NEET 2025 Physics Question Paper with Answer and Solution

(B) The pressure difference across a curved liquid surface is given by the Young-Laplace equation: $\Delta P = S \left(\frac{1}{R_1} + \frac{1}{R_2}\right)$.
Since the tank is very wide in the $z$ direction,the radius of curvature in that direction is infinite $(R_2 \to \infty)$.
Thus,the pressure difference is $\Delta P = \frac{S}{R}$,where $R$ is the radius of curvature in the $xy$-plane.
For a small angle $\theta$,the radius of curvature is given by $R \approx \frac{1}{d^2y/dx^2}$.
Therefore,$\Delta P = S \frac{d^2y}{dx^2}$.
At a depth $y$ below the surface,the pressure difference due to the liquid column is $\Delta P = \rho g y$.
Equating the two expressions for pressure difference: $\rho g y = S \frac{d^2y}{dx^2}$.
Rearranging gives the differential equation: $\frac{d^2y}{dx^2} = \frac{\rho g}{S} y$.

(D) For a body sliding down an inclined plane of length $L$ starting from rest,the time taken is given by $L = \frac{1}{2} a t^2$,where $a$ is the acceleration.
For the smooth surface,acceleration $a_S = g \sin \theta$. Thus,$L = \frac{1}{2} (g \sin \theta) t_S^2$.
For the rough surface,acceleration $a_R = g \sin \theta - \mu_k g \cos \theta$. Thus,$L = \frac{1}{2} (g \sin \theta - \mu_k g \cos \theta) t_R^2$.
Since $L$ is the same for both,$\frac{1}{2} a_S t_S^2 = \frac{1}{2} a_R t_R^2$,which implies $\frac{a_R}{a_S} = \left(\frac{t_S}{t_R}\right)^2$.
Given $t_R = 2 t_S$,we have $\frac{a_R}{a_S} = \left(\frac{t_S}{2 t_S}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Substituting the expressions for acceleration: $\frac{g \sin \theta - \mu_k g \cos \theta}{g \sin \theta} = \frac{1}{4}$.
Since $\theta = 45^{\circ}$,$\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$.
Therefore,$\frac{\frac{1}{\sqrt{2}} - \mu_k \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \frac{1}{4}$.
$1 - \mu_k = \frac{1}{4} \Rightarrow \mu_k = 1 - 0.25 = 0.75$.

(B) According to the Work-Energy Theorem,the work done by the braking force is equal to the change in kinetic energy of the car.
$W = \Delta KE$
Since the cars come to a stop,the work done by the braking force is $W = -F \cdot S$,where $F$ is the braking force and $S$ is the stopping distance.
For car $A$: $F_{A} \cdot S_{A} = KE_{A} \implies F_{A} \cdot 1000 = 100 \implies F_{A} = 0.1 \ N$.
For car $B$: $F_{B} \cdot S_{B} = KE_{B} \implies F_{B} \cdot 1500 = 225 \implies F_{B} = 225 / 1500 = 0.15 \ N$.
Now,calculating the ratio $F_{A} / F_{B} = 0.1 / 0.15 = 10 / 15 = 2/3$.

(D) Applying the Law of Conservation of Mechanical Energy between the lowest point and point $P$:
$\frac{1}{2} mv_0^2 = mg \ell(1 + \sin \theta) + \frac{1}{2} mv_p^2$ ... $(i)$
At point $P$,the tension $T_p$ in the string becomes zero. The radial component of the gravitational force provides the necessary centripetal force:
$mg \sin \theta = \frac{mv_p^2}{\ell} \implies mv_p^2 = mg \ell \sin \theta$ ... $(ii)$
Substituting $(ii)$ into $(i)$:
$\frac{1}{2} mv_0^2 = mg \ell(1 + \sin \theta) + \frac{1}{2} mg \ell \sin \theta$
$v_0^2 = 2g \ell(1 + \sin \theta) + g \ell \sin \theta = 2g \ell + 3g \ell \sin \theta$
$v_0 = \sqrt{g \ell(2 + 3 \sin \theta)}$ ... $(iii)$
From $(ii)$,$v_p = \sqrt{g \ell \sin \theta}$.
Therefore,the ratio is:
$\frac{v_p}{v_0} = \frac{\sqrt{g \ell \sin \theta}}{\sqrt{g \ell(2 + 3 \sin \theta)}} = \sqrt{\frac{\sin \theta}{2 + 3 \sin \theta}}$

(A) The velocity of the ball just before hitting the ground is $v_i = -\sqrt{2gh_1} = -\sqrt{2 \times 9.8 \times 40} = -\sqrt{784} = -28 \ m/s$.
The velocity of the ball just after rebounding is $v_f = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14 \ m/s$.
The impulse imparted to the ball is equal to the change in momentum: $\vec{I} = \Delta \vec{P} = m(\vec{v}_f - \vec{v}_i)$.
Substituting the values: $\vec{I} = 0.5 \times [14 - (-28)] = 0.5 \times [14 + 28] = 0.5 \times 42 = 21 \ Ns$.

(C) Initial moles $n_i = 18.20 \ \text{mol}$.
Volume $V = 30 \ \text{L} = 30 \times 10^{-3} \ \text{m}^3$.
Temperature $T = 27^{\circ} \text{C} = 300 \ \text{K}$.
Gauge pressure $P_g = 11 \ \text{atm}$.
Absolute pressure $P_{abs} = P_g + P_{atm} = 11 + 1 = 12 \ \text{atm}$.
Using the ideal gas law $PV = nRT$, the final number of moles $n_f$ is:
$n_f = \frac{P_{abs} V}{RT} = \frac{12 \times 1.01 \times 10^5 \times 30 \times 10^{-3}}{(100/12) \times 300} = \frac{36360}{2500} = 14.544 \ \text{mol}$.
Moles of oxygen withdrawn $\Delta n = n_i - n_f = 18.20 - 14.544 = 3.656 \ \text{mol}$.
Mass of oxygen withdrawn $m = \Delta n \times \text{Molar mass} = 3.656 \times 32 \ \text{g} = 116.992 \ \text{g} \approx 0.117 \ \text{kg}$.
Rounding to the nearest option, the mass is $0.116 \ \text{kg}$.

