AIIMS 2001 Physics Question Paper with Answer and Solution

(C) The dimensions of the given pairs are as follows:
$1$. Stress and pressure: Both have dimensions $[ML^{-1}T^{-2}]$.
$2$. Angle and strain: Both are dimensionless quantities $[M^0L^0T^0]$.
$3$. Tension and surface tension: Tension is a force with dimensions $[MLT^{-2}]$,while surface tension is force per unit length with dimensions $[MT^{-2}]$. Thus,they do not have similar dimensions.
$4$. Planck's constant and angular momentum: Both have dimensions $[ML^2T^{-1}]$.
Therefore,the pair that does not have similar dimensions is Tension and surface tension.

(A) The distance traveled by an object in the $n$th second is given by $S_n = u + \frac{a}{2}(2n - 1)$.
For body $A$,the distance traveled in the $5$th second is $S_{A,5} = 0 + \frac{a_1}{2}(2 \times 5 - 1) = \frac{9a_1}{2}$.
Body $B$ starts $2$ seconds after $A$. Therefore,the $5$th second after the start of $A$ corresponds to the $(5 - 2) = 3$rd second of motion for body $B$.
The distance traveled by body $B$ in its $3$rd second is $S_{B,3} = 0 + \frac{a_2}{2}(2 \times 3 - 1) = \frac{5a_2}{2}$.
Given that these distances are equal: $\frac{9a_1}{2} = \frac{5a_2}{2}$.
Simplifying this,we get $9a_1 = 5a_2$,which implies $\frac{a_1}{a_2} = \frac{5}{9}$.

(D) Given: Initial velocity $u = 200\,m/s$,Final velocity $v = 100\,m/s$,Displacement $s = 10\,cm = 0.1\,m$.
Using the third equation of motion: $v^2 = u^2 + 2as$.
Substituting the values: $(100)^2 = (200)^2 + 2 \times a \times 0.1$.
$10000 = 40000 + 0.2a$.
$0.2a = 10000 - 40000 = -30000$.
$a = -30000 / 0.2 = -150000\,m/s^2 = -15 \times 10^4\,m/s^2$.
The retardation is the magnitude of negative acceleration,which is $15 \times 10^4\,m/s^2$.

(C) At the lowest position of a vertical circle,two forces act on the mass $m$: the tension $T$ acting upwards towards the center and the weight $mg$ acting downwards.
According to Newton's second law,the net force providing the centripetal acceleration is $T - mg = \frac{mv^2}{r}$.
Rearranging this equation to solve for tension $T$,we get $T = \frac{mv^2}{r} + mg$.

(B) When a lubricant is applied between two surfaces,it forms a thin layer that fills the irregularities of the surfaces.
This reduces the direct contact between the surfaces and decreases the coefficient of friction.
As a result,the surfaces can easily slide over each other with minimal resistance.

(D) The work done $W$ by a constant force $\overrightarrow{F}$ during a displacement $\overrightarrow{s}$ is given by the dot product of the force and displacement vectors:
$W = \overrightarrow{F} \cdot \overrightarrow{s}$
Given:
$\overrightarrow{F} = (3\,\hat{i} + 4\,\hat{j}) \, N$
$\overrightarrow{s} = (3\,\hat{i} + 4\,\hat{j}) \, m$
Substituting the values:
$W = (3\,\hat{i} + 4\,\hat{j}) \cdot (3\,\hat{i} + 4\,\hat{j})$
Using the dot product property $\hat{i} \cdot \hat{i} = 1$,$\hat{j} \cdot \hat{j} = 1$,and $\hat{i} \cdot \hat{j} = 0$:
$W = (3 \times 3) + (4 \times 4)$
$W = 9 + 16$
$W = 25 \, J$

(C) According to the law of conservation of linear momentum,the total momentum before firing is equal to the total momentum after firing.
Initial momentum of the system (gun + bullet) = $0$.
Final momentum = $m_G v_G + m_B v_B = 0$.
Therefore,the magnitude of the momentum of the gun equals the magnitude of the momentum of the bullet: $m_G v_G = m_B v_B$.
Given:
Mass of bullet $(m_B)$ = $50 \, g = 0.05 \, kg$.
Velocity of bullet $(v_B)$ = $30 \, m/s$.
Velocity of gun $(v_G)$ = $1 \, m/s$.
$m_G = \frac{m_B v_B}{v_G} = \frac{0.05 \times 30}{1} = 1.5 \, kg$.

