KCET 2021 Physics Question Paper with Answer and Solution

(B) Given: $m_1 = 1 \,kg$, $u_1 = 12 \,ms^{-1}$, $m_2 = 2 \,kg$, $u_2 = -24 \,ms^{-1}$ (negative sign indicates opposite direction).
Coefficient of restitution $e = 2/3$.
The formula for the coefficient of restitution is $e = \frac{v_2 - v_1}{u_1 - u_2}$.
Substituting the values: $\frac{2}{3} = \frac{v_2 - v_1}{12 - (-24)} = \frac{v_2 - v_1}{36}$.
Thus, $v_2 - v_1 = 24$ ... $(i)$.
By the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.
$(1)(12) + (2)(-24) = (1)v_1 + (2)v_2$.
$12 - 48 = v_1 + 2v_2 \Rightarrow v_1 + 2v_2 = -36$ ... (ii).
From $(i)$, $v_2 = v_1 + 24$. Substituting into (ii):
$v_1 + 2(v_1 + 24) = -36$.
$v_1 + 2v_1 + 48 = -36 \Rightarrow 3v_1 = -84 \Rightarrow v_1 = -28 \,ms^{-1}$.
Then $v_2 = -28 + 24 = -4 \,ms^{-1}$.
The velocities are $-28 \,ms^{-1}$ and $-4 \,ms^{-1}$.

(C) According to the law of conservation of linear momentum,if the net external force on a system is zero,the total momentum of the system remains conserved.
When a ball hits the floor,the interaction between the ball and the Earth (floor) involves internal forces. Since there is no external force acting on the system of the ball and the Earth,the total momentum of the system is conserved.
In an inelastic collision,kinetic energy is not conserved as some energy is dissipated as heat,sound,or deformation energy.
Therefore,the mechanical energy of the ball is not conserved,and the total mechanical energy of the ball and the Earth is also not conserved due to the energy loss during the inelastic collision.

(C) The total kinetic energy of an object in pure rolling motion is given by:
$E = \frac{1}{2} m v_{cm}^{2} \left(1 + \frac{k^{2}}{R^{2}}\right)$
where $k$ is the radius of gyration,$R$ is the radius,and $m$ is the mass of the object.
Given $m = 1 \ kg$,the expression becomes:
$\frac{E}{v_{cm}^{2}} = \frac{1}{2} \left(1 + \frac{k^{2}}{R^{2}}\right) \quad ...(i)$
From the given graph,the slope of the line is $\frac{E}{v_{cm}^{2}} = \frac{3}{4}$.
Substituting this value into Eq. $(i)$:
$\frac{3}{4} = \frac{1}{2} \left(1 + \frac{k^{2}}{R^{2}}\right)$
$\frac{3}{2} = 1 + \frac{k^{2}}{R^{2}}$
$\frac{k^{2}}{R^{2}} = \frac{3}{2} - 1 = \frac{1}{2}$
We know that for a disc,the moment of inertia $I = \frac{1}{2} m R^{2}$,so $k^{2} = \frac{1}{2} R^{2}$,which means $\frac{k^{2}}{R^{2}} = \frac{1}{2}$.
Thus,the object is a disc.

(B) The gravitational force between two masses is given by $F = \frac{G m_1 m_2}{r^2}$.
Given $m_A = m_B = 8 \,kg$, $m_G = 2 \,kg$, and $AG = BG = 1 \,m$.
The force exerted by mass at $A$ on mass at $G$ is $F_A = \frac{G \times 8 \times 2}{1^2} = 16G$.
The force exerted by mass at $B$ on mass at $G$ is $F_B = \frac{G \times 8 \times 2}{1^2} = 16G$.
The angle between $F_A$ and $F_B$ is $120^{\circ}$.
The resultant force $F_{AB}$ is given by $F_{AB} = \sqrt{F_A^2 + F_B^2 + 2 F_A F_B \cos 120^{\circ}}$.
Since $F_A = F_B = 16G$, $F_{AB} = \sqrt{(16G)^2 + (16G)^2 + 2(16G)(16G)(-0.5)} = \sqrt{3(16G)^2 - (16G)^2} = 16G$.
This resultant force $F_{AB}$ acts along the line $GC$ directed towards $C$.
To make the net force on the $2 \,kg$ body zero, a fourth body of mass $m_C = 4 \,kg$ must be placed at a distance $x$ from $G$ along the line $GC$ such that the gravitational force $F_C$ exerted by it balances $F_{AB}$.
$F_C = \frac{G \times 4 \times 2}{x^2} = 16G$.
$\frac{8G}{x^2} = 16G \Rightarrow x^2 = \frac{8}{16} = 0.5$.
$x = \frac{1}{\sqrt{2}} \,m$.
Thus, the fourth body should be placed at a point $P$ on the line $CG$ such that $PG = \frac{1}{\sqrt{2}} \,m$.

(B) The height of capillary rise $h$ of a liquid in a capillary tube is given by the formula:
$h = \frac{2 T \cos \theta}{r \rho g}$
Since $T$,$\theta$,$\rho$,and $g$ are constant for the same liquid and tube material,we have:
$h \propto \frac{1}{r}$
Since the diameter $D = 2r$,we can also write $h \propto \frac{1}{D}$.
Given that $h_P = \frac{2}{3} h_Q$,we have $\frac{h_P}{h_Q} = \frac{2}{3}$.
Using the inverse proportionality $h \propto \frac{1}{D}$,we get:
$\frac{h_P}{h_Q} = \frac{D_Q}{D_P} = \frac{2}{3}$
Therefore,the ratio of their diameters is $\frac{D_P}{D_Q} = \frac{3}{2}$ or $3: 2$.

(A) In transverse waves,the particle motion is perpendicular to the direction of wave propagation. This requires the medium to resist shearing stress,which is characterized by the shear modulus.
In longitudinal waves,the particles oscillate along the direction of wave propagation,causing volume changes. This requires the medium to resist compressive stress,which is characterized by the bulk modulus.
Therefore,for a medium to support both types of waves,it must possess both bulk modulus and shear modulus.

(B) The displacement of an object is given by the area under the $v-t$ graph.
$s = \text{Area under } v-t \text{ graph}$
In the given graph, the area under the uniform motion part (a rectangle) is greater than the area under the uniform acceleration part (a trapezoid) for the same time interval.
Therefore, the displacement during uniform acceleration is less than that during uniform motion.