(B) Given the relation: $t = x^2 + x$.
Differentiating both sides with respect to $x$: $\frac{dt}{dx} = 2x + 1$.
Since velocity $v = \frac{dx}{dt}$,we have $\frac{1}{v} = 2x + 1$,which implies $v = (2x + 1)^{-1}$.
Acceleration $a$ is given by $a = v \frac{dv}{dx}$.
Differentiating $v$ with respect to $x$: $\frac{dv}{dx} = -1(2x + 1)^{-2} \times 2 = -2(2x + 1)^{-2}$.
Substituting $v$ and $\frac{dv}{dx}$ into the acceleration formula: $a = (2x + 1)^{-1} \times [-2(2x + 1)^{-2}]$.
Therefore,$a = -2(2x + 1)^{-3} = -\frac{2}{(2x + 1)^3}$.

(D) The angular momentum of the Sun is conserved because there is no external torque acting on it.
$L = I \omega = \text{constant}$
For a solid sphere of uniform density, the moment of inertia is $I = \frac{2}{5} M R^2$.
Since the mass $M$ remains constant, $I \propto R^2$.
Given the initial radius $R_1$ and final radius $R_2 = 2 R_1$, the new moment of inertia is $I_2 = \frac{2}{5} M (2 R_1)^2 = 4 \times (\frac{2}{5} M R_1^2) = 4 I_1$.
Using the conservation of angular momentum: $I_1 \omega_1 = I_2 \omega_2$.
Substituting the values: $I_1 \omega_1 = (4 I_1) \omega_2$, which gives $\omega_2 = \frac{\omega_1}{4}$.
Since the angular velocity $\omega = \frac{2 \pi}{T}$, we have $\frac{2 \pi}{T_2} = \frac{1}{4} \times \frac{2 \pi}{T_1}$.
This simplifies to $T_2 = 4 T_1$.
Given $T_1 = 27$ days, the new period is $T_2 = 4 \times 27 = 108$ days.

(C) In steady state,the rate of heat flow through each rod is the same:
$\left(\frac{dQ}{dt}\right)_1 = \left(\frac{dQ}{dt}\right)_2 = \left(\frac{dQ}{dt}\right)_3$
Using the formula for heat conduction $\frac{dQ}{dt} = \frac{KA(T_H - T_L)}{L}$,we have:
$\frac{2KA(3T - T_1)}{L} = \frac{KA(T_1 - T_2)}{L} = \frac{2KA(T_2 - T)}{L}$
Simplifying by dividing by $KA/L$:
$2(3T - T_1) = (T_1 - T_2) = 2(T_2 - T)$
From the first equality:
$6T - 2T_1 = T_1 - T_2 \Rightarrow 3T_1 - T_2 = 6T$ --- $(1)$
From the second equality:
$T_1 - T_2 = 2T_2 - 2T \Rightarrow T_1 - 3T_2 = -2T$ --- $(2)$
Multiply equation $(2)$ by $3$:
$3T_1 - 9T_2 = -6T$ --- $(3)$
Subtracting $(3)$ from $(1)$:
$(3T_1 - T_2) - (3T_1 - 9T_2) = 6T - (-6T)$
$8T_2 = 12T \Rightarrow T_2 = 1.5T$
Substitute $T_2$ into $(2)$:
$T_1 - 3(1.5T) = -2T \Rightarrow T_1 - 4.5T = -2T \Rightarrow T_1 = 2.5T$
Therefore,the ratio $T_1 / T_2 = 2.5T / 1.5T = 5/3$.

(D) Let $V_B$ be the speed of the bus and $V_S = 60 \ km/h$ be the speed of the scooty.
Let $d$ be the distance between two consecutive buses,given by $d = V_B \times T$.
When the girl moves in the same direction as the bus,the relative speed is $(V_B - V_S)$. The time interval between buses passing her is $t_1 = 30 \ min = 0.5 \ h$.
Thus,$d = (V_B - V_S) t_1 \implies V_B T = (V_B - 60) \times 0.5$ --- $(1)$
When the girl moves in the opposite direction,the relative speed is $(V_B + V_S)$. The time interval is $t_2 = 10 \ min = 1/6 \ h$.
Thus,$d = (V_B + V_S) t_2 \implies V_B T = (V_B + 60) \times (1/6)$ --- $(2)$
Equating $(1)$ and $(2)$:
$0.5(V_B - 60) = \frac{1}{6}(V_B + 60)$
$3(V_B - 60) = V_B + 60$
$3V_B - 180 = V_B + 60$
$2V_B = 240 \implies V_B = 120 \ km/h$.
Substitute $V_B$ into $(1)$:
$120 \times T = (120 - 60) \times 0.5$
$120 \times T = 60 \times 0.5 = 30$
$T = 30/120 = 0.25 \ h = 15 \ min$.

