AIIMS 2012 Physics Question Paper with Answer and Solution

(D) The weight of the passenger is the net gravitational force acting on them. Let $M_e$ and $M_m$ be the masses of Earth and Mars,and $R_e$ and $R_m$ be their radii. The gravitational force $F$ at a distance $x$ from Earth along the path is given by $F = |F_e - F_m| = |\frac{G M_e m}{x^2} - \frac{G M_m m}{(d-x)^2}|$,where $d$ is the distance between Earth and Mars.
At the surface of Earth $(x = R_e)$,the weight is $W_e = m g_e = 60 \times 10 = 600 \, N$.
At the surface of Mars $(x = d - R_m)$,the weight is $W_m = m g_m = 60 \times 4 = 240 \, N$.
As the spaceship moves from Earth to Mars,the gravitational pull of Earth decreases and the gravitational pull of Mars increases. At some point between the two planets,the gravitational forces from Earth and Mars will be equal in magnitude,making the net gravitational force (weight) zero.
Since the spaceship moves with constant velocity,the distance $x$ is proportional to time $t$. Therefore,the graph of weight versus time must decrease from $600 \, N$,reach $0 \, N$ at some intermediate time,and then increase to $240 \, N$ at the arrival time $t_0$. Curve $D$ is the only one that shows the weight becoming zero and then increasing to $240 \, N$.

(B) When a particle moves in a circle with uniform speed,it covers equal distances in equal intervals of time along the circumference.
Since it returns to the same position after a fixed time interval (the time period),the motion is periodic.
However,simple harmonic motion $(SHM)$ requires a restoring force proportional to the displacement from the mean position $(F = -kx)$,which is not the case for uniform circular motion.
Therefore,the motion is periodic but not simple harmonic.

(D) The time period of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{l}{g}}$,where $l$ is the effective length of the pendulum (distance from the point of suspension to the center of mass of the oscillating body).
When the chimpanzee stands up,the center of mass of the system shifts upward,closer to the point of suspension.
This results in a decrease in the effective length $l$ of the swing.
Since $T \propto \sqrt{l}$,a decrease in $l$ leads to a decrease in the time period $T$.

(C) The height of liquid in a capillary tube is given by $h = \frac{2T \cos \theta}{rdg}$,where $T$ is surface tension,$r$ is the radius,$d$ is density,$g$ is acceleration due to gravity,and $\theta$ is the angle of contact.
Rearranging for surface tension: $T = \frac{hrdg}{2 \cos \theta}$.
For water $(1)$ and mercury $(2)$: $\frac{T_1}{T_2} = \frac{h_1}{h_2} \times \frac{d_1}{d_2} \times \frac{\cos \theta_2}{\cos \theta_1}$.
Given: $h_1 = 10 \text{ cm}$,$h_2 = -3.112 \text{ cm}$ (depression),$d_1 = 1 \text{ g/cm}^3$,$d_2 = 13.6 \text{ g/cm}^3$,$\theta_1 = 0^o$,$\theta_2 = 135^o$.
Note: For capillary rise/fall,we consider the magnitude of height: $|h_2| = 3.112 \text{ cm}$.
$\frac{T_1}{T_2} = \frac{10}{3.112} \times \frac{1}{13.6} \times \frac{\cos(135^o)}{\cos(0^o)} = \frac{10}{3.112 \times 13.6} \times \frac{1}{\sqrt{2}} \approx \frac{10}{42.32} \times 0.707 \approx \frac{1}{6}$.

(C) The sound level $L$ in decibels $(dB)$ is given by $L = 10 \log_{10} \left( \frac{I}{I_0} \right)$,where $I$ is the intensity and $I_0$ is the reference intensity.
The change in sound level is given by $\Delta L = L_1 - L_2 = 20 \ dB$.
Substituting the formula,we get $\Delta L = 10 \log_{10} \left( \frac{I_1}{I_0} \right) - 10 \log_{10} \left( \frac{I_2}{I_0} \right) = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.
Given $\Delta L = 20$,we have $20 = 10 \log_{10} \left( \frac{I_1}{I_2} \right)$.
Dividing by $10$,we get $2 = \log_{10} \left( \frac{I_1}{I_2} \right)$.
Taking the antilog,we get $\frac{I_1}{I_2} = 10^2 = 100$.
Therefore,the intensity decreases by a factor of $100$.

