1$
Since the distance between the spheres is large compared to their diameters,the induced effects may be ignored.

(A) In a thermoelectric series,the direction of current flow at the cold junction is from the metal that appears later in the series to the metal that appears earlier.
For an Antimony-Bismuth thermocouple,the thermoelectric series order is Bismuth followed by Antimony.
Therefore,at the cold junction,the current flows from Antimony to Bismuth.
This creates a continuous loop where the current flows from Bismuth to Antimony at the hot junction.

(C) The balanced condition for a Wheatstone bridge is given by $\frac{P}{Q} = \frac{R}{S}$,where $S$ is the equivalent resistance of the fourth arm.
Since $S_1$ and $S_2$ are connected in parallel in the fourth arm,their equivalent resistance $S$ is given by $\frac{1}{S} = \frac{1}{S_1} + \frac{1}{S_2} = \frac{S_1 + S_2}{S_1 S_2}$.
Therefore,$S = \frac{S_1 S_2}{S_1 + S_2}$.
Substituting this value into the bridge balance condition,we get $\frac{P}{Q} = \frac{R}{\left( \frac{S_1 S_2}{S_1 + S_2} \right)}$.
This simplifies to $\frac{P}{Q} = \frac{R(S_1 + S_2)}{S_1 S_2}$.

(C) The circuit can be simplified by identifying series and parallel combinations.
Looking at the circuit,the $10 \Omega$ and $20 \Omega$ resistors are in series,giving a total of $30 \Omega$.
The $5 \Omega$ and $10 \Omega$ resistors are also in series,giving a total of $15 \Omega$.
These two branches ($30 \Omega$ and $15 \Omega$) are in parallel with each other.
The equivalent resistance $R_{eq}$ is given by:
$R_{eq} = \frac{30 \times 15}{30 + 15} = \frac{450}{45} = 10 \Omega$.
Using Ohm's law,the current $I$ drawn from the $5 \text{ V}$ source is:
$I = \frac{V}{R_{eq}} = \frac{5}{10} = 0.5 \text{ A}$.

(C) The resistance at temperature $T$ is given by $R_T = R_0(1 + \alpha \Delta T)$,where $R_0$ is the resistance at $0\,^{\circ}C$.
Given:
$R_1 = 100\,\Omega$ at $T_1 = 100\,^{\circ}C$
$R_2 = 200\,\Omega$ at $T_2 = T$
$\alpha = 0.005\,^{\circ}C^{-1}$
Using the formula $R = R_0(1 + \alpha T)$:
$100 = R_0(1 + 0.005 \times 100) = R_0(1 + 0.5) = 1.5 R_0$
$R_0 = \frac{100}{1.5} = \frac{200}{3}\,\Omega$
Now,for $R_2 = 200\,\Omega$:
$200 = R_0(1 + 0.005 \times T)$
$200 = \frac{200}{3}(1 + 0.005 T)$
$3 = 1 + 0.005 T$
$2 = 0.005 T$
$T = \frac{2}{0.005} = 400\,^{\circ}C$.

(A) Given: $\rho_B = 2\rho_A$ and $d_B = 2d_A$.
Since the wires are circular,the area of cross-section is $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.
For the resistances to be equal,$R_B = R_A$.
Using the formula $R = \frac{\rho l}{A}$,we have $\frac{\rho_B l_B}{A_B} = \frac{\rho_A l_A}{A_A}$.
Substituting the values: $\frac{2\rho_A l_B}{\frac{\pi (2d_A)^2}{4}} = \frac{\rho_A l_A}{\frac{\pi d_A^2}{4}}$.
Simplifying the equation: $\frac{2\rho_A l_B}{4 A_A} = \frac{\rho_A l_A}{A_A} \Rightarrow \frac{2 l_B}{4} = l_A$.
Therefore,$\frac{l_B}{l_A} = \frac{4}{2} = 2$.

(C) The charged particle is released from rest,so its initial velocity $v = 0$.
Since the magnetic force is given by $F_m = q(v \times B)$,if $v = 0$,then $F_m = 0$.
The electric force is given by $F_e = qE$.
Since the particle is released from rest,it will be accelerated by the electric field $E$ in the direction of the field (if $q > 0$) or opposite to it (if $q < 0$).
Because the electric and magnetic fields are parallel,the particle moves along the direction of the electric field lines.
Since the magnetic field exerts no force on a particle moving parallel to it,the particle continues to move in a straight line.