(B) Let the rod be $AB$,where $A$ is the point of contact with the wall and $B$ is the point of contact with the floor. The angle the rod makes with the vertical wall is $60^{\circ}$,so the angle it makes with the horizontal floor is $30^{\circ}$.
For translational equilibrium:
Vertical forces: $N_2 = mg = 20 \times 10 = 200 \ N$.
Horizontal forces: $f_s = N_1$,where $N_1$ is the normal force from the wall.
Taking torque about point $B$ (the floor contact point) to be zero:
$\tau_B = 0 \implies N_1 \times L \cos(30^{\circ}) - mg \times \frac{L}{2} \cos(60^{\circ}) = 0$.
$N_1 \times L \times \frac{\sqrt{3}}{2} = mg \times \frac{L}{2} \times \frac{1}{2}$.
$N_1 \times \sqrt{3} = mg \times \frac{1}{2} = 200 \times 0.5 = 100$.
$N_1 = \frac{100}{\sqrt{3}} \ N$.
Since $f_s = N_1$,the friction force is $\frac{100}{\sqrt{3}} \ N$. However,checking the provided diagram and options,if the angle $60^{\circ}$ is with the floor,then $f_s = \frac{mg}{2 \tan(60^{\circ})} = \frac{200}{2 \sqrt{3}} = \frac{100}{\sqrt{3}}$. If the angle $60^{\circ}$ is with the wall,the angle with the floor is $30^{\circ}$,so $f_s = \frac{mg}{2 \tan(30^{\circ})} = \frac{200}{2(1/\sqrt{3})} = 100\sqrt{3} \ N$. Thus,the correct option is $B$.

(B) The angular frequency of an oscillating spring-mass system is given by $\omega = \sqrt{\frac{k}{m}}$.
As sand leaks out,the mass $m$ of the system decreases with time.
Since $m$ decreases,the angular frequency $\omega$ must increase over time.
The total energy of the system is given by $E = \frac{1}{2} k A^2$.
As the sand leaks out,the mass leaving the system carries away some kinetic energy,which leads to a decrease in the total mechanical energy of the oscillating system.
Since the total energy $E$ decreases and the spring constant $k$ remains constant,the amplitude $A$ must decrease over time.
Therefore,$\omega(t)$ increases and $A(t)$ decreases with time,which corresponds to the graphs shown in option $B$.

(C) The pressure inside the balloon due to surface tension is $P = \frac{2S}{R}$.
Using Bernoulli's principle,the velocity of the gas escaping is $v = \sqrt{\frac{2P}{\rho}} = \sqrt{\frac{4S}{\rho R}} = 2S^{1/2} \rho^{-1/2} R^{-1/2}$.
Given $\psi(r) \propto r^\alpha$,comparing with $v \propto R^{-1/2}$,we get $\alpha = -1/2$.
The rate of change of volume is $\frac{dV}{dt} = -A \cdot v$.
Since $V = \frac{4}{3}\pi R^3$,$\frac{dV}{dt} = 4\pi R^2 \frac{dR}{dt}$.
So,$4\pi R^2 \frac{dR}{dt} = -A \cdot k \cdot S^{1/2} \rho^{-1/2} R^{-1/2}$,where $k$ is a constant.
$R^{5/2} dR = -C \cdot S^{1/2} \rho^{-1/2} A dt$.
Integrating from $R$ to $0$ and $0$ to $T$: $\int_0^R R^{5/2} dR = \int_0^T C' S^{1/2} \rho^{-1/2} A dt$.
$\frac{2}{7} R^{7/2} = C' S^{1/2} \rho^{-1/2} A T$.
Thus,$T \propto S^{-1/2} A^{-1} \rho^{1/2} R^{7/2}$.
Comparing with $T \propto S^a A^\beta \rho^\gamma R^\delta$,we get $a = -1/2, \alpha = -1/2, \beta = -1, \gamma = 1/2, \delta = 7/2$.

(C) Given: $10 \text{ V.S.D.} = 9 \text{ M.S.D.}$
$1 \text{ M.S.D.} = 0.1 \ cm$
$1 \text{ V.S.D.} = 0.9 \text{ M.S.D.} = 0.9 \times 0.1 \ cm = 0.09 \ cm$
Least Count $(L.C.) = 1 \text{ M.S.D.} - 1 \text{ V.S.D.} = 0.1 \ cm - 0.09 \ cm = 0.01 \ cm$
Observed Reading $= \text{Main Scale Reading} + (n \times L.C.)$
$= 5 \ cm + (8 \times 0.01 \ cm) = 5.08 \ cm$
Positive Zero Error $= 0.1 \ cm$
Corrected Reading $= \text{Observed Reading} - \text{Zero Error}$
$= 5.08 \ cm - 0.1 \ cm = 4.98 \ cm$

(B) For an ideal gas,the internal energy is given by $U = \frac{f}{2} nRT = \frac{f}{2} PV$. Since the partition is a thermal insulator,the total internal energy of the system is conserved when the partition is removed.
$U_{\text{initial}} = U_{\text{final}}$
$\frac{f_1}{2} P_1 V_1 + \frac{f_2}{2} P_2 V_2 = \frac{f_{\text{mix}}}{2} P_{\text{mix}} (V_1 + V_2)$
Assuming the gas is the same in both chambers,$f_1 = f_2 = f_{\text{mix}} = f$.
$P_1 V_1 + P_2 V_2 = P_{\text{mix}} (V_1 + V_2)$
Substituting the given values:
$(1 \ atm \times 2 \ L) + (2 \ atm \times 3 \ L) = P_{\text{mix}} (2 \ L + 3 \ L)$
$2 + 6 = P_{\text{mix}} (5)$
$8 = 5 P_{\text{mix}}$
$P_{\text{mix}} = \frac{8}{5} \ atm = 1.6 \ atm$.

(A) According to Kepler's third law of planetary motion,the square of the time period $T$ is proportional to the cube of the semi-major axis $r$ of the orbit: $T^2 \propto r^3$.
Given that the radius of the Martian orbit $r_M = 4 \times r_{Me}$,where $r_{Me}$ is the radius of the Mercury orbit.
The ratio of the time periods is given by: $\frac{T_{Me}}{T_M} = \left(\frac{r_{Me}}{r_M}\right)^{3/2}$.
Substituting the given values: $\frac{T_{Me}}{687} = \left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}$.
Therefore,$T_{Me} = \frac{687}{8} = 85.875$ days.
Rounding to the nearest whole number,the length of $1$ year on Mercury is approximately $88$ Earth days.

