NEET 2024 Physics Question Paper with Answer and Solution

(C) In uniform circular motion,the speed of the particle remains constant,but the direction of motion changes at every point along the circular path.
Since velocity is a vector quantity (having both magnitude and direction),a change in direction implies a change in velocity. Therefore,the velocity is varying.
Furthermore,the centripetal acceleration,given by $a_c = v^2/r$,is directed towards the center of the circle. As the particle moves,the direction of this acceleration vector changes continuously to remain pointing towards the center.
Thus,both velocity and acceleration are varying.

(D) Given:
Displacement $x = 2t - 1 \,m$
Force $F = 5 \,N$
Step $1$: Find the velocity of the particle.
Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(2t - 1) = 2 \,m/s$
Step $2$: Calculate the instantaneous power.
Power $P = F \cdot v$
$P = 5 \,N \times 2 \,m/s = 10 \,W$
Thus, the instantaneous power is $10 \,W$.

(D) The formula for the moment of inertia $(I)$ of a thin rod of mass $m$ and length $\ell$ about an axis passing through its midpoint and perpendicular to its length is given by:
$I = \frac{m \ell^2}{12}$
Given:
$I = 2400 \,g \,cm^2$
$m = 400 \,g$
Substituting the values into the formula:
$2400 = \frac{400 \times \ell^2}{12}$
Simplify the equation:
$2400 = \frac{100 \times \ell^2}{3}$
$2400 \times 3 = 100 \times \ell^2$
$7200 = 100 \times \ell^2$
$\ell^2 = 72$
Taking the square root:
$\ell = \sqrt{72} \approx 8.485 \,cm$
Rounding to the nearest value, we get:
$\ell \approx 8.5 \,cm$

(A) For a bob of mass $m$ rotating in a horizontal circle of radius $\ell$ with angular speed $\omega$,the centripetal force is provided by the tension $T$ in the string.
$T = m\ell\omega^2$ --- $(1)$
When the angular speed is increased to $2\omega$ while keeping the radius $\ell$ constant,the new tension $T'$ is given by:
$T' = m\ell(2\omega)^2$
$T' = m\ell(4\omega^2)$
$T' = 4(m\ell\omega^2)$
Substituting equation $(1)$ into this expression,we get:
$T' = 4T$

(D) The solid angle is defined as $d\Omega = \frac{dA}{r^2}$. Since it is the ratio of area to the square of distance,its dimensions are $[M^0 L^0 T^0]$,making it a dimensionless quantity.
Strain is defined as the ratio of change in length to original length,$\text{Strain} = \frac{\Delta l}{l}$,which is also dimensionless $[M^0 L^0 T^0]$.
An angle $\theta = \frac{l}{r}$ is the ratio of arc length to radius,which is also dimensionless $[M^0 L^0 T^0]$.
Therefore,both strain and angle have the same dimensions as a solid angle.

(A) In the case of pure rolling motion,the velocity of any point on the rim of the wheel is the vector sum of the translational velocity of the center of mass $(v)$ and the tangential velocity due to rotation $(v = r\omega)$.
For the topmost point $P$,the translational velocity and the rotational velocity are in the same direction. Thus,the resultant velocity is $v + v = 2v$.
For the lowest point $Q$,the translational velocity is in the forward direction $(v)$ and the rotational velocity is in the backward direction $(v)$. Thus,the resultant velocity is $v - v = 0$.
Since the velocity of point $P$ is $2v$ and the velocity of point $Q$ is $0$,point $P$ moves faster than point $Q$.

(D) The excess force required to pull a circular disc of radius $R$ from the surface of a liquid with surface tension $T$ is given by the force due to surface tension acting along the circumference of the disc.
Excess force $F = T \times (2 \pi R)$
Given:
Radius $R = 4.5 \,cm = 4.5 \times 10^{-2} \,m$
Surface tension $T = 0.07 \,N \,m^{-1}$
Substituting the values:
$F = 0.07 \times 2 \times 3.14 \times 4.5 \times 10^{-2}$
$F = 0.07 \times 28.26 \times 10^{-2}$
$F = 1.9782 \times 10^{-2} \,N$
$F \approx 19.8 \times 10^{-3} \,N$
$F = 19.8 \,mN$

(D) For maximum elongation,the stress applied must be equal to the elastic limit of the material.
Given:
Length of the wire,$L = 1 \,m$
Elastic limit (maximum stress),$\sigma = 8 \times 10^8 \,N \,m^{-2}$
Young's modulus,$Y = 2 \times 10^{11} \,N \,m^{-2}$
Using the formula for Young's modulus: $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\sigma}{\Delta L / L}$
Rearranging for elongation $\Delta L$:
$\Delta L = \frac{\sigma \times L}{Y}$
Substituting the values:
$\Delta L = \frac{8 \times 10^8 \times 1}{2 \times 10^{11}}$
$\Delta L = 4 \times 10^{-3} \,m$
Converting to millimeters:
$\Delta L = 4 \,mm$
Thus,the maximum elongation is $4 \,mm$.

