NEET 2022 Physics Question Paper with Answer and Solution

(A) The formula for angular velocity under uniform angular acceleration is $\omega = \omega_{0} + \alpha t$.
Here, the initial angular speed $\omega_{0} = 1200\,rpm$ and the final angular speed $\omega = 3120\,rpm$.
The time interval is $t = 16\,s$.
First, convert the angular speeds from $rpm$ to $rad/s$ using the relation $1\,rpm = \frac{2\pi}{60}\,rad/s$.
The change in angular speed is $\Delta\omega = 3120 - 1200 = 1920\,rpm$.
Converting this to $rad/s$: $\Delta\omega = 1920 \times \frac{2\pi}{60} = 32 \times 2\pi = 64\pi\,rad/s$.
The angular acceleration $\alpha$ is given by $\alpha = \frac{\Delta\omega}{t} = \frac{64\pi}{16} = 4\pi\,rad/s^{2}$.

(A) The radius of gyration $k$ is given by the formula $k = \sqrt{\frac{I}{m}}$,where $I$ is the moment of inertia and $m$ is the mass.
For a thin uniform disc of mass $m$ and radius $R$:
$1$. The moment of inertia about an axis passing through its centre and normal to its plane is $I_1 = \frac{mR^2}{2}$.
$2$. The moment of inertia about its diameter is $I_2 = \frac{mR^2}{4}$.
Let $k_1$ and $k_2$ be the radii of gyration corresponding to $I_1$ and $I_2$ respectively.
Then,$\frac{k_1}{k_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{mR^2 / 2}{mR^2 / 4}} = \sqrt{\frac{4}{2}} = \sqrt{2} = \sqrt{2}: 1$.

(A) In a $P-V$ diagram,the slope of an adiabatic process is given by $\frac{dP}{dV} = -\gamma \frac{P}{V}$,while the slope of an isothermal process is given by $\frac{dP}{dV} = -\frac{P}{V}$.
Since the adiabatic index $\gamma > 1$ for all gases,the magnitude of the slope of the adiabatic curve is greater than that of the isothermal curve.
Therefore,the adiabatic curve is steeper than the isothermal curve.
Looking at the graph:
$1$ represents an isochoric process (constant volume).
$4$ represents an isobaric process (constant pressure).
Between $2$ and $3$,curve $2$ is steeper than curve $3$.
Thus,curve $2$ represents the adiabatic process and curve $3$ represents the isothermal process.

(B) Since the lift is moving with a constant velocity,the net acceleration $a = 0$.
According to Newton's second law,the tension $T$ in the cable must balance both the gravitational force and the frictional force.
$T = mg + f$
Given $m = 2000\,kg$,$g = 10\,m/s^2$,and $f = 3000\,N$:
$T = (2000 \times 10) + 3000 = 20000 + 3000 = 23000\,N$.
The power delivered by the motor is given by $P = T \times v$.
Given $v = 1.5\,m/s$:
$P = 23000 \times 1.5 = 34500\,W$.

(D) Plane angle is defined as the ratio of arc length to radius $(d\theta = ds/r)$. Since both arc length and radius have the dimension of length $(L)$,the dimension of plane angle is $[L^1/L^1] = [M^0 L^0 T^0]$,which is dimensionless. However,it has a unit,which is radian $(rad)$.
Similarly,solid angle is defined as the ratio of the area of the spherical surface to the square of the radius $(d\Omega = dA/r^2)$. Since both area and the square of the radius have the dimension of length squared $(L^2)$,the dimension of solid angle is $[L^2/L^2] = [M^0 L^0 T^0]$,which is also dimensionless. However,it has a unit,which is steradian $(sr)$.
Therefore,plane angle and solid angle have units but no dimensions.

(B) When a spherical ball is dropped in a highly viscous liquid, it experiences three forces: gravitational force acting downwards, and buoyant force and viscous drag force acting upwards.
The net force on the ball is $F_{net} = mg - F_B - F_v$, where $F_v = 6\pi\eta rv$ is the viscous drag force.
Initially, the speed $(v)$ is zero, so the viscous drag is zero, and the ball accelerates downwards. As the speed increases, the viscous drag force increases.
According to Newton's second law, $ma = mg - F_B - 6\pi\eta rv$. As $v$ increases, the acceleration $a$ decreases.
Eventually, the ball reaches a constant terminal velocity when the net force becomes zero $(a = 0)$.
This behavior is represented by a curve that starts from the origin, has a decreasing slope (decreasing acceleration), and becomes horizontal (constant velocity) as time increases. Curve $B$ matches this description.

(B) The distance travelled by a body in the $n^{\text{th}}$ second is given by the formula: $S_{n} = u + \frac{a}{2}(2n - 1)$.
Since the body is freely falling,the initial velocity $u = 0$ and acceleration $a = g$.
Thus,$S_{n} = \frac{g}{2}(2n - 1)$.
This implies that $S_{n} \propto (2n - 1)$.
For the $1^{\text{st}}, 2^{\text{nd}}, 3^{\text{rd}}$,and $4^{\text{th}}$ seconds,the ratio is:
$S_{1} : S_{2} : S_{3} : S_{4} = (2(1) - 1) : (2(2) - 1) : (2(3) - 1) : (2(4) - 1)$.
$S_{1} : S_{2} : S_{3} : S_{4} = 1 : 3 : 5 : 7$.

(D) The excess pressure inside a soap bubble is given by the formula $P_{in} = P_{0} + \frac{4T}{R}$, where $P_{0}$ is the atmospheric pressure, $T$ is the surface tension, and $R$ is the radius of the bubble.
As the soap bubble expands, its radius $R$ increases.
Since the term $\frac{4T}{R}$ is inversely proportional to $R$, as $R$ increases, the value of $\frac{4T}{R}$ decreases.
Therefore, the total pressure inside the bubble $P_{in}$ decreases.

(A) The gravitational field intensity $I_g$ at a point is defined as the gravitational force $F$ per unit mass $m$ placed at that point.
$I_g = \frac{F}{m}$
Given:
Mass $m = 60 \, g = 60 \times 10^{-3} \, kg = 0.06 \, kg$
Force $F = 3.0 \, N$
Substituting the values:
$I_g = \frac{3.0 \, N}{0.06 \, kg} = 50 \, N/kg$
Therefore,the magnitude of the gravitational field intensity is $50 \, N/kg$.

(B) The speed of a transverse wave on a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $\mu$ remains constant,the speed is directly proportional to the square root of the tension: $v \propto \sqrt{T}$.
Let the initial tension be $T_i = T$ and the final tension be $T_f = 2T$.
The ratio of the initial speed $v_i$ to the final speed $v_f$ is given by:
$\frac{v_i}{v_f} = \sqrt{\frac{T_i}{T_f}}$
Substituting the values,we get:
$\frac{v_i}{v_f} = \sqrt{\frac{T}{2T}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$
Thus,the ratio is $1 : \sqrt{2}$.

(C) The velocity of a particle is given by the slope of the displacement-time graph.
$V = \frac{dx}{dt} = \tan \theta$
Given the angles $\theta_1 = 30^{\circ}$ and $\theta_2 = 45^{\circ}$,the ratio of their velocities is:
$\frac{V_1}{V_2} = \frac{\tan 30^{\circ}}{\tan 45^{\circ}}$
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$ and $\tan 45^{\circ} = 1$,we have:
$\frac{V_1}{V_2} = \frac{1/\sqrt{3}}{1} = \frac{1}{\sqrt{3}}$
Thus,the ratio is $1: \sqrt{3}$.