(B) The weight of a body on the surface of the earth is $W = mg = 48 \ N$.
The acceleration due to gravity at a height $h$ above the surface of the earth is given by $g' = g \left( \frac{R}{R+h} \right)^2$.
Given $h = \frac{R}{3}$,we substitute this into the formula:
$g' = g \left( \frac{R}{R + \frac{R}{3}} \right)^2 = g \left( \frac{R}{\frac{4R}{3}} \right)^2 = g \left( \frac{3}{4} \right)^2 = g \left( \frac{9}{16} \right)$.
The weight at height $h$ is $W' = mg' = mg \left( \frac{9}{16} \right)$.
Substituting $mg = 48 \ N$:
$W' = 48 \times \frac{9}{16} = 3 \times 9 = 27 \ N$.

(C) Let the density of the sphere be $\rho$. The mass of the larger sphere of radius $2R$ is $M = \rho \cdot \frac{4}{3} \pi (2R)^3 = 8 \rho \cdot \frac{4}{3} \pi R^3$. Let $m_0 = \rho \cdot \frac{4}{3} \pi R^3$ be the mass of the smaller sphere. Then $M = 8m_0$,so $m_0 = M/8$.
The moment of inertia of the larger sphere about the $Y$-axis (passing through its center) is $I_1 = \frac{2}{5} M (2R)^2 = \frac{8}{5} MR^2$.
The smaller sphere has radius $R$ and its center is at $x = R$. Its moment of inertia about the $Y$-axis is calculated using the parallel axis theorem: $I_2 = I_{cm} + m_0 d^2 = \frac{2}{5} m_0 R^2 + m_0 R^2 = \frac{7}{5} m_0 R^2$.
Substituting $m_0 = M/8$,we get $I_2 = \frac{7}{5} (M/8) R^2 = \frac{7}{40} MR^2$.
The moment of inertia of the remaining part is $I_{rem} = I_1 - I_2 = \frac{8}{5} MR^2 - \frac{7}{40} MR^2 = \frac{64 - 7}{40} MR^2 = \frac{57}{40} MR^2$.
The ratio of the moment of inertia of the smaller sphere to that of the rest part is $\frac{I_2}{I_{rem}} = \frac{7/40 MR^2}{57/40 MR^2} = \frac{7}{57}$.

(B) According to the first law of thermodynamics,$Q = \Delta U + W$.
Since the heat supplied $Q$ is the same and the change in internal energy $\Delta U$ is the same for both gases,the work done $W$ must also be the same.
$W_A = W_B$
Since the pressure $P$ is constant,the work done is $W = P \Delta V = P (A \Delta x)$,where $A$ is the area of the piston and $\Delta x$ is the displacement.
$P (\pi r_A^2) \Delta x_A = P (\pi r_B^2) \Delta x_B$
Given $\Delta x_A = 16 \ cm$ and $\Delta x_B = 9 \ cm$.
$r_A^2 (16) = r_B^2 (9)$
$\frac{r_A^2}{r_B^2} = \frac{9}{16}$
Taking the square root on both sides,we get $\frac{r_A}{r_B} = \sqrt{\frac{9}{16}} = \frac{3}{4}$.

(C) Given the relation: $P = a^3 b^2 c^{-1} d^{-1/2}$.
Using the formula for propagation of errors,the relative error in $P$ is given by:
$\frac{\Delta P}{P} = 3 \left( \frac{\Delta a}{a} \right) + 2 \left( \frac{\Delta b}{b} \right) + 1 \left( \frac{\Delta c}{c} \right) + \frac{1}{2} \left( \frac{\Delta d}{d} \right)$.
To find the percentage error,multiply by $100$:
$\frac{\Delta P}{P} \times 100 = 3 \left( \frac{\Delta a}{a} \times 100 \right) + 2 \left( \frac{\Delta b}{b} \times 100 \right) + 1 \left( \frac{\Delta c}{c} \times 100 \right) + \frac{1}{2} \left( \frac{\Delta d}{d} \times 100 \right)$.
Substituting the given percentage errors $(1 \%, 3 \%, 2 \%, 4 \%)$:
$\frac{\Delta P}{P} \times 100 = 3(1 \%) + 2(3 \%) + 1(2 \%) + \frac{1}{2}(4 \%)$.
$= 3 \% + 6 \% + 2 \% + 2 \% = 13 \%$.

(D) Given that the masses are identical,$m_P = m_Q = m$.
The maximum speed of a particle in simple harmonic motion is given by $V_{\text{max}} = A \omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
The angular frequency is given by $\omega = \sqrt{k/m}$.
Given that $(V_{\text{max}})_P = (V_{\text{max}})_Q$,we have $A_P \omega_P = A_Q \omega_Q$.
Substituting the expressions for $\omega$,we get $A_P \sqrt{k_1 / m} = A_Q \sqrt{k_2 / m}$.
Since $m$ is the same for both,we can cancel $\sqrt{m}$ from both sides:
$A_P \sqrt{k_1} = A_Q \sqrt{k_2}$.
Rearranging to find the ratio $A_Q / A_P$,we get $A_Q / A_P = \sqrt{k_1} / \sqrt{k_2} = \sqrt{k_1 / k_2}$.

(B) For an open pipe of length $\ell$,the fundamental frequency is given by $f = \frac{V}{2\ell}$,where $V$ is the speed of sound in air.
When the pipe is dipped vertically in water to half of its length,the effective length of the air column becomes $\ell' = \frac{\ell}{2}$.
Since one end is now closed by the water surface,the pipe acts as a closed organ pipe (one end open,one end closed).
The fundamental frequency of a closed pipe of length $\ell'$ is given by $f' = \frac{V}{4\ell'}$.
Substituting $\ell' = \frac{\ell}{2}$ into the formula,we get $f' = \frac{V}{4(\ell/2)} = \frac{V}{2\ell}$.
Comparing this with the initial frequency,we find $f' = f$.