(B) The change in velocity of a particle is equal to the area under the acceleration-time $(a-t)$ graph.
Given that the particle starts from rest,the initial velocity $u = 0$.
The area under the $a-t$ graph is a triangle with base $b = 10 \ s$ and height $h = 8 \ m/s^2$.
Area $= \frac{1}{2} \times \text{base} \times \text{height}$
Area $= \frac{1}{2} \times 10 \times 8 = 40 \ m/s$.
Since $\Delta v = v - u = \text{Area}$,and $u = 0$,we have $v = 40 \ m/s$.
Thus,the maximum speed of the particle is $40 \ m/s$.

(C) Given: Maximum range $R_{max} = 16 \,km = 16,000 \,m$ and acceleration due to gravity $g = 10 \,ms^{-2}$.
We know that the formula for the maximum range of a projectile is $R_{max} = \frac{u^2}{g}$,where $u$ is the muzzle velocity.
Substituting the given values into the formula:
$16,000 = \frac{u^2}{10}$
$u^2 = 16,000 \times 10$
$u^2 = 160,000$
$u = \sqrt{160,000}$
$u = 400 \,ms^{-1}$.
Thus,the muzzle velocity of the shell is $400 \,ms^{-1}$.

(D) The trajectory of a projectile is a parabola only when air resistance is neglected.
In the presence of air resistance,the projectile experiences a drag force that opposes its motion,causing both the range and the maximum height to decrease.
Consequently,the path deviates from a perfect parabola,making the statement 'a parabola always' incorrect.
Therefore,the trajectory is a parabola only in the absence of air resistance.

(A) In projectile motion,the acceleration is constant and directed vertically downwards (gravity).
Let the velocity vector be $\vec{v} = v_x \hat{i} + v_y \hat{j}$ and the acceleration vector be $\vec{a} = -g \hat{j}$.
The angle $\theta$ between $\vec{v}$ and $\vec{a}$ is given by $\cos \theta = \frac{\vec{v} \cdot \vec{a}}{|\vec{v}| |\vec{a}|} = \frac{-g v_y}{\sqrt{v_x^2 + v_y^2} \cdot g} = \frac{-v_y}{\sqrt{v_x^2 + v_y^2}}$.
As the projectile moves,$v_y$ changes from positive (upward) to negative (downward).
When $v_y > 0$,$\cos \theta$ is negative,meaning $\theta$ is obtuse $(> 90^{\circ})$.
When $v_y < 0$,$\cos \theta$ is positive,meaning $\theta$ is acute $(< 90^{\circ})$.
As $v_y$ becomes more negative,$\cos \theta$ increases,meaning $\theta$ decreases.
Thus,the angle is acute throughout the downward journey and reaches its minimum value at the point of impact (where $v_y$ is most negative).
However,in the context of standard physics problems,the angle is considered acute during the entire descent. The question asks for the point where it is minimum and acute; this occurs at the final point of the trajectory.

(D) Given: Initial velocity $\vec{u} = 10 \hat{j} \text{ ms}^{-1}$, Acceleration $\vec{a} = 8 \hat{i} + 2 \hat{j} \text{ ms}^{-2}$.
Using the kinematic equation $\vec{s} = \vec{u}t + \frac{1}{2} \vec{a}t^2$:
$\vec{s} = (10 \hat{j})t + \frac{1}{2} (8 \hat{i} + 2 \hat{j})t^2$
$\vec{s} = (4t^2) \hat{i} + (10t + t^2) \hat{j}$.
Comparing the components with $\vec{s} = x \hat{i} + y \hat{j}$, we get $x = 4t^2$ and $y = 10t + t^2$.
Given $x = 16 \text{ m}$, so $4t^2 = 16 \Rightarrow t^2 = 4 \Rightarrow t = 2 \text{ s}$.
Substituting $t = 2 \text{ s}$ into the expression for $y$:
$y = 10(2) + (2)^2 = 20 + 4 = 24 \text{ m}$.

(A) simple pendulum,in practice,consists of a heavy but small-sized metallic bob suspended by a light,inextensible,and flexible string.
It oscillates simple harmonically if and only if:
$(I)$ The size of the bob is negligible compared to the length of the string of the pendulum,which allows us to treat the bob as a point mass.
$(II)$ The angular amplitude (the angle between the vertical mean position and the string at the extreme point) is small,typically less than $10^{\circ}$,so that the approximation $\sin \theta \approx \theta$ holds true.
Thus,both statements are correct.

(A) When a coin is placed on a rotating turntable,the necessary centripetal force for circular motion is provided by the static frictional force between the coin and the surface of the turntable.
For the coin to just slip,the centripetal force must equal the maximum static frictional force:
$m r \omega^2 = \mu m g$
where $m$ is the mass of the coin,$r$ is the distance from the centre,$\omega$ is the angular velocity,$\mu$ is the coefficient of friction,and $g$ is the acceleration due to gravity.
Since $m, \mu$,and $g$ are constant,we have:
$r \omega^2 = \text{constant}$
This implies $r \propto \frac{1}{\omega^2}$.
Therefore,the ratio of distances for two different angular velocities is:
$\frac{r_2}{r_1} = \left( \frac{\omega_1}{\omega_2} \right)^2$
Given $r_1 = 4 \ cm$ and $\omega_2 = 2 \omega_1$,we substitute these values:
$\frac{r_2}{4} = \left( \frac{\omega_1}{2 \omega_1} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$
$r_2 = 4 \times \frac{1}{4} = 1 \ cm$.

(B) Volume expansion is the expansion of the volume of a gas due to an increase in its temperature.
$\therefore$ The coefficient of volume expansion $\alpha_{V}$ is defined as:
$\alpha_{V} = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_{p} ... (i)$
From the ideal gas equation,$pV = nRT$,we have $V = \frac{nRT}{p}$.
Differentiating with respect to $T$ at constant pressure $p$:
$\left( \frac{\partial V}{\partial T} \right)_{p} = \frac{nR}{p}$
Substituting this into equation $(i)$:
$\alpha_{V} = \frac{1}{V} \left( \frac{nR}{p} \right) = \frac{1}{V} \left( \frac{pV}{T} \cdot \frac{1}{p} \right) = \frac{1}{T}$
Thus,$\alpha_{V} = \frac{1}{T}$.
This implies that $\alpha_{V}$ is directly proportional to $\frac{1}{T}$. Therefore,a graph of $\alpha_{V}$ versus $\frac{1}{T}$ will be a straight line passing through the origin. This is correctly depicted in option $(b)$.