(B) Initial velocity of the first ball,$u_1 = 36 \,km/h = 36 \times \frac{5}{18} = 10 \,m/s$.
Initial velocity of the second ball,$u_2 = 0 \,m/s$.
Masses are $m_1 = 2 \,kg$ and $m_2 = 3 \,kg$.
By the law of conservation of linear momentum,$m_1 u_1 + m_2 u_2 = (m_1 + m_2)V$,where $V$ is the common velocity after the collision.
$2 \times 10 + 3 \times 0 = (2 + 3)V \implies 20 = 5V \implies V = 4 \,m/s$.
Initial kinetic energy,$K_i = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} \times 2 \times (10)^2 + 0 = 100 \,J$.
Final kinetic energy,$K_f = \frac{1}{2} (m_1 + m_2) V^2 = \frac{1}{2} \times 5 \times (4)^2 = \frac{1}{2} \times 5 \times 16 = 40 \,J$.
Loss in kinetic energy,$\Delta K = K_i - K_f = 100 - 40 = 60 \,J$.

(A) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$.
Since mass $M = \text{density} (\rho) \times \text{volume} = \rho \times \frac{4}{3}\pi R^3$,we can substitute $M$ into the formula:
$v_e = \sqrt{\frac{2G(\rho \cdot \frac{4}{3}\pi R^3)}{R}} = \sqrt{\frac{8}{3}\pi G \rho} \cdot R$.
Given that the mean density $\rho$ is constant,we find that $v_e \propto R$.
If the radius of the planet is twice that of the earth $(R' = 2R)$,then the new escape velocity $v_e'$ will be:
$v_e' = 2 \times v_e = 2 \times 11 \, km/s = 22 \, km/s$.

(D) The energy stored per unit volume $(u)$ in a stretched wire is given by the formula:
$u = \frac{1}{2} \times \text{Stress} \times \text{Strain}$
Since Young's modulus $y = \frac{\text{Stress}}{\text{Strain}}$,we have $\text{Stress} = y \times \text{Strain}$.
Substituting the given strain $x$ into the equation:
$u = \frac{1}{2} \times (y \times x) \times x$
$u = \frac{1}{2} y x^2$

(C) The excess pressure $\Delta P$ inside a spherical drop is given by the formula $\Delta P = \frac{2T}{R}$.
Given:
Surface tension $T = 70 \times 10^{-3}\, N/m$
Radius $R = 1\, mm = 1 \times 10^{-3}\, m$
Substituting the values:
$\Delta P = \frac{2 \times 70 \times 10^{-3}}{1 \times 10^{-3}}$
$\Delta P = 2 \times 70 = 140\, N/m^2$.
Therefore,the correct option is $C$.

(D) Thermodynamic coordinates are variables that define the state of a thermodynamic system,such as pressure $(P)$,volume $(V)$,and temperature $(T)$.
$R$ is the universal gas constant,which is a fundamental physical constant and not a variable coordinate of a system's state.
Therefore,the correct option is $(d)$.

(C) According to Wien's displacement law,the wavelength $\lambda_m$ corresponding to the maximum intensity of emission is inversely proportional to the absolute temperature $T$ of the body,given by $\lambda_m T = b$,where $b$ is Wien's constant. Since the colour of a star depends on the wavelength of the light it emits most intensely,the observed colour is a direct indicator of the star's surface temperature.

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{l}{g}}$.
This implies $T \propto \sqrt{l}$.
Let the initial length be $l_1 = 100$ units. Then the new length $l_2 = 100 + 21 = 121$ units.
The ratio of the time periods is $\frac{T_2}{T_1} = \sqrt{\frac{l_2}{l_1}} = \sqrt{\frac{121}{100}} = \frac{11}{10} = 1.1$.
Thus,$T_2 = 1.1 T_1$.
The percentage increase in the time period is given by $\frac{T_2 - T_1}{T_1} \times 100 = \frac{1.1 T_1 - T_1}{T_1} \times 100 = 0.1 \times 100 = 10\%$.
Therefore,the correct option is $A$.

