AIPMT 2009 Physics Question Paper with Answer and Solution

(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta Q$ is the heat absorbed,$\Delta U$ is the change in internal energy,and $\Delta W$ is the work done by the system.
Given:
Heat absorbed,$\Delta Q = 2 \; kcal = 2 \times 10^3 \times 4.2 \; J = 8400 \; J$.
Work done,$\Delta W = 500 \; J$.
Substituting these values into the equation:
$\Delta U = \Delta Q - \Delta W$
$\Delta U = 8400 \; J - 500 \; J = 7900 \; J$.
Therefore,the internal energy change is $7900 \; J$.

(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) The energy gained by an electron in an electric field $E$ over a mean free path $\lambda$ is given by the work done: $W = qE\lambda$.
Given the energy $W = 2 \;eV$,the charge $q = e$,and the mean free path $\lambda = 4 \times 10^{-8} \;m$.
Substituting these values into the equation: $2 \;eV = e \times E \times (4 \times 10^{-8} \;m)$.
Since $1 \;eV$ is the energy gained by an electron moving through a potential difference of $1 \;V$,we have $2 \;e \;V = e \times E \times 4 \times 10^{-8} \;m$.
Dividing both sides by $e$ and $4 \times 10^{-8} \;m$,we get: $E = \frac{2}{4 \times 10^{-8}} \;V/m$.
$E = 0.5 \times 10^8 \;V/m = 5 \times 10^7 \;V/m$.

(D) The dimensional formula for pressure is given by $P = \frac{\text{Force}}{\text{Area}} = \frac{[M^1 L^1 T^{-2}]}{[L^2]}$.
Simplifying this,we get $[P] = [M^1 L^{-1} T^{-2}]$.
Comparing this with the given form $M^a L^b T^c$,we find $a=1, b=-1, c=-2$.
Thus,the physical quantity is pressure when $a=1, b=-1, c=-2$.

(C) Given that the particle starts from rest,the initial velocity $u = 0$.
Since the force is constant,the acceleration $a$ is also constant.
The distance $S$ covered in time $t$ is given by the equation of motion: $S = ut + \frac{1}{2}at^2$.
For $t = 10 \ s$,the distance $S_1 = 0(10) + \frac{1}{2}a(10)^2 = 50a$.
For $t = 20 \ s$,the distance $S_2 = 0(20) + \frac{1}{2}a(20)^2 = 200a$.
Comparing the two,$S_2 = 200a = 4(50a) = 4S_1$.
Therefore,$S_2 = 4S_1$.

(D) Let $v_s$ be the velocity of the scooter.
The initial distance between the scooter and the bus is $d = 1\; km = 1000\; m$.
The velocity of the bus is $v_b = 10\; m/s$.
The time taken to overtake is $t = 100\; s$.
The relative velocity of the scooter with respect to the bus is $v_{rel} = v_s - v_b = v_s - 10$.
Using the formula for relative motion,$d = v_{rel} \times t$:
$1000 = (v_s - 10) \times 100$
$10 = v_s - 10$
$v_s = 20\; m/s$.
Thus,the scooterist should chase the bus at a speed of $20\; m/s$.

(C) Given force vector $\vec F = 6\hat i - 8\hat j + 10\hat k$.
The magnitude of the force is given by $|\vec F| = \sqrt{6^2 + (-8)^2 + 10^2} = \sqrt{36 + 64 + 100} = \sqrt{200} = 10\sqrt 2 \,N$.
According to Newton's second law of motion,$F = ma$,where $m$ is the mass and $a$ is the acceleration.
Given acceleration $a = 1\, m/s^2$.
Therefore,$m = \frac{F}{a} = \frac{10\sqrt 2}{1} = 10\sqrt 2 \,kg$.

(B) Given:
Mass of the lift,$M = 2000 \, kg$
Tension in the cable,$T = 28000 \, N$
Acceleration due to gravity,$g = 10 \, m/s^2$
According to Newton's second law,the net force acting on the lift is $T - Mg = Ma$.
Substituting the values:
$28000 - (2000 \times 10) = 2000 \times a$
$28000 - 20000 = 2000 \times a$
$8000 = 2000 \times a$
$a = \frac{8000}{2000} = 4 \, m/s^2$
Since the tension is greater than the weight,the acceleration is directed upwards.