(B) The magnetic field at the centre of a long solenoid is given by $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the current.
For the first solenoid: $B_1 = \mu_0 n_1 i_1 = 6.28 \times 10^{-2} \ Wb/m^2$,where $n_1 = 200 \ turns/cm$ and $i_1 = i$.
For the second solenoid: $n_2 = 100 \ turns/cm$ and $i_2 = i/3$.
Taking the ratio: $\frac{B_2}{B_1} = \frac{\mu_0 n_2 i_2}{\mu_0 n_1 i_1} = \frac{n_2 i_2}{n_1 i_1}$.
Substituting the values: $\frac{B_2}{6.28 \times 10^{-2}} = \frac{100 \times (i/3)}{200 \times i} = \frac{100}{200 \times 3} = \frac{1}{6}$.
Therefore,$B_2 = \frac{6.28 \times 10^{-2}}{6} \approx 1.05 \times 10^{-2} \ Wb/m^2$.

(A) Ferromagnetic substances are strongly attracted by a magnetic field because they possess permanent magnetic domains that align easily with the external field.
Paramagnetic substances are weakly attracted by a magnetic field because they possess permanent magnetic dipoles that align weakly with the external field.
Diamagnetic substances are weakly repelled by a magnetic field because they develop an induced magnetic moment in a direction opposite to the applied field due to the orbital motion of electrons.
Therefore,the magnet will attract $N_1$ strongly,$N_2$ weakly,and repel $N_3$ weakly.

(B) The angle of minimum deviation $D$ for a prism is given by the formula $D = (\mu - 1)A$,where $\mu$ is the refractive index and $A$ is the prism angle.
Given that the refractive index for red light is $\mu_r = 1.520$ and for blue light is $\mu_b = 1.525$.
Since $\mu_b > \mu_r$,it follows that $(\mu_b - 1) > (\mu_r - 1)$.
Therefore,$D_2 > D_1$,which can also be written as $D_1 < D_2$.

(C) The photoelectric current $I$ depends on the number of photoelectrons emitted per unit time,which is proportional to the intensity of the incident light,provided the frequency is above the threshold frequency. However,if the anode voltage is fixed and not high enough to collect all emitted electrons,the current depends on the kinetic energy of the photoelectrons. As the wavelength $\lambda$ decreases,the energy of the incident photons $(E = hc/\lambda)$ increases. This results in an increase in the maximum kinetic energy of the emitted photoelectrons. With higher kinetic energy,a greater number of photoelectrons are able to overcome the potential barrier and reach the anode,thereby increasing the plate current $I$. Therefore,as $\lambda$ decreases,$I$ increases. This relationship is represented by a curve where $I$ decreases as $\lambda$ increases,which corresponds to Graph $C$.

(A) Initially,when steady state is achieved,the current in the circuit is $i_0 = \frac{E}{R} = \frac{100 \ V}{100 \ \Omega} = 1 \ A$.
When the battery is disconnected and points $A$ and $B$ are short-circuited,the circuit becomes an $LR$ decay circuit.
The equation for the decay of current is given by $i(t) = i_0 e^{-\frac{R}{L}t}$.
Given $R = 100 \ \Omega$,$L = 100 \ mH = 0.1 \ H$,and $t = 1 \ ms = 10^{-3} \ s$.
The time constant of the circuit is $\tau = \frac{L}{R} = \frac{0.1 \ H}{100 \ \Omega} = 10^{-3} \ s$.
Substituting the values into the decay equation:
$i(t) = 1 \times e^{-\frac{10^{-3}}{10^{-3}}} = 1 \times e^{-1} = \frac{1}{e} \ A$.

(A) The magnetic flux $\phi$ linked with the coil at any time $t$ is given by $\phi = N B A \cos(\omega t)$.
According to Faraday's law of electromagnetic induction,the induced emf $e$ is given by $e = -\frac{d\phi}{dt}$.
Substituting the expression for $\phi$: $e = -\frac{d}{dt}(N B A \cos(\omega t))$.
$e = -N B A \frac{d}{dt}(\cos(\omega t)) = -N B A (-\omega \sin(\omega t)) = N B A \omega \sin(\omega t)$.
The maximum value of emf $(e_{\max})$ occurs when $\sin(\omega t) = 1$.
Therefore,$e_{\max} = N B A \omega$.

(C) The dimension of inductance $L$ is given by the formula $L = \frac{\phi}{I}$,where $\phi$ is the magnetic flux and $I$ is the current.
Magnetic flux $\phi$ has the dimension of $[M L^2 T^{-1} Q^{-1}]$.
Current $I$ has the dimension of $[Q T^{-1}]$.
Therefore,the dimension of inductance is $[L] = \frac{[M L^2 T^{-1} Q^{-1}]}{[Q T^{-1}]} = [M L^2 Q^{-2}]$.
The unit of inductance is $Henry$ $(H)$.
Thus,the dimension $\frac{M L^2}{Q^2}$ corresponds to the unit $Henry$ $(H)$.