(C) Given the formula for the time period of a simple pendulum: $T = 2\pi \sqrt{\frac{l}{g}}$.
Squaring both sides,we get $T^2 = 4\pi^2 \frac{l}{g}$.
Rearranging for $g$,we get $g = 4\pi^2 \frac{l}{T^2}$.
Taking the relative error,we use the rule for multiplication and division: $\frac{\Delta g}{g} = \frac{\Delta l}{l} + 2 \frac{\Delta T}{T}$.
Given that the fractional error in $l$ is $\frac{\Delta l}{l} = y$ and the fractional error in $T$ is $\frac{\Delta T}{T} = x$.
Substituting these values,we get $\frac{\Delta g}{g} = y + 2x$ or $2x + y$.

(A) The capacitance $C$ is defined as $C = \frac{Q}{V}$,where $Q$ is charge and $V$ is potential difference.
Since potential difference $V = \frac{W}{Q}$,where $W$ is work done,we can write $C = \frac{Q^2}{W}$.
The dimensional formula for work $W$ is $[ML^2T^{-2}]$.
Substituting the dimensions,we get $[C] = \frac{[Q^2]}{[ML^2T^{-2}]}$.
Therefore,$[C] = [M^{-1}L^{-2}T^2Q^2]$.

(B) The velocity component along the $X$-axis is given by $v_x = \frac{dx}{dt} = \frac{d}{dt}(at^2) = 2at$.
The velocity component along the $Y$-axis is given by $v_y = \frac{dy}{dt} = \frac{d}{dt}(bt^2) = 2bt$.
The speed $v$ of the particle is the magnitude of the velocity vector,given by $v = \sqrt{v_x^2 + v_y^2}$.
Substituting the values,$v = \sqrt{(2at)^2 + (2bt)^2} = \sqrt{4a^2t^2 + 4b^2t^2} = \sqrt{4t^2(a^2 + b^2)}$.
Therefore,$v = 2t\sqrt{a^2 + b^2}$.

(A) The total height $h$ of the tower is given by the equation of motion for a body falling from rest: $h = \frac{1}{2}gT^2$.
At time $t = T/3$,the distance $h'$ covered by the ball from the top is: $h' = \frac{1}{2}g(T/3)^2 = \frac{1}{2}g(T^2/9) = \frac{1}{9}(\frac{1}{2}gT^2) = \frac{h}{9}$.
The position of the ball from the ground is the total height minus the distance covered from the top: $h_{ground} = h - h' = h - \frac{h}{9} = \frac{8h}{9}$ meters.

(C) Given radius $r = 25\, cm = 0.25\, m$.
Frequency $f = 2\, rev/s$.
The angular velocity is given by $\omega = 2\pi f = 2\pi \times 2 = 4\pi\, rad/s$.
The centripetal acceleration $a_c$ is given by the formula $a_c = r\omega^2$.
Substituting the values: $a_c = 0.25 \times (4\pi)^2$.
$a_c = 0.25 \times 16\pi^2$.
$a_c = 4\pi^2\, m/s^2$.

(C) Let the mass of block $A$ be $m$ and the mass of block $B$ be $2m$.
The total mass of the system is $M = m + 2m = 3m$.
The applied force is $F = 15 \, N$.
The acceleration $a$ of the system is given by $a = \frac{F}{M} = \frac{15}{3m} = \frac{5}{m} \, m/s^2$.
The force on block $B$ is the force exerted by block $A$ on block $B$,which causes block $B$ to accelerate. This is given by $F_B = m_B \times a$.
Substituting the values,$F_B = (2m) \times \left(\frac{5}{m}\right) = 10 \, N$.

(B) According to the Law of Conservation of Energy, the total mechanical energy of the system remains constant.
For ball $A$ thrown upwards with speed $u$ from height $h$: $\frac{1}{2}mu^2 + mgh = \frac{1}{2}mv_A^2$.
For ball $B$ thrown downwards with speed $u$ from height $h$: $\frac{1}{2}mu^2 + mgh = \frac{1}{2}mv_B^2$.
Since the initial kinetic energy and potential energy are identical for both balls, the final kinetic energy at the ground must be the same.
Therefore, $v_A^2 = v_B^2$, which implies $v_A = v_B$.

(D) The power $P$ delivered by a force $F$ moving an object with velocity $v$ is given by $P = F \cdot v$.
Since the force $F$ is constant,the acceleration $a = F/m$ is also constant.
Assuming the body starts from rest $(u = 0)$,the velocity at time $t$ is $v = a \cdot t = (F/m) \cdot t$.
Substituting this into the power equation,we get $P = F \cdot (F/m) \cdot t = (F^2/m) \cdot t$.
Since $F$ and $m$ are constants,$P \propto t$.
This represents a straight line passing through the origin,which corresponds to the graph in option $D$.

