NEET 2018 Physics Question Paper with Answer and Solution

(C) The diameter of the ball is calculated using the formula: $\text{Diameter} = \text{MSR} + (\text{CSR} \times \text{LC}) - \text{Zero Error}$.
Given: $\text{MSR} = 5 \, mm = 0.5 \, cm$,$\text{CSR} = 25$,$\text{LC} = 0.001 \, cm$,and $\text{Zero Error} = -0.004 \, cm$.
Substituting the values:
$\text{Diameter} = 0.5 \, cm + (25 \times 0.001 \, cm) - (-0.004 \, cm)$.
$\text{Diameter} = 0.5 \, cm + 0.025 \, cm + 0.004 \, cm$.
$\text{Diameter} = 0.529 \, cm$.

(C) To complete a vertical circle of radius $R$,the minimum speed at the lowest point $A$ must be $v_A = \sqrt{5gR}$.
Given that the diameter of the circle is $D$,the radius is $R = \frac{D}{2}$.
Substituting $R$ into the velocity equation,we get $v_A = \sqrt{5g \left(\frac{D}{2}\right)} = \sqrt{\frac{5gD}{2}}$.
Using the principle of conservation of mechanical energy,the potential energy at height $h$ is converted into kinetic energy at point $A$:
$mgh = \frac{1}{2}mv_A^2$
$gh = \frac{1}{2}v_A^2$
Substituting the value of $v_A^2 = \frac{5gD}{2}$:
$gh = \frac{1}{2} \left(\frac{5gD}{2}\right)$
$h = \frac{5D}{4}$

(C) To keep the block stationary on the wedge,we analyze the forces in the non-inertial frame of the wedge.
$1$. The forces acting on the block are: gravitational force $mg$ (downwards),normal reaction $N$ (perpendicular to the surface),and pseudo force $ma$ (towards the left).
$2$. Resolving the forces into components parallel and perpendicular to the inclined surface:
- Perpendicular to the surface: $N = mg \cos \theta + ma \sin \theta$. However,for the block to be stationary relative to the wedge,the net force along the incline must be zero.
- Parallel to the incline: $ma \cos \theta = mg \sin \theta$.
$3$. From the parallel component equation:
$ma \cos \theta = mg \sin \theta$
$a \cos \theta = g \sin \theta$
$a = g \frac{\sin \theta}{\cos \theta}$
$a = g \tan \theta$

(C) The coefficient of friction $(\mu)$ is defined by the relation $f = \mu N$,where $f$ is the frictional force and $N$ is the normal reaction force.
Rearranging the formula,we get $\mu = \frac{f}{N}$.
Since both $f$ and $N$ are forces,their units are Newtons $(N)$. Therefore,the coefficient of friction is a ratio of two forces,making it a dimensionless quantity.
Thus,the statement that the coefficient of sliding friction has dimensions of length is incorrect.

(D) According to $Wien's$ displacement law,$\lambda_{\max} T = \text{constant}$.
Let the initial temperature be $T$ and the final temperature be $T'$.
Given $\lambda_{\max, 1} = \lambda_0$ and $\lambda_{\max, 2} = \frac{3}{4}\lambda_0$.
Using the law: $\lambda_0 T = \left(\frac{3}{4}\lambda_0\right) T' \Rightarrow T' = \frac{4}{3}T$.
According to the $Stefan-Boltzmann$ law,the power radiated by a black body is $P = A \sigma T^4$,which implies $P \propto T^4$.
Therefore,$\frac{P_2}{P_1} = \left(\frac{T'}{T}\right)^4$.
Substituting the values: $n = \left(\frac{4/3 T}{T}\right)^4 = \left(\frac{4}{3}\right)^4 = \frac{256}{81}$.

(B) The escape velocity from the Earth's surface is $v_{\text{escape}} = 11200 \, m/s$.
For the oxygen molecules to escape,their $rms$ speed must be equal to the escape velocity:
$v_{\text{rms}} = \sqrt{\frac{3 k_B T}{m}} = v_{\text{escape}}$
Substituting the given values:
$11200 = \sqrt{\frac{3 \times 1.38 \times 10^{-23} \times T}{2.76 \times 10^{-26}}}$
Squaring both sides:
$(11200)^2 = \frac{3 \times 1.38 \times 10^{-23} \times T}{2.76 \times 10^{-26}}$
$1.2544 \times 10^8 = \frac{4.14 \times 10^{-23} \times T}{2.76 \times 10^{-26}}$
$1.2544 \times 10^8 = 1.5 \times 10^3 \times T$
Solving for $T$:
$T = \frac{1.2544 \times 10^8}{1.5 \times 10^3} \approx 8.360 \times 10^4 \, K$.