(D) Let the velocity of water be $v$ and the mass per unit length be $m$.
The mass of water flowing per unit time (mass flow rate) is given by the product of mass per unit length and velocity:
$\text{Mass flow rate} = m \times v$
The kinetic energy $K$ of a mass $M$ moving with velocity $v$ is $K = \frac{1}{2} Mv^2$.
The rate at which kinetic energy is imparted is the power $P$,which is the kinetic energy per unit time:
$P = \frac{1}{2} (\text{mass flow rate}) v^2$
$P = \frac{1}{2} (mv) v^2 = \frac{1}{2} mv^3$.

(D) According to the law of conservation of linear momentum,the initial momentum of the rock is zero. Therefore,the vector sum of the momenta of the three parts must be zero.
Let the momenta of the three parts be $\vec{p}_1$,$\vec{p}_2$,and $\vec{p}_3$.
$\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 \implies \vec{p}_3 = -(\vec{p}_1 + \vec{p}_2)$.
The magnitude of the momentum of the first part is $p_1 = m_1 v_1 = 1 \, kg \times 12 \, m s^{-1} = 12 \, kg \, m s^{-1}$.
The magnitude of the momentum of the second part is $p_2 = m_2 v_2 = 2 \, kg \times 8 \, m s^{-1} = 16 \, kg \, m s^{-1}$.
Since the two parts move at right angles to each other,the magnitude of the resultant momentum of these two parts is $p_{12} = \sqrt{p_1^2 + p_2^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, kg \, m s^{-1}$.
The third part must have a momentum equal in magnitude and opposite in direction to this resultant momentum to satisfy the conservation law.
Thus,$p_3 = p_{12} = 20 \, kg \, m s^{-1}$.
Given the speed of the third part $v_3 = 4 \, m s^{-1}$,its mass $m_3$ is:
$m_3 = \frac{p_3}{v_3} = \frac{20 \, kg \, m s^{-1}}{4 \, m s^{-1}} = 5 \, kg$.

(D) Initial kinetic energy $(K_i)$ of the body is given by: $K_i = \frac{1}{2}mv^2 = \frac{1}{2} \times 1 \times (20)^2 = 200\,J$.
At the maximum height $h$ reached in the absence of air friction,the kinetic energy would be completely converted into potential energy: $mgh = 200\,J$.
Using $m = 1\,kg$ and $g = 10\,m/s^2$,the theoretical maximum height is $h = \frac{200}{1 \times 10} = 20\,m$.
However,the body only reaches a height of $h' = 18\,m$ due to air friction.
The potential energy at this height is $U_f = mgh' = 1 \times 10 \times 18 = 180\,J$.
The energy lost due to air friction is the difference between the initial kinetic energy and the final potential energy: $\Delta E = K_i - U_f = 200\,J - 180\,J = 20\,J$.

(A) Let the maximum extension in the spring be $x$.
According to the law of conservation of energy,the loss in gravitational potential energy of the block is equal to the gain in elastic potential energy of the spring.
Loss in gravitational potential energy = $Mgx$.
Gain in elastic potential energy = $\frac{1}{2} k x^2$.
Equating the two: $Mgx = \frac{1}{2} k x^2$.
Solving for $x$ (where $x \neq 0$): $x = \frac{2Mg}{k}$.

(C) In a steady state,the rate of heat flow through a rod is analogous to the flow of electric current in a conductor,where the temperature difference corresponds to potential difference and thermal resistance corresponds to electrical resistance.
The formula for the rate of heat transfer $\frac{dQ}{dt}$ is given by Fourier's law of heat conduction:
$\frac{dQ}{dt} = \frac{kA(T_1 - T_2)}{L}$
Here,$k$ is the thermal conductivity of the material of the rod,$A$ is the cross-sectional area,$L$ is the length of the rod,and $(T_1 - T_2)$ is the temperature difference between the two ends.

(B) According to Stefan-Boltzmann law,the rate of heat radiation $E$ is proportional to the fourth power of the absolute temperature $T$,i.e.,$E \propto T^4$.
Given:
Initial temperature $T_1 = 227^o C = (227 + 273) K = 500 K$.
Initial rate of radiation $E_1 = 7 \; cal/cm^2 s$.
Final temperature $T_2 = 727^o C = (727 + 273) K = 1000 K$.
Using the ratio formula:
$\frac{E_2}{E_1} = \left( \frac{T_2}{T_1} \right)^4$
Substituting the values:
$\frac{E_2}{7} = \left( \frac{1000}{500} \right)^4$
$\frac{E_2}{7} = (2)^4$
$\frac{E_2}{7} = 16$
$E_2 = 7 \times 16 = 112 \; cal/cm^2 s$.
Therefore,the rate of heat radiated at $727^o C$ is $112 \; cal/cm^2 s$.