(A) The Vernier Constant $(V.C.)$ is defined as the difference between one Main Scale Division $(MSD)$ and one Vernier Scale Division $(VSD)$: $V.C. = 1 \text{ MSD} - 1 \text{ VSD}$.
Given that $(N+1)$ divisions of the vernier scale coincide with $N$ divisions of the main scale:
$(N+1) \text{ VSD} = N \text{ MSD}$
$1 \text{ VSD} = \frac{N}{N+1} \text{ MSD}$.
Substituting this into the $V.C.$ formula:
$V.C. = 1 \text{ MSD} - \left(\frac{N}{N+1}\right) \text{ MSD}$
$V.C. = \text{ MSD} \left(1 - \frac{N}{N+1}\right) = \frac{1}{N+1} \text{ MSD}$.
Given $1 \text{ MSD} = 0.1 \text{ mm}$. Since $1 \text{ cm} = 10 \text{ mm}$,$1 \text{ mm} = 0.1 \text{ cm}$.
Therefore,$1 \text{ MSD} = 0.1 \times 0.1 \text{ cm} = 0.01 \text{ cm}$.
Thus,$V.C. = \frac{0.01}{N+1} \text{ cm} = \frac{1}{100(N+1)} \text{ cm}$.

(A) The general equation for simple harmonic motion is given by $x = A \sin(\omega t + \phi)$.
Comparing the given equation $x = 5 \sin \left(\pi t + \frac{\pi}{3}\right) \text{ m}$ with the general equation:
Amplitude $A = 5 \text{ m}$.
Angular frequency $\omega = \pi \text{ rad/s}$.
The time period $T$ is related to angular frequency by the formula $T = \frac{2\pi}{\omega}$.
Substituting the value of $\omega$: $T = \frac{2\pi}{\pi} = 2 \text{ s}$.
Thus,the amplitude is $5 \text{ m}$ and the time period is $2 \text{ s}$.

(B) Given: Force $F = 10 \,N$, mass of block $A$ $(m_A) = 2 \,kg$, mass of block $B$ $(m_B) = 3 \,kg$.
Since the blocks are in contact and moving together, the acceleration $(a)$ of the system is given by Newton's second law:
$F = (m_A + m_B) a$
$10 = (2 + 3) a$
$10 = 5a$
$a = 2 \,m/s^2$
Now, consider the free body diagram of block $B$. The only horizontal force acting on block $B$ is the force exerted by block $A$ $(F_{AB})$, which causes it to accelerate at $2 \,m/s^2$:
$F_{AB} = m_B \times a$
$F_{AB} = 3 \,kg \times 2 \,m/s^2 = 6 \,N$
Therefore, the force exerted by block $A$ on block $B$ is $6 \,N$.

(A) Let the mass of both bodies $A$ and $B$ be $m$.
Before the collision,body $A$ has velocity $v_1$ and body $B$ is at rest $(v_B = 0)$.
After the collision,since it is a completely inelastic collision,both bodies stick together and move with a common velocity $v_2$.
According to the law of conservation of linear momentum:
Initial momentum = Final momentum
$m v_1 + m(0) = (m + m) v_2$
$m v_1 = 2m v_2$
Dividing both sides by $m v_2$,we get:
$\frac{v_1}{v_2} = \frac{2}{1}$
Therefore,the ratio $v_1: v_2$ is $2: 1$.

(D) In a $P-V$ diagram,the work done by a gas is given by the area under the curve,which is $\int P \ dV$.
Along the path $b \to c$,the volume $V$ remains constant at $400 \ cm^3$.
Since there is no change in volume $(dV = 0)$,the process is an isochoric process.
Therefore,the work done by the gas along the path $b \to c$ is $W = P \Delta V = P \times 0 = 0 \ J$.

(C) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.
Let the mass and radius of the earth be $M$ and $R$ respectively. Then $g = \frac{GM}{R^2} = 9.8 \ m \ s^{-2}$.
For the given planet,the mass $M' = \frac{M}{10}$ and the radius $R' = \frac{R}{2}$ (since diameter is half,radius is also half).
The acceleration due to gravity on the planet is $g' = \frac{GM'}{R'^2}$.
Substituting the values: $g' = \frac{G(M/10)}{(R/2)^2} = \frac{GM/10}{R^2/4} = \frac{4}{10} \frac{GM}{R^2}$.
Since $\frac{GM}{R^2} = 9.8 \ m \ s^{-2}$,we have $g' = 0.4 \times 9.8 \ m \ s^{-2} = 3.92 \ m \ s^{-2}$.