(D) In the given figure,the two springs with spring constants $K_1$ and $K_2$ are connected in series.
For springs connected in series,the equivalent spring constant $K_{eq}$ is given by the formula:
$\frac{1}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2} = \frac{K_1 + K_2}{K_1 K_2}$
Therefore,$K_{eq} = \frac{K_1 K_2}{K_1 + K_2}$.
The frequency of oscillation $f$ for a spring-mass system is given by:
$f = \frac{1}{2\pi} \sqrt{\frac{K_{eq}}{m}}$
Substituting the value of $K_{eq}$ into the frequency formula,we get:
$f = \frac{1}{2\pi} \sqrt{\frac{K_1 K_2}{m(K_1 + K_2)}}$
Thus,the correct option is $(d)$.

(A) In a sinusoidal wave,the motion from maximum displacement (amplitude) to zero displacement corresponds to one-quarter of the total time period $(T)$.
Therefore,the time taken is $t = \frac{T}{4}$.
Since the frequency ($n$ or $f$) is the reciprocal of the time period,$T = \frac{1}{n}$.
Substituting this into the equation,we get $t = \frac{1}{4n}$.
Rearranging for frequency,$n = \frac{1}{4t}$.
Given $t = 0.170\,s$,we calculate $n = \frac{1}{4 \times 0.170} = \frac{1}{0.680} \approx 1.47\,Hz$.

(C) The relationship between path difference $(\Delta x)$ and phase difference $(\phi)$ is given by the formula: $\Delta x = \frac{\lambda}{2\pi} \times \phi$.
Given phase difference $\phi = 60^{\circ}$.
Convert the phase difference into radians: $\phi = 60^{\circ} \times \frac{\pi}{180^{\circ}} = \frac{\pi}{3} \text{ radians}$.
Substitute the value of $\phi$ into the formula: $\Delta x = \frac{\lambda}{2\pi} \times \frac{\pi}{3}$.
Simplifying the expression: $\Delta x = \frac{\lambda}{6}$.
Therefore,the path difference is $\lambda / 6$.

(C) The frequency $n$ of a stretched string is given by the formula $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Since $l$ and $\mu$ are constant,we have $n \propto \sqrt{T}$.
Therefore,$\frac{n_1}{n_2} = \sqrt{\frac{T_1}{T_2}}$.
Given $n_1 = n$,$n_2 = 2n$,and $T_1 = 10 \ N$.
Substituting these values: $\frac{n}{2n} = \sqrt{\frac{10}{T_2}}$.
$\frac{1}{2} = \sqrt{\frac{10}{T_2}}$.
Squaring both sides: $\frac{1}{4} = \frac{10}{T_2}$.
Thus,$T_2 = 40 \ N$.

(D) The energy $E$ of a sound wave is given by the relation $E \propto a^2 n^2$,where $a$ is the amplitude and $n$ is the frequency.
Since the energies of the two sounds are equal,we have $E_A = E_B$.
Therefore,$a_A^2 n_A^2 = a_B^2 n_B^2$.
Given that the frequency of $B$ is one-eighth the frequency of $A$,we have $n_B = \frac{1}{8} n_A$,or $\frac{n_A}{n_B} = 8$.
Rearranging the energy equation: $\frac{a_B^2}{a_A^2} = \frac{n_A^2}{n_B^2}$.
Taking the square root on both sides: $\frac{a_B}{a_A} = \frac{n_A}{n_B} = 8$.
Thus,$a_B = 8 a_A$.
The amplitude of the note of $B$ is eight times that of $A$.

(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: $n = \left(\frac{4/3 T}{T}\right)^4 = \left(\frac{4}{3}\right)^4 = \frac{256}{81}$.

(D) For an adiabatic process,the relationship between pressure $P$ and temperature $T$ is given by $T^\gamma P^{1-\gamma} = \text{constant}$.
Rearranging this,we get $P^{1-\gamma} \propto T^{-\gamma}$,which implies $P \propto T^{-\frac{\gamma}{1-\gamma}}$ or $P \propto T^{\frac{\gamma}{\gamma-1}}$.
Comparing this with the given relation $P \propto T^C$,we find $C = \frac{\gamma}{\gamma-1}$.
For a monoatomic gas,the adiabatic index $\gamma = 5/3$.
Substituting the value of $\gamma$:
$C = \frac{5/3}{5/3 - 1} = \frac{5/3}{2/3} = 5/2$.