(D) Let the side of the square be $a$. By the theorem of perpendicular axes,the moment of inertia about an axis passing through the center $O$ and perpendicular to the plane of the lamina is $I_z = I_x + I_y$.
For a square lamina,the moment of inertia about any axis passing through the center and lying in the plane of the lamina is the same.
Let $I_{EF}$ be the moment of inertia about the axis $EF$ (passing through the midpoints of sides $AB$ and $CD$). By symmetry,$I_{EF} = I_{GH}$ where $GH$ is the axis passing through the midpoints of sides $AD$ and $BC$.
Thus,$I_z = I_{EF} + I_{GH} = 2I_{EF}$.
Now,consider the diagonal $AC$. The moment of inertia about the diagonal $AC$ is $I_{AC}$. By symmetry,the moment of inertia about the other diagonal $BD$ is $I_{BD} = I_{AC}$.
Since the diagonals are also perpendicular axes in the plane of the lamina,$I_z = I_{AC} + I_{BD} = 2I_{AC}$.
Equating the two expressions for $I_z$,we get $2I_{EF} = 2I_{AC}$,which implies $I_{AC} = I_{EF}$.

(A) rolling motion can be considered as a combination of pure translation of the centre of mass $C$ with velocity $V_C = R\omega$ and pure rotation about the centre $C$ with angular velocity $\omega$.
For any point at a distance $r$ from the centre $C$,the velocity due to rotation is $v_{rot} = r\omega$.
The velocity of any point is the vector sum of the translational velocity $\vec{V}_C$ and the rotational velocity $\vec{v}_{rot}$.
For points $P$ and $Q$ on the horizontal diameter:
$1$. At point $Q$,the rotational velocity is in the same direction as the translational velocity. Thus,$V_Q = V_C + r\omega = R\omega + r\omega = (R+r)\omega$.
$2$. At point $P$,the rotational velocity is in the opposite direction to the translational velocity. Thus,$V_P = |V_C - r\omega| = |R\omega - r\omega| = (R-r)\omega$.
Since $R > r$,we have $V_Q > V_C > V_P$.

(A) The gravitational force $F$ acting on both masses is equal in magnitude and opposite in direction.
According to Newton's second law,$F = ma$,which implies $a = F/m$.
Since the force $F$ is the same for both,the acceleration $a$ is inversely proportional to the mass $m$ $(a \propto 1/m)$.
Given $m_1 < m_2$,it follows that $a_1 > a_2$. Thus,the acceleration of $m_1$ is greater than that of $m_2$.
Since the gravitational force is an internal force for the system,there is no external force acting on the system.
Therefore,the total energy of the system remains constant.

(B) The lift of an airplane wing is based on Bernoulli's principle.
An airplane wing is designed with a special shape called an airfoil.
Due to this shape,the air velocity above the wing is higher than the air velocity below the wing.
According to Bernoulli's theorem,where the velocity of a fluid is higher,the pressure is lower.
Therefore,the pressure below the wing is greater than the pressure above the wing,which creates an upward force known as lift.

(C) The heat supplied at a constant rate is given by $Q = K t$,where $K$ is the heating rate.
For a substance in a single phase,the temperature change is given by $Q = m c \Delta T$,which implies $K t = m c (T - T_0)$. Thus,$T = T_0 + (K / mc) t$. This shows that the temperature increases linearly with time during the heating of a single phase.
When the substance undergoes a phase change (e.g.,vaporization),the heat supplied is used to change the state at a constant temperature,given by $Q = m L$. During this process,the temperature remains constant,resulting in a horizontal line on the $T-t$ graph.
Since oxygen at $1$ atmospheric pressure undergoes a phase change (boiling) between $50\, K$ and $300\, K$,the graph will show a linear increase,followed by a horizontal segment during phase change,and then another linear increase.
Therefore,the correct graph is $C$.

(D) For an open organ pipe of length $\ell$,the frequency of the $p^{th}$ mode is given by $f_{open} = p \frac{v}{2\ell}$,where $v$ is the speed of sound.
For a closed organ pipe of length $\ell$,the frequency of the $p^{th}$ mode (where $p$ is the harmonic number) is given by $f_{closed} = (2p - 1) \frac{v}{4\ell}$.
To find the ratio of the $p^{th}$ mode frequency of the open pipe to the closed pipe,we divide the two expressions:
$\frac{f_{open}}{f_{closed}} = \frac{p \frac{v}{2\ell}}{(2p - 1) \frac{v}{4\ell}}$
Simplifying the expression:
$\frac{f_{open}}{f_{closed}} = \frac{p}{2\ell} \times \frac{4\ell}{2p - 1} = \frac{2p}{2p - 1}$
Thus,the ratio is $\frac{2p}{2p - 1}$.