(D) The total energy of the satellite at the surface of the Earth is $E_i = K_i + U_i = K_i - \frac{G M m}{R}$.
At an altitude of $2R$ from the surface,the distance from the center of the Earth is $r = R + 2R = 3R$.
The orbital velocity $v$ at distance $r = 3R$ is given by $v = \sqrt{\frac{G M}{3R}}$.
The total energy at the orbit is $E_f = K_f + U_f = \frac{1}{2} m v^2 - \frac{G M m}{3R} = \frac{1}{2} m \left(\frac{G M}{3R}\right) - \frac{G M m}{3R} = \frac{G M m}{6R} - \frac{G M m}{3R} = -\frac{G M m}{6R}$.
By the law of conservation of energy,$E_i = E_f$:
$K_i - \frac{G M m}{R} = -\frac{G M m}{6R}$.
$K_i = \frac{G M m}{R} - \frac{G M m}{6R} = \frac{5 G M m}{6R}$.

(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{\ell}{g}}$,where $\ell$ is the length of the pendulum and $g$ is the acceleration due to gravity. The time period is independent of the mass of the bob.
Given that the new length $\ell^{\prime} = \frac{\ell}{2}$.
The new time period $T^{\prime} = 2 \pi \sqrt{\frac{\ell^{\prime}}{g}} = 2 \pi \sqrt{\frac{\ell}{2g}} = \frac{1}{\sqrt{2}} \left( 2 \pi \sqrt{\frac{\ell}{g}} \right) = \frac{1}{\sqrt{2}} T$.
According to the problem,$T^{\prime} = \frac{x}{2} T$.
Therefore,$\frac{x}{2} = \frac{1}{\sqrt{2}}$.
$x = \frac{2}{\sqrt{2}} = \sqrt{2}$.

(A) According to the principle of homogeneity of dimensions,the dimensions of each term in an equation must be the same.
Therefore,$[F] = [\alpha t^2] = [\beta t]$.
From $[\alpha t^2] = [\beta t]$,we can write:
$\frac{[\alpha]}{[\beta]} = \frac{[t]}{[t^2]} = \frac{1}{[t]}$.
Multiplying both sides by $[t]$,we get:
$\frac{[\alpha][t]}{[\beta]} = 1$.
This implies that the quantity $\frac{\alpha t}{\beta}$ is dimensionless.

(A) The thermal strain produced in the bar is given by $\epsilon = \alpha \Delta T$.
Given $\alpha = 10^{-5} \,^{\circ}C^{-1}$ and $\Delta T = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C$.
So, $\epsilon = 10^{-5} \times 100 = 10^{-3}$.
Since the bar is prevented from expanding, the compressive stress $\sigma$ is given by $\sigma = Y \times \epsilon$, where $Y$ is Young's modulus.
$\sigma = (0.5 \times 10^{11} \,N \,m^{-2}) \times 10^{-3} = 0.5 \times 10^8 \,N \,m^{-2}$.
The compressive force $F$ is given by $F = \sigma \times A$, where $A$ is the area of cross-section.
$F = (0.5 \times 10^8 \,N \,m^{-2}) \times (10^{-3} \,m^2) = 0.5 \times 10^5 \,N = 50 \times 10^3 \,N$.

(C) From the ideal gas equation,$PV = nRT$,we have $T = (P/nR)V$.
For a fixed amount of gas ($n$ is constant),the slope of the $T-V$ graph is given by $m = P/nR$.
Since $n$ and $R$ are constants,the slope $m$ is directly proportional to the pressure $P$.
Therefore,a steeper slope in the $T-V$ graph corresponds to a higher pressure.
Looking at the graph,the slope of the curve for $P_1$ is the steepest,followed by $P_2$,and then $P_3$.
Thus,the correct relation is $P_1 > P_2 > P_3$.

(B) The acceleration $(a)$ is defined as the rate of change of velocity,which corresponds to the slope of the velocity-time $(v-t)$ graph,i.e.,$a = \frac{dv}{dt}$.
$1$. In the first interval,the velocity increases linearly with time. Since the slope is constant and positive,the acceleration is constant and positive.
$2$. In the second interval,the velocity is constant. Since the slope of a horizontal line is zero,the acceleration is zero.
$3$. In the third interval,the velocity decreases linearly with time. Since the slope is constant and negative,the acceleration is constant and negative.
Comparing this with the given options,the graph that shows a positive constant acceleration,followed by zero acceleration,and then a negative constant acceleration is represented by option $B$.

