KCET 2025 Physics Question Paper with Answer and Solution

(A) The coordinates of the vertices are $A(0, 0)$,$B(1, 0)$,and $C(1/2, \sqrt{3}/2)$.
The masses are $m_A = 1 \ kg$,$m_B = 2 \ kg$,and $m_C = 3 \ kg$.
The total mass $M = m_A + m_B + m_C = 1 + 2 + 3 = 6 \ kg$.
The $x$-coordinate of the centre of mass is given by:
$x_{cm} = \frac{m_A x_A + m_B x_B + m_C x_C}{M} = \frac{1(0) + 2(1) + 3(1/2)}{6} = \frac{0 + 2 + 1.5}{6} = \frac{3.5}{6} = \frac{7}{12}$.
The $y$-coordinate of the centre of mass is given by:
$y_{cm} = \frac{m_A y_A + m_B y_B + m_C y_C}{M} = \frac{1(0) + 2(0) + 3(\sqrt{3}/2)}{6} = \frac{3\sqrt{3}/2}{6} = \frac{3\sqrt{3}}{12}$.
Thus,the centre of mass is $\left(\frac{7}{12}, \frac{3\sqrt{3}}{12}\right)$.

(C) $I$. In an elastic collision,the total kinetic energy of the system is conserved,meaning the initial kinetic energy equals the final kinetic energy.
$II$. Linear momentum is conserved in all types of collisions (elastic or inelastic) provided no external force acts on the system.
$III$. During the collision interval $\Delta t$,the bodies deform,and part of the kinetic energy is temporarily converted into elastic potential energy. Therefore,the kinetic energy is not conserved during the collision process itself.
$\therefore$ All three statements are correct.

(B) According to Kepler's second law,the angular momentum of a planet revolving around the Sun is conserved because the gravitational force is a central force.
Therefore,$L_p = L_a$.
Perihelion is the point in the orbit closest to the Sun,and aphelion is the point farthest from the Sun.
Therefore,$r_p < r_a$.
Since angular momentum $L = mvr$ is conserved,we have $m v_p r_p = m v_a r_a$.
Since $r_p < r_a$,it follows that $v_p > v_a$.
Thus,the correct relation is $r_p < r_a, v_p > v_a, L_a = L_p$.

(A) The total energy $(E)$ of a satellite in a circular orbit is given by the formula:
$E = -\frac{GMm}{2r}$
Where:
- $G$ is the universal gravitational constant.
- $M$ is the mass of the Earth.
- $m$ is the mass of the satellite.
- $r = R + h$ is the distance from the center of the Earth.
Substituting $r = R + h$ into the energy equation,we get:
$E = -\frac{GMm}{2(R+h)}$
Since $G, M, m,$ and $2$ are constants,the total energy $E$ is proportional to $-\frac{1}{R+h}$.
Thus,the total energy varies as $-\frac{1}{R+h}$.

(C) The mean kinetic energy $(E)$ of an ideal gas is directly proportional to its absolute temperature ($T$ in Kelvin),given by the formula $E = \frac{3}{2} k_B T$.
Therefore,$E \propto T$.
Given initial temperature $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$.
Given final temperature $T_2 = 327^{\circ} C = 327 + 273 = 600 \ K$.
Using the proportionality ratio: $\frac{E_2}{E_1} = \frac{T_2}{T_1}$.
Substituting the values: $E_2 = E_1 \times \frac{600}{300}$.
$E_2 = 2 E_1$.

(A) Let $T_1$ be the tension in the horizontal string and $T_2$ be the tension in the inclined string at angle $\theta$.
For the hanging block of mass $M$,the vertical equilibrium gives $T_2 \sin \theta = Mg$ ... $(i)$.
For the point $O$,the horizontal equilibrium gives $T_1 = T_2 \cos \theta$ ... (ii).
For the block on the floor,the horizontal force is $T_1$ and the limiting friction is $f_s = \mu N = \mu Mg$.
For equilibrium,$T_1 = f_s = \mu Mg$ ... (iii).
From (ii) and (iii),$T_2 \cos \theta = \mu Mg$ ... (iv).
Dividing $(i)$ by (iv),we get $\frac{T_2 \sin \theta}{T_2 \cos \theta} = \frac{Mg}{\mu Mg}$.
Therefore,$\tan \theta = \frac{1}{\mu}$,which implies $\mu = \cot \theta$.

(D) From the velocity-time graph,the acceleration $a$ of the block is the slope of the line:
$a = \frac{v - u}{t} = \frac{20 - 0}{3} = \frac{20}{3} \text{ m/s}^2$
Since the block is moving,the friction acting on it is kinetic friction,$f_k = \mu_k N = \mu_k mg$.
Given $\mu_k = 0.25$ and $g = 10 \text{ m/s}^2$,the kinetic friction is $f_k = 0.25 \times m \times 10 = 2.5m$.
Applying Newton's second law of motion,$F - f_k = ma$:
$20 - 2.5m = m \times \frac{20}{3}$
$20 = m \left( \frac{20}{3} + 2.5 \right)$
$20 = m \left( \frac{20 + 7.5}{3} \right) = m \left( \frac{27.5}{3} \right)$
$m = \frac{20 \times 3}{27.5} = \frac{60}{27.5} \approx 2.18 \text{ kg}$.
Rounding to the nearest value,the mass is $2.2 \text{ kg}$.