(A) Let the $10\,kg$ mass be at the origin $(x_1 = 0)$ and the $20\,kg$ mass be at $x_2 = 10\,m$.
The formula for the center of mass $X_{CM}$ is given by:
$X_{CM} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$
Substituting the given values:
$X_{CM} = \frac{10 \times 0 + 20 \times 10}{10 + 20}$
$X_{CM} = \frac{200}{30} = \frac{20}{3}\,m$.
Thus,the distance of the center of mass from the $10\,kg$ mass is $\frac{20}{3}\,m$.

(B) Initially,the shell is at rest,so the initial momentum is $0$.
Let the masses of the three fragments be $m_1 = \frac{2m}{5}$,$m_2 = \frac{2m}{5}$,and $m_3 = \frac{m}{5}$.
The two fragments of mass $\frac{2m}{5}$ move with speed $v$ in mutually perpendicular directions. Let their velocity vectors be $\vec{v}_1 = -v \hat{i}$ and $\vec{v}_2 = -v \hat{j}$.
By the law of conservation of linear momentum:
$\vec{P}_{initial} = \vec{P}_{final}$
$0 = m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3$
$0 = \frac{2m}{5}(-v \hat{i}) + \frac{2m}{5}(-v \hat{j}) + \frac{m}{5} \vec{v}_3$
$\frac{m}{5} \vec{v}_3 = \frac{2m}{5} v \hat{i} + \frac{2m}{5} v \hat{j}$
$\vec{v}_3 = 2v \hat{i} + 2v \hat{j}$
The speed of the third fragment is the magnitude of $\vec{v}_3$:
$v_3 = |\vec{v}_3| = \sqrt{(2v)^2 + (2v)^2} = \sqrt{4v^2 + 4v^2} = \sqrt{8v^2} = 2\sqrt{2} v$.

(D) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.
Let $L_1 = 121 \ cm = 1.21 \ m$ and $L_2 = 100 \ cm = 1.0 \ m$.
Let $n_1$ be the number of oscillations of the longer pendulum and $n_2$ be the number of oscillations of the shorter pendulum.
For them to be in phase at the mean position again,the total time must be equal: $n_1 T_1 = n_2 T_2$.
$n_1 (2 \pi \sqrt{\frac{1.21}{g}}) = n_2 (2 \pi \sqrt{\frac{1.0}{g}})$.
$n_1 (1.1) = n_2 (1.0)$.
$1.1 n_1 = n_2$,which implies $\frac{n_2}{n_1} = \frac{1.1}{1} = \frac{11}{10}$.
Since $n_2$ and $n_1$ must be integers,the minimum number of vibrations for the shorter pendulum $(n_2)$ is $11$ and for the longer pendulum $(n_1)$ is $10$.

(C) The area of a rectangle is given by the formula: $Area = \text{Length} \times \text{Breadth}$.
Given,Length $= 55.3 \ m$ (which has $3$ significant figures) and Breadth $= 25 \ m$ (which has $2$ significant figures).
Calculating the product: $55.3 \times 25 = 1382.5 \ m^{2}$.
According to the rules of significant figures,the result of multiplication should have the same number of significant figures as the measurement with the least number of significant figures.
Here,the least number of significant figures is $2$ (from $25 \ m$).
Rounding $1382.5$ to $2$ significant figures,we get $1400 \ m^{2}$,which can be written as $14 \times 10^{2} \ m^{2}$.

(A) The angle of projection with the horizontal is $\theta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.
At the highest point of the trajectory,the vertical component of velocity becomes zero,and only the horizontal component of velocity remains.
The horizontal component of velocity is given by $v_x = u \cos \theta$.
Given $u = 10 \ ms^{-1}$ and $\theta = 30^{\circ}$,we have:
$v_x = 10 \cos 30^{\circ} = 10 \times \frac{\sqrt{3}}{2} = 5 \sqrt{3} \ ms^{-1}$.
Thus,the speed at the highest point is $5 \sqrt{3} \ ms^{-1}$.

(A) $1$. Gravitational constant $(G)$: From $F = G \frac{m_1 m_2}{r^2}$,we have $G = \frac{F r^2}{m_1 m_2}$. Dimensional formula is $[MLT^{-2}][L^2] / [M^2] = [M^{-1}L^3T^{-2}]$. Thus,$(a)-(ii)$.
$2$. Gravitational potential energy $(U)$: $U = -\frac{G M m}{r}$. Dimensional formula is $[MLT^{-2}][L] = [ML^2T^{-2}]$. Thus,$(b)-(iv)$.
$3$. Gravitational potential $(V)$: $V = \frac{U}{m}$. Dimensional formula is $[ML^2T^{-2}] / [M] = [L^2T^{-2}]$. Thus,$(c)-(i)$.
$4$. Gravitational intensity $(E)$: $E = \frac{F}{m}$. Dimensional formula is $[MLT^{-2}] / [M] = [LT^{-2}]$. Thus,$(d)-(iii)$.
Therefore,the correct match is $(a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)$.

(B) When a spring is stretched,the wire of the spring undergoes a change in shape (twisting/shearing) rather than just a change in length. Therefore,the stiffness of the spring is determined by the shear modulus $(G)$ of the material.
Assertion $(A)$ is true.
Regarding Reason $(R)$,steel has a much higher Young's modulus and tensile strength compared to copper. Therefore,a steel spring is stronger and more resistant to deformation than a copper spring of the same dimensions. Thus,the statement that a copper spring has more tensile strength than a steel spring is false.
Reason $(R)$ is false.
Therefore,the correct option is $(B)$.

(C) The number of moles $n$ is given by the mass divided by the molar mass of water $(H_{2}O)$.
Given mass $= 4.5 \, kg = 4.5 \times 10^{3} \, g$.
Molar mass of water $= 18 \, g/mol$.
$n = \frac{4.5 \times 10^{3}}{18} = 0.25 \times 10^{3} = 250 \, mol$.
At $STP$,the volume of $1 \, mole$ of an ideal gas is $22.4 \, L = 22.4 \times 10^{-3} \, m^{3}$.
Therefore,the total volume $V = n \times 22.4 \times 10^{-3} \, m^{3}$.
$V = 250 \times 22.4 \times 10^{-3} \, m^{3} = 5.6 \, m^{3}$.

(A) periodic function is one that repeats its values at regular intervals of time.
Functions like $\sin \omega t$,$\cos \omega t$,and their linear combinations are periodic because they repeat their values after a time period $T = 2\pi / \omega$.
The function $f(t) = e^{-\omega t}$ is an exponential decay function.
As $t$ increases,$e^{-\omega t}$ decreases monotonically and approaches zero as $t \to \infty$.
Since it never repeats its values,it represents a non-periodic motion.

(A) The formula for the maximum height $H$ attained by a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Given values are initial velocity $u = 20\,m/s$,angle of projection $\theta = 30^{\circ}$,and acceleration due to gravity $g = 10\,m/s^2$.
Substituting these values into the formula:
$H = \frac{(20)^2 \sin^2(30^{\circ})}{2 \times 10}$
$H = \frac{400 \times (1/2)^2}{20}$
$H = \frac{400 \times 1/4}{20}$
$H = \frac{100}{20} = 5\,m$.
Thus,the maximum height attained by the ball is $5\,m$.

(A) First,calculate the acceleration of the system. The total mass is $M = 3 + 2 + 1 = 6\,kg$.
The forces acting along the incline are $F_1 = 60\,N$ (upwards) and $F_2 = 18\,N$ (downwards).
The component of gravity acting downwards for all blocks is $Mg \sin 30^{\circ} = 6 \times 10 \times 0.5 = 30\,N$.
The net force is $F_{\text{net}} = 60 - 18 - 30 = 12\,N$.
The acceleration is $a = \frac{F_{\text{net}}}{M} = \frac{12}{6} = 2\,ms^{-2}$ (upwards).
Now,consider the $1\,kg$ block. Let $N$ be the normal reaction force between the $2\,kg$ and $1\,kg$ blocks.
The forces on the $1\,kg$ block are $N$ (upwards),$F_2 = 18\,N$ (downwards),and $mg \sin 30^{\circ} = 1 \times 10 \times 0.5 = 5\,N$ (downwards).
Applying Newton's second law: $N - 18 - 5 = m \times a \implies N - 23 = 1 \times 2 \implies N = 25\,N$.