(D) Let the frequency of fork $1$ be $n_1 = 200 \, Hz$ and the frequency of fork $2$ be $n_2$.
Initially,the beat frequency is $|n_1 - n_2| = 4 \, Hz$. This implies $n_2 = 200 \pm 4$,so $n_2 = 204 \, Hz$ or $n_2 = 196 \, Hz$.
When tape is attached to the prong of fork $2$,its frequency $n_2$ decreases.
If $n_2$ was $204 \, Hz$,decreasing it would make the beat frequency $|200 - n_2'|$ smaller than $4 \, Hz$.
If $n_2$ was $196 \, Hz$,decreasing it further (e.g.,to $194 \, Hz$) would make the beat frequency $|200 - 194| = 6 \, Hz$.
Since the beat frequency increased to $6 \, Hz$,the original frequency of fork $2$ must have been $196 \, Hz$.

(B) The circuit consists of two parallel branches connected to a battery of potential $V$.
Branch $1$ contains capacitors $C_1, C_2,$ and $C_3$ in series. The equivalent capacitance $C_{eq1}$ of this branch is given by:
$\frac{1}{C_{eq1}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{3C} = \frac{6+3+2}{6C} = \frac{11}{6C}$
Thus,$C_{eq1} = \frac{6C}{11}$.
The charge $Q_1$ on each capacitor in this series branch is $Q_1 = C_{eq1} V = \frac{6CV}{11}$.
Since $C_1, C_2,$ and $C_3$ are in series,the charge on $C_2$ is $Q_{C2} = Q_1 = \frac{6CV}{11}$.
Branch $2$ contains only capacitor $C_4$ connected directly across the battery. The charge on $C_4$ is $Q_{C4} = C_4 V = 4CV$.
The ratio of the charges on $C_2$ and $C_4$ is:
$\frac{Q_{C2}}{Q_{C4}} = \frac{\frac{6CV}{11}}{4CV} = \frac{6}{11 \times 4} = \frac{6}{44} = \frac{3}{22}$.

(A) The resistance of a wire is given by $R = \rho \frac{L}{A}$.
Since the wires are of the same material and length,$R \propto \frac{1}{A}$.
Given the ratio of cross-sectional areas is $A_1 : A_2 = 3 : 1$,where $A_1$ is the thicker wire and $A_2$ is the thinner wire.
Therefore,the ratio of resistances is $\frac{R_1}{R_2} = \frac{A_2}{A_1} = \frac{1}{3}$,which implies $R_2 = 3R_1$.
Given the resistance of the thicker wire $R_1 = 10\,\Omega$.
Thus,the resistance of the thinner wire $R_2 = 3 \times 10 = 30\,\Omega$.
When joined in series,the total resistance $R_{eq} = R_1 + R_2 = 10 + 30 = 40\,\Omega$.

(B) The Curie temperature $(T_C)$ is a characteristic property of ferromagnetic materials.
At temperatures below $T_C$,the material exhibits ferromagnetic properties due to the alignment of magnetic domains.
As the temperature increases and reaches $T_C$,the thermal agitation becomes strong enough to overcome the exchange coupling that maintains the alignment of these domains.
Consequently,above the Curie temperature,the material loses its spontaneous magnetization and transitions into a paramagnetic state.
Therefore,the correct option is $(B)$.

(A) The quality factor $(Q)$ of an $LCR$ circuit is defined as the ratio of the voltage across the inductor $(V_L)$ or capacitor $(V_C)$ to the voltage across the resistor $(V_R)$ at resonance.
Mathematically,$Q = \frac{V_L}{V_R} = \frac{I \cdot X_L}{I \cdot R} = \frac{X_L}{R}$.
Since the inductive reactance is given by $X_L = \omega L$,we substitute this into the expression.
Therefore,the quality factor is $Q = \frac{\omega L}{R}$.

(D) The resolving power $(R.P.)$ of an optical instrument is inversely proportional to the wavelength $(\lambda)$ of the light used,i.e.,$R.P. \propto \frac{1}{\lambda}$.
Therefore,the ratio of the resolving powers for $\lambda_1$ and $\lambda_2$ is given by:
$\frac{(R.P.)_1}{(R.P.)_2} = \frac{\lambda_2}{\lambda_1}$
Substituting the given values:
$\frac{(R.P.)_1}{(R.P.)_2} = \frac{5000 \; \mathring{A}}{4000 \; \mathring{A}} = \frac{5}{4}$
Thus,the ratio is $5:4$.