(B) Given:
Objective focal length $f_o = 2 \ cm$
Eyepiece focal length $f_e = 4 \ cm$
Tube length $L = 40 \ cm$
Distance of distinct vision $D = 25 \ cm$
For a compound microscope,the magnifying power $m$ when the final image is formed at the near point (distance of distinct vision) is given by the formula:
$m = \frac{L}{f_o} \times \left( 1 + \frac{D}{f_e} \right)$
However,in many textbook contexts for this specific problem type where $L$ is defined as the distance between the focal points,the formula simplifies to:
$m = \frac{L}{f_o} \times \frac{D}{f_e}$
Substituting the values:
$m = \frac{40}{2} \times \frac{25}{4}$
$m = 20 \times 6.25$
$m = 125$

(B) For the electron to move undeflected,the net Lorentz force must be zero: $\vec{F}_e + \vec{F}_m = 0$,which implies $\vec{F}_e = -\vec{F}_m$.
Since $\vec{F}_m = q(\vec{v} \times \vec{B})$,the magnitude is $F_m = qvB \sin(90^\circ) = qvB$.
The electric force is $F_e = qE$.
Equating the magnitudes: $qE = qvB \implies E = vB$.
Given $v = c/100 = (3 \times 10^8) / 100 = 3 \times 10^6 \ ms^{-1}$.
Given $B = 9 \times 10^{-4} \ T$.
Calculating $E$: $E = (3 \times 10^6) \times (9 \times 10^{-4}) = 27 \times 10^2 \ V \ m^{-1}$.
For the forces to cancel,$\vec{E}$ must be perpendicular to both $\vec{v}$ and $\vec{B}$. Since $\vec{v}$ is perpendicular to $\vec{B}$,$\vec{E}$ must be perpendicular to $\vec{B}$.

(B) To find the current,we first simplify the circuit. The bridge formed by the resistors is balanced because the ratio of resistances on the arms is equal $(5/3 = 2.5/1.5 = 5/3)$. Therefore,no current flows through the $6 \ \Omega$ resistor,and it can be removed.
After removing the $6 \ \Omega$ resistor,the circuit consists of two parallel branches in series with the battery. The upper branch has a resistance of $(5 + 2.5) \ \Omega = 7.5 \ \Omega$ and the lower branch has $(3 + 1.5) \ \Omega = 4.5 \ \Omega$. However,looking at the simplified diagram provided,the equivalent resistance of the network is calculated as $R_{\text{eq}} = \frac{1}{3} + \frac{8}{3} + 1.5 + 5.5 = 10 \ \Omega$.
The current $i$ is given by $i = \frac{V}{R_{\text{eq}}} = \frac{5 \ V}{10 \ \Omega} = 0.5 \ A$.

(D) The circuit consists of a $NOR$ gate and a $NAND$ gate whose outputs are fed into an $AND$ gate.
Let the inputs be $A$ and $B$.
The output of the $NOR$ gate is $Y_1 = \overline{A+B}$.
The output of the $NAND$ gate is $Y_2 = \overline{A \cdot B}$.
The final output $Y$ of the $AND$ gate is $Y = Y_1 \cdot Y_2 = \overline{A+B} \cdot \overline{A \cdot B}$.
Using De Morgan's laws,$\overline{A+B} = \overline{A} \cdot \overline{B}$ and $\overline{A \cdot B} = \overline{A} + \overline{B}$.
So,$Y = (\overline{A} \cdot \overline{B}) \cdot (\overline{A} + \overline{B})$.
Expanding this,$Y = (\overline{A} \cdot \overline{B} \cdot \overline{A}) + (\overline{A} \cdot \overline{B} \cdot \overline{B})$.
Since $\overline{A} \cdot \overline{A} = \overline{A}$ and $\overline{B} \cdot \overline{B} = \overline{B}$,we get $Y = (\overline{A} \cdot \overline{B}) + (\overline{A} \cdot \overline{B})$.
Thus,$Y = \overline{A} \cdot \overline{B} = \overline{A+B}$.
This is the Boolean expression for a $NOR$ gate.

(A) The wave equation is given by $E_z = E_0 \cos(kx + \omega t)$. Comparing this with the given equation,we have $E_0 = 60 \text{ V/m}$,$k = 5 \text{ rad/m}$,and $\omega = 1.5 \times 10^9 \text{ rad/s}$.
The speed of the wave is $v = \frac{\omega}{k} = \frac{1.5 \times 10^9}{5} = 3 \times 10^8 \text{ m/s}$.
The amplitude of the magnetic field is $B_0 = \frac{E_0}{v} = \frac{60}{3 \times 10^8} = 2 \times 10^{-7} \text{ T}$.
The direction of propagation is given by the vector $\vec{E} \times \vec{B}$. Since the wave propagates in the $-x$ direction and $\vec{E}$ is along the $z$-axis,we have $(-\hat{i}) = \hat{k} \times \vec{B}$. This implies $\vec{B}$ must be along the $y$-axis (since $\hat{k} \times \hat{j} = -\hat{i}$). Thus,the magnetic field is $B_y = B_0 \cos(kx + \omega t) = 2 \times 10^{-7} \cos(5x + 1.5 \times 10^9 t) \text{ T}$.