(A) The pressure exerted by a liquid column at the base of a vessel is given by the formula $P = h \rho g$,where $h$ is the height of the liquid column,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
Since both vessels $A$ and $B$ are filled to the same height $h$ with the same liquid (water),the pressure at the base depends only on the height $h$,the density $\rho$,and the constant $g$.
Because $h$,$\rho$,and $g$ are identical for both vessels,the pressure at the base of vessel $A$ must be equal to the pressure at the base of vessel $B$.

(A) According to Stokes' Law,the viscous drag force $F$ acting on a spherical object of radius $r$ moving with terminal velocity $v$ in a fluid of viscosity $\eta$ is given by:
$F = 6 \pi \eta r v$
Given values:
Radius $r = 5\,mm = 5 \times 10^{-3}\,m$
Velocity $v = 10\,cm\,s^{-1} = 10 \times 10^{-2}\,m\,s^{-1} = 0.1\,m\,s^{-1}$
Viscosity $\eta = 0.9\,kg\,m^{-1}s^{-1}$
Substituting the values into the formula:
$F = 6 \times 3.14 \times 0.9 \times (5 \times 10^{-3}) \times (0.1)$
$F = 6 \times 3.14 \times 0.9 \times 5 \times 10^{-4}$
$F = 84.78 \times 10^{-4}\,N$
$F = 8.478 \times 10^{-3}\,N \approx 8.48 \times 10^{-3}\,N$

(C) Given: $\overrightarrow{ F }=2 \hat{ i }+\hat{ j }-\hat{ k }$ and $\overrightarrow{ r }=3 \hat{ i }+2 \hat{ j }-2 \hat{ k }$.
$1$. Scalar Product (Dot Product):
$\overrightarrow{ F } \cdot \overrightarrow{ r } = (2)(3) + (1)(2) + (-1)(-2) = 6 + 2 + 2 = 10$.
$2$. Vector Product (Cross Product):
$\overrightarrow{ F } \times \overrightarrow{ r } = \begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 1 & -1 \\ 3 & 2 & -2 \end{vmatrix}$
$= \hat{ i }((1)(-2) - (-1)(2)) - \hat{ j }((2)(-2) - (-1)(3)) + \hat{ k }((2)(2) - (1)(3))$
$= \hat{ i }(-2 + 2) - \hat{ j }(-4 + 3) + \hat{ k }(4 - 3)$
$= 0 \hat{ i } + 1 \hat{ j } + 1 \hat{ k } = \hat{ j } + \hat{ k }$.
$3$. Magnitude of Vector Product:
$|\overrightarrow{ F } \times \overrightarrow{ r }| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.
Thus,the magnitudes are $10$ and $\sqrt{2}$.

(D) According to Hooke's law,the restoring force $F$ of a spring is directly proportional to the displacement $x$ from the equilibrium position and acts in the opposite direction.
This is expressed by the equation $F = -kx$,where $k$ is the spring constant.
Since the relationship is $F = -kx$,the graph of $F$ versus $x$ is a straight line passing through the origin with a negative slope $(-k)$.
Therefore,the correct representation is a straight line with a negative slope,which corresponds to the graph shown in option $D$.

(A) Given: Mass $m = 5\,g = 0.005\,kg$,Linear momentum $p = 0.3\,kg\,m/s$,Time $t = 5\,s$.
We know that linear momentum $p = mv$,where $v$ is the velocity.
Substituting the values: $0.005 \times v = 0.3$.
Solving for velocity: $v = \frac{0.3}{0.005} = \frac{300}{5} = 60\,m/s$.
Assuming the body moves with constant velocity,the distance covered $d = v \times t$.
$d = 60\,m/s \times 5\,s = 300\,m$.

(A) The gravitational force is a conservative force.
By definition,the work done by a conservative force in moving an object between two points is independent of the path taken.
It depends only on the initial and final positions of the object.
Since all three paths start at point $A$ and end at point $B$,the work done by the gravitational force along each path must be equal.
Therefore,$W_1 = W_2 = W_3$.

(D) Assertion $(A)$ is incorrect because it states that the fragments themselves follow the original path. In reality,only the centre of mass $(C.O.M.)$ of the fragments continues to follow the original parabolic trajectory.
Reason $(R)$ is correct because the explosion is caused by internal chemical forces,and the external force (gravity) remains unchanged during the explosion.
Since the assertion statement is factually incorrect in its phrasing,the correct choice is that $(A)$ is not correct but $(R)$ is correct.

(A) The matching is as follows:
$(a) \rightarrow (iv)$: The graph shows damped oscillations where the amplitude decreases over time due to air friction.
$(b) \rightarrow (iii)$: The graph represents a linear relationship $y = -kx$,which corresponds to the restoring force of a spring $(F = -kx)$.
$(c) \rightarrow (ii)$: The graph shows simple harmonic motion with constant amplitude,which corresponds to an ideal pendulum oscillating with negligible air friction.
$(d) \rightarrow (i)$: The graph shows the variation of Kinetic Energy and Potential Energy,where their sum (Total Mechanical Energy) remains constant.
Therefore,the correct matching is $(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)$.

(C) The given process equation is $PV^2 = C$.
Using the ideal gas law $PV = nRT$,we can write $P = \frac{nRT}{V}$.
Substituting this into the process equation: $(\frac{nRT}{V})V^2 = C$.
This simplifies to $nRTV = C$,or $TV = \text{constant}$ (since $nR$ and $C$ are constants).
Therefore,$T_1 V_1 = T_2 V_2$,which implies $\frac{T_1}{T_2} = \frac{V_2}{V_1}$.
If $V_2 > V_1$,then $\frac{V_2}{V_1} > 1$,which means $\frac{T_1}{T_2} > 1$,so $T_1 > T_2$ or $T_2 < T_1$.
Thus,option $C$ is correct.

(C) Pressure is defined as force per unit area,so its dimensional formula is $[M L^{-1} T^{-2}]$.
Young's modulus of elasticity $(Y)$ is defined as the ratio of stress to strain.
Since strain is a dimensionless quantity,the dimensional formula of $Y$ is the same as that of stress.
Stress is also defined as force per unit area,so its dimensional formula is $[M L^{-1} T^{-2}]$.
Therefore,the dimensional formula of Young's modulus of elasticity is the same as that of pressure.

(A) The initial angular velocity is $\omega_i = 60\,rpm = 60 \times \frac{2\pi}{60} = 2\pi\,rad/s$.
The final angular velocity is $\omega_f = 360\,rpm = 360 \times \frac{2\pi}{60} = 12\pi\,rad/s$.
The change in rotational kinetic energy is given by $\Delta K.E. = \frac{1}{2} I (\omega_f^2 - \omega_i^2) = 484\,J$.
Substituting the values: $\frac{1}{2} I ((12\pi)^2 - (2\pi)^2) = 484$.
$\frac{1}{2} I (144\pi^2 - 4\pi^2) = 484$.
$\frac{1}{2} I (140\pi^2) = 484$.
$70\pi^2 I = 484$.
Using $\pi^2 \approx 9.86$,we get $70 \times 9.86 \times I = 484$.
$690.2 I = 484$.
$I = \frac{484}{690.2} \approx 0.701\,kg\cdot m^2$.