(D) Consider one rod of the square frame,say rod $AB$. The moment of inertia of a rod of mass $M$ and length $l$ about an axis passing through its center and perpendicular to its length is $I_{cm} = \frac{Ml^2}{12}$.
The distance $d$ from the center of the rod $AB$ to the center of the square frame is $d = \frac{l}{2}$.
Using the parallel axis theorem,the moment of inertia of one rod about the axis passing through the center of the square and perpendicular to its plane is:
$I_{rod} = I_{cm} + Md^2 = \frac{Ml^2}{12} + M\left(\frac{l}{2}\right)^2 = \frac{Ml^2}{12} + \frac{Ml^2}{4} = \frac{Ml^2 + 3Ml^2}{12} = \frac{4Ml^2}{12} = \frac{Ml^2}{3}$.
Since there are four identical rods forming the square frame,the total moment of inertia $I$ is:
$I = 4 \times I_{rod} = 4 \times \frac{Ml^2}{3} = \frac{4}{3}Ml^2$.

(D) The torque $\vec{\tau}$ is defined as the cross product of the position vector $\vec{r}$ and the force vector $\vec{F}$,given by $\vec{\tau} = \vec{r} \times \vec{F}$.
By the definition of the cross product,the resulting vector $\vec{\tau}$ is always perpendicular to both the vectors $\vec{r}$ and $\vec{F}$ that form it.
Since the dot product of two perpendicular vectors is always zero,we have $\vec{r} \cdot \vec{\tau} = 0$ and $\vec{F} \cdot \vec{\tau} = 0$.

(B) According to Kepler's Second Law of Planetary Motion,a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
This means the areal velocity $\frac{dA}{dt}$ is constant.
Therefore,the time taken to sweep an area is directly proportional to the area swept.
Given that the area $SCD = 2 \times \text{Area } SAB$.
Since $t_1$ is the time taken to sweep area $SCD$ and $t_2$ is the time taken to sweep area $SAB$,we have:
$\frac{t_1}{t_2} = \frac{\text{Area } SCD}{\text{Area } SAB} = \frac{2 \times \text{Area } SAB}{\text{Area } SAB} = 2$.
Thus,$t_1 = 2t_2$.

(A) In an isochoric process,the volume of the system is kept constant.
If the pressure is kept constant,the process is known as an isobaric process.
Therefore,the statement that pressure remains constant in an isochoric process is incorrect.

(C) For simple harmonic motion,the velocity $v$ at position $x$ is given by $v = \omega \sqrt{a^2 - x^2}$.
Given the amplitude is $a$ and the time period is $T$,the angular frequency is $\omega = \frac{2\pi}{T}$.
At position $x = \frac{a}{2}$,the speed is:
$v = \omega \sqrt{a^2 - (\frac{a}{2})^2}$
$v = \omega \sqrt{a^2 - \frac{a^2}{4}}$
$v = \omega \sqrt{\frac{3a^2}{4}}$
$v = \omega \cdot \frac{a\sqrt{3}}{2}$
Substituting $\omega = \frac{2\pi}{T}$:
$v = \frac{2\pi}{T} \cdot \frac{a\sqrt{3}}{2}$
$v = \frac{\pi a\sqrt{3}}{T}$

(C) The defining condition for Simple Harmonic Motion $(S.H.M.)$ is that the acceleration $(a)$ is directly proportional to the negative of the displacement $(x)$ from the mean position.
Mathematically,this is expressed as $a = -\omega^2 x$,where $\omega$ is the angular frequency.
In option $(C)$,the equation is $a = -k(x+a)$. If we define a new coordinate $x' = x+a$,then the equation becomes $a = -k x'$.
This represents an $S.H.M.$ about the equilibrium position $x = -a$.
Since the acceleration is proportional to the displacement from the mean position and directed towards it,option $(C)$ correctly represents $S.H.M.$

(A) Given: Amplitude $A = 2\, cm = 0.02\, m$,speed $v = 128\, m/s$.
Since $5$ complete waves fit in $4\, m$,the wavelength $\lambda = \frac{4\, m}{5} = 0.8\, m$.
The frequency $f = \frac{v}{\lambda} = \frac{128}{0.8} = 160\, Hz$.
The angular frequency $\omega = 2\pi f = 2 \times 3.14159 \times 160 \approx 1005\, rad/s$.
The wave number $k = \frac{2\pi}{\lambda} = \frac{2 \times 3.14159}{0.8} \approx 7.85\, rad/m$.
Since the wave travels in the $+ve$ $x$-direction,the equation is $y = A \sin(kx - \omega t)$.
Substituting the values: $y = 0.02 \sin(7.85x - 1005t)$.