(D) The electric field $E$ increases along the positive $x-$ axis.
Let $E_1$ be the field at the position of the negative charge $(-q)$ and $E_2$ be the field at the position of the positive charge $(+q)$.
Since the field increases along the positive $x-$ axis,$E_2 > E_1$.
The force on the positive charge is $F_2 = qE_2$ (directed along the positive $x-$ axis).
The force on the negative charge is $F_1 = qE_1$ (directed along the negative $x-$ axis).
Since $E_2 > E_1$,the net force $F_{net} = F_2 - F_1 = q(E_2 - E_1)$ is directed along the positive $x-$ axis.
Wait,looking at the diagram,the negative charge is at a larger $x$ coordinate than the positive charge. Therefore,the field at the negative charge is stronger.
Let $x_{-q} > x_{+q}$,so $E(-q) > E(+q)$.
Force on $-q$ is $F_1 = qE(-q)$ (towards negative $x-$ axis).
Force on $+q$ is $F_2 = qE(+q)$ (towards positive $x-$ axis).
Since $E(-q) > E(+q)$,the net force is towards the negative $x-$ axis.
The torque $\tau = p \times E$ tends to rotate the dipole to align with the field. In this configuration,the torque causes an anticlockwise rotation.

(C) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{\text{enclosed}}}{\epsilon_0}$.
In the given figure,the charges enclosed by the Gaussian surface are $q_1$ and $q_2$. Therefore,the flux depends only on $q_1$ and $q_2$. Thus,the Assertion is correct.
However,the electric field at any point on the Gaussian surface is the vector sum of the electric fields produced by all the charges present in the space ($q_1, q_2, q_3$ and $q_4$). Therefore,the electric field at any point on the surface depends on all the charges,not just the enclosed ones. Thus,the Reason is incorrect.

(B) The dipole moment of the dipole is $\vec{p}$ and the uniform electric field is $\vec{E}$.
When an electric dipole is placed in a uniform electric field,each charge $q$ experiences a force $\vec{F} = q\vec{E}$ and $\vec{F} = -q\vec{E}$.
These two equal and opposite forces form a couple,which exerts a torque on the dipole.
The magnitude of the torque is given by the product of the magnitude of one of the forces and the perpendicular distance between them.
$\tau = F \times (d \sin \theta) = (qE) \times (a \sin \theta) = (qa) E \sin \theta$.
Since the dipole moment $p = qa$,we have $\tau = pE \sin \theta$.
In vector form,this is expressed as $\vec{\tau} = \vec{p} \times \vec{E}$.

(B) From the circuit diagram,the voltmeter is connected across the parallel combination of the two capacitors on the left.
Since the capacitors are connected in parallel,the potential difference across each capacitor is equal to the reading of the voltmeter,which is $V = 200\,V$.
The capacitance of each capacitor is $C = 25\,\mu F = 25 \times 10^{-6}\,F$.
The charge $Q$ on each plate of a capacitor is given by the formula $Q = CV$.
Substituting the given values:
$Q = (25 \times 10^{-6}\,F) \times (200\,V)$
$Q = 5000 \times 10^{-6}\,C$
$Q = 5 \times 10^{-3}\,C$.
Thus,the charge on each plate is $\pm 5 \times 10^{-3}\,C$.

(C) For capacitor $C_1$,the charges on the plates are $q_1 = 2 \mu C$ and $q_2 = 4 \mu C$. The potential difference across $C_1$ is $V_1 = (q_2 - q_1) / (2C_1) = (4 - 2) / (2C_1) = 1 / C_1$.
For capacitor $C_2$,the charges on the plates are $q_3 = 4 \mu C$ and $q_4 = 8 \mu C$. The potential difference across $C_2$ is $V_2 = (q_4 - q_3) / (2C_2) = (8 - 4) / (2 \times 2C_1) = 4 / (4C_1) = 1 / C_1$.
Since $V_1 = V_2$,there is no potential difference between the plates when the key $K$ is closed,so no charge flows.
The net charge on the upper plate of $C_1$ is $2 \mu C$ (positive) and on the lower plate is $4 \mu C$ (positive). Thus,the Assertion is correct.
The Reason is incorrect because,in an isolated capacitor,the plates carry equal and opposite charges,but when connected to external circuits or given arbitrary charges,the plates do not necessarily carry equal and opposite charges.

(B) The heat developed in a wire with constant voltage $V$ is given by $H = \frac{V^2}{R} t$. For the heat to be doubled,the resistance $R$ must be halved $(R' = R/2)$.
Resistance is given by $R = \frac{\rho \ell}{A} = \frac{\rho \ell}{\pi r^2}$.
If both length $\ell$ and radius $r$ are halved,$\ell' = \ell/2$ and $r' = r/2$.
Then $R' = \frac{\rho (\ell/2)}{\pi (r/2)^2} = \frac{\rho \ell / 2}{\pi r^2 / 4} = 2 \left( \frac{\rho \ell}{\pi r^2} \right) = 2R$.
If both length $\ell$ and radius $r$ are doubled,$\ell' = 2\ell$ and $r' = 2r$.
Then $R' = \frac{\rho (2\ell)}{\pi (2r)^2} = \frac{2\rho \ell}{4\pi r^2} = \frac{1}{2} \left( \frac{\rho \ell}{\pi r^2} \right) = \frac{R}{2}$.
Since $R' = R/2$,the heat $H' = \frac{V^2}{R'} t = \frac{V^2}{R/2} t = 2 \left( \frac{V^2}{R} t \right) = 2H$. Thus,the heat is doubled.