(B) The fundamental frequency of an organ pipe is given by $n = \frac{v}{4\ell}$ (for a closed pipe) or $n = \frac{v}{2\ell}$ (for an open pipe). In both cases,$n \propto v$.
Since the speed of sound in a gas is given by $v = \sqrt{\frac{\gamma RT}{M}}$,we have $v \propto \sqrt{T}$.
Therefore,the frequency $n$ is directly proportional to the square root of the absolute temperature $T$,i.e.,$n \propto \sqrt{T}$.
Given $T_1 = 27^{\circ}C = 300\,K$ and $T_2 = 90^{\circ}C = 363\,K$.
Using the ratio: $\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1}}$.
$\frac{n_2}{400} = \sqrt{\frac{363}{300}} = \sqrt{1.21} = 1.1$.
$n_2 = 400 \times 1.1 = 440\,Hz$.

(A) For an object moving with constant positive acceleration $(a > 0)$,the position-time equation is given by the kinematic equation: $x(t) = x_0 + v_0 t + \frac{1}{2} a t^2$.
If we assume the initial position $x_0 = 0$ and initial velocity $v_0 = 0$,the equation simplifies to $x = \frac{1}{2} a t^2$.
This equation represents a parabola opening upwards in the $x-t$ plane.
As $t$ increases,the slope of the tangent to the curve (which represents velocity $v = \frac{dx}{dt} = at$) also increases,indicating that the velocity is increasing over time,which is the definition of positive acceleration.
Therefore,the graph shown in option $A$ represents motion with positive acceleration.

(C) According to the ideal gas law $PV = nRT$,since $P$,$V$,and $T$ are the same for all vessels,the number of moles $n$ is the same. By Avogadro's law,the number of molecules is the same in all vessels.
The root mean square speed $(V_{rms})$ of gas molecules is given by the formula:
$V_{rms} = \sqrt{\frac{3RT}{M}}$
where $R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
Since $R$ and $T$ are constant,$V_{rms} \propto \frac{1}{\sqrt{M}}$.
The molar masses are:
$M_{He} = 4 \text{ g/mol}$
$M_{F_2} = 38 \text{ g/mol}$
$M_{SF_6} = 146 \text{ g/mol}$
Since helium has the smallest molar mass,it will have the largest root mean square speed.

(C) The gravitational field intensity $\vec{E}$ is related to the gravitational potential $V$ by the relation $\vec{E} = -\nabla V$.
Given $V = -\frac{K}{x}$,the field intensity component along the $x$-axis is $E_x = -\frac{dV}{dx}$.
$E_x = -\frac{d}{dx} \left( -\frac{K}{x} \right) = K \frac{d}{dx} (x^{-1}) = K (-1) x^{-2} = -\frac{K}{x^2}$.
Since the potential depends only on $x$,the components $E_y$ and $E_z$ are zero.
At the point $(2, 0, 3) \ m$,the $x$-coordinate is $2$.
Substituting $x = 2$ into the expression for $E_x$:
$E_x = -\frac{K}{2^2} = -\frac{K}{4}$.

(D) In a steady state,the rate of heat flow through both rods must be equal.
Let $\theta$ be the temperature at the junction.
The rate of heat flow is given by $H = \frac{KA(T_1 - T_2)}{\ell}$.
Since the rods are in series,the heat current $H$ is the same for both.
$H_{Cu} = H_{Steel}$
$\frac{K_{Cu} A (100 - \theta)}{\ell} = \frac{K_{Steel} A (\theta - 0)}{\ell}$
Given $K_{Cu} = 385 \, W \, m^{-1} \, K^{-1}$ and $K_{Steel} = 50 \, W \, m^{-1} \, K^{-1}$.
$385(100 - \theta) = 50(\theta - 0)$
Divide by $5$:
$77(100 - \theta) = 10\theta$
$7700 - 77\theta = 10\theta$
$87\theta = 7700$
$\theta = \frac{7700}{87} \approx 88.5^{\circ} \, C$.

(C) The formula for acceleration due to gravity is $g = \frac{4 \pi^2 L}{T^2}$.
The relative error in $g$ is given by $\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2 \frac{\Delta T}{T}$.
Substituting the given values: $L = 10 \, cm$,$\Delta L = 0.1 \, cm$,$T = 100 \, s$,and $\Delta T = 1 \, s$.
The percentage error in $g$ is $\left( \frac{\Delta g}{g} \times 100 \right) = \left( \frac{\Delta L}{L} \times 100 \right) + 2 \left( \frac{\Delta T}{T} \times 100 \right)$.
Percentage error $= \left( \frac{0.1}{10} \times 100 \right) + 2 \left( \frac{1}{100} \times 100 \right)$.
Percentage error $= 1\% + 2\% = 3\%$.
Therefore,the percentage error in the measurement of $g$ is $3\%$.

(D) Let $N_0$ be the initial number of radioactive atoms for both elements at $t = 0$.
According to the law of radioactive decay,the number of atoms remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For element $X_1$ with decay constant $\lambda_1 = 10\lambda$,the number of atoms is $N_1 = N_0 e^{-10\lambda t}$.
For element $X_2$ with decay constant $\lambda_2 = \lambda$,the number of atoms is $N_2 = N_0 e^{-\lambda t}$.
We are given that the ratio $\frac{N_1}{N_2} = \frac{1}{e} = e^{-1}$.
Substituting the expressions for $N_1$ and $N_2$:
$\frac{N_0 e^{-10\lambda t}}{N_0 e^{-\lambda t}} = e^{-1}$
$e^{-10\lambda t + \lambda t} = e^{-1}$
$e^{-9\lambda t} = e^{-1}$
Comparing the exponents:
$-9\lambda t = -1$
$t = \frac{1}{9\lambda}$.

(B) The Biot-Savart law is given by the expression: $d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \vec{r}}{r^3}$.
Statement $I$ is correct because the law specifically defines the magnetic field contribution from an infinitesimal current element $Id\vec{l}$.
Statement $II$ is incorrect because it swaps the nature of the sources. Biot-Savart's law involves a vector source $(Id\vec{l})$,whereas Coulomb's law involves a scalar source (charge $q$). The statement claims the opposite,making it false.
Therefore,Statement $I$ is correct and Statement $II$ is incorrect.

(B) Given: Radii of curvature $R_{1} = +20 \ cm = +0.2 \ m$ and $R_{2} = -20 \ cm = -0.2 \ m$ (by sign convention for a biconvex lens).
Refractive index $\mu = 1.5 = \frac{3}{2}$.
The lens maker's formula is given by $P = \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$.
Substituting the values: $P = (1.5 - 1) \left( \frac{1}{0.2} - \frac{1}{-0.2} \right)$.
$P = (0.5) \left( \frac{1}{0.2} + \frac{1}{0.2} \right) = (0.5) \left( \frac{2}{0.2} \right)$.
$P = (0.5) \times 10 = +5 \ D$.

(A) The electric potential $V$ at the surface of a hollow conducting sphere of radius $R$ carrying charge $Q$ is given by the formula: $V = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{Q}{R}$.
Here,$\frac{1}{4 \pi \epsilon_{0}}$ is a constant.
It is given that both spheres have the same charge $Q$.
Therefore,the potential $V$ is inversely proportional to the radius $R$,i.e.,$V \propto \frac{1}{R}$.
Since $R_{1} >> R_{2}$,the radius of the smaller sphere is $R_{2}$.
Because $R_{2} < R_{1}$,it follows that $V_{2} > V_{1}$.
Thus,the potential is higher on the smaller sphere.

(B) For conductors,the temperature coefficient of resistance $\alpha$ is positive,meaning resistance increases with temperature.
For semiconductors,the temperature coefficient of resistance $\alpha$ is negative,meaning resistance decreases as temperature increases due to the increase in the number of charge carriers.