(D) Statement $A$: $A$ solar cell converts light energy into electrical energy. In the $I-V$ characteristic curve of a solar cell,the voltage is positive and the current is negative (as it delivers power to an external load). This corresponds to the $IV$ quadrant of the $I-V$ graph. Thus,statement $A$ is correct.
Statement $B$: In a reverse-biased $pn$ junction diode,the current is extremely small (in the order of $\mu A$) and is caused by the drift of minority charge carriers across the junction. The majority charge carriers are pushed away from the junction,preventing them from contributing to the current. Thus,statement $B$ is incorrect.

(A) Let the prism angle at the top be $A = 90^{\circ}$.
In the prism,the sum of the angles of refraction at the two faces is equal to the prism angle,so $r_1 + r_2 = A = 90^{\circ}$.
Since the ray travels parallel to the base $BC$,the angle of refraction at the second face is the critical angle $c$,so $r_2 = c$.
Thus,$r_1 = 90^{\circ} - c$.
Applying Snell's law at the first surface:
$1 \cdot \sin 30^{\circ} = \mu \cdot \sin r_1$
$1 \cdot \frac{1}{2} = \mu \cdot \sin(90^{\circ} - c)$
$\frac{1}{2} = \mu \cdot \cos c$.
We know that $\sin c = \frac{1}{\mu}$,so $\cos c = \sqrt{1 - \sin^2 c} = \sqrt{1 - \frac{1}{\mu^2}} = \frac{\sqrt{\mu^2 - 1}}{\mu}$.
Substituting this into the equation:
$\frac{1}{2} = \mu \cdot \frac{\sqrt{\mu^2 - 1}}{\mu}$
$\frac{1}{2} = \sqrt{\mu^2 - 1}$.
Squaring both sides:
$\frac{1}{4} = \mu^2 - 1$
$\mu^2 = 1 + \frac{1}{4} = \frac{5}{4}$
$\mu = \frac{\sqrt{5}}{2}$.

(A) For an ideal transformer,the relationship between the voltage ratio and the turns ratio is given by the formula: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.
Given the turns ratio $\frac{N_p}{N_s} = \frac{1}{2}$,we can find the inverse ratio $\frac{N_s}{N_p} = \frac{2}{1}$.
Substituting this into the transformer equation,we get $\frac{V_s}{V_p} = \frac{2}{1}$.
Therefore,the ratio $V_s : V_p$ is $2 : 1$.

$(A)$ The energy difference for a transition is given by $\Delta E = \frac{hc}{\lambda}$.
Therefore, the wavelength is inversely proportional to the energy difference: $\lambda \propto \frac{1}{\Delta E}$.
As the energy difference increases, the wavelength decreases.
For the Balmer series $(n_1=2)$, the energy differences are:
$(\Delta E)_{3-2} < (\Delta E)_{4-2} < (\Delta E)_{5-2} < (\Delta E)_{6-2}$.
Consequently, the wavelengths follow the order:
$\lambda_{3-2} > \lambda_{4-2} > \lambda_{5-2} > \lambda_{6-2}$.
Matching the values:
$A$ ($n_2=3$ to $n_1=2$) corresponds to $656.3 \, nm$ $(III)$.
$B$ ($n_2=4$ to $n_1=2$) corresponds to $486.1 \, nm$ $(IV)$.
$C$ ($n_2=5$ to $n_1=2$) corresponds to $434.1 \, nm$ $(II)$.
$D$ ($n_2=6$ to $n_1=2$) corresponds to $410.2 \, nm$ $(I)$.
Thus, the correct match is $A-III, B-IV, C-II, D-I$.

(C) According to Brewster's law,when unpolarised light is incident on a transparent medium at Brewster's angle $(i_B)$,the reflected light is completely plane-polarised with its electric field vector perpendicular to the plane of incidence.
The refracted light,however,remains partially polarised because it contains both components of the electric field vector,although the intensity of the component parallel to the plane of incidence is greater than that of the perpendicular component.

(B) The current $i$ in the circuit is given by Ohm's law: $i = \frac{E}{R + r}$, where $E = 10 \, V$, $R = 4 \, \Omega$, and $r = 1 \, \Omega$.
$i = \frac{10}{4 + 1} = \frac{10}{5} = 2 \, A$.
The terminal voltage $V$ of the battery is given by the formula: $V = E - ir$.
Substituting the values: $V = 10 - (2 \times 1) = 10 - 2 = 8 \, V$.
Thus, the terminal voltage is $8 \, V$.