(B) Let $I_1$ be the current through the $5\,\Omega$ resistance and $I_2$ be the current through the $(6+9)\,\Omega$ branch.
Given that the heat developed in the $5\,\Omega$ resistance is $20.00\,cal/s$,we have:
$P_1 = I_1^2 R_1 = 20.00\,cal/s$
$I_1^2 \times 5 = 20 \implies I_1^2 = 4 \implies I_1 = 2\,A$.
The potential difference across points $C$ and $D$ is $V_{CD} = I_1 \times 5 = 2 \times 5 = 10\,V$.
The current through the upper branch $(6\,\Omega + 9\,\Omega)$ is $I_2 = \frac{V_{CD}}{6+9} = \frac{10}{15} = \frac{2}{3}\,A$.
The total current $I$ flowing through the $2\,\Omega$ resistance is $I = I_1 + I_2 = 2 + \frac{2}{3} = \frac{8}{3}\,A$.
The heat developed per second in the $2\,\Omega$ resistance is $P = I^2 R = \left(\frac{8}{3}\right)^2 \times 2 = \frac{64}{9} \times 2 = \frac{128}{9} \approx 14.22\,cal/s$.

(D) In circuit $I$,the switch (key) is open,which means the circuit is incomplete. Therefore,no current flows through the resistor $(I = 0 \ A)$. According to Ohm's law,the voltage drop across the resistor is $V = I \times R = 0 \times 2 = 0 \ V$. Thus,the voltmeter reads $0 \ V$.
In circuit $II$,the switch is closed,completing the circuit. The current flows through the resistor. Since the internal resistance of the battery is not mentioned,we assume it to be ideal. The entire electromotive force $(2 \ V)$ appears across the $2 \ \Omega$ resistor. Therefore,the voltmeter reads $2 \ V$.

(D) In the given circuit, the battery $E$, ammeter $A$, resistor $R$, and voltmeter $V$ are connected in series.
An ideal voltmeter has infinite resistance, which prevents any current from flowing through the circuit.
Since the circuit is effectively open due to the infinite resistance of the ideal voltmeter, the current $I$ flowing through the ammeter is $I = \frac{E}{R + R_V} = \frac{E}{R + \infty} = 0$.
Therefore, the ammeter reading is $0$.
The voltmeter is connected in series with the rest of the circuit. The potential difference across the ideal voltmeter will be the entire $EMF$ of the battery, $V = E$. Thus, the voltmeter reading is not zero.
Therefore, the Assertion is incorrect, and the Reason is also incorrect.

(D) The torque acting on a current-carrying loop in a uniform magnetic field is given by $\tau = NIAB \sin \theta$,where $N$ is the number of turns,$I$ is the current,$A$ is the area of the loop,$B$ is the magnetic field,and $\theta$ is the angle between the normal to the loop and the magnetic field.
For a given perimeter (length of wire $L = 2.0\,m$),the area $A$ enclosed by a loop is maximum for a circle.
Since the torque $\tau$ is directly proportional to the area $A$ $(\tau \propto A)$,the torque will be maximum for the loop with the largest area.
Among the given shapes,the circular loop $S$ encloses the maximum area for a fixed perimeter.
Therefore,the torque is maximum on the loop $S$.

(D) The magnetic field due to a straight wire of length $L$ at a perpendicular distance $d$ is given by $B = \frac{\mu_0 I}{4\pi d}(\sin \theta_1 + \sin \theta_2)$.
For a square coil of side $a$,the distance from the centre to any side is $d = a/2$.
The angles subtended by the corners at the centre are $\theta_1 = \theta_2 = 45^\circ$ (or $\pi/4$ radians).
The magnetic field due to one side is $B_{\text{side}} = \frac{\mu_0 I}{4\pi (a/2)}(\sin 45^\circ + \sin 45^\circ) = \frac{\mu_0 I}{2\pi a}(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) = \frac{\mu_0 I}{2\pi a}(\frac{2}{\sqrt{2}}) = \frac{\sqrt{2}\mu_0 I}{2\pi a} = \frac{\mu_0 I}{\sqrt{2}\pi a}$.
Since there are $4$ identical sides,the total magnetic field at the centre is $B_{\text{total}} = 4 \times B_{\text{side}} = 4 \times \frac{\mu_0 I}{\sqrt{2}\pi a} = \frac{4}{\sqrt{2}} \frac{\mu_0 I}{\pi a} = \frac{2\sqrt{2}\mu_0 I}{\pi a}$.