(C) The total internal energy of a gas is given by the formula:
$U = \frac{n f R T}{2}$
where $n$ is the number of moles and $f$ is the degree of freedom.
Given,$n_{\text{diatomic}} = n_{\text{monoatomic}} = 2$.
For a monoatomic gas,the degree of freedom $f_{\text{monoatomic}} = 3$.
For a diatomic gas,the degree of freedom $f_{\text{diatomic}} = 5$.
Calculating internal energy for each:
$U_{\text{monoatomic}} = \frac{2 \times 3 \times R T}{2} = 3 R T$
$U_{\text{diatomic}} = \frac{2 \times 5 \times R T}{2} = 5 R T$
Therefore,the total internal energy of the mixture is:
$U_{\text{total}} = U_{\text{monoatomic}} + U_{\text{diatomic}} = 3 R T + 5 R T = 8 R T$.

(B) The efficiency $(\eta)$ of a Carnot engine is given by the formula:
$\eta = 1 - \frac{T_{L}}{T_{H}}$
where $T_{L}$ is the temperature of the cold reservoir and $T_{H}$ is the temperature of the hot reservoir.
For a fixed cold reservoir temperature $T_{L}$,as the hot reservoir temperature $T_{H}$ increases,the ratio $\frac{T_{L}}{T_{H}}$ decreases.
Consequently,the efficiency $\eta = 1 - \frac{T_{L}}{T_{H}}$ increases as $T_{H}$ increases.
When $T_{H} = T_{L}$,the efficiency $\eta = 1 - \frac{T_{L}}{T_{L}} = 0$.
As $T_{H} \to \infty$,the efficiency $\eta \to 1$.
The graph of $\eta$ versus $T_{H}$ starts from $0$ at $T_{H} = T_{L}$ and increases with a decreasing slope (concave down) towards the value $1$.
This behavior is correctly represented by the graph in option $(b)$.

(B) The charge $q$ as a function of time $t$ is given by $q = q_0 \sin(\omega t)$,where $q_0$ is the maximum charge and $\omega$ is the angular frequency.
Differentiating with respect to $t$,the current $I$ is given by $I = \frac{dq}{dt} = \omega q_0 \cos(\omega t)$.
At $t = 0$,the current is $I = \omega q_0 \cos(0) = \omega q_0$.
Given $I = 2 \ A$ at $t = 0$,we have $2 = \omega q_0$.
Since $\omega = \frac{1}{\sqrt{LC}}$,we substitute this into the equation: $q_0 = I \sqrt{LC}$.
Given $L = 3 \times 10^{-3} \ H$ and $C = 2.7 \times 10^{-6} \ F$,we calculate:
$q_0 = 2 \times \sqrt{3 \times 10^{-3} \times 2.7 \times 10^{-6}}$
$q_0 = 2 \times \sqrt{8.1 \times 10^{-9}} = 2 \times \sqrt{81 \times 10^{-10}}$
$q_0 = 2 \times 9 \times 10^{-5} = 18 \times 10^{-5} \ C$.

(D) The induced emf in an inductor is given by the formula $|e| = L \left| \frac{dI}{dt} \right|$.
Given self-inductance $L = 6 \text{ mH} = 6 \times 10^{-3} \text{ H}$.
From the graph,at $t_1 = 20 \text{ s}$,the current $I_1 = 4 \text{ A}$,and at $t_2 = 40 \text{ s}$,the current $I_2 = 3 \text{ A}$.
The rate of change of current is $\left| \frac{dI}{dt} \right| = \left| \frac{I_2 - I_1}{t_2 - t_1} \right| = \left| \frac{3 - 4}{40 - 20} \right| = \left| \frac{-1}{20} \right| = 0.05 \text{ A/s}$.
Thus,the induced emf is $|e| = (6 \times 10^{-3} \text{ H}) \times (0.05 \text{ A/s}) = 0.3 \times 10^{-3} \text{ V} = 3 \times 10^{-4} \text{ V}$.
However,checking the options provided,$3 \times 10^{-4} \text{ V}$ is equivalent to $30 \times 10^{-5} \text{ V}$. Re-evaluating the calculation: $6 \times 10^{-3} \times (1/20) = 6/20 \times 10^{-3} = 0.3 \times 10^{-3} = 3 \times 10^{-4} \text{ V}$.
Given the options,$30 \times 10^{-5} \text{ V}$ is not listed,but $30 \times 10^{-4} \text{ V}$ is option $D$. Let's re-examine the graph. If the current at $t=40$ was $0$,the change would be $4/20 = 0.2$. $6 \times 10^{-3} \times 0.2 = 1.2 \times 10^{-3}$. If the current at $t=20$ was $4$ and $t=40$ was $3$,the slope is $1/20$. The calculation $3 \times 10^{-4} \text{ V}$ is correct. Since $30 \times 10^{-5} \text{ V}$ is not an option,and $3 \times 10^{-4} \text{ V}$ is the result,we select the closest magnitude if applicable,or note the discrepancy. Given the options,$30 \times 10^{-5} \text{ V}$ is $3 \times 10^{-4} \text{ V}$. Option $D$ is $30 \times 10^{-4} \text{ V} = 3 \times 10^{-3} \text{ V}$. There might be a typo in the question's options. Assuming the intended answer is $3 \times 10^{-4} \text{ V}$.

(B) The energy of an electron in the $n$th orbit of a hydrogen-like atom is given by the formula:
$E_{n} = -13.6 \text{ eV} \times \frac{Z^{2}}{n^{2}}$
This implies that $E_{n} \propto \frac{Z^{2}}{n^{2}}$.
For a hydrogen atom,the atomic number $Z = 1$. Thus,for the second orbit $(n = 2)$:
$E_{2} = k \times \frac{1^{2}}{2^{2}} = \frac{k}{4}$,where $k = -13.6 \text{ eV}$.
For a $He^{+}$ ion,the atomic number $Z = 2$. For the third orbit $(n = 3)$:
$E_{3} = k \times \frac{2^{2}}{3^{2}} = \frac{4k}{9}$.
Now,we find the ratio of $E_{3}$ to $E_{2}$:
$\frac{E_{3}}{E_{2}} = \frac{4k/9}{k/4} = \frac{4}{9} \times 4 = \frac{16}{9}$.
Therefore,$E_{3} = \frac{16}{9} E_{2}$.