(D) According to Stefan-Boltzmann Law,the energy radiated per unit time by a black body is proportional to the fourth power of its absolute temperature: $E \propto T^4$.
Given temperatures in Celsius: $T_1 = 27^oC$ and $T_2 = 127^oC$.
Convert these to Kelvin: $T_1 = 27 + 273 = 300 \ K$ and $T_2 = 127 + 273 = 400 \ K$.
The ratio of the energies emitted is given by: $\frac{E_1}{E_2} = \left( \frac{T_1}{T_2} \right)^4$.
Substituting the values: $\frac{E_1}{E_2} = \left( \frac{300}{400} \right)^4 = \left( \frac{3}{4} \right)^4 = \frac{81}{256}$.
Thus,the ratio is $81:256$.

(D) The $Assertion$ is incorrect because a rocket does not require surrounding air to move forward. In fact,rockets operate efficiently in the vacuum of space.
The rocket moves forward by ejecting its own fuel combustion products (exhaust gases) at high velocity backwards. According to Newton's third law of motion,the rocket exerts a force on the gases,and the gases exert an equal and opposite reaction force on the rocket,providing the necessary thrust.
Since the $Assertion$ is false and the $Reason$ is a scientifically correct statement regarding the principle of rocket propulsion (Newton's third law),the correct choice is that the $Assertion$ is incorrect but the $Reason$ is correct. However,based on the provided options,if we consider the $Assertion$ false,the most appropriate classification is that the $Assertion$ is incorrect.

(D) The relationship between torque $(\tau)$,moment of inertia $(I)$,and angular acceleration $(\alpha)$ is given by the formula: $\tau = I \alpha$.
Given:
Torque $(\tau)$ = $31.4 \, Nm$
Angular acceleration $(\alpha)$ = $4\pi \, rad/s^2$
Using the value of $\pi \approx 3.14$,we have $\alpha = 4 \times 3.14 = 12.56 \, rad/s^2$.
Substituting the values into the formula:
$31.4 = I \times 12.56$
$I = \frac{31.4}{12.56} = 2.5 \, kg \cdot m^2$.

(B) Bernoulli's principle is derived from the work-energy theorem for a flowing fluid.
It states that for an incompressible,non-viscous,and steady flow of a fluid,the sum of pressure energy,kinetic energy,and potential energy per unit volume remains constant along a streamline.
Therefore,it is based on the law of conservation of energy.

(D) In an adiabatic process,the system is thermally insulated from the surroundings,meaning $dQ = 0$.
According to the first law of thermodynamics,$dQ = dU + dW$.
For adiabatic compression,work is done on the system,so $dW < 0$.
This implies $dU = -dW > 0$,which means the internal energy $U$ increases.
Since internal energy of an ideal gas is a function of temperature $(U \propto T)$,the temperature of the system also increases.
Therefore,the $Assertion$ is incorrect.
Adiabatic processes are typically fast processes to ensure no heat exchange occurs with the surroundings.
Therefore,the $Reason$ is also incorrect.
Thus,both $Assertion$ and $Reason$ are incorrect.

(D) The $Assertion$ is incorrect. Two isothermal curves for different temperatures cannot intersect each other. If they were to intersect at a point,it would imply that at that specific $(P, V)$ state,the system has two different temperatures simultaneously,which is impossible.
The $Reason$ is correct. An isothermal process is a slow process that allows the system to remain in thermal equilibrium with its surroundings. The slope of an isothermal curve on a $P-V$ diagram is given by $-\frac{dP}{dV} = \frac{P}{V}$. Since $P$ and $V$ are positive,the slope is finite and relatively small compared to adiabatic curves (which have a slope of $\gamma \frac{P}{V}$).