(B) Translational kinetic energy is given by $K_t = \frac{1}{2}mv^2$.
Rotational kinetic energy is given by $K_r = \frac{1}{2}I\omega^2$.
For a solid sphere,the moment of inertia is $I = \frac{2}{5}mr^2$ and the relation between linear and angular velocity is $v = r\omega$,so $\omega = \frac{v}{r}$.
Substituting these into the rotational kinetic energy formula:
$K_r = \frac{1}{2} \left( \frac{2}{5}mr^2 \right) \left( \frac{v}{r} \right)^2 = \frac{1}{5}mv^2$.
The total kinetic energy is $K_{total} = K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \left( \frac{5+2}{10} \right)mv^2 = \frac{7}{10}mv^2$.
The ratio $K_t : (K_t + K_r)$ is $\frac{\frac{1}{2}mv^2}{\frac{7}{10}mv^2} = \frac{1}{2} \times \frac{10}{7} = \frac{5}{7}$.

(C) The moment of force (torque) is given by the cross product of the position vector relative to the point of rotation and the force vector:
$\overrightarrow{\tau} = (\overrightarrow{r} - \overrightarrow{r_0}) \times \overrightarrow{F}$
Here,the point of rotation is $\overrightarrow{r_0} = 2\hat{i} - 2\hat{j} - 2\hat{k}$ and the point of application of force is $\overrightarrow{r} = 2\hat{i} + 0\hat{j} - 3\hat{k}$.
First,calculate the relative position vector:
$\overrightarrow{r} - \overrightarrow{r_0} = (2\hat{i} + 0\hat{j} - 3\hat{k}) - (2\hat{i} - 2\hat{j} - 2\hat{k}) = 0\hat{i} + 2\hat{j} - 1\hat{k}$.
Now,calculate the cross product $\overrightarrow{\tau} = (0\hat{i} + 2\hat{j} - 1\hat{k}) \times (4\hat{i} + 5\hat{j} - 6\hat{k})$:
$\overrightarrow{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -1 \\ 4 & 5 & -6 \end{vmatrix}$
$\overrightarrow{\tau} = \hat{i}(2(-6) - (-1)(5)) - \hat{j}(0(-6) - (-1)(4)) + \hat{k}(0(5) - 2(4))$
$\overrightarrow{\tau} = \hat{i}(-12 + 5) - \hat{j}(0 + 4) + \hat{k}(0 - 8)$
$\overrightarrow{\tau} = -7\hat{i} - 4\hat{j} - 8\hat{k}$.

(A) The work done required to bring an object to rest is equal to its rotational kinetic energy: $W = \Delta KE = \frac{1}{2} I \omega^2$.
Since all objects have the same angular speed $\omega$,the work done is directly proportional to the moment of inertia: $W \propto I$.
The moments of inertia for the given objects about their symmetry axes are:
$1$. Solid sphere $(A)$: $I_A = \frac{2}{5} MR^2 = 0.4 MR^2$
$2$. Thin circular disk $(B)$: $I_B = \frac{1}{2} MR^2 = 0.5 MR^2$
$3$. Circular ring $(C)$: $I_C = MR^2 = 1.0 MR^2$
Comparing these values,we get $I_C > I_B > I_A$.
Therefore,the work required satisfies the relation $W_C > W_B > W_A$.

(C) According to the principle of conservation of angular momentum,if the net external torque acting on a system is zero,the total angular momentum of the system remains constant.
In this case,the sphere is rotating freely in free space,meaning there is no external torque acting on it $({\tau_{ext}} = 0)$.
Since ${\tau_{ext}} = \frac{dL}{dt} = 0$,the angular momentum $(L)$ must remain constant.
When the radius of the sphere increases,its moment of inertia $(I = \frac{2}{5}MR^2)$ increases. Since $L = I\omega$ is constant,the angular velocity $(\omega)$ will decrease,and the rotational kinetic energy $(K = \frac{L^2}{2I})$ will also change. Therefore,only the angular momentum remains constant.

(B) According to Kepler's second law of planetary motion,the areal velocity of a planet is constant. This implies that the planet moves faster when it is closer to the Sun and slower when it is farther away.
Point $A$ is the perihelion (closest to the Sun),and point $C$ is the aphelion (farthest from the Sun). Point $B$ is at an intermediate distance.
Therefore,the orbital speeds at these positions follow the relation: $v_A > v_B > v_C$.
Since the kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$,it is directly proportional to the square of the speed $(K \propto v^2)$.
Thus,the kinetic energies at these positions follow the relation: $K_A > K_B > K_C$.