(B) The car acts as the source and the hill acts as the observer.
First,we calculate the frequency $f_1$ heard by the hill:
$f_1 = f_0 \times \frac{v}{v - v_s} = 600 \times \frac{330}{330 - 30} = 600 \times \frac{330}{300} = 660 \, Hz$.
Now,the hill reflects this sound,acting as the source,and the driver acts as the observer moving towards the source.
The frequency $f_2$ heard by the driver is:
$f_2 = f_1 \times \frac{v + v_o}{v} = 660 \times \frac{330 + 30}{330} = 660 \times \frac{360}{330} = 2 \times 360 = 720 \, Hz$.

(A) Given: Lengths $l_1 = 0.516 \, m$ and $l_2 = 0.491 \, m$. Tension $T = 20 \, N$. Mass per unit length $\mu = 1 \, g/m = 0.001 \, kg/m$.
The fundamental frequency of a vibrating string is given by $v = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$.
Calculate the wave speed $v_w = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{20}{0.001}} = \sqrt{20000} = 100\sqrt{2} \approx 141.42 \, m/s$.
Frequency $v_1 = \frac{141.42}{2 \times 0.516} \approx 137.04 \, Hz$.
Frequency $v_2 = \frac{141.42}{2 \times 0.491} \approx 144.02 \, Hz$.
The number of beats is the difference in frequencies: $|v_2 - v_1| = 144.02 - 137.04 = 6.98 \approx 7 \, Hz$.

(D) The position vector of the centre of mass is given by $\vec{r}_{com} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$.
Substituting the given values: $\vec{r}_{com} = \frac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3}$.
$\vec{r}_{com} = \frac{\hat{i} + 2\hat{j} + \hat{k} - 9\hat{i} - 6\hat{j} + 3\hat{k}}{4} = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4} = -2\hat{i} - \hat{j} + \hat{k}$.
The magnitude of the centre of mass position vector is $|\vec{r}_{com}| = \sqrt{(-2)^2 + (-1)^2 + (1)^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.
Now,check the magnitude of the vector in option $D$: $|-2\hat{i} - \hat{j} + \hat{k}| = \sqrt{(-2)^2 + (-1)^2 + (1)^2} = \sqrt{6}$.
Thus,the magnitudes are equal.

(B) Diamagnetic substances are materials that develop a weak magnetization in the direction opposite to the applied magnetic field. When a diamagnetic substance is placed in a non-uniform magnetic field (such as near the poles of a bar magnet),it experiences a force that pushes it from the stronger part of the field to the weaker part. Since the magnetic field is strongest at the poles of a bar magnet,the diamagnetic substance is repelled by both the north and south poles.

(B) When $n$ capacitors of capacitance $C$ are connected in series,the equivalent capacitance $C_{\text{eq}}$ is given by $\frac{1}{C_{\text{eq}}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} = \frac{3}{C}$.
Thus,$C_{\text{eq}} = \frac{C}{3}$.
In a series combination,the potential difference across each capacitor is $V$. Since the capacitors are identical,the total breakdown voltage $V_{\text{total}}$ of the combination is the sum of the individual breakdown voltages.
$V_{\text{total}} = V + V + V = 3V$.
Therefore,the combination has a capacitance of $C/3$ and a breakdown voltage of $3V$.