(C) Using the equation of continuity,$A_1 V_1 = A_2 V_2$.
Given $A_1 = 10 \text{ cm}^2$,$V_1 = 1 \text{ ms}^{-1}$,and $A_2 = 5 \text{ cm}^2$.
$V_2 = (A_1 V_1) / A_2 = (10 \times 1) / 5 = 2 \text{ ms}^{-1}$.
Using Bernoulli's equation for a horizontal pipe,$P_1 + 0.5 \rho V_1^2 = P_2 + 0.5 \rho V_2^2$.
$P_2 = P_1 + 0.5 \rho (V_1^2 - V_2^2)$.
$P_2 = 2000 + 0.5 \times 1000 \times (1^2 - 2^2)$.
$P_2 = 2000 + 500 \times (1 - 4) = 2000 - 1500 = 500 \text{ Pa}$.

(A) First,calculate the average time $t_{\text{avg}}$ taken by the ball to sink: $t_{\text{avg}} = \frac{2.75 + 2.65 + 2.70}{3} = 2.7 \,s$.
The terminal velocity $v_t$ is given by $v_t = \frac{h}{t_{\text{avg}}} = \frac{0.9}{2.7} = \frac{1}{3} \,m/s$.
According to Stokes' Law,the terminal velocity is $v_t = \frac{2r^2g(\rho_s - \rho_l)}{9\eta}$,where $\eta$ is the coefficient of viscosity.
Rearranging for $\eta$: $\eta = \frac{2r^2g(\rho_s - \rho_l)}{9v_t}$.
Given $r = \sqrt{3} \times 10^{-3} \,m$,so $r^2 = 3 \times 10^{-6} \,m^2$.
Given $(\rho_s - \rho_l) = 7000 \,kg/m^3$ and $g = 10 \,m/s^2$.
Substituting the values: $\eta = \frac{2 \times (3 \times 10^{-6}) \times 10 \times 7000}{9 \times (1/3)} = \frac{6 \times 10^{-5} \times 7000}{3} = 2 \times 10^{-5} \times 7000 = 0.14 \,Pa \cdot s$.

(D) The formula for Young's modulus $Y$ is given by $Y = \frac{F/A}{\Delta \ell / \ell}$,where $F$ is the force,$A$ is the cross-sectional area,$\ell$ is the original length,and $\Delta \ell$ is the change in length.
Rearranging for $\Delta \ell$,we get $\Delta \ell = \frac{F \ell}{AY}$.
Since the area $A = \pi r^2 = \pi (d/2)^2$,where $d$ is the diameter,we have $\Delta \ell = \frac{F \ell}{\pi (d/2)^2 Y} = \frac{4F \ell}{\pi d^2 Y}$.
Given that the wires are made of the same material ($Y$ is constant) and are stretched by the same force ($F$ is constant),the change in length is proportional to $\frac{\ell}{d^2}$.
Therefore,$\frac{\Delta \ell_A}{\Delta \ell_B} = \left( \frac{\ell_A}{\ell_B} \right) \times \left( \frac{d_B}{d_A} \right)^2$.
Given $\frac{\ell_A}{\ell_B} = \frac{1}{3}$ and $\frac{d_A}{d_B} = \frac{1}{2}$ (which implies $\frac{d_B}{d_A} = 2$),we substitute these values:
$\frac{\Delta \ell_A}{\Delta \ell_B} = \left( \frac{1}{3} \right) \times (2)^2 = \frac{1}{3} \times 4 = \frac{4}{3}$.
Thus,the ratio of the increase in their lengths is $4: 3$.

(C) Let the first stone fall for time $t_0$. Its distance covered is $S_1 = \frac{1}{2} g t_0^2$.
The second stone starts $\Delta t$ seconds later,so it falls for time $(t_0 - \Delta t)$. Its distance covered is $S_2 = \frac{1}{2} g (t_0 - \Delta t)^2$.
The separation distance $H$ is given by $H = S_1 - S_2$.
Substituting the expressions: $H = \frac{1}{2} g t_0^2 - \frac{1}{2} g (t_0 - \Delta t)^2$.
Expanding the term: $H = \frac{1}{2} g [t_0^2 - (t_0^2 - 2 t_0 \Delta t + \Delta t^2)]$.
$H = \frac{1}{2} g [2 t_0 \Delta t - \Delta t^2]$.
$H = g t_0 \Delta t - \frac{1}{2} g \Delta t^2$.
Rearranging for $t_0$: $g t_0 \Delta t = H + \frac{1}{2} g \Delta t^2$.
Dividing by $g \Delta t$: $t_0 = \frac{H}{g \Delta t} + \frac{\Delta t}{2}$.

(B) In projectile motion,the horizontal component of velocity $(v_x)$ remains constant with respect to time and position.
This is because there is no horizontal acceleration $(a_x = 0)$ acting on the projectile,assuming air resistance is neglected.
The vertical component of velocity $(v_y)$ changes due to the acceleration due to gravity $(g)$.
Therefore,the horizontal component of velocity is the only quantity among the options that remains constant throughout the motion.