(C) The acceleration due to gravity on the surface of the Earth is given by $g = \frac{GM}{R^2}$,where $M$ is the mass of the Earth and $R$ is the radius of the Earth.
Note that the mass of the Sun does not affect the value of $g$ on the Earth's surface.
If the universal gravitational constant $G$ becomes $10G$,then the new acceleration due to gravity $g'$ becomes $g' = \frac{(10G)M}{R^2} = 10g$.
Since $g' = 10g$,the acceleration due to gravity increases significantly.
$1$. Raindrops will fall faster because the acceleration due to gravity is higher.
$2$. Walking on the ground becomes more difficult because the effective weight of a person increases.
$3$. The time period of a simple pendulum is $T = 2\pi \sqrt{\frac{l}{g}}$. Since $g$ increases,$T$ will decrease.
$4$. The statement '$g$ on the Earth will not change' is incorrect because $g$ increases by a factor of $10$.

(A) Young's modulus is given by $Y = \frac{Fl}{A\Delta l}$.
Since the volume $V = A \times L$ is the same for both wires,and the cross-sectional areas are $A_1 = A$ and $A_2 = 3A$,their lengths must be $L_1 = 3l$ and $L_2 = l$ respectively.
For the first wire:
$\Delta l = \frac{F \cdot (3l)}{A \cdot Y} = \frac{3Fl}{AY} \quad ...(i)$
For the second wire,let the required force be $F'$. The extension is the same $\Delta l$:
$\Delta l = \frac{F' \cdot l}{(3A) \cdot Y} = \frac{F'l}{3AY} \quad ...(ii)$
Equating $(i)$ and $(ii)$:
$\frac{3Fl}{AY} = \frac{F'l}{3AY}$
$3F = \frac{F'}{3}$
$F' = 9F$

(D) The viscous drag force is given by Stokes' Law: $F = 6\pi \eta rv$,where $v$ is the terminal velocity.
The rate of production of heat is equal to the power dissipated by the viscous force:
$P = F \cdot v = (6\pi \eta rv) \cdot v = 6\pi \eta r v^2$.
The terminal velocity $v$ of a sphere falling in a viscous liquid is given by:
$v = \frac{2r^2(\rho - \sigma)g}{9\eta}$,which implies $v \propto r^2$.
Substituting $v \propto r^2$ into the power equation:
$P \propto r \cdot (r^2)^2 = r \cdot r^4 = r^5$.
Therefore,the rate of production of heat is proportional to $r^5$.

(A) From the graph,$V \propto T$,which implies the process is isobaric (constant pressure).
For an isobaric process,the heat absorbed is given by $dQ = nC_p dT$,where $C_p$ is the molar heat capacity at constant pressure.
For a monatomic gas,the degrees of freedom $f = 3$. Thus,$C_p = \frac{f+2}{2}R = \frac{3+2}{2}R = \frac{5}{2}R$.
Therefore,$dQ = n \left( \frac{5}{2}R \right) dT$.
The work done by the gas is $dW = PdV$. Since $PV = nRT$,for a constant pressure process,$PdV = nRdT$.
The required ratio is $\frac{dW}{dQ} = \frac{nRdT}{n(\frac{5}{2}R)dT} = \frac{1}{5/2} = \frac{2}{5}$.

(A) The efficiency of an ideal (Carnot) heat engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$.
Here,$T_1$ is the temperature of the source (boiling point of water) and $T_2$ is the temperature of the sink (freezing point of water).
Freezing point of water,$T_2 = 0^{\circ}C = 273\,K$.
Boiling point of water,$T_1 = 100^{\circ}C = (100 + 273)\,K = 373\,K$.
Substituting these values into the efficiency formula:
$\eta = 1 - \frac{273}{373} = \frac{373 - 273}{373} = \frac{100}{373}$.
To express this as a percentage:
$\% \eta = \left( \frac{100}{373} \right) \times 100 \approx 26.8\%$.

(B) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.
First,convert the heat energy from calories to Joules: $\Delta Q = 54 \, cal \times 4.18 \, J/cal = 225.72 \, J$.
Next,calculate the work done during expansion: $\Delta W = P \Delta V$.
Given $P = 1.013 \times 10^5 \, N/m^2$ and $\Delta V = 167.1 \, cc = 167.1 \times 10^{-6} \, m^3$.
$\Delta W = 1.013 \times 10^5 \times 167.1 \times 10^{-6} \approx 16.93 \, J$.
Now,calculate the change in internal energy: $\Delta U = \Delta Q - \Delta W$.
$\Delta U = 225.72 \, J - 16.93 \, J = 208.79 \, J$.
Rounding to the nearest option,$\Delta U \approx 208.7 \, J$.