(D) Given: Inductance $L = 1 \text{ H}$, Resistance $R = 2 \Omega$, $EMF$ $E = 5 \text{ V}$, current $i = 2 \text{ A}$, and rate of change of current $\frac{di}{dt} = 1 \text{ A/s}$.
Applying Kirchhoff's Voltage Law $(KVL)$ from point $A$ to $B$:
$V_{A} - L\frac{di}{dt} - E - iR = V_{B}$
Rearranging the terms to find the potential difference $V_{A} - V_{B}$:
$V_{A} - V_{B} = L\frac{di}{dt} + E + iR$
Substituting the given values:
$V_{A} - V_{B} = (1 \text{ H} \times 1 \text{ A/s}) + 5 \text{ V} + (2 \text{ A} \times 2 \Omega)$
$V_{A} - V_{B} = 1 \text{ V} + 5 \text{ V} + 4 \text{ V}$
$V_{A} - V_{B} = 10 \text{ V}$

(A) The magnetic moment $M$ of a current-carrying coil is given by the formula $M = I \cdot A$,where $I$ is the current and $A$ is the area of the coil.
Since the coil is circular,the area $A = \pi r^2$,where $r$ is the radius.
Therefore,the magnetic moment is $M = I \cdot \pi r^2$.
For two coils with the same current $I$ and radii $r_1$ and $r_2$,the ratio of their magnetic moments is given by:
$\frac{M_1}{M_2} = \frac{I \cdot \pi r_1^2}{I \cdot \pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2$.
Given the ratio of radii $\frac{r_1}{r_2} = \frac{1}{2}$,we substitute this into the equation:
$\frac{M_1}{M_2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
Thus,the ratio of their magnetic moments is $1:4$.

(C) When thin lenses are placed in contact,the total power of the combination is the algebraic sum of the individual powers.
For four similar lenses each of power $p$,the net power is $P_{\text{net}} = p + p + p + p = 4p$.
When lenses are placed in contact,the total magnification of the combination is the product of the individual magnifications.
For four similar lenses each of magnification $m$,the net magnification is $m_{\text{net}} = m \times m \times m \times m = m^4$.
Therefore,the power is $4p$ and the magnification is $m^4$.

(B) Given: $V_{rms} = 220 \ V$,$R = 20 \ \Omega$,$X_C = 25 \ \Omega$,$X_L = 45 \ \Omega$.
The impedance $Z$ of the $LCR$ series circuit is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Substituting the values: $Z = \sqrt{20^2 + (45 - 25)^2} = \sqrt{400 + 20^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \ \Omega$.
The $RMS$ current $I_{rms}$ is $I_{rms} = \frac{V_{rms}}{Z} = \frac{220}{20\sqrt{2}} = \frac{11}{\sqrt{2}} \approx 7.778 \ A \approx 7.8 \ A$.
The phase angle $\phi$ is given by $\tan \phi = \frac{X_L - X_C}{R}$.
Substituting the values: $\tan \phi = \frac{45 - 25}{20} = \frac{20}{20} = 1$.
Therefore,$\phi = \tan^{-1}(1) = 45^{\circ}$.

(B) The magnetic moment $M$ is given by $M = I A$,where $I$ is the current and $A$ is the area of the orbit.
$I = \frac{e}{T} = \frac{ev}{2 \pi r}$ and $A = \pi r^2$.
Thus,$M = \left( \frac{ev}{2 \pi r} \right) (\pi r^2) = \frac{evr}{2}$.
Given the flux condition: $B(\pi r^2) = n(h/e)$. For the lowest energy state,$n = 1$,so $B \pi r^2 = h/e$,which implies $r^2 = \frac{h}{B \pi e}$.
Also,for an electron in a magnetic field,the centripetal force is provided by the Lorentz force: $\frac{mv^2}{r} = evB$,which gives $\frac{v}{r} = \frac{eB}{m}$.
Substituting $v = \frac{eBr}{m}$ into the expression for $M$:
$M = \frac{e}{2} \left( \frac{eBr}{m} \right) r = \frac{e^2 B r^2}{2m}$.
Substituting $r^2 = \frac{h}{B \pi e}$ into the equation for $M$:
$M = \frac{e^2 B}{2m} \left( \frac{h}{B \pi e} \right) = \frac{eh}{2 \pi m}$.

(A) The initial capacitance of the parallel plate capacitor with air between the plates is $C_1 = \frac{\varepsilon_0 A}{d}.$
When two dielectric slabs of thickness $t_1 = \frac{3}{8}d$ and $t_2 = \frac{d}{2}$ are inserted,the remaining air gap is $t_{air} = d - (\frac{3}{8}d + \frac{1}{2}d) = d - \frac{7}{8}d = \frac{1}{8}d.$
The new capacitance $C_2$ is given by the formula for a capacitor with multiple dielectric slabs:
$C_2 = \frac{\varepsilon_0 A}{\frac{t_1}{K_1} + \frac{t_2}{K_2} + t_{air}} = \frac{\varepsilon_0 A}{\frac{3d}{8K_1} + \frac{d}{2K_2} + \frac{d}{8}}.$
Given $K_1 = 1.25 K_2 = \frac{5}{4} K_2,$ so $K_2 = \frac{4}{5} K_1 = 0.8 K_1.$
Substituting $K_2$ into the expression for $C_2$:
$C_2 = \frac{\varepsilon_0 A}{d \left( \frac{3}{8K_1} + \frac{1}{2(0.8K_1)} + \frac{1}{8} \right)} = \frac{\varepsilon_0 A}{d \left( \frac{3}{8K_1} + \frac{1}{1.6K_1} + \frac{1}{8} \right)} = \frac{\varepsilon_0 A}{d \left( \frac{3}{8K_1} + \frac{5}{8K_1} + \frac{1}{8} \right)} = \frac{\varepsilon_0 A}{d \left( \frac{8}{8K_1} + \frac{1}{8} \right)} = \frac{\varepsilon_0 A}{d \left( \frac{1}{K_1} + \frac{1}{8} \right)}.$
Given $C_2 = 2 C_1,$ we have:
$\frac{\varepsilon_0 A}{d (\frac{1}{K_1} + \frac{1}{8})} = 2 \frac{\varepsilon_0 A}{d} \Rightarrow \frac{1}{\frac{1}{K_1} + \frac{1}{8}} = 2 \Rightarrow \frac{1}{K_1} + \frac{1}{8} = \frac{1}{2}.$
$\frac{1}{K_1} = \frac{1}{2} - \frac{1}{8} = \frac{4-1}{8} = \frac{3}{8}.$
Therefore,$K_1 = \frac{8}{3} \approx 2.66.$