(B) According to Bohr's postulate, the angular momentum $L$ is given by $L = \frac{n h}{2 \pi}$.
Given $L = \frac{3 h}{2 \pi}$, we get $n = 3$.
The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.
From the angular momentum quantization, $mvr = \frac{nh}{2\pi} = \frac{3h}{2\pi}$, so $mv = \frac{3h}{2\pi r}$.
Substituting this into the wavelength formula: $\lambda = \frac{h}{mv} = \frac{h \cdot 2\pi r}{3h} = \frac{2}{3} \pi r$.
The radius of the $n$-th orbit for a hydrogen-like ion is $r = a_{0} \frac{n^2}{Z}$.
For $Li^{2+}$, $Z = 3$ and $n = 3$, so $r = a_{0} \frac{3^2}{3} = 3 a_{0}$.
Substituting $r$ into the expression for $\lambda$: $\lambda = \frac{2}{3} \pi (3 a_{0}) = 2 \pi a_{0}$.
Comparing this with the given form $p \pi a_{0}$, we find $p = 2$.

(A) Let $R$ and $r$ be the radii of the bigger and each smaller drop, respectively.
Volume of the bigger drop $= 8 \times$ Volume of a smaller drop.
$\frac{4}{3} \pi R^{3} = 8 \times \frac{4}{3} \pi r^{3}$
$R^{3} = 8r^{3} \Rightarrow R = 2r$ ... $(i)$
As the capacitance of a spherical conductor is given by $C = 4 \pi \varepsilon_{0} r$, we have $C \propto r$.
Therefore, the ratio of the capacitance of the bigger drop to the smaller drop is:
$\frac{C_{\text{bigger}}}{C_{\text{smaller}}} = \frac{R}{r} = \frac{2r}{r} = 2$.
Thus, the capacitance of the bigger drop is $2$ times that of each smaller drop.

(B) Let $t$ be the thickness of the dielectric slab and $K$ be the dielectric constant.
When a dielectric slab of thickness $t$ is introduced,the effective separation between the plates increases by an amount $x = t(1 - 1/K)$ to maintain the same capacitance.
Given,$x = 3.5 \times 10^{-3} \ m$ and $t = 4 \times 10^{-3} \ m$.
Substituting these values into the equation:
$3.5 \times 10^{-3} = 4 \times 10^{-3} \left(1 - \frac{1}{K}\right)$
Dividing both sides by $4 \times 10^{-3}$:
$\frac{3.5}{4} = 1 - \frac{1}{K}$
$0.875 = 1 - \frac{1}{K}$
$\frac{1}{K} = 1 - 0.875 = 0.125$
$K = \frac{1}{0.125} = 8$.
Thus,the dielectric constant of the material is $8$.

(B) The energy stored in a capacitor is given by the formula:
$U = \frac{1}{2} C V^{2} \dots (i)$
where $C$ is the capacitance and $V$ is the potential difference.
From the given graph,the slope is:
$\text{Slope} = \tan \theta = \frac{Q}{V} = \frac{120 \mu C}{10 \text{ V}} = 12 \mu F \dots (ii)$
Since the capacitance $C = \frac{Q}{V}$,we have:
$C = 12 \mu F = 12 \times 10^{-6} \text{ F}$
Now,substituting the values of $C = 12 \times 10^{-6} \text{ F}$ and $V = 10 \text{ V}$ into Eq. $(i)$:
$U = \frac{1}{2} \times (12 \times 10^{-6} \text{ F}) \times (10 \text{ V})^{2}$
$U = \frac{1}{2} \times 12 \times 10^{-6} \times 100$
$U = 6 \times 10^{-4} \text{ J} = 600 \times 10^{-6} \text{ J} = 600 \mu J$
Thus,the capacitance is $12 \mu F$ and the energy stored is $600 \mu J$.

(D) To withstand a potential difference of $1.5 kV$ $(1500 V)$ using capacitors that can withstand $500 V$ each,the number of capacitors $(m)$ in each series row is:
$m = \frac{1500 V}{500 V} = 3$
The equivalent capacitance of one such row of $3$ capacitors (each $2 \mu F$) is:
$C_{row} = \frac{2 \mu F}{3}$
If we connect $n$ such rows in parallel to achieve a total capacitance of $6 \mu F$,we have:
$C_{eff} = n \times C_{row} = n \times \frac{2}{3} \mu F = 6 \mu F$
$n = \frac{6 \times 3}{2} = 9$
The total number of capacitors required is $N = m \times n = 3 \times 9 = 27$.

(B) Given, $X_{L} = 4 \, \Omega$ and $X_{C} = 4 \, \Omega$.
In a series $LC$ circuit, the voltage across the inductor $(V_{L})$ and the capacitor $(V_{C})$ are $180^{\circ}$ out of phase.
Therefore, the net voltage across the series combination of the inductor and capacitor is $V_{\text{net}} = |V_{L} - V_{C}|$.
Since $X_{L} = X_{C}$, the magnitudes of the voltages are equal, i.e., $V_{L} = I X_{L}$ and $V_{C} = I X_{C}$.
Thus, $V_{\text{net}} = I(X_{L} - X_{C}) = I(4 - 4) = 0 \, V$.
The voltmeter is connected across this series $LC$ combination, so its reading is $0 \, V$.
The total impedance of the circuit is $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$.
Substituting the values, $Z = \sqrt{45^{2} + (4 - 4)^{2}} = \sqrt{45^{2}} = 45 \, \Omega$.
The current in the circuit is $I = \frac{V}{Z} = \frac{90 \, V}{45 \, \Omega} = 2 \, A$.
Therefore, the voltmeter reading is $0 \, V$ and the ammeter reading is $2 \, A$.

(D) Given: Length $l = 1 \,m$,Area $A = 5 \times 10^{-7} \,m^{2}$,Current $I = 1 \,A$,Electron density $n = 8 \times 10^{28} \,m^{-3}$,Charge of electron $e = 1.6 \times 10^{-19} \,C$.
The drift velocity $v_{d}$ is given by the formula $v_{d} = \frac{I}{neA}$.
Substituting the values: $v_{d} = \frac{1}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-7}} = \frac{1}{64 \times 10^{2}} = \frac{1}{6.4 \times 10^{3}} \,m/s$.
The time $T$ taken to drift across the length $l$ is $T = \frac{l}{v_{d}}$.
$T = \frac{1}{1 / (6.4 \times 10^{3})} = 6.4 \times 10^{3} \,s$.