(D) The potential at any point on a shell is the sum of potentials due to all shells. The charge on a shell of radius $r$ and surface charge density $\sigma$ is $q = 4\pi r^2 \sigma$.
Thus,$q_A = 4\pi a^2 \sigma$,$q_B = -4\pi b^2 \sigma$,and $q_C = 4\pi c^2 \sigma$.
The potentials are:
$V_A = \frac{1}{4\pi\epsilon_0} [\frac{q_A}{a} + \frac{q_B}{b} + \frac{q_C}{c}] = \frac{1}{\epsilon_0} [a\sigma - b\sigma + \frac{c^2\sigma}{c}] = \frac{\sigma}{\epsilon_0} [a - b + c]$
$V_B = \frac{1}{4\pi\epsilon_0} [\frac{q_A}{b} + \frac{q_B}{b} + \frac{q_C}{c}] = \frac{1}{\epsilon_0} [\frac{a^2\sigma}{b} - b\sigma + \frac{c^2\sigma}{c}] = \frac{\sigma}{\epsilon_0} [\frac{a^2}{b} - b + c]$
$V_C = \frac{1}{4\pi\epsilon_0} [\frac{q_A}{c} + \frac{q_B}{c} + \frac{q_C}{c}] = \frac{1}{\epsilon_0} [\frac{a^2\sigma}{c} - \frac{b^2\sigma}{c} + c\sigma] = \frac{\sigma}{\epsilon_0} [\frac{a^2 - b^2}{c} + c]$
Given $c = a + b$,then $c - b = a$ and $c - a = b$. Also $a^2 - b^2 = (a - b)(a + b) = (a - b)c$.
Substituting $c = a + b$ into $V_A$: $V_A = \frac{\sigma}{\epsilon_0} [a - b + (a + b)] = \frac{2a\sigma}{\epsilon_0}$.
Substituting $c = a + b$ into $V_C$: $V_C = \frac{\sigma}{\epsilon_0} [\frac{(a - b)(a + b)}{c} + c] = \frac{\sigma}{\epsilon_0} [\frac{(a - b)c}{c} + c] = \frac{\sigma}{\epsilon_0} [a - b + a + b] = \frac{2a\sigma}{\epsilon_0}$.
Since $V_A = V_C = \frac{2a\sigma}{\epsilon_0}$ and $V_B = \frac{\sigma}{\epsilon_0} [\frac{a^2}{b} - b + a + b] = \frac{\sigma}{\epsilon_0} [\frac{a^2}{b} + a]$,clearly $V_A = V_C \neq V_B$.

(D) The electric potential is given by $V = -x^2y - xz^3 + 4$.
The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\vec{\nabla} V = -\left( \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j} + \frac{\partial V}{\partial z}\hat{k} \right)$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(-x^2y - xz^3 + 4) = -2xy - z^3$
$\frac{\partial V}{\partial y} = \frac{\partial}{\partial y}(-x^2y - xz^3 + 4) = -x^2$
$\frac{\partial V}{\partial z} = \frac{\partial}{\partial z}(-x^2y - xz^3 + 4) = -3xz^2$
Substituting these into the expression for $\vec{E}$:
$\vec{E} = -[(-2xy - z^3)\hat{i} + (-x^2)\hat{j} + (-3xz^2)\hat{k}]$
$\vec{E} = (2xy + z^3)\hat{i} + x^2\hat{j} + 3xz^2\hat{k}$.

(D) Applying Kirchhoff's voltage law to the loop $ABFE$ in the clockwise direction:
Starting from point $A$ and moving towards $B$,the potential drop across resistor $R$ is $-(i_1 + i_2)R$.
Moving from $F$ to $E$ through the branch containing $\varepsilon_1$ and $r_1$,we encounter the negative terminal of the battery first,so we add $\varepsilon_1$,and then we pass through the resistor $r_1$ in the direction of current $i_1$,resulting in a potential drop of $-i_1 r_1$.
Equating the sum of potential changes to zero,we get:
$-(i_1 + i_2)R + \varepsilon_1 - i_1 r_1 = 0$
Rearranging the terms,we obtain:
$\varepsilon_1 - (i_1 + i_2)R - i_1 r_1 = 0$

(D) The circumference of the circle is $L = 2 \pi r = 2 \pi \times 0.1 \, m = 0.2 \pi \, m$.
The total resistance of the wire is $R_{total} = (12 \, \Omega/m) \times (0.2 \pi \, m) = 2.4 \pi \, \Omega$.
When the wire is bent into a circle and points $A$ and $B$ are diametrically opposite, the wire is divided into two equal semicircular parts, each with resistance $R' = R_{total} / 2 = 1.2 \pi \, \Omega$.
These two parts are connected in parallel between points $A$ and $B$.
The equivalent resistance $R_{eq}$ is given by $\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R'}$.
Therefore, $R_{eq} = \frac{R'}{2} = \frac{1.2 \pi \, \Omega}{2} = 0.6 \pi \, \Omega$.