(C) The equation of the trajectory is $(x-2)^2 + y^2 = 25$. This represents a circle with center at $(2, 0)$ and radius $R = 5 \text{ m}$.
In uniform circular motion,the acceleration is purely centripetal,directed towards the center of the circle.
The magnitude of centripetal acceleration is $a_c = \frac{v^2}{R}$.
Given $v = 2 \text{ m/s}$ and $R = 5 \text{ m}$,we have $a_c = \frac{2^2}{5} = \frac{4}{5} = 0.8 \text{ m/s}^2$.
The particle attains the lowest $y$ coordinate at the point $(2, -5)$.
At this point,the center of the circle $(2, 0)$ is directly above the particle (along the positive $y$-axis).
Therefore,the centripetal acceleration vector is directed along the positive $y$-axis,which is $0.8 \hat{j} \text{ m/s}^2$.

(B) In $SHM$,the kinetic energy $K(x)$ and potential energy $U(x)$ are given by:
$K(x) = \frac{1}{2}m\omega^2(A^2 - x^2)$
$U(x) = \frac{1}{2}m\omega^2x^2$
At the point $x = x_0$,the kinetic energy and potential energy are equal,i.e.,$K(x_0) = U(x_0)$.
$\frac{1}{2}m\omega^2(A^2 - x_0^2) = \frac{1}{2}m\omega^2x_0^2$
$A^2 - x_0^2 = x_0^2$
$A^2 = 2x_0^2$
$x_0^2 = \frac{A^2}{2}$
$|x_0| = \frac{A}{\sqrt{2}}$

(A) Given: $I_1 = 4 \ kg \ m^2$,$r_1 = 0.2 \ m$,$I_2 = 20 \ kg \ m^2$,$r_2 = 0.3 \ m$,$\tau_1 = 10 \ Nm$.
Since the belt is non-slipping,the linear acceleration of the rims is the same: $a = \alpha_1 r_1 = \alpha_2 r_2$.
For the smaller wheel: $\tau_1 = I_1 \alpha_1 \implies 10 = 4 \alpha_1 \implies \alpha_1 = 2.5 = 5/2 \ rad/s^2$.
For the larger wheel: $\alpha_2 = \alpha_1 (r_1 / r_2) = (5/2) \times (0.2 / 0.3) = (5/2) \times (2/3) = 5/3 \ rad/s^2$.
The tension $T$ in the belt provides the torque on the larger wheel: $\tau_2 = T r_2$ and $\tau_1 - T r_1 = I_1 \alpha_1$. However,using the relation $\tau_2 = I_2 \alpha_2$ is more direct: $\tau_2 = 20 \times (5/3) = 100/3 \ Nm$.
Matching: $a \to 3, b \to 2, c \to 1$.

(A) Let the thermal resistance of each rod be $R$ and the temperature of the junction be $T_X$.
According to the principle of conservation of energy,the total heat flowing into the junction must equal the total heat flowing out of the junction.
Here,heat flows from the two rods at $90^{\circ} C$ towards the junction $X$,and then from the junction $X$ towards the rod at $0^{\circ} C$.
Let $H_1$ and $H_2$ be the heat currents from the two rods at $90^{\circ} C$,and $H$ be the heat current flowing towards the $0^{\circ} C$ end.
$H = H_1 + H_2$
Using the formula for heat current $H = \frac{\Delta T}{R}$,we have:
$\frac{T_X - 0}{R} = \frac{90 - T_X}{R} + \frac{90 - T_X}{R}$
Since the rods are identical,the thermal resistance $R$ is the same for all.
$T_X = (90 - T_X) + (90 - T_X)$
$T_X = 180 - 2T_X$
$3T_X = 180$
$T_X = 60^{\circ} C$
Thus,the temperature at the junction $X$ is $60^{\circ} C$.

(B) According to the first law of thermodynamics,the change in internal energy $\Delta U$ is given by $\Delta U = Q - W$.
Since internal energy is a state function,it depends only on the initial and final states of the system.
For both paths $1$ and $2$,the system moves from state $A$ to state $B$,so the change in internal energy is the same for both paths: $\Delta U_1 = \Delta U_2$.
Therefore,$Q_1 - W_1 = Q_2 - W_2$.

(C) Dimensional analysis is a powerful tool for checking the consistency of physical equations,but it has limitations. It cannot determine dimensionless constants (like $6 \pi$) or the presence of transcendental functions (like trigonometric,exponential,or logarithmic functions) because their arguments must be dimensionless.
$1$. $x=A \cos (\omega t)$: While dimensionally consistent,the presence of the trigonometric function $\cos$ cannot be derived using dimensional analysis.
$2$. $N=N_0 e^{-\lambda t}$: While dimensionally consistent,the exponential form cannot be derived using dimensional analysis.
$3$. $F=6 \pi \eta r \nu$ (Stokes' Law): The dimensions of the right-hand side are $[M L^{-1} T^{-1}] \cdot [L] \cdot [L T^{-1}] = [M L T^{-2}]$,which is the dimension of force. Dimensional analysis can relate $F$ to $\eta$,$r$,and $\nu$ as $F \propto \eta^a r^b \nu^c$. By comparing dimensions,we can find $a=1, b=1, c=1$,leading to $F \propto \eta r \nu$. Thus,the functional dependence can be deduced.
Therefore,option $C$ is the correct choice.