(C) In circuit $(a)$,both p-n junctions are connected in the same forward-biased orientation. Since they are identical and in series,the potential drop across each junction is equal.
In circuit $(b)$,one junction is forward-biased while the other is reverse-biased. The reverse-biased junction offers a much higher resistance,so the potential drop across it will be significantly higher than the forward-biased junction.
In circuit $(c)$,both p-n junctions are connected in the same reverse-biased orientation. Since they are identical and in series,the potential drop across each junction is equal.
Therefore,the potential drop across the two p-n junctions is equal in both circuits $(a)$ and $(c)$.

(B) Method $(i)$:
Using Snell's law,$n_1 \sin i = n_2 \sin r$.
Given $n_1 = 1$ (air),$n_2 = \sqrt{3}$ (glass),and $i = 60^{\circ}$.
$1 \cdot \sin 60^{\circ} = \sqrt{3} \cdot \sin r$
$\frac{\sqrt{3}}{2} = \sqrt{3} \sin r$
$\sin r = \frac{1}{2} \implies r = 30^{\circ}$.
The angle between the reflected ray and the normal is $i = 60^{\circ}$.
The angle between the refracted ray and the normal is $r = 30^{\circ}$.
The angle between the reflected ray and the refracted ray is $180^{\circ} - (i + r) = 180^{\circ} - (60^{\circ} + 30^{\circ}) = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
Method $(ii)$:
Since the angle of incidence $i = 60^{\circ}$ satisfies the condition for Brewster's angle,where $\tan i_p = \mu = \sqrt{3}$,it implies $i_p = 60^{\circ}$.
At Brewster's angle,the reflected and refracted rays are perpendicular to each other,so the angle between them is $90^{\circ}$.

(D) The power $P$ of the transmitter is $100\,kW = 100 \times 10^{3}\,W = 10^{5}\,W$.
The time $t$ is $1\,hour = 3600\,s$.
The energy $E$ radiated is given by the formula $E = P \times t$.
Substituting the values, we get $E = 10^{5}\,W \times 3600\,s$.
$E = 3600 \times 10^{5}\,J = 36 \times 10^{7}\,J$.

(B) The magnetic force on a current-carrying conductor is given by $F = BIl \sin \theta$. Thus,the magnetic field $B = F / (Il)$.
The dimensions of force $F$ are $[MLT^{-2}]$.
The dimensions of current $I$ are $[A]$.
The dimensions of length $l$ are $[L]$.
Therefore,the dimensions of $B$ are $[MLT^{-2}] / ([A][L]) = [MT^{-2}A^{-1}]$.
The magnetic permeability $\mu$ is related to $B$ and $H$ by $B = \mu H$,where $H$ has dimensions $[IL^{-1}] = [AL^{-1}]$.
Thus,$\mu = B / H = [MT^{-2}A^{-1}] / [AL^{-1}] = [MLT^{-2}A^{-2}]$.
Hence,the dimensions $[MLT^{-2}A^{-2}]$ correspond to magnetic permeability.

(B) For an alternating current $(ac)$ source,the relationship between the peak voltage $(V_0)$ and the root mean square $(rms)$ voltage $(V_{rms})$ is given by the formula: $V_{rms} = \frac{V_0}{\sqrt{2}}$.
Rearranging this formula to solve for the peak voltage,we get: $V_0 = \sqrt{2} \times V_{rms}$.
Therefore,the peak voltage is $\sqrt{2}$ times the $rms$ value of the $ac$ source.

(A) The magnetic field $B$ at the centre of a long solenoid is given by the formula $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the current.
Given:
Number of turns per unit length $n = 100 \text{ turns/mm} = 100 \times 10^3 \text{ turns/m} = 10^5 \text{ turns/m}$.
Current $i = 1\,A$.
Permeability of free space $\mu_0 = 4\pi \times 10^{-7} \text{ T}\,\text{m/A}$.
Substituting the values:
$B = (4\pi \times 10^{-7}) \times (10^5) \times (1)$
$B = 4\pi \times 10^{-2} \text{ T}$
$B \approx 4 \times 3.14 \times 10^{-2} \text{ T} = 12.56 \times 10^{-2} \text{ T}$.

(C) The electromagnetic spectrum classifies waves based on their wavelength range:
$(a)$ $AM$ radio waves have long wavelengths, typically around $10^{2} \, m$ (Matches with $ii$).
$(b)$ Microwaves have wavelengths in the range of $10^{-2} \, m$ (Matches with $iii$).
$(c)$ Infrared radiations have wavelengths around $10^{-4} \, m$ (Matches with $iv$).
$(d)$ $X$-rays have very short wavelengths, typically around $10^{-10} \, m$ (Matches with $i$).
Therefore, the correct matching is $(a) - (ii), (b) - (iii), (c) - (iv), (d) - (i)$.

(A) Since both resistors are connected in parallel,the potential difference $(V)$ across both resistors is the same.
Thermal energy $(H)$ developed in a resistor is given by $H = \frac{V^2}{R} \times t$.
For a given time $(t)$ and constant potential difference $(V)$,the thermal energy is inversely proportional to the resistance: $H \propto \frac{1}{R}$.
Therefore,the ratio of thermal energy in $100\,\Omega$ $(H_1)$ to that in $200\,\Omega$ $(H_2)$ is:
$\frac{H_1}{H_2} = \frac{R_2}{R_1} = \frac{200\,\Omega}{100\,\Omega} = \frac{2}{1}$.
Thus,the ratio is $2: 1$.

(B) In a half-wave rectifier,the diode conducts only during the positive half-cycle of the input $AC$ signal.
Therefore,the output signal consists of one pulse for every complete cycle of the input signal.
As a result,the frequency of the output signal is equal to the frequency of the input signal.
Given that the input frequency $f_{\text{in}} = 60\,Hz$,the output frequency $f_{\text{out}}$ will also be $60\,Hz$.

(B) The work done in moving a charge $q$ along an equipotential surface is zero because the potential difference between any two points on the surface is zero.
Work done $W = \int \vec{F} \cdot d\vec{l} = q \int \vec{E} \cdot d\vec{l} = 0$.
Since $q \neq 0$ and $d\vec{l} \neq 0$,it implies that $\vec{E} \cdot d\vec{l} = 0$.
This means the electric field vector $\vec{E}$ must be perpendicular to the displacement vector $d\vec{l}$ on the equipotential surface.
Therefore,the angle between the electric lines of force and the equipotential surface is $90^{\circ}$.

(C) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = h\nu - h\nu_0$,where $\nu_0$ is the threshold frequency.
For frequency $\nu$,the stopping potential is $\frac{V_s}{2}$. Thus,$e(\frac{V_s}{2}) = h\nu - h\nu_0$ --- $(1)$
For frequency $\frac{\nu}{2}$,the stopping potential is $V_s$. Thus,$eV_s = h(\frac{\nu}{2}) - h\nu_0$ --- $(2)$
From equation $(2)$,we have $eV_s = \frac{h\nu}{2} - h\nu_0$. Substituting this into equation $(1)$:
$\frac{1}{2}(\frac{h\nu}{2} - h\nu_0) = h\nu - h\nu_0$
$\frac{h\nu}{4} - \frac{h\nu_0}{2} = h\nu - h\nu_0$
$h\nu_0 - \frac{h\nu_0}{2} = h\nu - \frac{h\nu}{4}$
$\frac{h\nu_0}{2} = \frac{3h\nu}{4}$
$\nu_0 = \frac{3}{2}\nu$.