(A) The terminal potential difference $(V)$ of a cell is given by the equation: $V = \varepsilon - Ir$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = V$ and $x = I$:
$V = (-r)I + \varepsilon$.
Here,the slope $(m)$ is equal to $-r$ and the $y$-intercept $(c)$ is equal to $\varepsilon$.
Therefore,the slope is $-r$ and the intercept is $\varepsilon$.

(C) To convert a galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with the galvanometer.
The formula for the shunt resistance is $i_g G = (I - i_g) S$,where $i_g$ is the full-scale deflection current,$G$ is the galvanometer resistance,and $I$ is the maximum current to be measured.
Given: $G = 60\,\Omega$,$i_g = 1.0\,A$,and $I = 5.0\,A$.
Substituting the values: $1.0 \times 60 = (5.0 - 1.0) \times S$.
$60 = 4 \times S$.
$S = \frac{60}{4} = 15\,\Omega$.
Therefore,a resistance of $15\,\Omega$ must be connected in parallel.

(D) The magnetic force $\vec{F}$ on a charged particle is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
Given:
Charge $q = -2 \times 10^{-6}\, C$
Velocity $\vec{v} = (2\hat{i} + 3\hat{j}) \times 10^6\, m/s$
Magnetic field $\vec{B} = 2\hat{j}\, T$
Substituting these values into the formula:
$\vec{F} = (-2 \times 10^{-6}) \times [(2\hat{i} + 3\hat{j}) \times 10^6] \times (2\hat{j})$
$\vec{F} = -2 \times 2 \times [ (2\hat{i} \times \hat{j}) + (3\hat{j} \times \hat{j}) ]$
Since $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{j} = 0$:
$\vec{F} = -4 \times (2\hat{k} + 0)$
$\vec{F} = -8\hat{k}\, N$
The negative sign indicates the force is in the $-z$ direction.
Thus,the force is $8\, N$ in the $-z$ direction.

(C) The work done $W$ in rotating a magnetic dipole in a magnetic field is given by the formula:
$W = MB(\cos \theta_1 - \cos \theta_2)$
Given:
Magnetic moment $M = 2 \times 10^4 \, JT^{-1}$
Magnetic field $B = 6 \times 10^{-4} \, T$
Initial angle $\theta_1 = 0^{\circ}$ (parallel to the field)
Final angle $\theta_2 = 60^{\circ}$
Substituting the values:
$W = (2 \times 10^4) \times (6 \times 10^{-4}) \times (\cos 0^{\circ} - \cos 60^{\circ})$
$W = 12 \times (1 - 0.5)$
$W = 12 \times 0.5 = 6 \, J$
Thus,the work done is $6 \, J$.

(A) Given: Wavelength $\lambda = 667 \, nm = 667 \times 10^{-9} \, m$. Power $P = 9 \, mW = 9 \times 10^{-3} \, W$.
Energy of a single photon $E = \frac{hc}{\lambda}$.
Using $h = 6.63 \times 10^{-34} \, J \cdot s$ and $c = 3 \times 10^8 \, m/s$:
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{667 \times 10^{-9}} \approx 2.98 \times 10^{-19} \, J$.
The number of photons per second $n = \frac{P}{E}$.
$n = \frac{9 \times 10^{-3}}{2.98 \times 10^{-19}} \approx 3.02 \times 10^{16} \approx 3 \times 10^{16}$ photons/sec.

(B) According to the photoelectric effect,the number of photoelectrons emitted per unit time is directly proportional to the number of incident photons per unit time.
Since the intensity of light is defined as the energy incident per unit area per unit time,and for a given frequency $\nu$,the energy of each photon is $E = h\nu$ (a constant),the intensity is directly proportional to the number of incident photons.
Therefore,the number of photoelectrons emitted is directly proportional to the intensity of the incident light,provided the frequency $\nu$ is greater than the threshold frequency $\nu_0$.

(D) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = B \cdot \pi r^2$,where $B$ is the magnetic field and $r$ is the radius of the loop.
According to Faraday's law,the induced $emf$ is $\varepsilon = -\frac{d\phi}{dt}$.
Since $B$ is constant,$\varepsilon = -\frac{d}{dt}(B \pi r^2) = -B \pi (2r) \frac{dr}{dt}$.
Given: $B = 0.04\, T$,$r = 2\, cm = 0.02\, m$,and $\frac{dr}{dt} = -2\, mm/s = -2 \times 10^{-3}\, m/s$ (negative because the radius is shrinking).
Substituting the values:
$\varepsilon = -(0.04) \cdot \pi \cdot 2 \cdot (0.02) \cdot (-2 \times 10^{-3})$
$\varepsilon = 0.04 \cdot \pi \cdot 0.04 \cdot 2 \times 10^{-3}$
$\varepsilon = 0.0032 \times 10^{-3} \cdot \pi\, V$
$\varepsilon = 3.2 \times 10^{-6} \cdot \pi\, V = 3.2\pi\, \mu V$.