(A) Magnetic field lines are continuous closed loops. Outside the magnet,they emerge from the North pole $(N)$ and enter the South pole $(S)$. Inside the magnet,they travel from the South pole $(S)$ to the North pole $(N)$. This ensures that the lines of magnetic induction form continuous paths. Looking at the provided figures,figure $(1)$ correctly depicts these lines emerging from the North pole,entering the South pole,and continuing through the interior of the magnet from $S$ to $N$.

(C) The magnetic flux $\phi$ is given by the formula $\phi = B A \cos \theta$,where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{n}$ (normal to the surface).
Given that the area makes an angle of $60^o$ with the magnetic field,the angle $\theta$ between the normal to the area and the magnetic field is $\theta = 90^o - 60^o = 30^o$.
Substituting the values: $\phi = 2.0 \times 0.5 \times \cos(30^o)$.
$\phi = 1.0 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} \, Wb$.

(C) In the given parallel $LCR$ circuit,at resonance,the inductive reactance $X_L$ equals the capacitive reactance $X_C$ $(X_L = X_C)$.
This means the impedance of the $LC$ branch becomes zero (assuming ideal components).
However,the current in the resistance $R$ is determined by the applied voltage $V$ and the resistance $R$ itself,as $I = V/R$.
Since $R$ is in parallel with the $LC$ combination,the voltage across $R$ remains equal to the source voltage $V$.
Therefore,the current in the resistance $R$ is $I_{max} = V/R$,which is the maximum possible value for this circuit and is finite.

(D) The peak current in a series $LCR$ circuit is given by $I_{\max} = \frac{\varepsilon_{\max}}{Z}$,where $Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}$.
In a purely resistive element,the current is independent of frequency,but in a series $LCR$ circuit,the impedance $Z$ depends on the angular frequency $\omega$.
As $\omega$ increases,the term $(\omega L - \frac{1}{\omega C})^2$ changes. Specifically,the current $I_{\max}$ reaches its maximum value at resonance $(\omega = \frac{1}{\sqrt{LC}})$ and decreases as we move away from resonance in either direction.
Therefore,the assertion that the current always increases with an increase in angular frequency is incorrect.
The reason provided is a correct formula for peak current,but it does not support the false assertion.
Thus,the Assertion is incorrect and the Reason is correct.

(C) According to Cauchy's equation,the refractive index $\mu$ of a material depends on the wavelength $\lambda$ of light,where $\mu$ is inversely proportional to $\lambda$. Since the wavelength of red light is greater than that of violet light $(\lambda_R > \lambda_V)$,the refractive index for red light is less than that for violet light $(\mu_R < \mu_V)$.
Using the lens maker's formula,$\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} - \frac{1}{R_2})$,we see that $\frac{1}{f} \propto (\mu - 1)$.
Since $\mu_V > \mu_R$,it follows that $(\mu_V - 1) > (\mu_R - 1)$.
Therefore,$\frac{1}{f_V} > \frac{1}{f_R}$,which implies $f_V < f_R$.

(C) For a convex mirror,the focal length $f$ is taken as positive $(f = +26 \ cm)$. The object distance $u$ is always taken as negative $(u = -26 \ cm)$.
Using the mirror formula: $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$.
Substituting the values: $\frac{1}{v} + \frac{1}{-26} = \frac{1}{26}$.
$\frac{1}{v} = \frac{1}{26} + \frac{1}{26} = \frac{2}{26} = \frac{1}{13}$.
Thus,$v = +13 \ cm$.
The image is formed at $13 \ cm$ behind the mirror,not at infinity. Therefore,the Assertion is correct.
The Reason states that the equation gives $v = \infty$,which is mathematically incorrect based on the sign convention. Thus,the Reason is incorrect.

(A) Fringe visibility $(V)$ is a measure of the contrast in an interference pattern.
It is defined as the ratio of the difference between the maximum and minimum intensities to the sum of the maximum and minimum intensities.
The formula is given by:
$V = \frac{I_{\max} - I_{\min}}{I_{\max} + I_{\min}}$

(C) The fringe width $\beta$ is given by the formula $\beta = \frac{\lambda' D'}{d'}$.
Given: $\lambda' = 3\lambda$,$D' = 3D$,and $d' = d/3$.
Substituting these values,we get $\beta = \frac{(3\lambda)(3D)}{d/3} = \frac{9\lambda D}{d/3} = 27 \frac{\lambda D}{d}$.
The number of fringes $n$ in a distance $x = 1/3 \ m$ is given by $n = \frac{x}{\beta}$.
Therefore,$n = \frac{1/3}{27 \lambda D / d} = \frac{d}{81 \lambda D}$.