(C) Let the inputs to the first $NAND$ gate be $A$ and $A$. The output is $Y_1 = \overline{A \cdot A} = \bar{A}$.
Let the inputs to the $NOR$ gate be $B$ and $B$. The output is $Y_2 = \overline{B+B} = \bar{B}$.
The final gate is a $NOR$ gate with inputs $Y_1$ and $Y_2$. The output $Y$ is given by:
$Y = \overline{Y_1 + Y_2}$
Substituting the values of $Y_1$ and $Y_2$:
$Y = \overline{\bar{A} + \bar{B}}$
Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
Thus,the output $Y = A \cdot B$,which is the output of an $AND$ gate.

(D) The given circuit is a balanced Wheatstone bridge because the ratio of capacitances in the arms is equal $(2 \mu F / 2 \mu F = 2 \mu F / 2 \mu F)$.
In a balanced Wheatstone bridge,the central capacitor does not store any charge,so it can be removed from the circuit.
After removing the central capacitor,the circuit consists of two parallel branches,each containing two $2 \mu F$ capacitors in series.
For the upper branch,the equivalent capacitance $C_1$ is given by:
$1/C_1 = 1/2 + 1/2 = 1 \implies C_1 = 1 \mu F$.
Similarly,for the lower branch,the equivalent capacitance $C_2$ is:
$1/C_2 = 1/2 + 1/2 = 1 \implies C_2 = 1 \mu F$.
Since these two branches are in parallel,the total equivalent capacitance $C_{AB}$ is:
$C_{AB} = C_1 + C_2 = 1 \mu F + 1 \mu F = 2 \mu F$.

(C) $1$. Alpha decay ($\alpha$): Mass number decreases by $4$, atomic number decreases by $2$.
${ }_{82}^{290} X \xrightarrow{\alpha} { }_{80}^{286} Y$
$2$. Positron emission ($e^{+}$): Mass number remains same, atomic number decreases by $1$.
${ }_{80}^{286} Y \xrightarrow{e^{+}} { }_{79}^{286} Z$
$3$. Beta decay ($\beta^{-}$): Mass number remains same, atomic number increases by $1$.
${ }_{79}^{286} Z \xrightarrow{\beta^{-}} { }_{80}^{286} P$
$4$. Electron emission ($e^{-}$): This is equivalent to $\beta^{-}$ decay. Mass number remains same, atomic number increases by $1$.
${ }_{80}^{286} P \xrightarrow{e^{-}} { }_{81}^{286} Q$
Thus, for the final product $Q$, the mass number $A = 286$ and the atomic number $Z = 81$.

(A) The resistance of the original wire is $R = 100 \Omega$.
When the wire is divided into $10$ equal parts,the resistance of each part is $r = \frac{R}{10} = \frac{100 \Omega}{10} = 10 \Omega$.
For the first $5$ parts connected in series,the equivalent resistance is $R_S = 5 \times r = 5 \times 10 \Omega = 50 \Omega$.
For the next $5$ parts connected in parallel,the equivalent resistance $R_P$ is given by $\frac{1}{R_P} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} = \frac{5}{r}$.
Thus,$R_P = \frac{r}{5} = \frac{10 \Omega}{5} = 2 \Omega$.
Since these two combinations are connected in series,the final equivalent resistance is $R_{eq} = R_S + R_P = 50 \Omega + 2 \Omega = 52 \Omega$.

(B) The magnitude of the magnetic field at the centre of a circular coil with $N$ turns is given by the formula:
$B_c = \frac{\mu_0 i N}{2 R}$
Given:
Number of turns $N = 100$
Radius $R = 10 \,cm = 0.1 \,m$
Current $i = 7 \,A$
Permeability of free space $\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A$
Substituting the values:
$B_c = \frac{4 \pi \times 10^{-7} \times 7 \times 100}{2 \times 0.1}$
$B_c = \frac{4 \times 3.14159 \times 10^{-7} \times 700}{0.2}$
$B_c = \frac{8796.46 \times 10^{-7}}{0.2} \approx 4.398 \times 10^{-3} \,T$
Rounding to two significant figures,we get:
$B_c = 4.4 \times 10^{-3} \,T = 4.4 \,mT$

(B) By observing the truth table:
When $B = 0$,$Y = 1$ (regardless of $A$).
When $B = 1$,$Y = 0$ (regardless of $A$).
This shows that the output $Y$ depends only on the input $B$ and is the inverse of $B$.
Therefore,the Boolean expression for the output is $Y = \bar{B}$.