(D) In a series $LCR$ circuit,the current $I$ flowing through the circuit is given by $I = \frac{V_R}{R}$.
Given $V_R = 100 \ V$ and $R = 1 \ k\Omega = 1000 \ \Omega$,we have $I = \frac{100}{1000} = 0.1 \ A$.
At resonance,the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$,where $X_C = \frac{1}{\omega C}$.
Given $\omega = 200 \ rad/s$ and $C = 2 \ \mu F = 2 \times 10^{-6} \ F$,we calculate $X_L = X_C = \frac{1}{200 \times 2 \times 10^{-6}} = \frac{1}{400 \times 10^{-6}} = \frac{10^6}{400} = 2500 \ \Omega$.
The voltage across the inductor $L$ is given by $V_L = I X_L$.
Substituting the values,$V_L = 0.1 \times 2500 = 250 \ V$.

(A) The average total energy density $u$ of an electromagnetic wave is the sum of the electric energy density and the magnetic energy density.
$u = u_E + u_B = \frac{1}{2} \varepsilon_0 E_{rms}^2 + \frac{1}{2\mu_0} B_{rms}^2$
Since $B_{rms} = \frac{E_{rms}}{c}$,we have $u_B = \frac{1}{2\mu_0} \left(\frac{E_{rms}}{c}\right)^2 = \frac{1}{2\mu_0} (E_{rms}^2 \varepsilon_0 \mu_0) = \frac{1}{2} \varepsilon_0 E_{rms}^2$.
Thus,the total energy density is $u = \frac{1}{2} \varepsilon_0 E_{rms}^2 + \frac{1}{2} \varepsilon_0 E_{rms}^2 = \varepsilon_0 E_{rms}^2$.
Given $\varepsilon_0 = 8.85 \times 10^{-12} \; F/m$ and $E_{rms} = 720 \; N/C$:
$u = (8.85 \times 10^{-12}) \times (720)^2$
$u = 8.85 \times 10^{-12} \times 518400$
$u = 4.58784 \times 10^{-6} \; J/m^3 \approx 4.58 \times 10^{-6} \; J/m^3$.

(D) In $\beta$-decay,the energy released is shared between the $\beta$-particle and the antineutrino (or neutrino). Because the energy is shared in a continuous manner,the $\beta$-particles are emitted with a continuous range of kinetic energies,from zero up to a maximum value $E_0$. The distribution curve $N(E)$ versus $E$ starts from zero at $E=0$,reaches a peak at an intermediate energy,and drops to zero at the maximum energy $E_0$. This corresponds to the graph shown in option $D$.

(D) The nuclear reaction is given by: $_3^7Li + _1^1p \to _4^8Be + X$.
Applying the law of conservation of mass number: $7 + 1 = 8 + A \implies A = 0$.
Applying the law of conservation of atomic number: $3 + 1 = 4 + Z \implies Z = 0$.
The particle with mass number $0$ and atomic number $0$ is a gamma photon ($\gamma$).
Therefore, the emitted particle is a $\gamma$-particle.

(C) The binding energy of a nucleus is given by the product of the number of nucleons and the binding energy per nucleon.
For $_3^7Li$,the total binding energy is $7 \times 5.60 \, MeV = 39.20 \, MeV$.
For $_2^4He$,the total binding energy of one nucleus is $4 \times 7.06 \, MeV = 28.24 \, MeV$.
In the reaction $p + {}_3^7Li \to 2 {}_2^4He$,the total energy must be conserved.
The energy of the proton $(E_p)$ plus the binding energy of the reactant nucleus must equal the total binding energy of the product nuclei.
$E_p + 39.20 \, MeV = 2 \times (28.24 \, MeV)$.
$E_p + 39.20 = 56.48$.
$E_p = 56.48 - 39.20 = 17.28 \, MeV$.

(C) The '$rad$' (radiation absorbed dose) is a unit of absorbed radiation dose.
It is defined as the absorption of $100 \ erg$ of energy per gram of matter (usually tissue).
Therefore,it measures the energy delivered by radiation to a target material.
One '$rad$' is equivalent to $0.01 \ J/kg$ or $0.01 \ Gray$ $(Gy)$.