(B) In a transverse wave,the particles of the medium oscillate in a direction perpendicular to the direction of wave propagation.
Since the wave velocity is along the direction of propagation and the particle velocity is perpendicular to it,the angle between them is $\frac{\pi}{2}$ radian.
This holds true for all positions of the particle except at the mean position,where the particle velocity is momentarily zero.

(D) According to the Work-Energy Theorem,the net work done on a body is equal to the change in its kinetic energy.
$W_{\text{net}} = \Delta KE = KE_f - KE_i$
Given mass $m = 0.25 \ kg$,speed $v = k x^{3/2}$,and $k = 2$.
At $x = 0$,$v_i = 2(0)^{3/2} = 0 \ m/s$,so $KE_i = 0 \ J$.
At $x = 2 \ m$,$v_f = 2(2)^{3/2} = 2 \times 2 \sqrt{2} = 4\sqrt{2} \ m/s$.
$KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 0.25 \times (4\sqrt{2})^2$
$KE_f = \frac{1}{2} \times 0.25 \times 16 \times 2 = 0.25 \times 16 = 4 \ J$.
Therefore,$W_{\text{net}} = 4 \ J - 0 \ J = 4 \ J$.

(A) Given: $X_L = 24 \ \Omega$,$X_C = 16 \ \Omega$,$R = 6 \ \Omega$,$V = 100 \ V$.
In a series $LCR$ circuit,the impedance $Z$ is given by:
$Z = \sqrt{R^2 + (X_L - X_C)^2}$
$Z = \sqrt{6^2 + (24 - 16)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \ \Omega$.
The current $i$ in the circuit is:
$i = \frac{V}{Z} = \frac{100}{10} = 10 \ A$.
The potential difference across the series combination of the inductor and capacitor is given by:
$V_{LC} = i |X_L - X_C|$
$V_{LC} = 10 \times |24 - 16| = 10 \times 8 = 80 \ V$.

(D) In domestic electric mains supply,the power is transmitted as Alternating Current $(AC)$.
This means that both the voltage and the current oscillate sinusoidally with time.
Therefore,the domestic supply consists of $AC$ voltage and $AC$ current.

(C) The given equation is $V = V_0 \sin(\omega t)$,where $V_0 = 311 \ V$ and $\omega = 314 \ rad/s$.
The $rms$ value of voltage is given by $V_{rms} = \frac{V_0}{\sqrt{2}}$.
$V_{rms} = \frac{311}{1.414} \approx 220 \ V$.
The angular frequency is $\omega = 2\pi f$.
$314 = 2 \times 3.14 \times f$.
$314 = 6.28 \times f$.
$f = \frac{314}{6.28} = 50 \ Hz$.
Thus,the $rms$ voltage is $220 \ V$ and the frequency is $50 \ Hz$.

(B) The radius of the $n^{th}$ Bohr orbit is given by the formula $r_n = n^2 r_1$, where $r_1$ is the radius of the first orbit $(r_1 = r)$.
For the second Bohr orbit, $n = 2$.
Substituting the value of $n$ into the formula:
$r_2 = (2)^2 \times r$
$r_2 = 4 r$
Therefore, the radius of the second Bohr orbit is $4 r$.

(B) The circuit consists of a cell with $EMF$ $E = 2 \ V$ and internal resistance $r = 0.1 \ \Omega$,connected in series with an external resistor $R = 3.9 \ \Omega$.
First,calculate the total resistance of the circuit: $R_{total} = R + r = 3.9 \ \Omega + 0.1 \ \Omega = 4.0 \ \Omega$.
Next,calculate the current $i$ flowing through the circuit using Ohm's law: $i = \frac{E}{R_{total}} = \frac{2 \ V}{4.0 \ \Omega} = 0.5 \ A$.
The terminal voltage $V$ across the cell is given by the formula: $V = E - i \times r$.
Substituting the values: $V = 2 \ V - (0.5 \ A \times 0.1 \ \Omega) = 2 \ V - 0.05 \ V = 1.95 \ V$.

(D) The equivalent emf of two cells connected in parallel is given by the formula:
$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}$
This can be rewritten as:
$E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2} = \frac{E_1 r_2 + E_1 r_1 - E_1 r_1 + E_2 r_1}{r_1 + r_2} = \frac{E_1(r_1 + r_2) + r_1(E_2 - E_1)}{r_1 + r_2} = E_1 + \frac{r_1}{r_1 + r_2}(E_2 - E_1)$
Since $E_2 > E_1$ and $r_1, r_2 > 0$,it is clear that $E_{eq} > E_1$.
Similarly,$E_{eq} = E_2 - \frac{r_2}{r_1 + r_2}(E_2 - E_1)$. Since $E_2 > E_1$ and $r_1, r_2 > 0$,it is clear that $E_{eq} < E_2$.
Thus,$E_1 < E_{eq} < E_2$.
Given $r_2 > r_1$,the term $\frac{r_1}{r_1 + r_2} < \frac{1}{2}$.
Therefore,$E_{eq}$ is closer to $E_1$ than to $E_2$.