(A) The properties of a photon are as follows:
$1$. The energy of a photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency. Thus,statement $A$ is correct.
$2$. $A$ photon travels with the speed of light in free space,which is $c$. Thus,statement $B$ is correct.
$3$. The momentum $p$ of a photon is related to its energy $E$ by $p = E/c$. Substituting $E = h\nu$,we get $p = h\nu/c$. Thus,statement $C$ is correct.
$4$. In any collision,including a photon-electron collision (Compton scattering),both total energy and total momentum are conserved. Thus,statement $D$ is correct.
$5$. $A$ photon is an electrically neutral particle,meaning it has zero charge. Thus,statement $E$ is incorrect.
Therefore,statements $A, B, C,$ and $D$ are correct.

(C) For a uniformly charged thin spherical shell of radius $R$ and charge $q$,the electric potential at any point inside the shell (including the center) is constant and equal to the potential at the surface.
The potential at any point $r \leq R$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{R}$.
Since point $C$ is at the center $(r = 0)$ and point $P$ is on the surface $(r = R)$,both points are at the same potential.
Therefore,$V_C = V_P = \frac{1}{4 \pi \varepsilon_0} \frac{q}{R}$.
The potential difference between points $C$ and $P$ is $V_C - V_P = 0$.

(A) According to Lenz's Law,the induced current in a coil opposes the change in magnetic flux that produces it.
For solenoid-$1$: The North pole of the magnet is moving away from it. To oppose this,the end $B$ of solenoid-$1$ must act as a South pole. Using the right-hand thumb rule,the current flows from $B$ to $A$.
For solenoid-$2$: The South pole of the magnet is approaching it. To oppose this,the end $C$ of solenoid-$2$ must act as a South pole. Using the right-hand thumb rule,the current flows from $C$ to $D$.
Therefore,the directions of induced current are $B A$ and $C D$ respectively.

(C) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $E$ is its kinetic energy.
Squaring both sides,we get $\lambda^2 = \frac{h^2}{2mE}$.
Rearranging the terms,we find $\frac{1}{\lambda^2} = \left(\frac{2m}{h^2}\right) E$.
This equation is of the form $y = mx$,where $y = \frac{1}{\lambda^2}$,$x = E$,and the slope $m = \frac{2m}{h^2}$ is a constant.
Therefore,the graph of $\frac{1}{\lambda^2}$ versus $E$ is a straight line passing through the origin.

(C) The time period of oscillation for a magnetic needle in a uniform magnetic field is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$, where $I$ is the moment of inertia, $M$ is the magnetic moment, and $B$ is the magnetic field strength.
Given: $B = 0.049 \ T$, $I = 9.8 \times 10^{-5} \ kg \ m^2$, and the needle performs $20$ oscillations in $5 \ s$.
The time period $T = \frac{5 \ s}{20} = 0.25 \ s = \frac{1}{4} \ s$.
Substituting the values into the formula: $\frac{1}{4} = 2 \pi \sqrt{\frac{9.8 \times 10^{-5}}{M \times 0.049}}$.
Squaring both sides: $\frac{1}{16} = 4 \pi^2 \times \frac{9.8 \times 10^{-5}}{M \times 0.049}$.
Rearranging for $M$: $M = \frac{4 \pi^2 \times 9.8 \times 10^{-5} \times 16}{0.049} = \frac{4 \pi^2 \times 9.8 \times 10^{-5} \times 16}{0.049}$.
Since $9.8 / 0.049 = 200$, we get $M = 4 \pi^2 \times 200 \times 16 \times 10^{-5} = 12800 \pi^2 \times 10^{-5} \ A \ m^2$.
Wait, re-calculating: $M = \frac{4 \pi^2 \times 9.8 \times 10^{-5} \times 16}{49 \times 10^{-3}} = 4 \pi^2 \times 0.2 \times 16 \times 10^{-2} = 12.8 \pi^2 \times 10^{-2} = 1280 \pi^2 \times 10^{-5} \ A \ m^2$.
Thus, $x = 1280 \pi^2$.

(D) The magnetic susceptibility $(\chi)$ characterizes the magnetic properties of materials:
$1$. Diamagnetic materials: These materials are weakly repelled by a magnetic field, and their susceptibility is small and negative, typically $0 > \chi \geq -1$. Thus, $A-II$.
$2$. Ferromagnetic materials: These materials are strongly attracted by a magnetic field, and their susceptibility is large and positive, $\chi >> 1$. Thus, $B-III$.
$3$. Paramagnetic materials: These materials are weakly attracted by a magnetic field, and their susceptibility is small and positive, $0 < \chi < \varepsilon$. Thus, $C-IV$.
$4$. Non-magnetic materials: These materials do not interact with magnetic fields, so their susceptibility is zero, $\chi = 0$. Thus, $D-I$.
Therefore, the correct matching is $A-II, B-III, C-IV, D-I$.