(C) The property of increasing conductivity with an increase in temperature is a characteristic feature of semiconductors.
Semiconductors are typically formed by covalent bonding (e.g.,Silicon,Germanium).
Metals (metallic bonding) show a decrease in conductivity with an increase in temperature.
Ionic solids are generally transparent to visible light and are insulators.
Therefore,the solid described is formed by covalent bonding.

(D) The current in a semiconductor is given by $I = n e A v_d$,where $n$ is the concentration,$e$ is the charge,$A$ is the cross-sectional area,and $v_d$ is the drift velocity.
For electrons and holes,the ratio of currents is given by:
$\frac{I_e}{I_h} = \frac{n_e e A v_e}{n_h e A v_h} = \frac{n_e}{n_h} \times \frac{v_e}{v_h}$
Given $\frac{n_e}{n_h} = \frac{7}{5}$ and $\frac{I_e}{I_h} = \frac{7}{4}$.
Substituting these values:
$\frac{7}{4} = \frac{7}{5} \times \frac{v_e}{v_h}$
$\frac{v_e}{v_h} = \frac{7}{4} \times \frac{5}{7} = \frac{5}{4}$.

(B) Given: Collector current $I_C = 5.488 \ mA$ and Emitter current $I_E = 5.60 \ mA$.
First,calculate the base current $I_B$ using the relation $I_B = I_E - I_C$.
$I_B = 5.60 \ mA - 5.488 \ mA = 0.112 \ mA$.
The base current amplification factor $\beta$ is defined as the ratio of collector current to base current: $\beta = \frac{I_C}{I_B}$.
Substituting the values: $\beta = \frac{5.488}{0.112} = 49$.
Thus,the value of $\beta$ is $49$.

(C) In a semiconductor crystal,the lattice constant represents the spacing between atoms.
When the lattice constant decreases,the atoms are brought closer together,which increases the overlap of atomic orbitals.
This increased overlap leads to a broadening of the energy bands,meaning the widths of the conduction band $(E_c)$ and the valence band $(E_v)$ increase.
Simultaneously,the increased interaction between atoms results in a decrease in the band gap $(E_g)$.
Therefore,$E_c$ and $E_v$ increase,while $E_g$ decreases.

(A) diode is reverse biased when its $p$-side is connected to a lower potential and its $n$-side is connected to a higher potential.
Let us analyze the options:
$A$: The $p$-side is at $0 \ V$ (ground) and the $n$-side is at $+5 \ V$. Since $0 \ V < +5 \ V$,this is reverse biased.
$B$: The $p$-side is at $+5 \ V$ and the $n$-side is at $+10 \ V$. Since $5 \ V < 10 \ V$,this is also reverse biased.
$C$: The $p$-side is at $-12 \ V$ and the $n$-side is at $-5 \ V$. Since $-12 \ V < -5 \ V$,this is also reverse biased.
$D$: The $p$-side is at $0 \ V$ (ground) and the $n$-side is at $-10 \ V$. Since $0 \ V > -10 \ V$,this is forward biased.
Note: In standard multiple-choice questions of this type,usually only one option is intended to be correct. However,based on the provided diagrams,options $A$,$B$,and $C$ all represent reverse-biased configurations. Given the typical structure,$A$ is the most standard representation of a reverse-biased diode.

(C) In the given circuit,the positive terminal of the $12 \ V$ battery is connected such that diode $D_2$ is forward-biased and diode $D_1$ is reverse-biased.
Since $D_1$ is reverse-biased,it acts as an open circuit and no current flows through the $3 \ \Omega$ resistor branch.
Since $D_2$ is forward-biased and ideal,it acts as a short circuit. The current flows through the $4 \ \Omega$ resistor and the $2 \ \Omega$ resistor branch.
The total resistance of the circuit is $R = 4 \ \Omega + 2 \ \Omega = 6 \ \Omega$.
Using Ohm's law,the current $i$ in the circuit is given by:
$i = \frac{V}{R} = \frac{12 \ V}{6 \ \Omega} = 2 \ A$.

(B) The magnetic flux linked with the coil is given by $\phi = 10t^2 - 50t + 250$.
According to Faraday's law of electromagnetic induction,the induced $emf$ $(e)$ is given by $e = -\frac{d\phi}{dt}$.
Differentiating the flux with respect to time $t$:
$\frac{d\phi}{dt} = \frac{d}{dt}(10t^2 - 50t + 250) = 20t - 50$.
Therefore,the induced $emf$ is $e = -(20t - 50) = 50 - 20t$.
At $t = 3 \ s$,the induced $emf$ is:
$e = 50 - 20(3) = 50 - 60 = -10 \ V$.