(B) The magnitude of acceleration $a$ of a particle performing $SHM$ is given by $|a| = \omega^2 y$,where $\omega$ is the angular frequency and $y$ is the displacement from the mean position.
Given: $|a| = 20 \; m/s^2$ and $y = 5 \; m$.
Substituting these values into the formula: $20 = \omega^2 (5)$.
$\omega^2 = 4 \Rightarrow \omega = 2 \; rad/s$.
The time period $T$ is given by $T = \frac{2\pi}{\omega}$.
$T = \frac{2\pi}{2} = \pi \; s$.

(A) The fundamental frequency of an open organ pipe of length $l'$ is given by $f_{open} = \frac{v}{2l'}$.
The frequencies of a closed organ pipe of length $l$ are given by $f_n = \frac{nv}{4l}$,where $n = 1, 3, 5, \dots$ (odd harmonics).
The third harmonic of a closed organ pipe corresponds to $n = 3$,so $f_{closed, 3} = \frac{3v}{4l}$.
According to the problem,$f_{open} = f_{closed, 3}$.
Therefore,$\frac{v}{2l'} = \frac{3v}{4l}$.
Simplifying for $l'$,we get $l' = \frac{4l}{6} = \frac{2l}{3}$.
Given $l = 20 \ cm$,we have $l' = \frac{2 \times 20}{3} = \frac{40}{3} \approx 13.33 \ cm$.

(B) In a resonance tube experiment,the distance between two successive resonance positions is equal to half the wavelength of the sound wave,i.e.,$\frac{\lambda}{2} = L_2 - L_1$.
Given,$L_1 = 20\,cm = 0.20\,m$ and $L_2 = 73\,cm = 0.73\,m$.
Therefore,$\frac{\lambda}{2} = 0.73\,m - 0.20\,m = 0.53\,m$.
This implies $\lambda = 2 \times 0.53\,m = 1.06\,m$.
The velocity of sound $v$ is given by the formula $v = f \lambda$,where $f$ is the frequency of the tuning fork.
Given $f = 320\,Hz$.
$v = 320 \times 1.06 = 339.2\,m/s$.
Rounding to the nearest integer,the velocity of sound is $339\,m/s$.

(B) Let the final velocity of the block of mass $4m$ be $v'$.
Initial velocity of the block of mass $m = v$.
Initial velocity of the block of mass $4m = 0$.
Final velocity of the block of mass $m = 0$.
According to the law of conservation of linear momentum:
$mv + 4m(0) = m(0) + 4mv'$
$mv = 4mv'$
$v' = v/4$
Coefficient of restitution $(e)$ is defined as the ratio of relative velocity of separation to the relative velocity of approach:
$e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}}$
$e = \frac{v' - 0}{v - 0} = \frac{v/4}{v} = 0.25$

(A) For an isolated parallel plate capacitor,the charge $Q$ on the plates remains constant.
The electric field $E$ produced by one plate at the location of the other plate is given by $E = \frac{\sigma}{2\varepsilon_0}$,where $\sigma = \frac{Q}{A}$ is the surface charge density.
The electrostatic force $F$ experienced by one plate due to the electric field of the other is $F = Q \times E$.
Substituting the expression for $E$,we get $F = Q \times \frac{Q}{2A\varepsilon_0} = \frac{Q^2}{2A\varepsilon_0}$.
Since $Q$,$A$,and $\varepsilon_0$ are constants,the force $F$ is independent of the distance $d$ between the plates.

(A) The equation of motion for a particle starting from rest under a constant electric force is given by $h = \frac{1}{2} a t^2$,where $a = \frac{qE}{m}$.
Substituting $a$,we get $h = \frac{1}{2} \left( \frac{eE}{m} \right) t^2$.
Therefore,the time of fall is $t = \sqrt{\frac{2hm}{eE}}$.
Since $h$,$e$,and $E$ are constant for both particles,we have $t \propto \sqrt{m}$.
Because the mass of an electron $(m_e \approx 9.1 \times 10^{-31} \text{ kg})$ is much smaller than the mass of a proton $(m_p \approx 1.67 \times 10^{-27} \text{ kg})$,the time taken by the electron will be smaller than the time taken by the proton.