(D) When a charge $\Delta q$ moves through a potential difference $V$,the change in its electric potential energy is given by $\Delta U = \Delta q \cdot V$. Since $I = \frac{\Delta q}{\Delta t}$,we have $\Delta q = I \Delta t$. Thus,the potential energy decreases by $I V \Delta t$.
In a steady current,the drift velocity of the charge carriers remains constant,meaning the kinetic energy of the charge does not increase significantly. Instead,the lost potential energy is dissipated as heat due to collisions with the lattice ions of the conductor.
Therefore,the thermal energy of the conductor increases by $I V \Delta t$.
Statements $II$ and $III$ are correct.

(A) In a meter bridge,the condition for null deflection is given by the Wheatstone bridge principle: $\frac{R_1}{R_2} = \frac{R_{AD}}{R_{DB}}$.
Here,$R_{AD}$ is the resistance of the wire segment $AD$ and $R_{DB}$ is the resistance of the wire segment $DB$.
Let $l$ be the length of the wire $AD$,then the length of $DB$ is $(100 - l)$.
The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $\rho$ is resistivity,$L$ is length,and $A$ is the cross-sectional area.
Substituting this into the balance condition: $\frac{R_1}{R_2} = \frac{\rho l / A}{\rho (100 - l) / A} = \frac{l}{100 - l}$.
Since the cross-sectional area $A$ cancels out from the numerator and denominator,the balancing length $l$ (which is $X$) is independent of the radius (and thus the area of cross-section) of the wire.
Therefore,if the radius of the wire $AB$ is doubled,the balancing length will remain $X$.

(D) Given,initial resistance,$R_{1} = 3 \Omega$.
Let the original length of the wire be $l$. When the wire is stretched to twice its length,the new length becomes $l^{\prime} = 2l$.
Since the volume of the wire remains constant during stretching,$V = A \times l = A^{\prime} \times l^{\prime}$.
Therefore,$A^{\prime} = \frac{A \times l}{l^{\prime}} = \frac{A \times l}{2l} = \frac{A}{2}$.
The resistance of a wire is given by $R = \rho \frac{l}{A}$.
The new resistance $R_{2}$ is $R_{2} = \rho \frac{l^{\prime}}{A^{\prime}} = \rho \frac{2l}{A/2} = 4 \left( \rho \frac{l}{A} \right) = 4 R_{1}$.
Substituting the value of $R_{1}$,we get $R_{2} = 4 \times 3 \Omega = 12 \Omega$.

(D) For a proton,the total energy $E_{total} = E_k + mc^2 = \frac{1}{8}mc^2 + mc^2 = \frac{9}{8}mc^2$.
Using the relativistic relation $E_{total}^2 = (pc)^2 + (mc^2)^2$:
$(\frac{9}{8}mc^2)^2 = (p_1c)^2 + (mc^2)^2$
$\frac{81}{64}m^2c^4 = p_1^2c^2 + m^2c^4$
$p_1^2c^2 = (\frac{81}{64} - 1)m^2c^4 = \frac{17}{64}m^2c^4$
$p_1 = \frac{\sqrt{17}}{8}mc$.
For a photon,energy $E = p_2c = E_k = \frac{1}{8}mc^2$,so $p_2 = \frac{mc}{8}$.
The ratio $\frac{p_1 - p_2}{p_1} = 1 - \frac{p_2}{p_1} = 1 - \frac{mc/8}{\sqrt{17}mc/8} = 1 - \frac{1}{\sqrt{17}}$.
Note: If we assume non-relativistic kinetic energy $E_k = \frac{p_1^2}{2m} = \frac{1}{8}mc^2$,then $p_1^2 = \frac{1}{4}m^2c^2$,so $p_1 = \frac{mc}{2}$.
Then $p_2 = \frac{E_k}{c} = \frac{mc}{8}$.
The ratio $\frac{p_1 - p_2}{p_1} = \frac{mc/2 - mc/8}{mc/2} = \frac{3/8}{1/2} = \frac{3}{4}$.

(D) According to the de-Broglie hypothesis,the circumference of a stationary orbit must be an integral number of wavelengths:
$n \lambda = 2 \pi r_n$
Also,the angular momentum of the electron is given by:
$m v_n r_n = \frac{n h}{2 \pi}$
From the Bohr model,the velocity of an electron in the $n^{th}$ orbit is $v_n = \frac{e^2}{2 n h \varepsilon_0}$.
Substituting this into the angular momentum equation:
$m \left( \frac{e^2}{2 n h \varepsilon_0} \right) r_n = \frac{n h}{2 \pi}$
$r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2}$
In the given figure,counting the number of standing wave loops (or wavelengths),we find $n = 6$.
Therefore,the radius is:
$r_n = \frac{6^2 h^2 \varepsilon_0}{\pi m e^2} = \frac{36 h^2 \varepsilon_0}{\pi m e^2}$

(D) Given: Work-function, $\phi_{0} = 1 eV = 1.6 \times 10^{-19} J$.
Wavelength, $\lambda = 3000 \text{Å} = 3000 \times 10^{-10} m$.
The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3000 \times 10^{-10}} = 6.63 \times 10^{-19} J$.
According to Einstein's photoelectric equation, $E = \phi_{0} + KE_{max}$.
$KE_{max} = E - \phi_{0} = 6.63 \times 10^{-19} - 1.6 \times 10^{-19} = 5.03 \times 10^{-19} J$.
Since $KE_{max} = \frac{1}{2}mv^{2}$, we have $v = \sqrt{\frac{2 \times KE_{max}}{m}}$.
$v = \sqrt{\frac{2 \times 5.03 \times 10^{-19}}{9.1 \times 10^{-31}}} = \sqrt{1.105 \times 10^{12}} \approx 1.05 \times 10^{6} m/s$.
Thus, the velocity is approximately $1 \times 10^{6} m/s$.