(A) The induced electromotive force (emf) is given by $\varepsilon = B l v$,where $l$ is the length of the conductor cutting the magnetic field lines perpendicular to the velocity vector.
For a rectangular or square loop,as it moves out of the magnetic field,the length $l$ of the side cutting the magnetic field lines remains constant until the entire loop exits the field. Thus,the induced emf remains constant.
For a circular or elliptical loop,the width of the loop perpendicular to the velocity vector changes continuously as the loop moves out of the magnetic field. Since the length $l$ is not constant,the induced emf $\varepsilon = B l v$ will not remain constant.
Therefore,the induced emf will not remain constant for the circular and elliptical loops.

(D) The average power dissipated in an $LCR$ series circuit is given by $P = E_{rms} I_{rms} \cos \phi$.
The impedance of the circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2}$,where $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$.
The power factor is $\cos \phi = \frac{R}{Z}$.
The root-mean-square current is $I_{rms} = \frac{E_{rms}}{Z}$.
Substituting these into the power formula: $P = E_{rms} \cdot \frac{E_{rms}}{Z} \cdot \frac{R}{Z} = \frac{E_{rms}^2 R}{Z^2}$.
Substituting $Z^2 = R^2 + (\omega L - \frac{1}{\omega C})^2$ and $E_{rms} = \varepsilon$,we get $P = \frac{\varepsilon^2 R}{R^2 + (\omega L - \frac{1}{\omega C})^2}$.

(B) The given electric field is $E_y = E_0 \cos(\omega t - kx)$.
Comparing this with the given equation $E_y = 2.5 \cos[(2\pi \times 10^6)t - (\pi \times 10^{-2})x]$,we identify the angular frequency $\omega = 2\pi \times 10^6 \text{ rad/s}$ and the wave number $k = \pi \times 10^{-2} \text{ rad/m}$.
Since the term inside the cosine is $(\omega t - kx)$,the wave is propagating in the positive $x$-direction.
The frequency $f$ is given by $\omega = 2\pi f$,so $f = \frac{2\pi \times 10^6}{2\pi} = 10^6 \text{ Hz}$.
The wavelength $\lambda$ is given by $k = \frac{2\pi}{\lambda}$,so $\lambda = \frac{2\pi}{k} = \frac{2\pi}{\pi \times 10^{-2}} = 2 \times 10^2 = 200 \text{ m}$.
Thus,the wave is moving along the $x$-direction with a frequency of $10^6 \text{ Hz}$ and a wavelength of $200 \text{ m}$.

(C) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula $N = \frac{n(n-1)}{2}$.
Given $N = 6$,we have $\frac{n(n-1)}{2} = 6$,which implies $n^2 - n - 12 = 0$. Solving this quadratic equation,we get $(n-4)(n+3) = 0$,so $n = 4$.
The energy of the emitted photon is given by $\Delta E = E_i - E_f = \frac{hc}{\lambda}$.
Since $\lambda = \frac{hc}{\Delta E}$,the wavelength $\lambda$ is maximum when the energy difference $\Delta E$ is minimum.
The possible transitions from $n=4$ are: $(4 \to 3), (4 \to 2), (4 \to 1), (3 \to 2), (3 \to 1), (2 \to 1)$.
Comparing the energy gaps,the transition $n=4 \to n=3$ has the smallest energy difference,and therefore corresponds to the maximum wavelength of the emitted radiation.

(A) In a Rutherford scattering experiment,at the distance of closest approach $(r_0)$,the entire initial kinetic energy $(K)$ of the projectile is converted into electrostatic potential energy $(U)$.
The electrostatic potential energy is given by the formula:
$U = \frac{1}{4 \pi \varepsilon_0} \frac{(Z_1 e)(Z_2 e)}{r_0}$
Since the kinetic energy $K$ is equal to the potential energy $U$ at the point of closest approach:
$K = \frac{1}{4 \pi \varepsilon_0} \frac{Z_1 Z_2 e^2}{r_0}$
From this expression,it is clear that the energy of the projectile is directly proportional to the product of the charges of the projectile and the target nucleus,i.e.,$K \propto Z_1 Z_2$.