(A) Given:
Side length of the square loop,$l = 1\,m$.
Area of the loop,$A = l^2 = (1\,m)^2 = 1\,m^2$.
Magnetic field,$B = 0.5\,T$.
The plane of the loop is perpendicular to the magnetic field,which means the area vector $\overrightarrow{A}$ is parallel to the magnetic field vector $\overrightarrow{B}$.
Therefore,the angle $\theta$ between $\overrightarrow{B}$ and $\overrightarrow{A}$ is $0^\circ$.
The magnetic flux $\phi$ is given by the formula:
$\phi = B A \cos\theta$
Substituting the values:
$\phi = 0.5 \times 1 \times \cos(0^\circ)$
Since $\cos(0^\circ) = 1$,
$\phi = 0.5 \times 1 \times 1 = 0.5\,Wb$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
The first excited state corresponds to $n = 2$.
Therefore,$T_1 = -\frac{13.6}{2^2} = -\frac{13.6}{4} \text{ eV}$.
The second excited state corresponds to $n = 3$.
Therefore,$T_2 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \text{ eV}$.
The ratio $T_1 : T_2$ is calculated as:
$\frac{T_1}{T_2} = \frac{-13.6/4}{-13.6/9} = \frac{9}{4}$.
Thus,the ratio is $9:4$.

(C) According to the de Broglie hypothesis,the wavelength $(\lambda)$ associated with a particle of momentum $(p)$ is given by the relation: $\lambda = \frac{h}{p}$,where $h$ is Planck's constant.
This equation shows that $\lambda$ is inversely proportional to $p$ (i.e.,$\lambda \propto \frac{1}{p}$).
As the momentum $(p)$ increases,the de Broglie wavelength $(\lambda)$ decreases.
This relationship represents a rectangular hyperbola,which is correctly depicted in option $C$.

(C) The width of the segment on the screen is given by $y = n \beta$,where $\beta = \frac{\lambda D}{d}$ is the fringe width.
Thus,$y = n \lambda \left(\frac{D}{d}\right)$.
Since the segment length $y$ and the experimental setup parameters $D$ and $d$ remain constant,we have $n_1 \lambda_1 = n_2 \lambda_2$.
Given $n_1 = 8$,$\lambda_1 = 600 \ nm$,and $\lambda_2 = 400 \ nm$.
Substituting the values: $8 \times 600 = n_2 \times 400$.
$n_2 = \frac{8 \times 600}{400} = \frac{4800}{400} = 12$.
Therefore,the student will observe $12$ fringes.

(B) The given reaction is ${ }_{11}^{22} Na \rightarrow X + e ^{+} + \nu$.
This represents a $\beta^{+}$ decay (positron emission).
In $\beta^{+}$ decay,the atomic number $Z$ decreases by $1$ while the mass number $A$ remains constant.
For the parent nucleus ${ }_{11}^{22} Na$,the atomic number $Z = 11$ and the mass number $A = 22$.
After emitting a positron $(e^{+})$,the new atomic number $Z' = 11 - 1 = 10$.
The mass number remains $A' = 22$.
The element with atomic number $10$ is Neon $(Ne)$.
Therefore,the reaction is ${ }_{11}^{22} Na \rightarrow { }_{10}^{22} Ne + e ^{+} + \nu$.
Thus,$X$ is ${ }_{10}^{22} Ne$.

(C) Given: Length $L = 10 \, m$,Radius $r = \frac{10^{-2}}{\sqrt{\pi}} \, m$,Resistance $R = 10 \, \Omega$,Electric field $E = 10 \, V/m$.
First,calculate the cross-sectional area $A$:
$A = \pi r^{2} = \pi \left( \frac{10^{-2}}{\sqrt{\pi}} \right)^{2} = \pi \cdot \frac{10^{-4}}{\pi} = 10^{-4} \, m^{2}$.
We know that current density $J = \sigma E$,where $\sigma$ is conductivity.
Since $\sigma = \frac{1}{\rho}$ and $R = \rho \frac{L}{A}$,we have $\rho = \frac{RA}{L}$.
Thus,$\sigma = \frac{L}{RA}$.
Substituting this into the expression for $J$:
$J = \left( \frac{L}{RA} \right) E = \frac{E \cdot L}{R \cdot A}$.
Plugging in the values:
$J = \frac{10 \times 10}{10 \times 10^{-4}} = \frac{100}{10^{-3}} = 10^{5} \, A/m^{2}$.

(C) The refractive index $n$ of a medium is related to its relative permittivity $\varepsilon_{r}$ and relative permeability $\mu_{r}$ by the formula: $n = \sqrt{\varepsilon_{r}\mu_{r}}$.
Since the refractive index is defined as the ratio of the speed of light in vacuum $c$ to the speed of light in the medium $v$,we have $n = \frac{c}{v}$.
Rearranging this for $v$,we get $v = \frac{c}{n}$.
Substituting the expression for $n$,we obtain the velocity of light in the medium as $v = \frac{c}{\sqrt{\varepsilon_{r}\mu_{r}}}$.

(B) According to Ampere's circuital law,for a long straight wire of radius $R$ carrying a steady current $I$:
$1$. Inside the wire $(r < R)$: The magnetic field $B$ is given by $B = \frac{\mu_0 I r}{2 \pi R^2}$. Thus,$B \propto r$,which means the magnetic field increases linearly with distance $r$ from the center.
$2$. Outside the wire $(r > R)$: The magnetic field $B$ is given by $B = \frac{\mu_0 I}{2 \pi r}$. Thus,$B \propto 1/r$,which means the magnetic field decreases as $1/r$ with distance $r$ from the center.
Therefore,the magnetic field increases linearly up to the boundary $(r = R)$ and then decreases as $1/r$ for the outside region.

(A) Given: $L = 10\,H$,$C = 10 \times 10^{-6}\,F$,$R = 50\,\Omega$,and $V = 200 \sin(100t)$.
Comparing with $V = V_{m} \sin(\omega t)$,we get angular frequency $\omega = 100\,rad/s$.
The frequency of the $AC$ source is $\nu = \frac{\omega}{2\pi} = \frac{100}{2\pi} = \frac{50}{\pi}\,Hz$.
The resonant frequency $\nu_{0}$ of the $LCR$ circuit is given by $\nu_{0} = \frac{1}{2\pi\sqrt{LC}}$.
Substituting the values: $\nu_{0} = \frac{1}{2\pi\sqrt{10 \times 10 \times 10^{-6}}} = \frac{1}{2\pi\sqrt{10^{-4}}} = \frac{1}{2\pi \times 10^{-2}} = \frac{100}{2\pi} = \frac{50}{\pi}\,Hz$.
Thus,$\nu_{0} = \nu = \frac{50}{\pi}\,Hz$.

(B) The circuit consists of two $NAND$ gates whose outputs are fed into an $AND$ gate. Let the inputs be $A$ and $B$. The upper $NAND$ gate receives inputs $A$ and $B$,so its output is $\overline{A \cdot B}$. The lower $NAND$ gate receives inputs $\bar{A}$ (from the $NOT$ gate) and $B$,so its output is $\overline{\bar{A} \cdot B}$. The final output $C$ is the $AND$ operation of these two outputs: $C = (\overline{A \cdot B}) \cdot (\overline{\bar{A} \cdot B})$.
Using De Morgan's theorem,$\overline{A \cdot B} = \bar{A} + \bar{B}$ and $\overline{\bar{A} \cdot B} = A + \bar{B}$.
Thus,$C = (\bar{A} + \bar{B}) \cdot (A + \bar{B})$.
Using the distributive law,$C = \bar{A}A + \bar{A}\bar{B} + \bar{B}A + \bar{B}\bar{B}$.
Since $\bar{A}A = 0$ and $\bar{B}\bar{B} = \bar{B}$,we get $C = 0 + \bar{B}(\bar{A} + A) + \bar{B} = \bar{B}(1) + \bar{B} = \bar{B} + \bar{B} = \bar{B}$.
Therefore,the output $C$ is the $NOT$ of $B$ $(\bar{B})$.
| $A$ | $B$ | $C$ |
|---|---|---|
| $0$ | $0$ | $1$ |
| $0$ | $1$ | $0$ |
| $1$ | $0$ | $1$ |
| $1$ | $1$ | $0$ |

(B) The radius $R$ of a nucleus with mass number $A$ is given by the formula: $R = R_{0} A^{1/3}$, where $R_{0}$ is a constant.
For the two daughter nuclei with mass numbers $A_{1} = 125$ and $A_{2} = 64$, their radii $R_{1}$ and $R_{2}$ are:
$R_{1} = R_{0} (125)^{1/3} = R_{0} \times 5$
$R_{2} = R_{0} (64)^{1/3} = R_{0} \times 4$
The ratio of the radii is:
$\frac{R_{1}}{R_{2}} = \frac{R_{0} (125)^{1/3}}{R_{0} (64)^{1/3}} = \frac{5}{4}$
Thus, the ratio is $5: 4$.