(C) The correct answer is current only.
According to the principle of continuity for steady current,the current $I$ flowing through any cross-section of a conductor remains constant,regardless of the area of the cross-section.
Since $I = nAev_d$,where $n$ is the number density of electrons,$A$ is the area of cross-section,$e$ is the charge of an electron,and $v_d$ is the drift speed,if the area $A$ changes,the drift speed $v_d$ must change to keep the current $I$ constant.
Similarly,the electric field $E$ depends on the current density $J = I/A$,so it also varies with the cross-sectional area.

(C) galvanometer is converted into an ammeter by connecting a small shunt resistance $S$ in parallel with it.
For a galvanometer with resistance $G$ and full-scale deflection current $i_g$,if it is to measure a maximum current $i$,the shunt resistance $S$ is given by:
$S = \frac{i_g G}{i - i_g}$
From this formula,we can see that $S \propto \frac{1}{i - i_g}$.
Since an ammeter is designed to measure a larger current $(i_{ammeter})$ compared to a milliammeter $(i_{milliammeter})$,the denominator $(i - i_g)$ will be larger for the ammeter.
Therefore,the shunt resistance $S$ required for the ammeter will be less than the shunt resistance required for the milliammeter.
Thus,the shunt resistance of the ammeter is less than that of the milliammeter.

(C) In the half-deflection method,the resistance of the galvanometer $G$ is given by the formula:
$G = \frac{S \cdot R}{R - S}$
Given:
Resistance in series $R = 5200 \ \Omega$
Shunt resistance $S = 90 \ \Omega$
Initial deflection $\theta = 26 \ \text{div}$
Final deflection $\theta' = \theta/2 = 13 \ \text{div}$
Substituting the values:
$G = \frac{90 \times 5200}{5200 - 90}$
$G = \frac{468000}{5110}$
$G \approx 91.58 \ \Omega$
Rounding to the nearest option,$G \approx 91.6 \ \Omega$.

(A) In the given graph:
$1$. For material $Z$,resistivity $\rho$ decreases exponentially with an increase in temperature $T$. This is a characteristic property of semiconductors.
$2$. For material $Y$,resistivity $\rho$ increases linearly with temperature $T$. This is a characteristic property of alloys like nichrome,which have a very small temperature coefficient of resistance.
$3$. For material $X$,resistivity $\rho$ increases non-linearly with temperature $T$. This is a characteristic property of metals like copper.
Therefore,$X$ is copper,$Y$ is nichrome,and $Z$ is a semiconductor.

(C) The resistance $(R)$ of a wire is given by the formula: $R = \frac{\rho \ell}{A}$,where $\rho$ is resistivity,$\ell$ is length,and $A$ is the cross-sectional area.
Since the wire is cylindrical,the cross-sectional area $A = \pi r^2 = \pi (D/2)^2 = \frac{\pi D^2}{4}$.
Substituting this into the resistance formula,we get: $R = \frac{\rho \ell}{\pi D^2 / 4} = \frac{4 \rho \ell}{\pi D^2}$.
Since $\rho$,$\ell$,and $\pi$ are constants,we have $R \propto \frac{1}{D^2}$.
This relationship represents an inverse square law,which corresponds to a curve that decreases as $D$ increases. Among the given options,Graph $C$ represents this behavior.

(D) Electrical conductivity $(\sigma)$ is the reciprocal of electrical resistivity $(\rho)$,i.e.,$\sigma = 1/\rho$.
For metals,the resistivity $(\rho)$ is very low,typically in the range of $10^{-8} \Omega \text{ m}$ to $10^{-6} \Omega \text{ m}$.
Consequently,the electrical conductivity $(\sigma)$ is very high,typically in the range of $10^6 \text{ S m}^{-1}$ to $10^8 \text{ S m}^{-1}$.
Comparing this with the given options,option $D$ correctly represents these ranges.

(D) The de-Broglie wavelength $\lambda$ is related to the momentum $p$ of a particle by the equation $\lambda = \frac{h}{p}$,where $h$ is Planck's constant.
Since the de-Broglie wavelength $\lambda$ is the same for both the electron and the photon,their momenta $p$ must also be the same.
Therefore,they will have the same momentum.

(A) In the photoelectric effect,the photoelectric current $I$ is directly proportional to the intensity of the incident light,provided the frequency $\nu$ of the incident light is greater than the threshold frequency $\nu_0$.
If the intensity of the incident light is kept constant while the frequency $\nu$ is increased,the number of photons incident per unit time remains constant (since $P = n h \nu$,if power $P$ is constant,$n$ decreases as $\nu$ increases). However,in practical experimental setups,the photoelectric current is independent of the frequency of light as long as $\nu > \nu_0$.
Therefore,the graph of photoelectric current $I$ versus frequency $\nu$ is a horizontal line for $\nu > \nu_0$ and zero for $\nu < \nu_0$. This corresponds to the graph shown in option $A$.