(B) The acceleration $a$ is given by $a = \frac{v - u}{t} = \frac{6 - 0}{1} = 6\, m s^{-2}$.
For the interval $t = 0$ to $t = 1\, s$:
The displacement $S_1 = u t + \frac{1}{2} a t^2 = 0 + \frac{1}{2} \times 6 \times (1)^2 = 3\, m$.
At $t = 1\, s$,the field is reversed,so the new acceleration is $a' = -6\, m s^{-2}$.
For the interval $t = 1\, s$ to $t = 2\, s$:
The displacement $S_2 = v_1 t + \frac{1}{2} a' t^2 = 6 \times 1 + \frac{1}{2} \times (-6) \times (1)^2 = 6 - 3 = 3\, m$.
For the interval $t = 2\, s$ to $t = 3\, s$:
The velocity at $t = 2\, s$ is $v_2 = v_1 + a' t = 6 + (-6) \times 1 = 0\, m s^{-1}$.
The displacement $S_3 = v_2 t + \frac{1}{2} a' t^2 = 0 \times 1 + \frac{1}{2} \times (-6) \times (1)^2 = -3\, m$.
Total displacement $S = S_1 + S_2 + S_3 = 3 + 3 - 3 = 3\, m$.
Average velocity = $\frac{\text{Total displacement}}{\text{Total time}} = \frac{3}{3} = 1\, m s^{-1}$.
Total distance = $|S_1| + |S_2| + |S_3| = 3 + 3 + 3 = 9\, m$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{9}{3} = 3\, m s^{-1}$.

(A) When $n$ resistors of resistance $R$ are connected in series,the equivalent resistance is $n R$. The total resistance of the circuit including the internal resistance $R$ is $(n R + R)$.
Thus,the current $I$ is given by $I = \frac{E}{n R + R} = \frac{E}{R(n + 1)}$ .....$(i)$
When $n$ resistors are connected in parallel,the equivalent resistance is $R/n$. The total resistance of the circuit is $(R/n + R)$.
Thus,the new current $10I$ is given by $10I = \frac{E}{R/n + R} = \frac{E}{R(\frac{1+n}{n})} = \frac{nE}{R(n+1)}$ .....$(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{10I}{I} = \frac{nE / R(n+1)}{E / R(n+1)}$
$10 = n$
Therefore,the value of $n$ is $10$.

(A) When $n$ identical cells,each with electromotive force $\varepsilon$ and internal resistance $r$,are connected in series,the total electromotive force of the battery is $n\varepsilon$ and the total internal resistance is $nr$.
When the terminals are short-circuited,the current $I$ flowing through the circuit is given by Ohm's law:
$I = \frac{\text{Total EMF}}{\text{Total Resistance}} = \frac{n\varepsilon}{nr} = \frac{\varepsilon}{r}$
Since $\varepsilon$ and $r$ are constants for the given cells,the current $I$ is independent of the number of cells $n$. Thus,the graph of $I$ versus $n$ is a horizontal straight line,which corresponds to Graph $A$.

(D) Current sensitivity $(I_s)$ is defined as the deflection per unit current: $I_s = \frac{\theta}{I}$.
Voltage sensitivity $(V_s)$ is defined as the deflection per unit voltage: $V_s = \frac{\theta}{V}$.
Since $V = IR$,where $R$ is the resistance of the galvanometer $(R_G)$,we have $V_s = \frac{\theta}{I R_G} = \frac{I_s}{R_G}$.
Therefore,the resistance of the galvanometer is $R_G = \frac{I_s}{V_s}$.
Given: $I_s = 5\,div/mA = 5\,div/(10^{-3}\,A) = 5000\,div/A$.
Given: $V_s = 20\,div/V$.
Substituting the values: $R_G = \frac{5000}{20} = 250\,\Omega$.

(A) When the current in the electromagnet is switched on,it creates a non-uniform magnetic field. Diamagnetic materials are weakly repelled by magnetic fields and tend to move from regions of stronger magnetic field to regions of weaker magnetic field. As the rod is pushed upward against gravity,it gains gravitational potential energy. This energy is supplied by the external current source that maintains the magnetic field in the electromagnet.