(B) Statement $I$ is correct. Atoms are electrically neutral because the number of protons (positive charge) in the nucleus is equal to the number of electrons (negative charge) orbiting the nucleus.
Statement $II$ is incorrect. While many atoms are stable,there are many radioactive elements whose atoms are unstable and undergo decay. Furthermore,while atoms do emit characteristic spectra,the statement implies that 'each' element's atom is stable,which is a generalization that ignores radioactive instability.

(B) In Young's double slit experiment,the path difference at the central point on the screen is zero for all wavelengths present in white light.
Since the condition for constructive interference is $\Delta x = n\lambda$,for $n=0$,the path difference is zero regardless of the wavelength $\lambda$.
Therefore,all colors overlap at the central point,resulting in a central bright white fringe.
For other fringes,the fringe width is given by $\beta = \frac{\lambda D}{d}$. Since $\beta$ depends on the wavelength $\lambda$,different colors will form fringes at different positions,resulting in a few coloured fringes surrounding the central white fringe.

(D) The electric potential $V$ at a point due to an electric dipole at a distance $r$ from its center is given by $V = \frac{1}{4 \pi \epsilon_0} \frac{P \cos \theta}{r^2}$.
For an axial point, the angle $\theta$ is either $0^{\circ}$ or $180^{\circ}$.
Thus, the potential is $V = \pm \frac{1}{4 \pi \epsilon_0} \frac{P}{r^2}$.
Given $P = 4 \times 10^{-6} \ C \ m$, $r = 2 \ m$, and $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \ SI$ units.
Substituting these values: $V = \pm \frac{9 \times 10^9 \times 4 \times 10^{-6}}{2^2} = \pm \frac{36 \times 10^3}{4} = \pm 9 \times 10^3 \ V$.
Both Assertion $A$ and Reason $R$ are true, and $R$ correctly explains $A$.

(A) Given:
Focal length of the objective lens,$f_o = 140 \ cm$.
Focal length of the eyepiece,$f_e = 5.0 \ cm$.
For a telescope viewing a distant object (at normal adjustment),the magnifying power $m$ is given by the formula:
$m = \frac{f_o}{f_e}$
Substituting the values:
$m = \frac{140}{5.0} = 28$
Therefore,the magnifying power of the telescope is $28$.

(A) The power rating $P$ of a heater is given by $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.
Since $V$ is constant,$R = \frac{V^2}{P}$.
For heater $A$,$P_A = 1 \text{ kW}$,so $R_A = \frac{V^2}{1}$.
For heater $B$,$P_B = 2 \text{ kW}$,so $R_B = \frac{V^2}{2}$.
Thus,$R_A = 2R_B$.
In series combination,the equivalent resistance is $R_S = R_A + R_B = 2R_B + R_B = 3R_B$. The power consumed is $P_S = \frac{V^2}{R_S} = \frac{V^2}{3R_B}$.
In parallel combination,the equivalent resistance is $R_P = \frac{R_A R_B}{R_A + R_B} = \frac{(2R_B)(R_B)}{2R_B + R_B} = \frac{2R_B^2}{3R_B} = \frac{2}{3}R_B$. The power consumed is $P_P = \frac{V^2}{R_P} = \frac{V^2}{(2/3)R_B} = \frac{3V^2}{2R_B}$.
The ratio of power outputs is $\frac{P_S}{P_P} = \frac{V^2 / 3R_B}{3V^2 / 2R_B} = \frac{1}{3} \times \frac{2}{3} = \frac{2}{9}$.

$(A)$ The capacitive reactance $X_C$ is given by $X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$.
Substituting the given values: $X_C = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} = \frac{1}{3140 \times 10^{-6}} = \frac{1000}{3.14} \approx 318.47 \Omega$.
The $RMS$ voltage is $V_{\text{rms}} = 210 \text{ V}$.
The $RMS$ current is $I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = \frac{210}{1000 / 3.14} = \frac{210 \times 3.14}{1000} = 0.6594 \text{ A}$.
The peak current $I_0$ is given by $I_0 = \sqrt{2} \times I_{\text{rms}}$.
$I_0 = 1.414 \times 0.6594 \approx 0.932 \text{ A}$.
Thus, the peak current is approximately $0.93 \text{ A}$.