(C) The formula for the maximum height $H$ attained by a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Since both projectiles are projected with the same speed $u$,the ratio of their maximum heights $H_1$ and $H_2$ is given by $\frac{H_1}{H_2} = \frac{\sin^2 \theta_1}{\sin^2 \theta_2}$.
Given $\theta_1 = 30^{\circ}$ and $\theta_2 = 60^{\circ}$,we have $\sin 30^{\circ} = \frac{1}{2}$ and $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$.
Substituting these values,we get $\frac{H_1}{H_2} = \frac{(\sin 30^{\circ})^2}{(\sin 60^{\circ})^2} = \frac{(1/2)^2}{(\sqrt{3}/2)^2} = \frac{1/4}{3/4} = \frac{1}{3}$.
Thus,the ratio is $1:3$.

(B) Hooke in $1679$ showed experimentally that if strain is small,then the stress is proportional to strain.
The ratio of stress to strain is constant for a given material and is called the modulus of elasticity $E$.
Thus,$E = \frac{\text{stress}}{\text{strain}} = \text{constant}$.
Hence,if stress is increased (within the elastic limit),the ratio of stress to strain remains constant.

(A) According to Gauss's theorem,the total electric flux $\phi_{net}$ through a closed surface is given by $\phi_{net} = \frac{q}{\varepsilon_0}$.
Since the charge $q$ is placed at the centre of the cube,the flux is distributed symmetrically through all $6$ faces of the cube.
Therefore,the electric flux through one face of the cube is $\phi_{face} = \frac{\phi_{net}}{6} = \frac{q}{6\varepsilon_0}$.

(B) The given arrangement can be considered as two capacitors connected in parallel,each having an area $A/2$ and plate separation $t$.
For the first capacitor with dielectric constant $k_1$,the capacitance is $C_1 = \frac{k_1 \varepsilon_0 (A/2)}{t} = \frac{k_1 \varepsilon_0 A}{2t}$.
For the second capacitor with dielectric constant $k_2$,the capacitance is $C_2 = \frac{k_2 \varepsilon_0 (A/2)}{t} = \frac{k_2 \varepsilon_0 A}{2t}$.
Since they are in parallel,the equivalent capacitance $C = C_1 + C_2$.
$C = \frac{k_1 \varepsilon_0 A}{2t} + \frac{k_2 \varepsilon_0 A}{2t} = \frac{\varepsilon_0 A}{2t}(k_1 + k_2) = \frac{\varepsilon_0 A}{t} \cdot \frac{k_1 + k_2}{2}$.
Thus,the correct option is $B$.

(C) The electromotive force $(E)$ of the cell is $2\, V$.
The internal resistance $(r)$ of the cell is $0.1\,\Omega$.
The external resistance $(R)$ connected is $3.9\,\Omega$.
The total resistance of the circuit is $R_{total} = R + r = 3.9\,\Omega + 0.1\,\Omega = 4.0\,\Omega$.
The current $(I)$ flowing through the circuit is given by Ohm's law: $I = \frac{E}{R + r} = \frac{2\, V}{4.0\,\Omega} = 0.5\, A$.
The terminal voltage $(V)$ across the cell is given by $V = I \times R$.
Substituting the values: $V = 0.5\, A \times 3.9\,\Omega = 1.95\, V$.

(C) The internal resistance of a cell is defined as the opposition offered by the electrolyte and the electrodes of the cell to the flow of current through it.
However,the primary contribution to the internal resistance comes from the electrolyte present between the two electrodes.
Therefore,the correct option is $(c)$.

(C) The thermo e.m.f. $E$ is given by $E = AT - \frac{1}{2}BT^2$.
At the temperature of inversion $(T_i)$,the thermo e.m.f. becomes zero.
Setting $E = 0$:
$0 = AT_i - \frac{1}{2}BT_i^2$
$AT_i = \frac{1}{2}BT_i^2$
$T_i = \frac{2A}{B}$
Given $A = 16$ and $B = 0.08$:
$T_i = \frac{2 \times 16}{0.08} = \frac{32}{0.08} = \frac{3200}{8} = 400\,^oC$.
Therefore,the temperature of inversion is $400\,^oC$.