(B) The given decay sequence is: ${}_Z^AX \to {}_{Z + 1}^AY \to {}_{Z - 1}^{A - 4}K \to {}_{Z - 1}^{A - 4}K$.
$1$. In the first step,${}_Z^AX \to {}_{Z + 1}^AY$: The atomic number increases by $1$ while the mass number remains the same. This corresponds to the emission of a $\beta^-$ particle.
$2$. In the second step,${}_{Z + 1}^AY \to {}_{Z - 1}^{A - 4}K$: The atomic number decreases by $2$ and the mass number decreases by $4$. This corresponds to the emission of an $\alpha$ particle.
$3$. In the third step,${}_{Z - 1}^{A - 4}K \to {}_{Z - 1}^{A - 4}K$: There is no change in the atomic number or mass number,which indicates the emission of a $\gamma$ ray (de-excitation of the nucleus).
Therefore,the sequence of particles emitted is $\beta, \alpha, \gamma$.

(C) Let the parent nucleus be represented as ${}_Z^AX$.
When one $\alpha$ particle $({}_2^4He)$ is emitted,the atomic number $Z$ decreases by $2$ and the mass number $A$ decreases by $4$.
When one $\beta^-$ particle $({}_{-1}^0e)$ is emitted,the atomic number $Z$ increases by $1$ and the mass number $A$ remains unchanged.
Given that the number of $\beta$ particles emitted is twice the number of $\alpha$ particles,let the number of $\alpha$ particles be $n$. Then the number of $\beta$ particles is $2n$.
The change in atomic number $Z'$ of the daughter nucleus is given by: $Z' = Z - 2(n) + 1(2n) = Z - 2n + 2n = Z$.
The change in mass number $A'$ of the daughter nucleus is given by: $A' = A - 4(n) + 0(2n) = A - 4n$.
Since the atomic number $Z$ remains the same,the resulting daughter nucleus is an isotope of the parent nucleus.

(A) By observing the given logic gate symbols:
$(i)$ represents a $NAND$ gate.
(ii) represents a $NOT$ gate.
(iii) represents an $AND$ gate.
(iv) represents an $OR$ gate.
Therefore,the symbols for $OR$,$NOT$,and $NAND$ gates are $(iv)$,$(ii)$,and $(i)$ respectively.

(C) The energy of a photon required to excite an electron across the band gap is given by $E_g = 2.5 \, eV$.
For a photodiode to detect a signal,the energy of the incident photon $(E = \frac{hc}{\lambda})$ must be greater than or equal to the band gap energy $(E_g)$.
Thus,$\frac{12400 \, eV \cdot \mathring{A}}{\lambda} \geq 2.5 \, eV$.
Calculating the threshold wavelength $\lambda_{max} = \frac{12400}{2.5} \, \mathring{A} = 4960 \, \mathring{A}$.
Any signal with a wavelength less than or equal to $4960 \, \mathring{A}$ can be detected.
Among the given options,$4000 \, \mathring{A}$ is the only wavelength smaller than $4960 \, \mathring{A}$.

(C) In a common-emitter configuration,the current gain $\beta$ is defined as the ratio of the change in collector current to the change in base current at a constant collector-emitter voltage $V_{CE}$.
$\beta = \left( \frac{\Delta I_C}{\Delta I_B} \right)_{V_{CE}}$
Given:
$\Delta I_C = 10\, mA - 5\, mA = 5\, mA = 5 \times 10^{-3}\, A$
$\Delta I_B = 200\,\mu A - 100\,\mu A = 100\,\mu A = 100 \times 10^{-6}\, A$
Substituting these values into the formula:
$\beta = \frac{5 \times 10^{-3}}{100 \times 10^{-6}}$
$\beta = \frac{5 \times 10^{-3}}{10^{-4}}$
$\beta = 5 \times 10^1 = 50$
Thus,the current gain is $50$.

(B) In the photoelectric effect,the stopping potential depends on the frequency of the incident radiation,while the saturation current depends on the intensity of the incident radiation.
From the graph,curves $(a)$ and $(b)$ have the same stopping potential (the point where the curve intersects the negative x-axis),which implies they have the same frequency.
However,they have different saturation currents (the plateau value of the photo current),which implies they have different intensities.
Therefore,curves $(a)$ and $(b)$ represent incident radiations of the same frequency but different intensities.