(A) According to Lenz's law,the induced current in the coil always opposes the change in magnetic flux that produces it.
When the north pole of the magnet is pushed towards the coil,the magnetic flux through the coil increases,and the induced current flows in a direction to oppose this increase.
When the magnet is pulled away from the coil,the magnetic flux through the coil decreases. To oppose this decrease,the induced current flows in the opposite direction compared to the first case.
Since the galvanometer pointer deflected towards $X$ when the magnet was pushed in,it will deflect in the opposite direction,i.e.,towards $X^1$,when the magnet is pulled away.

(A) To match the electromagnetic waves with their respective wavelength ranges,we refer to the standard electromagnetic spectrum:
$1$. Microwave: The wavelength range is approximately $0.1 \ m$ to $1 \ mm$ (corresponds to $c$).
$2$. Visible light: The wavelength range is approximately $700 \ nm$ to $400 \ nm$ (corresponds to $a$).
$3$. Ultraviolet: The wavelength range is approximately $400 \ nm$ to $1 \ nm$ (corresponds to $d$).
$4$. $X$-rays: The wavelength range is approximately $1 \ nm$ to $10^{-3} \ nm$ (corresponds to $b$).
Thus,the correct matching is $i-c, ii-a, iii-d, iv-b$.

(A) When a light wave travels from a rarer medium to a non-reflecting and non-absorbing medium,the total energy it carries remains the same.
Energy in a light wave is proportional to the frequency of the wave and the number of photons present.
Since the frequency of light remains constant when it changes mediums and the medium is non-absorbing (no energy loss) and non-reflecting (no energy diverted),the total energy is conserved.

(D) The potential energy $(U)$ of an electric dipole in a uniform electric field is given by the formula: $U = -\vec{p} \cdot \vec{E} = -pE \cos \theta$.
For each case:
$(a)$ At $\theta = 180^{\circ}$,$U = -pE \cos 180^{\circ} = -pE(-1) = pE$. This matches $(ii)$.
$(b)$ At $\theta = 120^{\circ}$,$U = -pE \cos 120^{\circ} = -pE(-1/2) = \frac{1}{2} pE$. This matches $(iii)$.
$(c)$ At $\theta = 90^{\circ}$,$U = -pE \cos 90^{\circ} = -pE(0) = 0$. This matches $(iv)$.
Therefore,the correct matching is $a-ii, b-iii, c-iv$.

(A) For a conducting sphere of radius $R$ carrying a total charge $Q$:
$1$. Inside the sphere $(r < R)$,the electric field $E$ is zero because all charges reside on the surface.
$2$. Outside the sphere $(r \geq R)$,the sphere acts as a point charge located at its center,so the electric field is given by $E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}$,which means $E \propto \frac{1}{r^2}$.
$3$. Therefore,the electric field is zero for $r < R$ and decreases as $1/r^2$ for $r \geq R$. The graph that represents this behavior is Graph $A$,where $E=0$ up to $r=R$ and then follows an inverse-square decay.

(B) According to the properties of conductors in electrostatic equilibrium,the electric field inside the material of a metallic conductor is always zero.
Since the electric field $E$ is zero everywhere inside the metallic sphere of radius $R$,the electric flux $\phi$ through any closed surface drawn inside the material of this sphere is given by $\phi = \oint E \cdot dA = 0$.
Therefore,the electric flux at any point inside the metallic sphere of radius $R$ due to the external charge $Q$ on the other sphere is zero.

(D) According to Gauss's Law,the electric flux $\phi$ through any closed surface is given by $\phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}$,where $q_{\text{enclosed}}$ is the total charge enclosed by the surface.
In the given figure,sphere '$A$' has a radius ' $R$ ' and is centered at the midpoint of the dipole. The distance between the charges $+q$ and $-q$ is $2R$. Therefore,the distance of each charge from the center is $R$.
Since sphere '$A$' has a radius ' $R$ ',it encloses only the charge $-q$ located at the center of the sphere.
Thus,the charge enclosed by sphere '$A$' is $q_{\text{enclosed}} = -q$.
Therefore,the electric flux through sphere '$A$' is $\phi_A = \frac{-q}{\varepsilon_0}$.

(D) Gauss's law states that the total electric flux through a closed surface is equal to $\frac{1}{\epsilon_0}$ times the total charge enclosed by the surface.
Therefore,Gauss's law is valid for any closed surface,regardless of the shape or the distribution of charges inside it.
Charges outside the surface do not contribute to the net flux through the surface.
Thus,option $D$ is the correct statement.

(A) The work done $W$ in moving a charge $q$ from point $B$ to point $A$ is given by the formula: $W = q(V_A - V_B)$.
Given:
Charge $q = 5 \ mC = 5 \times 10^{-3} \ C$.
Potential at $A$,$V_A = -3 \ V$.
Potential at $B$,$V_B = 5 \ V$.
Substituting the values:
$W = 5 \times 10^{-3} \times (-3 - 5)$
$W = 5 \times 10^{-3} \times (-8)$
$W = -40 \times 10^{-3} \ J$
$W = -0.04 \ J$.