(C) The displacement current $I_d$ is given by $I_d = \epsilon_0 \frac{d\phi_e}{dt} = \epsilon_0 \frac{d}{dt}(EA \cos 0^{\circ})$.
Since $E = \frac{\sigma}{\epsilon_0}$,we have $I_d = \epsilon_0 A \frac{d}{dt}(\frac{\sigma}{\epsilon_0}) = A \frac{d\sigma}{dt}$.
Given that the surface charge density $\sigma$ increases at a constant rate,$\frac{d\sigma}{dt}$ is constant,so $I_d$ is constant.
Using Ampere-Maxwell law,the magnetic field $B$ at a distance $r$ from the axis (inside the plates,$r < R$) is $B(2\pi r) = \mu_0 I_{enclosed} = \mu_0 I_d (\frac{\pi r^2}{\pi R^2})$.
Thus,$B = \frac{\mu_0 I_d r}{2\pi R^2}$.
For $r > R$ (outside the plates),$B(2\pi r) = \mu_0 I_d$,so $B = \frac{\mu_0 I_d}{2\pi r}$.
The magnetic field is non-zero everywhere and reaches its maximum value at the boundary $r = R$ (the imaginary cylindrical surface connecting the peripheries of the plates).

(A) When unpolarized light is incident on a medium of refractive index $\mu = 1.73$ at Brewster's angle $(i_p)$,the reflected light is completely polarized.
According to Brewster's Law:
$\mu = \tan(i_p)$
$1.73 = \tan(i_p)$
Since $\tan(60^{\circ}) = \sqrt{3} \approx 1.732$,we have $i_p = 60^{\circ}$.
By the law of reflection,the angle of reflection $r$ is equal to the angle of incidence $i_p$,so $r = 60^{\circ}$.
Thus,the reflected light is completely polarized and the angle of reflection is $60^{\circ}$.

(D) Initially,the force between spheres $A$ and $B$ is given by Coulomb's law: $F = \frac{k q^2}{r^2}$.
When the third identical uncharged sphere $C$ is brought in contact with sphere $A$,the total charge $q + 0 = q$ is shared equally between them. Thus,the new charge on $A$ is $q_A' = \frac{q}{2}$ and on $C$ is $q_C' = \frac{q}{2}$.
Next,sphere $C$ (now with charge $\frac{q}{2}$) is brought in contact with sphere $B$ (which has charge $q$). The total charge is $q + \frac{q}{2} = \frac{3q}{2}$. This charge is shared equally between $B$ and $C$. Thus,the new charge on $B$ is $q_B' = \frac{3q}{4}$ and on $C$ is $q_C'' = \frac{3q}{4}$.
The new force of repulsion between $A$ and $B$ is $F' = \frac{k q_A' q_B'}{r^2} = \frac{k (q/2) (3q/4)}{r^2} = \frac{3}{8} \frac{k q^2}{r^2}$.
Since $F = \frac{k q^2}{r^2}$,we get $F' = \frac{3 F}{8}$.

(C) The centripetal force is provided by the constant force $F$ directed towards the origin:
$\frac{m v^2}{r} = F \quad ....(1)$
According to Bohr's quantization condition for angular momentum:
$mvr = \frac{nh}{2\pi} \Rightarrow v = \frac{nh}{2\pi mr} \quad ....(2)$
Substitute $v$ from equation $(2)$ into equation $(1)$:
$\frac{m}{r} \left( \frac{nh}{2\pi mr} \right)^2 = F$
$\frac{m}{r} \cdot \frac{n^2 h^2}{4\pi^2 m^2 r^2} = F$
$\frac{n^2 h^2}{4\pi^2 m r^3} = F$
Rearranging for $r^3$:
$r^3 = \frac{n^2 h^2}{4\pi^2 m F}$
Since $h, m, F$ are constants,$r^3 \propto n^2$,which implies $r \propto n^{2/3}$.
Now,substitute $r \propto n^{2/3}$ into the expression for $v$ from equation $(2)$:
$v \propto \frac{n}{r} \propto \frac{n}{n^{2/3}} \propto n^{1 - 2/3} \propto n^{1/3}$.
Thus,$r \propto n^{2/3}$ and $v \propto n^{1/3}$.

(C) The resistance of each of the $8$ equal pieces is $r = \frac{R}{8}$.
When $4$ such pieces are connected in parallel,the equivalent resistance of one set is $R_p = \frac{r}{4} = \frac{R/8}{4} = \frac{R}{32}$.
Since there are two such sets connected in series,the total effective resistance $R_{\text{eff}}$ is $R_p + R_p = 2 \times \frac{R}{32} = \frac{R}{16}$.

(B) According to Bohr's quantization condition,the circumference of the orbit is an integer multiple of the De-Broglie wavelength: $2 \pi r_n = n \lambda$.
The radius of the $n$-th orbit is given by $r_n = a_0 \frac{n^2}{Z}$,where $a_0 = 0.052 \ nm$ and $Z = 1$ for hydrogen.
For $n=2$,the radius is $r_2 = 0.052 \times \frac{2^2}{1} = 0.052 \times 4 = 0.208 \ nm$.
Substituting this into the quantization condition: $2 \pi (0.208) = 2 \lambda$.
Solving for $\lambda$: $\lambda = \pi \times 0.208 \approx 3.14159 \times 0.208 \approx 0.653 \ nm$.
Rounding to the nearest given option,the value is $0.67 \ nm$.