(A) In a Wheatstone bridge,the condition for balance is given by $\frac{P}{Q} = \frac{X}{Y}$.
To achieve the most precise measurement of the unknown resistance $X$,the sensitivity of the bridge must be maximized.
The sensitivity of a Wheatstone bridge is maximum when all four resistances are of the same order of magnitude.
Specifically,if $P$ and $Q$ are approximately equal and small,the relative error in the measurement of $X$ is minimized,leading to the most precise result.

(B) $1$. Initial charge on the first capacitor: $Q = C V = 900 \times 10^{-12} \, F \times 100 \, V = 9 \times 10^{-8} \, C$.
$2$. When connected to an uncharged capacitor of the same capacitance $C$,the charge $Q$ is shared equally between the two capacitors because they are in parallel.
$3$. The common potential $V'$ is given by: $V' = \frac{Q_{total}}{C_{total}} = \frac{Q}{C + C} = \frac{9 \times 10^{-8} \, C}{1800 \times 10^{-12} \, F} = 50 \, V$.
$4$. The total electrostatic energy stored in the system is: $U = \frac{1}{2} (C + C) (V')^2 = \frac{1}{2} (1800 \times 10^{-12} \, F) (50 \, V)^2$.
$5$. $U = 900 \times 10^{-12} \times 2500 = 225 \times 10^{-8} \, J = 2.25 \times 10^{-6} \, J$.

(A) The refractive index $\mu$ is inversely proportional to the speed of light $v$ in a medium,given by $\mu = \frac{c}{v}$,where $c$ is the speed of light in vacuum.
The critical angle $i_c$ for light traveling from a denser medium to a rarer medium is given by the formula $\sin i_c = \frac{\mu_R}{\mu_D}$,where $\mu_R$ is the refractive index of the rarer medium and $\mu_D$ is the refractive index of the denser medium.
Since $\mu \propto \frac{1}{v}$,we have $\frac{\mu_R}{\mu_D} = \frac{v_D}{v_R}$.
Given $v_A = 1.5 \times 10^8 \ m/s$ and $v_B = 2.0 \times 10^8 \ m/s$,medium $A$ is the denser medium $(D)$ and medium $B$ is the rarer medium $(R)$.
Therefore,$\sin i_c = \frac{v_A}{v_B} = \frac{1.5 \times 10^8}{2.0 \times 10^8} = \frac{1.5}{2.0} = \frac{3}{4} = 0.750$.
Thus,$i_c = \sin^{-1}(0.750)$.

(A) The given system consists of two equal and opposite point charges separated by a small distance $L$,which forms an electric dipole.
For an electric dipole,the electric field intensity $E$ at a large distance $R$ $(R \gg L)$ from the center of the dipole is given by the general formula:
$E = \frac{1}{4\pi\epsilon_0} \frac{p}{R^3} \sqrt{1 + 3 \cos^2 \theta}$
where $p$ is the dipole moment and $\theta$ is the angle between the position vector and the dipole axis.
Since $p = qL$ is constant,we can observe that $E \propto \frac{1}{R^3}$.
Therefore,the magnitude of the electric field intensity varies as $1/R^3$.

(B) The magnetic flux $\phi$ through the coil rotating in a magnetic field is given by $\phi = NBA \cos(\omega t)$.
The induced electromotive force $(EMF)$ is $E = -\frac{d\phi}{dt} = NBA\omega \sin(\omega t)$.
The maximum $EMF$ is $E_{\max} = NBA\omega$.
The maximum induced current is $i_{\max} = \frac{E_{\max}}{R} = \frac{NBA\omega}{R}$.
Given: $N = 1000$,$B = 2 \times 10^{-5} \, T$,$r = 10 \, m$,$\omega = 2 \, rad \cdot s^{-1}$,$R = 12.56 \, \Omega$.
Area $A = \pi r^2 = \pi (10)^2 = 100\pi \, m^2$.
Substituting the values:
$i_{\max} = \frac{1000 \times (2 \times 10^{-5}) \times (100\pi) \times 2}{12.56}$.
Since $100\pi \approx 314$,$i_{\max} = \frac{1000 \times 2 \times 10^{-5} \times 314 \times 2}{12.56} = \frac{12.56}{12.56} = 1 \, A$.

(D) Given the magnetic field $\overrightarrow{B} = B_0 \cos(kx + \omega t) \hat{j}$,where $B_0 = 3 \times 10^{-8} \text{ T}$.
The relationship between the amplitude of the electric field $(E_0)$ and the magnetic field $(B_0)$ is $E_0 = c B_0$,where $c = 3 \times 10^8 \text{ m/s}$.
$E_0 = (3 \times 10^8 \text{ m/s}) \times (3 \times 10^{-8} \text{ T}) = 9 \text{ V/m}$.
The wave propagates in the negative $x$-direction (since the argument is $kx + \omega t$).
The direction of propagation is given by the direction of $\overrightarrow{E} \times \overrightarrow{B}$.
Since $\overrightarrow{B}$ is along $\hat{j}$ and the wave travels along $-\hat{i}$,we have $\hat{k} \times \hat{j} = -\hat{i}$.
Thus,the electric field must be along the $\hat{k}$ direction.
Therefore,$\overrightarrow{E} = 9 \cos (1.6 \times 10^3 x + 48 \times 10^{10} t) \hat{k} \text{ V/m}$.

(C) Zener diode is a special type of diode designed to operate reliably in the reverse breakdown region.
$1$. Doping: Zener diodes are heavily doped,which results in a very thin depletion region.
$2$. Depletion Region: Due to heavy doping,the depletion region is extremely thin (typically $10^{-6} \ m$),which allows for a high electric field even at low reverse voltages.
$3$. Operation: It is specifically designed to operate in the reverse bias region.
$4$. Voltage: Once the breakdown voltage is reached,the Zener voltage $(V_z)$ remains constant despite changes in the current.
Therefore,the statement that the depletion region is very wide is incorrect.

(A) Given:
$emf (E) = 4\,V$
Internal resistance $(r) = 0.5\,\Omega$
External resistance $(R) = 7.5\,\Omega$
First,calculate the current $(I)$ flowing through the circuit using Ohm's law:
$I = \frac{E}{R + r} = \frac{4}{7.5 + 0.5} = \frac{4}{8} = 0.5\,A$
The terminal potential difference $(V)$ is given by the formula:
$V = E - Ir$
$V = 4 - (0.5 \times 0.5)$
$V = 4 - 0.25$
$V = 3.75\,V$
Alternatively,$V = IR = 0.5 \times 7.5 = 3.75\,V$.

(C) Statement-$I$: In a purely capacitive $AC$ circuit,the current leads the voltage by a phase angle of $\pi/2$ radians $(90^{\circ})$. Thus,statement-$I$ is correct.
Statement-$II$: In a purely capacitive $AC$ circuit,the phase difference between the current and the voltage is $\pi/2$,not $\pi$. Thus,statement-$II$ is incorrect.
Therefore,statement-$I$ is correct but statement-$II$ is incorrect.