(A) An equipotential surface is defined as a surface where the electric potential is the same at every point.
Since the potential difference $(V_B - V_A)$ between any two points on an equipotential surface is zero,the work done $(W)$ in moving a charge $(q)$ from one point to another is given by $W = q(V_B - V_A) = 0$.
Therefore,the statement that 'Work done to move a charge on an equipotential surface is not zero' is false.

(B) The magnetic force per unit length between two long parallel wires carrying currents $I_1$ and $I_2$ separated by a distance $r$ is given by the formula:
$\frac{F}{\ell} = \frac{\mu_0 I_1 I_2}{2 \pi r}$
In this case,both wires carry the same current $I$,so $I_1 = I_2 = I$. Thus,the force per unit length is:
$\frac{F}{\ell} = \frac{\mu_0 I^2}{2 \pi r}$
According to the right-hand rule,parallel wires carrying currents in the same direction attract each other,while wires carrying currents in opposite directions repel each other.
Since the currents are in opposite directions,the wires will repel each other.

(B) Given: Length of solenoid $\ell = 1 \ m$,Diameter $d = 4 \ cm$,Radius $r = 2 \ cm = 0.02 \ m$.
Total number of turns $N = 5 \times 1000 = 5000$.
Current $I = 7 \ A$.
The magnetic field at the centre of a finite solenoid is given by $B = \frac{\mu_0 N I}{2\ell} (\cos \theta_1 + \cos \theta_2)$.
Since the solenoid is long compared to its radius,we can approximate the field using the formula for an ideal solenoid $B = \mu_0 n I$,where $n = N/\ell$.
$B = \frac{\mu_0 N I}{\ell} = \frac{4\pi \times 10^{-7} \times 5000 \times 7}{1}$.
$B = 4\pi \times 10^{-7} \times 35000 = 14\pi \times 10^{-3} \ T$.
$B \approx 14 \times 3.14159 \times 10^{-3} \ T = 43.98 \times 10^{-3} \ T = 4.398 \times 10^{-2} \ T$.
Rounding to the given options,the correct value is $4.396 \times 10^{-2} \ T$.

(B) According to Oersted's experiment and the Biot-Savart law,a straight current-carrying conductor produces a magnetic field. The magnetic field lines around a long straight current-carrying wire are concentric circles with the wire as the center. Therefore,option $B$ is correct. Fleming's Left-Hand Rule is used to determine the direction of force on a current-carrying conductor in a magnetic field,not the field itself. The magnetic field inside an ideal solenoid is uniform.

(B) The magnetic field $B$ at a perpendicular distance $r$ from an infinitely long,straight current-carrying conductor is given by the formula:
$B = \frac{\mu_0 I}{2 \pi r}$
From this expression,it is clear that $B \propto \frac{1}{r}$.
This relationship represents a rectangular hyperbola,where $B$ decreases as $r$ increases. Therefore,the graph in option $B$ correctly represents this variation.

(A) The area vector $\vec{A}$ of a square loop of side $2 \ m$ lying in the $Y-Z$ plane is directed along the $X$-axis.
$\vec{A} = (2 \times 2) \hat{i} = 4 \hat{i} \ m^2$.
The magnetic field is given by $\vec{B} = (5 \hat{i} + 3 \hat{j} - 4 \hat{k}) \ T$.
The magnetic flux $\phi$ is given by the dot product of the magnetic field and the area vector:
$\phi = \vec{B} \cdot \vec{A}$
$\phi = (5 \hat{i} + 3 \hat{j} - 4 \hat{k}) \cdot (4 \hat{i})$
$\phi = (5 \times 4) + (3 \times 0) + (-4 \times 0)$
$\phi = 20 \ Wb$.
Thus,the magnitude of the magnetic flux is $20 \ Wb$.

(D) Diamagnetic substances are materials that develop a weak magnetization in the direction opposite to the applied magnetic field.
Their magnetic susceptibility $(\chi)$ is small and negative, typically ranging from $-10^{-5}$ to $-10^{-9}$.
Unlike paramagnetic or ferromagnetic substances, the magnetic susceptibility of diamagnetic substances is independent of temperature.
Therefore, the statement that susceptibility is small and negative is correct.

(D) $1$. Isotopes: Atoms with the same atomic number $(Z)$ but different mass numbers $(A)$. Example: $_1H^1$ and $_1H^2$ have $Z=1$. Thus,$A-iii$.
$2$. Isobars: Atoms with the same mass number $(A)$ but different atomic numbers $(Z)$. Example: $Li^7$ and $Be^7$ both have $A=7$. Thus,$B-i$.
$3$. Isotones: Atoms with the same number of neutrons $(N = A-Z)$. For $_8O^{18}$,$N = 18-8 = 10$. For $_9F^{19}$,$N = 19-9 = 10$. Thus,$C-ii$.
Therefore,the correct matching is $A-iii, B-i, C-ii$.

(C) The strong nuclear force is not always attractive. It is strongly attractive at distances larger than $0.8 \ fm$ and becomes strongly repulsive at distances less than $0.8 \ fm$ to prevent the collapse of the nucleus. Therefore,the statement that the nuclear force is 'always attractive' is incorrect.