(A) For a Wheatstone bridge to be balanced,the ratio of resistances in the two arms must be equal,i.e.,$\frac{R_1}{R_2} = \frac{R_3}{R_4}$.
In the given circuit,the left arm has resistances $10 \Omega$ and $15 \Omega$,so the ratio is $\frac{10}{15} = \frac{2}{3}$.
For the right arm,we need the effective resistance $R_{eff}$ such that $\frac{10}{R_{eff}} = \frac{2}{3}$,which implies $R_{eff} = 15 \Omega$.
The right arm consists of a $5 \Omega$ resistor in series with a diode $D$. If the diode is forward-biased and has a dynamic resistance $R_D = 10 \Omega$,the total resistance becomes $R_{eff} = 5 \Omega + 10 \Omega = 15 \Omega$.
Thus,the bridge is balanced when the diode is in series with the $5 \Omega$ resistor and is forward-biased. Looking at the options,the circuit where the diode is in series with the $5 \Omega$ resistor and oriented to be forward-biased allows for this balance.

(A) Given that the capacitor is connected to a battery,the potential difference $V$ remains constant.
$(i)$ Capacitance $C = \frac{\varepsilon_0 A}{d}$. As the plates are moved closer,the distance $d$ decreases,so the capacitance $C$ increases. Thus,statement $C$ is correct.
$(ii)$ Charge $Q = CV$. Since $C$ increases and $V$ is constant,the charge $Q$ increases. Thus,statement $A$ is correct.
$(iii)$ Energy stored $U = \frac{1}{2}CV^2$. Since $C$ increases and $V$ is constant,the energy $U$ increases. Thus,statement $B$ is incorrect.
$(iv)$ The ratio of charge to potential is $\frac{Q}{V} = C$. Since $C$ increases,the ratio $\frac{Q}{V}$ changes. Thus,statement $D$ is incorrect.
$(v)$ The product of charge and voltage is $QV = CV^2$. Since $C$ increases and $V$ is constant,the product $QV$ increases. Thus,statement $E$ is correct.
Therefore,statements $A, C,$ and $E$ are correct.

(A) According to the modified Ampere's circuital law,the total current is the sum of conduction current $(I_C)$ and displacement current $(I_D)$: $\oint B \cdot dl = \mu_0(I_C + I_D)$.
In the wire,there is only conduction current $I_C = I$,and the displacement current $I_D = 0$.
In the gap between the plates,there is no conduction current $(I_C = 0)$,but the changing electric field produces a displacement current $I_D$.
By the principle of continuity of current,the displacement current in the gap must be equal to the conduction current in the wire,i.e.,$I_D = I_C = I$.
This displacement current flows in the same direction as the conduction current to maintain the continuity of the total current in the circuit.

(C) Electromagnetic $(EM)$ waves are produced by accelerating charges.
According to the theory of electromagnetism,a charge moving with a uniform velocity produces a steady magnetic field and does not radiate energy in the form of $EM$ waves.
Therefore,the statement that they originate from charges moving with uniform speed is incorrect.
All other options are standard properties of $EM$ waves:
$1$. The energy density of the electric field $(u_E = \frac{1}{2} \varepsilon_0 E^2)$ is equal to the energy density of the magnetic field $(u_B = \frac{1}{2\mu_0} B^2)$.
$2$. The speed of $EM$ waves in free space is $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
$3$. $EM$ waves are transverse in nature,meaning the oscillations of electric and magnetic fields are perpendicular to the direction of wave propagation.

(A) . $A$ magnetic pole will exert an attractive or repulsive force on a magnetic sheet,so a force is required to hold it in place.
$B$. If the sheet is non-magnetic,there is no magnetic interaction,so no force is needed to hold it.
$C$. If the sheet is conducting,moving it through a magnetic field induces eddy currents. According to Lenz's Law,these currents create a force that opposes the motion. Therefore,an external force is required to move the sheet with uniform velocity.
$D$. $A$ non-conducting and non-polar sheet does not interact with the magnetic field of the pole,so no force is required to move it.

(A) Let the pole strength of the original iron bar be $m$. The magnetic moment is given by $M = m L$.
When the bar is bent at the middle,each arm has a length of $L/2$.
The new magnetic moment $M'$ is the product of the pole strength $m$ and the effective length (the distance between the two poles).
The effective length $L'$ is the distance between the two ends of the bent bar. Since the two arms of length $L/2$ make an angle of $60^{\circ}$,the effective length forms the third side of an isosceles triangle with two sides of length $L/2$ and an included angle of $60^{\circ}$.
This triangle is equilateral,so the effective length $L' = L/2$.
The new magnetic moment is $M' = m L' = m (L/2) = M/2$.