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ of an ejected photoelectron is given by:
$K_{max} = h\nu - \phi_0$
where $h$ is Planck's constant,$\nu$ is the frequency of incident radiation,and $\phi_0$ is the work function of the metal.
Since $\phi_0 = h\nu_0$,where $\nu_0$ is the threshold frequency,we can write:
$K_{max} = h\nu - h\nu_0 = h(\nu - \nu_0)$
This equation represents a straight line of the form $y = mx + c$,where the slope is $h$ and the intercept on the frequency axis is $\nu_0$.
For frequencies $\nu < \nu_0$,the kinetic energy is zero because no photoemission occurs. For $\nu \geq \nu_0$,the kinetic energy increases linearly with frequency. This behavior is correctly represented by graph $D$.

(D) The self-inductance $(L)$ and mutual inductance $(M)$ both have the same unit.
Both are defined as the magnetic flux $(\phi)$ per unit current $(I)$.
Mathematically,$M = L = \frac{\phi}{I}$.
The unit of magnetic flux $(\phi)$ is Weber $(\text{wb})$ and the unit of current $(I)$ is Ampere $(\text{A})$.
Therefore,the unit of both self-inductance and mutual inductance is $\text{wb A}^{-1}$ (also known as Henry $(\text{H})$).
Thus,both $(a)$ and $(b)$ are correct.

(C) Given: $E_{0}=120 \text{ NC}^{-1}$,$f=50 \text{ MHz} = 50 \times 10^{6} \text{ Hz}$.
$(a)$ Magnetic field amplitude: $B_{0} = \frac{E_{0}}{c} = \frac{120}{3 \times 10^{8}} = 40 \times 10^{-8} \text{ T} = 400 \text{ nT}$. (Correct)
$(b)$ Angular frequency: $\omega = 2\pi f = 2\pi \times 50 \times 10^{6} = \pi \times 10^{8} \text{ rad/s}$. (Correct)
$(c)$ Propagation constant: $k = \frac{\omega}{c} = \frac{\pi \times 10^{8}}{3 \times 10^{8}} = \frac{\pi}{3} \approx 1.047 \text{ rad/m}$. The given value $2.1 \text{ rad/m}$ is incorrect.
$(d)$ Wavelength: $\lambda = \frac{c}{f} = \frac{3 \times 10^{8}}{50 \times 10^{6}} = 6 \text{ m}$. (Correct)

(B) According to the principles of electromagnetism,electromagnetic waves are produced by accelerated charges.
An electric charge at rest produces only a static electric field.
$A$ charge moving with a constant velocity produces a steady electric current,which creates a constant magnetic field,but does not radiate energy as electromagnetic waves.
$A$ charge moving in a circular orbit is undergoing centripetal acceleration,which is a form of accelerated motion.
Therefore,a charge moving in a circular orbit acts as a source of electromagnetic waves.

(B) The electric field intensity $E$ at a perpendicular distance $r$ from an infinitely long,straight,uniformly charged wire with linear charge density $\lambda$ is given by Gauss's Law as:
$E = \frac{\lambda}{2 \pi \varepsilon_{0} r}$
Here,$\lambda$ is the linear charge density,$\varepsilon_{0}$ is the permittivity of free space,and $\pi$ is a constant.
Since $\lambda$,$\pi$,and $\varepsilon_{0}$ are constants,the relationship between the electric field and distance is:
$E \propto \frac{1}{r}$
Therefore,the electric field varies inversely with the distance $r$.

(C) Polar molecules are molecules in which the centres of positive and negative charges are separated even when there is no external electric field.
Such molecules possess a permanent electric dipole moment.
Examples include $HCl$,$H_{2}O$,etc.
Therefore,the statement in option $(C)$ is false because polar molecules do possess permanent dipole moments.

(A) In circular motion,the net centripetal force on an electron of mass $m$ at a distance $x$ from the axis of rotation is given by $F_{c} = m \omega^{2} x$.
When the rod rotates,the electrons within the rod also rotate,experiencing this centripetal force.
This centripetal force is provided by the induced electric field $E$ acting on the electron,such that $F_{e} = e E$.
Equating the two forces,we have $e E = m \omega^{2} x$.
Solving for the electric field $E$,we get $E = \frac{m \omega^{2} x}{e}$.

(C) Given: Mass $m = 2 \ g = 2 \times 10^{-3} \ kg$.
Electric field $E = 300 \ NC^{-1}$.
Initial kinetic energy $(KE)_1 = 0$ at $x = 0$.
Final kinetic energy $(KE)_2 = 0.12 \ J$ at $x = 0.5 \ m$.
According to the work-energy theorem,the work done by the electric force is equal to the change in kinetic energy.
Work done $W = F \cdot d = (QE) \cdot d$.
Change in kinetic energy $\Delta KE = KE_2 - KE_1 = 0.12 \ J - 0 \ J = 0.12 \ J$.
Equating the two: $Q \times 300 \times 0.5 = 0.12$.
$Q \times 150 = 0.12$.
$Q = \frac{0.12}{150} = 0.0008 \ C$.
$Q = 800 \times 10^{-6} \ C = 800 \ \mu C$.
Since the object gains kinetic energy moving in the direction of the electric field,the charge $Q$ must be positive.

(A) Given:
$r = 10 \,cm = 0.1 \,m$
$I_1 = I_2 = 10 \,A$
The force per unit length between two parallel current-carrying wires is given by the formula:
$\frac{F}{l} = \frac{\mu_0 I_1 I_2}{2 \pi r}$
Substituting the values:
$\frac{F}{l} = \frac{(4 \pi \times 10^{-7} \,T \cdot m/A) \times (10 \,A) \times (10 \,A)}{2 \pi \times 0.1 \,m}$
$\frac{F}{l} = \frac{2 \times 10^{-7} \times 100}{0.1} = 2 \times 10^{-4} \,N/m$
Since the currents are in the same direction,the force is attractive.

(C) The magnetic field $B$ due to a toroid is determined using Ampere's circuital law:
$(i)$ For $r < R_{1}$ (inside the hollow space of the toroid),the net current enclosed is zero,so $B = 0$.
(ii) For $R_{1} < r < R_{2}$ (inside the core of the toroid),the magnetic field is given by $B = \frac{\mu_{0} N I}{2 \pi r}$.
(iii) For $r > R_{2}$ (outside the toroid),the net current enclosed by the Amperian loop is zero,so $B = 0$.
Since the question asks for the variation of the magnetic field with radial distance $r$ within the core,and the options provided represent the behavior across the cross-section,the correct representation for the field inside the toroid is a curve proportional to $1/r$,which is shown in graph $C$.