(D) $1$. When a lens is cut along its principal axis,the focal length of each half remains the same as the original lens,so the power of each half remains $P$. Thus,the power of $L_1$ is $P$.
$2$. When a lens is cut perpendicular to its principal axis,the focal length of each part doubles,which means the power of each part becomes half of the original power.
$3$. Since $L_2$ and $L_3$ are obtained by cutting the upper half of the original lens perpendicular to the principal axis,the power of $L_2$ is $\frac{P}{2}$ and the power of $L_3$ is $\frac{P}{2}$.
$4$. Comparing these with the given options,the incorrect statement is that the power of $L_1$ is $\frac{P}{2}$.

(B) In a compound microscope,the objective lens is placed close to the object. The object is placed just beyond the focal point of the objective lens. This results in the formation of a real,inverted,and enlarged image. This image acts as an object for the eyepiece,which further magnifies it to produce the final virtual image.

(D) For a prism,the sum of the angles of refraction at the two faces is equal to the angle of the prism $(A)$:
$r + r^1 = A$
Given that the angle of the prism $A = 50^{\circ}$ and $r = 10^{\circ} + t^2$,we can substitute these values into the equation:
$10^{\circ} + t^2 + r^1 = 50^{\circ}$
Solving for $r^1$:
$r^1 = 50^{\circ} - 10^{\circ} - t^2$
$r^1 = 40^{\circ} - t^2$

(B) According to Snell's law,$n_1 \sin i = n_2 \sin r$.
Here,$n_1 = 1$ (vacuum),$n_2 = n$,$i$ is the angle of incidence,and $r$ is the angle of refraction.
Given that $i = 2r$,we have $r = \frac{i}{2}$.
Substituting these into Snell's law:
$1 \times \sin i = n \sin \left(\frac{i}{2}\right)$
Using the trigonometric identity $\sin i = 2 \sin \left(\frac{i}{2}\right) \cos \left(\frac{i}{2}\right)$:
$2 \sin \left(\frac{i}{2}\right) \cos \left(\frac{i}{2}\right) = n \sin \left(\frac{i}{2}\right)$
Dividing both sides by $\sin \left(\frac{i}{2}\right)$ (assuming $i \neq 0$):
$2 \cos \left(\frac{i}{2}\right) = n$
$\cos \left(\frac{i}{2}\right) = \frac{n}{2}$
$\frac{i}{2} = \cos^{-1} \left(\frac{n}{2}\right)$
$i = 2 \cos^{-1} \left(\frac{n}{2}\right)$

(A) The refractive index $n$ is given by the ratio of real depth to apparent depth.
Real depth $(RD)$ is the actual thickness of the glass slab,which is the difference between the reading for the chalk dust on top of the slab and the reading for the ink mark at the bottom: $RD = 8.123 \ cm - 5.123 \ cm = 3.000 \ cm$.
Apparent depth $(AD)$ is the depth of the ink mark as seen through the glass slab,which is the difference between the reading for the chalk dust on top and the reading for the ink mark through the slab: $AD = 8.123 \ cm - 6.123 \ cm = 2.000 \ cm$.
The refractive index is $n = \frac{RD}{AD} = \frac{3.000}{2.000} = 1.5$.

(C) In the given circuit,the positive terminal of the $3 \text{ V}$ battery is connected to the cathode of diode $D_1$ and the anode of diode $D_2$.
$1$. Diode $D_1$: The positive terminal of the battery is connected to the n-side (cathode) of $D_1$. Thus,$D_1$ is reverse-biased and acts as an open circuit (no current flows through this branch).
$2$. Diode $D_2$: The positive terminal of the battery is connected to the p-side (anode) of $D_2$. Thus,$D_2$ is forward-biased and acts as a closed switch (ideal diode has zero resistance).
$3$. Equivalent Resistance: Since $D_1$ is an open circuit,the current flows only through the branch containing $D_2$ and the $30 \Omega$ resistor,which is in series with the $70 \Omega$ resistor.
$R_{eq} = 30 \Omega + 70 \Omega = 100 \Omega$
$4$. Current Calculation: Using Ohm's law,$I = \frac{V}{R_{eq}}$
$I = \frac{3 \text{ V}}{100 \Omega} = 0.03 \text{ A}$

(D) In an $n$-type semiconductor,donor impurities are added to the intrinsic semiconductor. These impurities create discrete energy levels known as donor energy levels. These levels are located within the forbidden energy gap,very close to the bottom of the conduction band. This proximity allows electrons from the donor level to be easily excited into the conduction band at room temperature,thereby increasing the conductivity of the material.

(A) When a plane wave front is incident on a convex lens,the central part of the wave front travels through the thickest part of the lens,while the edges travel through thinner parts.
Since the refractive index of the lens $n_2$ is greater than the surrounding medium $n_1$,the speed of light inside the lens is lower than outside.
Consequently,the central part of the wave front is delayed more than the edges.
This causes the initially plane wave front to become spherical and converge towards the focal point of the lens.
Therefore,the refracted wave front is a spherical wave front that is concave towards the direction of propagation (or concave towards the lens as it converges).
Looking at the options,the shape that represents a converging spherical wave front is a concave curve.