(B) Using the mirror formula,$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$.
For a concave mirror,$f = -15 \, cm$.
Initially,the object distance $u_1 = -40 \, cm$.
$\frac{1}{-15} = \frac{1}{v_1} + \frac{1}{-40} \Rightarrow \frac{1}{v_1} = \frac{1}{40} - \frac{1}{15} = \frac{3 - 8}{120} = \frac{-5}{120} = -\frac{1}{24}$.
So,$v_1 = -24 \, cm$ (image is $24 \, cm$ in front of the mirror).
When the object is displaced by $20 \, cm$ towards the mirror,the new object distance $u_2 = -(40 - 20) = -20 \, cm$.
$\frac{1}{-15} = \frac{1}{v_2} + \frac{1}{-20} \Rightarrow \frac{1}{v_2} = \frac{1}{20} - \frac{1}{15} = \frac{3 - 4}{60} = -\frac{1}{60}$.
So,$v_2 = -60 \, cm$ (image is $60 \, cm$ in front of the mirror).
The displacement of the image is $|v_2 - v_1| = |-60 - (-24)| = |-36| = 36 \, cm$.
Since the magnitude of the image distance increased,the image shifts $36 \, cm$ away from the mirror.

(B) For the light ray to retrace its path,it must strike the silvered surface normally (at an angle of $90^{\circ}$ to the surface).
Let the angle of incidence at the first surface be $i$ and the angle of refraction be $r_1$. The angle of the prism is $A = 30^{\circ}$.
Since the ray strikes the second surface normally,the angle of refraction at the second surface is $r_2 = 0^{\circ}$.
From the prism formula,$A = r_1 + r_2$,we have $30^{\circ} = r_1 + 0^{\circ}$,so $r_1 = 30^{\circ}$.
Applying Snell's law at the first surface:
$\mu = \frac{\sin i}{\sin r_1}$
$\sqrt{2} = \frac{\sin i}{\sin 30^{\circ}}$
$\sin i = \sqrt{2} \times \sin 30^{\circ} = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$
Therefore,$i = 45^{\circ}$.

(D) For an astronomical telescope,the angular magnification $M$ is given by $M = \frac{f_o}{f_e}$,where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece.
To achieve a large angular magnification,the objective lens must have a large focal length $f_o$.
The angular resolution of a telescope is given by $\theta = \frac{1.22 \lambda}{D}$,where $D$ is the diameter of the objective lens and $\lambda$ is the wavelength of light.
To achieve high angular resolution,the value of $\theta$ must be small,which implies that the diameter $D$ of the objective lens must be large.
Therefore,an astronomical refracting telescope requires an objective lens with a large focal length and a large diameter.

(A) The force acting on the electron due to the electric field is $\vec{F} = q\vec{E} = (-e)(-E_0 \hat{i}) = eE_0 \hat{i}$.
The acceleration produced in the electron is $\vec{a} = \frac{\vec{F}}{m} = \frac{eE_0}{m} \hat{i}$.
The velocity of the electron at time $t$ is given by $\vec{v}_t = \vec{v} + \vec{a}t = \left(V_0 + \frac{eE_0}{m}t\right) \hat{i}$.
The magnitude of the velocity is $v_t = V_0 + \frac{eE_0}{m}t = V_0 \left(1 + \frac{eE_0}{mV_0}t\right)$.
The de-Broglie wavelength at time $t$ is $\lambda_t = \frac{h}{mv_t}$.
Substituting $v_t$,we get $\lambda_t = \frac{h}{m V_0 \left(1 + \frac{eE_0}{mV_0}t\right)}$.
Since the initial de-Broglie wavelength is $\lambda_0 = \frac{h}{mV_0}$,we have $\lambda_t = \frac{\lambda_0}{\left(1 + \frac{eE_0}{mV_0}t\right)}$.

(A) According to Einstein's photoelectric equation,$K_{max} = h\nu - W_0 = \frac{1}{2}mv^2$,where $W_0 = h\nu_0$.
For frequency $2\nu_0$:
$h(2\nu_0) = h\nu_0 + \frac{1}{2}mv_1^2$
$h\nu_0 = \frac{1}{2}mv_1^2$ ..... $(i)$
For frequency $5\nu_0$:
$h(5\nu_0) = h\nu_0 + \frac{1}{2}mv_2^2$
$4h\nu_0 = \frac{1}{2}mv_2^2$ ..... $(ii)$
Dividing equation $(i)$ by $(ii)$:
$\frac{h\nu_0}{4h\nu_0} = \frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2}$
$\frac{1}{4} = \frac{v_1^2}{v_2^2}$
Taking the square root on both sides:
$\frac{v_1}{v_2} = \frac{1}{2}$
Thus,the ratio is $1:2$.