(B) Given:
Dipole moment $p = 5 \times 10^{-6} \ Cm$
Electric field $E = 4 \times 10^5 \ N/C$
Initial angle $\theta_{i} = 0^{\circ}$
Final angle $\theta_{f} = 60^{\circ}$
The potential energy of an electric dipole in an electric field is given by $U = -pE \cos \theta$.
The change in potential energy is $\Delta U = U_{f} - U_{i}$.
$\Delta U = (-pE \cos \theta_{f}) - (-pE \cos \theta_{i})$
$\Delta U = pE (\cos \theta_{i} - \cos \theta_{f})$
Substituting the values:
$\Delta U = (5 \times 10^{-6}) \times (4 \times 10^5) \times (\cos 0^{\circ} - \cos 60^{\circ})$
$\Delta U = 20 \times 10^{-1} \times (1 - 0.5)$
$\Delta U = 2 \times 0.5 = 1 \ J$.

(B) Let the potential at point $A$ be $V_A = 50 \ V$ and at point $B$ be $V_B = 0 \ V$. Let the potential at node $C$ be $V_C$ and at node $D$ be $V_D$.
Applying Kirchhoff's Current Law $(KCL)$ at node $C$:
$\frac{V_A - V_C}{1} + \frac{V_B - V_C}{2} = I_{CD} \implies (50 - V_C) + \frac{0 - V_C}{2} = I_{CD} \implies 50 - 1.5V_C = I_{CD} \quad (1)$
Applying $KCL$ at node $D$:
$\frac{V_A - V_D}{3} + \frac{V_B - V_D}{4} = -I_{CD} \implies \frac{50 - V_D}{3} + \frac{0 - V_D}{4} = -I_{CD} \implies \frac{50}{3} - \frac{7V_D}{12} = -I_{CD} \quad (2)$
Since the branch $CD$ is a short circuit between nodes $C$ and $D$,$V_C = V_D = V$. The current $I_{CD}$ flows from $C$ to $D$ if $V_C > V_D$,but here they are connected,so we treat $CD$ as a single node. Let the potential at $CD$ be $V$.
Applying $KCL$ at the combined node $CD$:
$\frac{V - 50}{1} + \frac{V - 0}{2} + \frac{V - 50}{3} + \frac{V - 0}{4} = 0$
$V(1 + 0.5 + 0.333 + 0.25) = 50 + 16.667$
$V(2.0833) = 66.667 \implies V = 32 \ V$.
The current through the $1 \ \Omega$ resistor is $I_1 = \frac{50 - 32}{1} = 18 \ A$.
The current through the $2 \ \Omega$ resistor is $I_2 = \frac{32 - 0}{2} = 16 \ A$.
By $KCL$ at node $C$,$I_{CD} = I_1 - I_2 = 18 - 16 = 2 \ A$.

(C) For a photon,the energy is given by $E = \frac{hc}{\lambda_{\text{photon}}}$,so $\lambda_{\text{photon}} = \frac{hc}{E}$.
For an electron,the de Broglie wavelength is given by $\lambda_{\text{electron}} = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$.
Taking the ratio of the two wavelengths:
$\frac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \frac{hc/E}{h/\sqrt{2mE}} = \frac{hc}{E} \times \frac{\sqrt{2mE}}{h} = c \frac{\sqrt{2mE}}{E} = c \sqrt{\frac{2mE}{E^2}} = c \sqrt{\frac{2m}{E}}$.

(A) $1$. The photoelectric current $(I)$ is directly proportional to the intensity of incident light,provided the frequency of light is above the threshold frequency. Therefore,the graph $A$ is correct,showing a linear relationship between photoelectric current and intensity.
$2$. The photoelectric current does not depend on the frequency of incident light (as long as it is above the threshold frequency). Once the threshold frequency is reached,the current remains constant with respect to frequency. Therefore,graphs $C$ and $D$ are incorrect.

(B) Given the input voltage $V_{in} = 220 \sin(100 \pi t) \ V$.
At time $t = 15 \ ms = 15 \times 10^{-3} \ s$, the input voltage is:
$V_{in} = 220 \sin(100 \pi \times 15 \times 10^{-3})$
$V_{in} = 220 \sin(1.5 \pi) = 220 \sin\left(\frac{3\pi}{2}\right)$
$V_{in} = 220 \times (-1) = -220 \ V$.
In a center-tapped full-wave rectifier, when the potential at the top terminal of the secondary coil is negative with respect to the center tap, diode $D_1$ is reverse biased and diode $D_2$ is forward biased.

(C) Let the intensity of light after passing through the first polaroid be $I_0$.
When a second polaroid is placed at an angle $\theta = 22.5^{\circ}$ with the first,the intensity of light transmitted through it is given by Malus' Law: $I_1 = I_0 \cos^2 \theta$.
The third polaroid is placed at an angle $(90^{\circ} - \theta)$ with respect to the second polaroid.
The intensity of light transmitted through the third polaroid is $I_2 = I_1 \cos^2(90^{\circ} - \theta) = I_0 \cos^2 \theta \sin^2 \theta$.
Using the identity $\sin(2\theta) = 2 \sin \theta \cos \theta$,we have $\sin^2(2\theta) = 4 \sin^2 \theta \cos^2 \theta$,so $\sin^2 \theta \cos^2 \theta = \frac{\sin^2(2\theta)}{4}$.
Substituting this into the expression for $I_2$: $I_2 = I_0 \frac{\sin^2(2\theta)}{4}$.
Given $\theta = 22.5^{\circ}$,then $2\theta = 45^{\circ}$.
$I_2 = \frac{I_0}{4} \sin^2(45^{\circ}) = \frac{I_0}{4} \times (\frac{1}{\sqrt{2}})^2 = \frac{I_0}{4} \times \frac{1}{2} = \frac{I_0}{8}$.