(C) The number of radioactive nuclei $N$ at any time $t$ is given by the radioactive decay law:
$N(t) = N_0 e^{-\lambda t}$
Where $N_0$ is the initial number of radioactive nuclei at $t = 0$ and $\lambda$ is the decay constant.
This equation represents an exponential decay function,which is of the form $y = a e^{-kx}$.
As $t$ increases,$N$ decreases exponentially from the initial value $N_0$ and approaches zero as $t \to \infty$.
Therefore,the correct graph is the one showing an exponential decay curve starting from a positive value on the $N$-axis,which corresponds to option $C$.

(B) The particle ${}_{-1}e^{0}$ is known as a $\beta^{-}$ particle (electron),and $\bar{\nu}$ is known as an antineutrino.
In this reaction,a neutron is converted into a proton,emitting an electron and an antineutrino.
Since the reaction involves the emission of a $\beta^{-}$ particle,it is classified as $\beta^{-}$ decay.

(A) In a nuclear decay process, energy is released because the total mass of the products is less than the mass of the parent nucleus. This mass difference, known as the mass defect $(\Delta m)$, is converted into energy according to Einstein's mass-energy equivalence principle, $E = \Delta m c^2$.
Since energy is released, the total rest mass energy of the products must be less than the rest mass energy of the parent nucleus to satisfy the law of conservation of energy.
Therefore, both the Assertion and the Reason are correct, and the Reason provides the correct explanation for the Assertion.

(A) The nucleus of a hydrogen atom $(_{1}^{1}H)$ consists of only a single proton.
Nuclear binding energy is defined as the energy required to disassemble a nucleus into its constituent protons and neutrons (nucleons).
Since a hydrogen nucleus contains only one nucleon (a proton),there are no other nucleons to bind with,and thus no nuclear force is acting to hold it together.
Therefore,the binding energy and mass defect of a hydrogen nucleus are zero.
Since the Assertion is true and the Reason correctly explains why the binding energy is zero (due to the presence of only one nucleon),the correct option is $A$.

(D) $NAND$ gate is formed by connecting a $NOT$ gate to the output of an $AND$ gate.
In the given options,option $D$ shows an $AND$ gate followed by a $NOT$ gate,which is the standard representation of a $NAND$ gate.

(B) $NAND$ gate produces an output of $0$ only when both inputs are $1$. Otherwise,it produces an output of $1$.
Based on the input waveforms for $A$ and $B$:
- For $t = 0$ to $2 \ s$: $A=0, B=0 \implies Y=1$.
- For $t = 2$ to $4 \ s$: $A=1, B=0 \implies Y=1$.
- For $t = 4$ to $6 \ s$: $A=0, B=0 \implies Y=1$.
- For $t = 6$ to $8 \ s$: $A=1, B=1 \implies Y=0$.
- For $t > 8 \ s$: $A=0, B=0 \implies Y=1$.
Comparing this sequence $(1, 1, 1, 0, 1)$ with the given options,the waveform in option $B$ matches this output.

(A) photo-diode is a reverse-biased $p-n$ junction diode. At the $p-n$ junction,there exists a junction electric field which,at equilibrium,does not permit the flow of charge carriers across the junction.
When such a $p-n$ diode is illuminated with light photons having energy $h
u > E_{g}$,electron-hole pairs are generated in the depletion layer (or near the junction). These charge carriers are separated by the junction field and made to flow across the junction,resulting in a change in the reverse saturation current.
Because the reverse current is highly sensitive to the generation of these photo-excited carriers,the change in current is directly proportional to the incident light intensity. Thus,the device acts as a photo-detector.

(D) The bandwidth of each signal is $5\, kHz$. Since there are $12$ signals,the total signal bandwidth is $12 \times 5\, kHz = 60\, kHz$.
For $N$ signals,the number of guard bands required between them is $N-1$.
Here,$N = 12$,so the number of guard bands required is $12 - 1 = 11$.
Each guard band has a bandwidth of $1\, kHz$.
Therefore,the total guard bandwidth is $11 \times 1\, kHz = 11\, kHz$.
The total bandwidth of the multiplexed signal is the sum of the signal bandwidths and the guard bandwidths:
Total Bandwidth $= 60\, kHz + 11\, kHz = 71\, kHz$.