(C) The magnetic potential energy $U$ stored in an inductor of inductance $L$ carrying current $I$ is given by the formula:
$U = \frac{1}{2} L I^{2}$
Given:
$U = 25\, mJ = 25 \times 10^{-3}\, J$
$I = 60\, mA = 60 \times 10^{-3}\, A = 6 \times 10^{-2}\, A$
Substituting the values into the formula:
$25 \times 10^{-3} = \frac{1}{2} \times L \times (6 \times 10^{-2})^{2}$
$25 \times 10^{-3} = \frac{1}{2} \times L \times 36 \times 10^{-4}$
$25 \times 10^{-3} = L \times 18 \times 10^{-4}$
$L = \frac{25 \times 10^{-3}}{18 \times 10^{-4}}$
$L = \frac{250}{18} = 13.888...\, H \approx 13.89\, H$.

(C) The force due to gravity acting down the incline is $F_g = mg \sin 30^{\circ}$.
The magnetic force acting on the rod is $F_m = IlB$,which acts horizontally. The component of this magnetic force acting up the incline is $F_{m, \text{up}} = IlB \cos 30^{\circ}$.
For the rod to remain stationary,the forces along the incline must balance:
$mg \sin 30^{\circ} = IlB \cos 30^{\circ}$
Rearranging for current $I$:
$I = \frac{mg}{lB} \tan 30^{\circ}$
Given mass per unit length $\frac{m}{l} = 0.5\; kg\; m^{-1}$,$B = 0.25\; T$,and $g = 9.8\; m/s^2$:
$I = \frac{0.5 \times 9.8}{0.25} \times \tan 30^{\circ}$
$I = 2 \times 9.8 \times \frac{1}{\sqrt{3}}$
$I = \frac{19.6}{1.732} \approx 11.32\; A$

(A) The given values are $L = 20 \, mH = 20 \times 10^{-3} \, H$,$C = 100 \, \mu F = 100 \times 10^{-6} \, F$,$R = 50 \, \Omega$,and $V = 10 \sin(314 \, t)$.
Comparing with $V = V_0 \sin(\omega t)$,we get $V_0 = 10 \, V$ and $\omega = 314 \, rad/s$.
The inductive reactance is $X_L = \omega L = 314 \times 20 \times 10^{-3} = 6.28 \, \Omega$.
The capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{314 \times 100 \times 10^{-6}} = \frac{1}{0.0314} \approx 31.85 \, \Omega$.
The impedance $Z$ is given by $Z = \sqrt{R^2 + (X_C - X_L)^2} = \sqrt{50^2 + (31.85 - 6.28)^2} = \sqrt{2500 + (25.57)^2} = \sqrt{2500 + 653.8} = \sqrt{3153.8} \approx 56.16 \, \Omega$.
The $RMS$ voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, V$.
The power loss $P_{av}$ is given by $P_{av} = I_{rms}^2 R = \left(\frac{V_{rms}}{Z}\right)^2 R = \left(\frac{10}{\sqrt{2} \times 56.16}\right)^2 \times 50$.
$P_{av} = \left(\frac{10}{79.42}\right)^2 \times 50 = (0.1259)^2 \times 50 = 0.01585 \times 50 \approx 0.79 \, W$.

(B) The direction of propagation of an electromagnetic wave is given by the direction of the Poynting vector,which is $\vec{S} \propto \vec{E} \times \vec{B}$.
Given that the velocity $\vec{v}$ is along the $+x$ axis $(\hat{i})$ and the electric field $\vec{E}$ is along the $+y$ axis $(\hat{j})$.
We know that $\vec{v} \parallel (\vec{E} \times \vec{B})$.
Substituting the known directions: $\hat{i} = \hat{j} \times \vec{B}_{dir}$.
Using the properties of unit vectors in a Cartesian coordinate system,we know that $\hat{j} \times \hat{k} = \hat{i}$.
Therefore,the direction of the magnetic field $\vec{B}$ must be along the $+z$ axis $(\hat{k})$.
Thus,the oscillating magnetic field is along the $+z$ direction.

(B) When the reflected and refracted rays are perpendicular to each other,the angle of incidence $i$ is known as the Brewster angle $(i_B)$.
According to Brewster's law,the refractive index $\mu$ is given by $\mu = \tan(i_B)$.
In this condition,the reflected light is completely plane-polarised,and its electric field vector is perpendicular to the plane of incidence.

(B) The angular width $\theta$ of the fringes in Young's double slit experiment is given by the formula: $\theta = \frac{\lambda}{d}$.
Given the initial conditions: $\theta_1 = 0.20^\circ$ and $d_1 = 2 \ mm$.
So,$0.20^\circ = \frac{\lambda}{2 \ mm}$ .... $(i)$.
For the new condition: $\theta_2 = 0.21^\circ$ and $d_2 = d$.
So,$0.21^\circ = \frac{\lambda}{d}$ .... $(ii)$.
Dividing equation $(i)$ by equation $(ii)$,we get:
$\frac{0.20}{0.21} = \frac{d}{2 \ mm}$.
Solving for $d$:
$d = \frac{0.20 \times 2}{0.21} \ mm = \frac{0.40}{0.21} \ mm \approx 1.9047 \ mm$.
Rounding to the nearest provided option,$d = 1.9 \ mm$.