(B) The magnetic field inside a long solenoid is uniform and directed along its axis,given by $B = \mu_{0} n I$.
When a particle of charge $q$ and mass $m$ is projected perpendicular to the axis,it experiences a Lorentz force $F = q(v \times B)$. Since $v \perp B$,the force is $F = qvB$,which acts as the centripetal force.
The particle moves in a circular path of radius $R_{c} = \frac{mv}{qB}$.
For the particle not to strike the solenoid wall,the diameter of its circular path must be less than or equal to the radius of the solenoid $r$.
Thus,$2R_{c} \leq r$,which implies $R_{c} \leq \frac{r}{2}$.
Substituting $R_{c} = \frac{mv}{qB}$,we get $\frac{mv}{qB} \leq \frac{r}{2}$.
Solving for $v$,we get $v \leq \frac{qBr}{2m}$.
Substituting $B = \mu_{0} n I$,the maximum speed is $v_{max} = \frac{\mu_{0} n I q r}{2m}$.

(C) The magnetic force $F$ acting on a charged particle moving with velocity $v$ in a magnetic field $B$ is given by the Lorentz force formula: $F = q(v \times B)$.
Since the electron is stationary,its velocity $v = 0$.
Substituting this into the formula,we get $F = q(0 \times B) = 0$.
Since the net force acting on the electron is zero,it will remain stationary.

(B) At the magnetic poles,the angle of dip (inclination),$\delta = 90^{\circ}$.
The horizontal component of the Earth's magnetic field $(B_{H})$ is given by the formula:
$B_{H} = B \cos \delta$
where $B$ is the net magnetic field of the Earth.
Substituting the value of $\delta$ at the poles:
$B_{H} = B \cos 90^{\circ}$
Since $\cos 90^{\circ} = 0$,we get:
$B_{H} = 0$
Therefore,at the magnetic poles,the horizontal component of the Earth's magnetic field is zero.

(D) Option $(A)$ and $(B)$ represent electric field lines due to isolated positive and negative charges,respectively. However,magnetic monopoles do not exist in nature,so these patterns are not valid for magnetic field lines.
Option $(C)$ represents circular magnetic field lines around a current-carrying conductor. Electric field lines cannot form closed loops,so this is not valid for an electric field.
Option $(D)$ represents the field lines of a dipole. This pattern is valid for an electric dipole (electric field lines originating from positive and terminating at negative charge) and also for a magnetic dipole (magnetic field lines forming continuous loops from North to South pole outside and South to North inside).
Thus,option $(D)$ is valid for both electric and magnetic fields.

(D) The potential energy $U$ of a pair of nucleons as a function of their separation $r$ is characterized by a deep potential well.
For separations $r < r_{0}$ (where $r_{0} \approx 0.8 \ \text{fm}$), the nuclear force is strongly repulsive, causing the potential energy to rise sharply.
For separations $r > r_{0}$, the nuclear force is attractive, and the potential energy increases towards zero as $r$ increases.
The minimum potential energy occurs at the equilibrium separation $r_{0}$, where the net force between the nucleons is zero.
Graph $D$ correctly depicts this behavior with $U$ in $\text{MeV}$ and $r$ in $\text{fm}$.

(B) The moderator used in a nuclear reactor must have light nuclei (like protons or deuterium).
When neutrons undergo a perfectly elastic collision with light nuclei,they transfer a significant portion of their kinetic energy to the nuclei,effectively slowing down the neutrons.
In contrast,if neutrons collide with heavy nuclei,the mass of the heavy nucleus is much larger than that of the neutron.
According to the laws of conservation of momentum and energy,in an elastic collision with a very heavy object,the velocity of the incident particle (neutron) remains nearly unchanged.
Therefore,heavy nuclei cannot effectively slow down neutrons and are not suitable as moderators.

(C) For $LC$-oscillations,the differential equation is given by $L \frac{di}{dt} + \frac{q}{C} = 0$.
Differentiating with respect to time $t$,we get $L \frac{d^2i}{dt^2} + \frac{1}{C} \frac{dq}{dt} = 0$. Since $i = \frac{dq}{dt}$,this becomes $L \frac{d^2i}{dt^2} + \frac{1}{C} i = 0$ ... $(i)$.
For a mechanical spring-block system,the equation of motion is $m \frac{d^2x}{dt^2} + kx = 0$ ... $(ii)$.
Comparing equation $(i)$ and $(ii)$,we observe that the mass $m$ is analogous to inductance $L$,and the force constant $k$ is analogous to the reciprocal of capacitance,i.e.,$k \propto \frac{1}{C}$.

(D) According to the Cartesian sign convention:
Focal length of the first biconvex lens $= f_{1}$
Focal length of the second biconvex lens $= f_{2}$
Focal length of the third biconcave lens $= -f_{3}$
For a combination of thin lenses in contact,the total power $P = P_{A} + P_{B} = \frac{1}{f_{A}} + \frac{1}{f_{B}}$.
For the first and second lenses: $P_{1} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{f_{1} + f_{2}}{f_{1} f_{2}}$.
For the first and third lenses: $P_{2} = \frac{1}{f_{1}} + \frac{1}{-f_{3}} = \frac{1}{f_{1}} - \frac{1}{f_{3}} = \frac{f_{3} - f_{1}}{f_{1} f_{3}}$.
For the second and third lenses: $P_{3} = \frac{1}{f_{2}} + \frac{1}{-f_{3}} = \frac{1}{f_{2}} - \frac{1}{f_{3}} = \frac{f_{3} - f_{2}}{f_{2} f_{3}}$.

(D) Given,refractive index of glass with respect to air,${ }_{a} \mu_{g} = \frac{3}{2}$.
Refractive index of water with respect to air,${ }_{a} \mu_{w} = \frac{4}{3}$.
Using the Lens Maker's formula: $\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a lens in air: $\frac{1}{f_a} = ({ }_{a} \mu_{g} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.
For a lens in water: $\frac{1}{f_w} = ({ }_{w} \mu_{g} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,where ${ }_{w} \mu_{g} = \frac{{ }_{a} \mu_{g}}{{ }_{a} \mu_{w}} = \frac{3/2}{4/3} = \frac{9}{8}$.
Taking the ratio $\frac{f_w}{f_a}$:
$\frac{f_w}{f_a} = \frac{{ }_{a} \mu_{g} - 1}{{ }_{w} \mu_{g} - 1} = \frac{3/2 - 1}{9/8 - 1} = \frac{1/2}{1/8} = \frac{1}{2} \times 8 = 4$.
Thus,the ratio is $4: 1$.