(C) Let the equivalent resistance of the infinite network be $x$.
Since the network is infinite,adding one more section to the front does not change the total resistance. Thus,the network can be represented as two $1\,\Omega$ resistors in series with the parallel combination of a $1\,\Omega$ resistor and the equivalent resistance $x$.
The equivalent resistance $x$ is given by:
$x = 1 + \left( \frac{1 \times x}{1 + x} \right) + 1$
$x = 2 + \frac{x}{1 + x}$
$x - 2 = \frac{x}{1 + x}$
$(x - 2)(x + 1) = x$
$x^2 + x - 2x - 2 = x$
$x^2 - 2x - 2 = 0$
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-2)}}{2(1)}$
$x = \frac{2 \pm \sqrt{4 + 8}}{2} = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}$
Since resistance cannot be negative,we take the positive value:
$x = (1 + \sqrt{3})\,\Omega$

(D) The magnetic field $B$ at the centre of a circular coil with $N$ turns,radius $R$,and current $I$ is given by the formula:
$B = \frac{\mu_0 NI}{2R}$
Given values:
$N = 1000$
$I = 1\,A$
$R = 62.8\,cm = 0.628\,m = 62.8 \times 10^{-2}\,m$
$\mu_0 = 4\pi \times 10^{-7}\,T\cdot m/A$
Substituting the values into the formula:
$B = \frac{4 \times 3.14 \times 10^{-7} \times 1000 \times 1}{2 \times 62.8 \times 10^{-2}}$
$B = \frac{12.56 \times 10^{-4}}{125.6 \times 10^{-2}}$
$B = \frac{12.56 \times 10^{-4}}{1.256}$
$B = 10 \times 10^{-4} = 10^{-3}\,T$
Thus,the magnetic field produced at the centre of the coil is $10^{-3}\,T$.

(C) For an $AC$ source,the inductive reactance is given by $X_L = \omega L = 2 \pi f L$.
Given $L = 2 \, mH = 2 \times 10^{-3} \, H$ and $f = 50 \, Hz$.
$X_1 = 2 \times \pi \times 50 \times 2 \times 10^{-3} = 100 \pi \times 2 \times 10^{-3} = 0.2 \pi \, \Omega$.
Using $\pi \approx 3.14$,we get $X_1 = 0.2 \times 3.14 = 0.628 \, \Omega$.
For a $DC$ source,the frequency $f = 0$,so the angular frequency $\omega = 2 \pi f = 0$.
The inductive reactance $X_2 = \omega L = 0 \times L = 0 \, \Omega$.
Thus,$X_1 = 0.628 \, \Omega$ and $X_2 = 0$.

(C) primary rainbow is formed by a single internal reflection of sunlight within the raindrops,and it appears at an angular range of $40^{\circ}$ to $42^{\circ}$.
$A$ secondary rainbow is formed by two internal reflections of sunlight within the raindrops,and it appears at an angular range of $51^{\circ}$ to $54^{\circ}$.
Because the secondary rainbow has a larger angular radius,it is always formed above the primary rainbow.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE_{max})$ of the emitted photoelectrons is given by:
$KE_{max} = h\nu - \phi$
Where $h\nu$ is the energy of the incident photon and $\phi$ is the work function of the metal.
Given:
Energy of incident photon $(h\nu)$ = $4.2\,eV$
Work function $(\phi)$ = $2.2\,eV$
We know that the maximum kinetic energy is related to the stopping potential $(V_0)$ by the equation:
$KE_{max} = eV_0$
Substituting the values:
$eV_0 = 4.2\,eV - 2.2\,eV$
$eV_0 = 2.0\,eV$
Therefore,the stopping potential $V_0 = 2\,V$.

(A) In the given circuit,two switches are connected in parallel with the $LED$. The $LED$ glows when current flows through it,which happens when at least one of the switches is closed $(1)$. If both switches are open $(0)$,the circuit is incomplete and the $LED$ does not glow $(0)$.

$A$	$B$	$LED$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$1$

This truth table corresponds to the $OR$ gate.

(D) The initial capacitance of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 A}{d}$,where $A$ is the area of the plates and $d$ is the distance between them.
According to the problem,the new distance $d' = 2d$ and the new area $A' = A / 2$.
The final capacitance $C'$ is given by $C' = \frac{\varepsilon_0 A'}{d'}$.
Substituting the new values,we get $C' = \frac{\varepsilon_0 (A / 2)}{2d} = \frac{\varepsilon_0 A}{4d}$.
Since $C = \frac{\varepsilon_0 A}{d}$,we can write $C' = \frac{C}{4}$.

(C) According to Malus's law,the intensity of light transmitted through an analyser is given by $I_2 = I_1 \cos^2 \theta$.
Here,$I_1 = I$ is the intensity of the incident linearly polarised light.
The angle between the transmission axes of the polariser and the analyser is $\theta = 30^{\circ}$.
Substituting these values into the formula:
$I_2 = I \cos^2(30^{\circ})$
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have:
$I_2 = I \left( \frac{\sqrt{3}}{2} \right)^2 = I \left( \frac{3}{4} \right) = \frac{3I}{4}$.

(C) The linear fringe width $\beta$ in a Young's double slit experiment is given by the formula $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
When the screen is moved away from the plane of the slits,the distance $D$ increases.
Since $\beta \propto D$,an increase in $D$ leads to an increase in the linear fringe width $\beta$.
Therefore,the linear separation of the fringes increases.

(C) Let the capacitances of the two capacitors be $C_1$ and $C_2$.
When connected in series,the effective capacitance is given by $\frac{C_1 C_2}{C_1 + C_2} = 3\,\mu F$. (Equation $1$)
When connected in parallel,the effective capacitance is given by $C_1 + C_2 = 16\,\mu F$. (Equation $2$)
Substituting Equation $2$ into Equation $1$:
$C_1 C_2 = 3 \times 16 = 48$.
We have the sum $C_1 + C_2 = 16$ and the product $C_1 C_2 = 48$. These are the roots of the quadratic equation $x^2 - 16x + 48 = 0$.
Solving the quadratic equation:
$(x - 12)(x - 4) = 0$.
Thus,$x = 12$ or $x = 4$.
Therefore,the capacitances are $12\,\mu F$ and $4\,\mu F$.

(D) The reciprocal of resistance is defined as conductance.
Mathematically, $\text{Conductance} = \frac{1}{\text{Resistance}}$.
The $SI$ unit of conductance is Siemens $(S)$ or $\Omega^{-1}$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE)_{\max}$ of emitted electrons is given by:
$(KE)_{\max} = h\nu - \phi_0$
where $\nu$ is the frequency of incident light and $\phi_0$ is the work function of the metal.
Given that the threshold frequency is $\nu_0$,the work function is $\phi_0 = h\nu_0$.
The frequency of incident light is $\nu = 4\nu_0$.
Substituting these values into the equation:
$(KE)_{\max} = h(4\nu_0) - h\nu_0$
$(KE)_{\max} = 4h\nu_0 - h\nu_0$
$(KE)_{\max} = 3h\nu_0$.
Thus,the maximum kinetic energy is $3h\nu_0$.

(B) For a plane electromagnetic wave in free space,the relationship between the magnitude of the electric field $(E_0)$ and the magnetic field $(B_0)$ is given by $E_0 = c B_0$,where $c$ is the speed of light in free space.
Therefore,the ratio of the magnitude of the magnetic field to the electric field intensity is $\frac{B_0}{E_0} = \frac{1}{c}$.
Since the speed of light in free space is defined as $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$,we can also express this ratio as $\frac{B_0}{E_0} = \sqrt{\mu_0 \varepsilon_0}$.
However,in terms of the given options,the correct ratio is $\frac{1}{c}$.

(B) According to the Right-Hand Thumb Rule,when a current flows through an infinitely long,straight conductor,it generates a magnetic field around it. The magnetic field lines are concentric circles centered on the wire,with the plane of these circles being perpendicular to the length of the conductor.