(B) In a Bohr orbit of the hydrogen atom,the total energy $E$ is given by $E = -K$,where $K$ is the kinetic energy.
This relationship arises because the total energy is the sum of kinetic energy and potential energy,$E = K + U$,and for a Bohr orbit,$U = -2K$.
Therefore,$E = K + (-2K) = -K$.
Thus,the ratio of kinetic energy to total energy is $K/E = K/(-K) = 1:-1$.

(A) The number of nuclei remaining after disintegration is given by $N = N_{0} - N_{\text{disintegrated}}$.
Given $N_{0} = 600$ and $N_{\text{disintegrated}} = 450$,the remaining nuclei $N = 600 - 450 = 150$.
According to the law of radioactive decay,$\frac{N}{N_{0}} = \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$,where $T_{1/2} = 10$ minutes.
Substituting the values: $\frac{150}{600} = \left(\frac{1}{2}\right)^{\frac{t}{10}}$.
This simplifies to $\frac{1}{4} = \left(\frac{1}{2}\right)^{\frac{t}{10}}$.
Since $\frac{1}{4} = \left(\frac{1}{2}\right)^{2}$,we have $\left(\frac{1}{2}\right)^{2} = \left(\frac{1}{2}\right)^{\frac{t}{10}}$.
Equating the exponents: $2 = \frac{t}{10}$,which gives $t = 20$ minutes.

(C) Given: $V_{BE} = 0$,$V_{CE} = 0$,$V_i = 20\, V$,$R_B = 500\, k\Omega$,$R_C = 4\, k\Omega$.
For the output circuit (collector-emitter loop):
$V_{CC} = I_C R_C + V_{CE}$
Since $V_{CE} = 0$,we have:
$20\, V = I_C \times (4 \times 10^3\,\Omega) + 0$
$I_C = \frac{20}{4 \times 10^3} = 5 \times 10^{-3}\, A = 5\, mA$.
For the input circuit (base-emitter loop):
$V_i = I_B R_B + V_{BE}$
Since $V_{BE} = 0$,we have:
$20\, V = I_B \times (500 \times 10^3\,\Omega) + 0$
$I_B = \frac{20}{500 \times 10^3} = 40 \times 10^{-6}\, A = 40\,\mu A$.
The current gain $\beta$ is given by:
$\beta = \frac{I_C}{I_B} = \frac{5 \times 10^{-3}\, A}{40 \times 10^{-6}\, A} = \frac{5000}{40} = 125$.

(C) When the temperature of a $p-n$ junction diode increases,the thermal energy causes more covalent bonds to break,resulting in an increase in the number of electron-hole pairs.
This increase in charge carriers leads to a decrease in the resistance of the semiconductor material.
Consequently,both the forward current and the reverse saturation current are significantly affected.
Therefore,the overall $V-I$ characteristics of the $p-n$ junction diode are altered by the change in temperature.

(B) The given circuit consists of two $AND$ gates,two $NOT$ gates,and one $OR$ gate.
$1$. The upper $AND$ gate receives inputs $A$ and $\overline{B}$ (since $B$ passes through a $NOT$ gate). Its output is $A \cdot \overline{B}$.
$2$. The lower $AND$ gate receives inputs $\overline{A}$ (since $A$ passes through a $NOT$ gate) and $B$. Its output is $\overline{A} \cdot B$.
$3$. These two outputs are fed into an $OR$ gate. Therefore,the final output $Y$ is the sum of these two expressions: $Y = A \cdot \overline{B} + \overline{A} \cdot B$.
This is the Boolean expression for an $XOR$ (Exclusive-$OR$) gate.

(C) The given resistance is $(47 \pm 4.7) \; k\Omega = 47 \times 10^3 \; \Omega \pm 10 \%$.
According to the carbon resistor color code:
$1$. The first digit is $4$, which corresponds to the color Yellow.
$2$. The second digit is $7$, which corresponds to the color Violet.
$3$. The multiplier is $10^3$, which corresponds to the color Orange.
$4$. The tolerance is $10 \%$, which corresponds to the color Silver.
Therefore, the color code sequence is Yellow - Violet - Orange - Silver.