(B) Given: Focal length of convex lens $f_{1} = 30 \,cm$, focal length of concave lens $f_{2} = -20 \,cm$, and initial image size $h_{o} = 2 \,cm$.
Since the object is at infinity, the image formed by the convex lens is at its focus, $v_{1} = 30 \,cm$.
The concave lens is placed at a distance of $26 \,cm$ from the convex lens. Thus, the image formed by the convex lens acts as a virtual object for the concave lens.
The distance of this virtual object from the concave lens is $u_{2} = v_{1} - 26 = 30 - 26 = 4 \,cm$.
Using the lens formula for the concave lens: $\frac{1}{v_{2}} - \frac{1}{u_{2}} = \frac{1}{f_{2}}$.
Substituting the values: $\frac{1}{v_{2}} - \frac{1}{4} = \frac{1}{-20}$.
$\frac{1}{v_{2}} = \frac{1}{4} - \frac{1}{20} = \frac{5-1}{20} = \frac{4}{20} = \frac{1}{5}$.
So, $v_{2} = 5 \,cm$.
The magnification $m$ produced by the concave lens is $m = \frac{v_{2}}{u_{2}} = \frac{5}{4} = 1.25$.
The new size of the image $h_{i} = m \times h_{o} = 1.25 \times 2 \,cm = 2.5 \,cm$.

(B) The bending of light when passing from one medium to another is due to the change in the speed of light.
Refraction occurs because the refractive index of the second medium is different from that of the first medium.
The refractive index $( \mu )$ is defined as the ratio of the speed of light in a vacuum $( c )$ to the speed of light in the medium $( v )$,i.e.,$ \mu = c/v $.
When light travels from one medium to another,its frequency remains constant,but its speed changes,which causes the light ray to deviate from its original path (bending).
Therefore,the correct reason for the bending of light is that the speed of light is different in the second medium.

(D) The correct answer is $D$. An $LED$ (Light Emitting Diode) is a heavily doped $p-n$ junction diode,not a lightly doped one. When it is forward biased,the recombination of electrons and holes releases energy in the form of photons (light). Options $A$,$B$,and $C$ are correct statements regarding semiconductor diodes.

(B) The given circuit consists of two $NAND$ gates connected in series,where the second $NAND$ gate acts as a $NOT$ gate because its inputs are shorted together.
Let the inputs to the first $NAND$ gate be $A$ and $B$. The output of the first $NAND$ gate is $\overline{A \cdot B}$.
This output is fed as the input to the second $NAND$ gate. Since both inputs of the second $NAND$ gate are connected to the same signal,its output $Y$ is given by:
$Y = \overline{(\overline{A \cdot B}) \cdot (\overline{A \cdot B})}$
Using the property of Boolean algebra $\overline{X \cdot X} = \overline{X}$,we get:
$Y = \overline{(\overline{A \cdot B})} = A \cdot B$
Thus,the circuit performs the operation of an $AND$ gate.

(C) The energy of a photon corresponding to a wavelength $\lambda$ is given by $E = \frac{1240}{\lambda (nm)} eV$.
For $\lambda = 600 nm$,the energy of the incident photon is $E = \frac{1240}{600} \approx 2.07 eV$.
$A$ photodiode can detect light only if the energy of the incident photon is greater than or equal to the band gap energy $(E_{g})$ of the semiconductor material $(E \ge E_{g})$.
Comparing the photon energy $(2.07 eV)$ with the band gaps:
For $D_{1}$: $E_{g} = 2.5 eV$. Since $2.07 eV < 2.5 eV$,$D_{1}$ cannot detect this light.
For $D_{2}$: $E_{g} = 2 eV$. Since $2.07 eV > 2 eV$,$D_{2}$ can detect this light.
For $D_{3}$: $E_{g} = 3 eV$. Since $2.07 eV < 3 eV$,$D_{3}$ cannot detect this light.
Therefore,only $D_{2}$ will be able to detect the light of wavelength $600 nm$.

(B) Given: Wavelength $\lambda = 6500 Å = 6500 \times 10^{-10} \text{ m}$,Angle $\theta = 30^{\circ}$.
For the diffraction pattern of a single slit,the condition for the $n^{\text{th}}$ minimum is given by $a \sin \theta = n \lambda$.
For the first diffraction minimum,$n = 1$.
Substituting the values: $a \sin 30^{\circ} = 1 \times (6500 \times 10^{-10} \text{ m})$.
Since $\sin 30^{\circ} = 0.5$,we have $a \times 0.5 = 6500 \times 10^{-10} \text{ m}$.
$a = 2 \times 6500 \times 10^{-10} \text{ m} = 13000 \times 10^{-10} \text{ m}$.
$a = 1.3 \times 10^{-6} \text{ m} = 1.3 \text{ micron}$.

(D) In a single slit diffraction pattern,the width of the central maxima is given by $w = 2\lambda D/a$,while the width of secondary maxima is $\lambda D/a$.
Thus,the central maxima is twice as wide as the secondary maxima (Statement $I$ is correct).
The intensity of the secondary maxima decreases rapidly as we move away from the central maxima (Statement $II$ is correct).
The width of the central maxima is inversely proportional to the slit width $a$ (Statement $III$ is incorrect).
The intensity of the central maxima is much higher than that of the secondary maxima (Statement $IV$ is incorrect).
Therefore,statements $I$ and $II$ are correct.

(B) The path difference between the waves reaching point $P$ from the two slits is given by $\Delta x = \frac{xd}{D}$.
The corresponding phase difference $\phi$ is given by $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \left(\frac{xd}{D}\right)$.
The resultant intensity $I_P$ at point $P$ for two coherent sources of equal intensity $I_0$ is given by the formula $I_P = 4I_0 \cos^2\left(\frac{\phi}{2}\right)$.
Substituting the value of $\phi$:
$I_P = 4I_0 \cos^2\left(\frac{1}{2} \cdot \frac{2\pi xd}{\lambda D}\right)$
$I_P = 4I_0 \cos^2\left(\frac{\pi dx}{\lambda D}\right)$.