(C) The magnetic field $B$ at the centre of a circular coil is given by the formula: $B = \frac{\mu_0 n i}{2r}$.
Given:
Number of turns $n = 25$.
Diameter $D = 10\, cm$,so radius $r = 5\, cm = 5 \times 10^{-2}\, m$.
Current $i = 4\, A$.
Permeability of free space $\mu_0 = 4\pi \times 10^{-7}\, T\cdot m/A$.
Substituting the values:
$B = \frac{(4\pi \times 10^{-7}) \times 25 \times 4}{2 \times 5 \times 10^{-2}}$
$B = \frac{4\pi \times 10^{-7} \times 100}{10 \times 10^{-2}}$
$B = 4\pi \times 10^{-7} \times 10^3 = 4\pi \times 10^{-4}\, T$
$B \approx 4 \times 3.14159 \times 10^{-4} = 12.566 \times 10^{-4} = 1.257 \times 10^{-3}\, T$.

(C) cyclotron is a device used to accelerate positively charged particles such as protons,deuterons,and alpha particles.
It works on the principle that a positively charged particle can be accelerated to a sufficiently high energy by repeatedly crossing an oscillating electric field,while being guided in a circular path by a strong magnetic field.
Electrons are not accelerated in a cyclotron because their mass is very small,causing them to reach relativistic speeds quickly,which leads to a loss of synchrony with the oscillating electric field.

(B) The correct answer is $(b)$.
In an $AC$ circuit,a choke coil is essentially an inductor with high inductance and negligible resistance.
It is used to limit or decrease the current in the circuit without significant loss of electrical energy.
Unlike a resistor,which dissipates energy as heat ($I^2R$ loss),an ideal choke coil consumes no power because the phase difference between voltage and current is $90^{\circ}$,resulting in a power factor of $\cos(90^{\circ}) = 0$.

(A) In an atomic bomb,the energy is released due to the uncontrolled chain reaction of neutrons and $_{92}U^{235}$.
The nuclear fission reaction taking place is as follows:
$_{92}U^{235} + _{0}n^{1} \rightarrow _{56}Ba^{141} + _{36}Kr^{92} + 3_{0}n^{1} + Q \text{ (approx. } 200 \text{ MeV energy)}$.
The three neutrons generated in each fission event can react with three other $_{92}U^{235}$ nuclei,which leads to a self-sustaining chain reaction,releasing a massive amount of energy in a very short time.

(B) When a radioactive nucleus emits an $\alpha$-particle $(_{2}He^{4})$,the atomic number $Z$ of the nucleus decreases by $2$ and the mass number $A$ decreases by $4$.
The decay equation is given by: $_{Z}X^{A} \rightarrow _{Z-2}Y^{A-4} + _{2}He^{4}$.
Since the atomic number $Z$ decreases by $2$,the element shifts two places to the left in the periodic table.

(A) The Fraunhofer lines are observed in the solar spectrum.
These lines are formed when the atoms in the cooler outer layers of the Sun (the chromosphere) absorb specific wavelengths of light emitted by the hotter inner layers (the photosphere).
Since these atoms absorb discrete wavelengths,they produce dark lines in the continuous spectrum of the Sun.
Therefore,the Fraunhofer spectrum is a line absorption spectrum.

(A) The resolving limit of a healthy human eye is defined as the minimum angle that two distinct points must subtend at the eye to be seen as separate. For a healthy human eye,this limit is approximately $1$ minute of arc,which is denoted as $1'$.
Since $60' = 1^\circ$,it follows that $1' = \left( \frac{1}{60} \right)^\circ$.

(A) The magnification $M$ of an astronomical telescope in normal adjustment is given by the formula $M = -\frac{f_o}{f_e}$,where $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eye-piece.
To obtain the largest magnification $M$,the ratio $\frac{f_o}{f_e}$ must be maximized.
This implies that for a given objective lens,the focal length of the eye-piece $f_e$ should be the smallest available value.
Among the given focal lengths $(+ 15\, cm, + 20\, cm, + 150\, cm, + 250\, cm)$,the smallest value is $+ 15\, cm$.
Therefore,the focal length of the eye-piece should be $+ 15\, cm$.

(D) The dual nature of light implies that light exhibits both wave-like and particle-like properties.
Diffraction is a phenomenon that demonstrates the wave nature of light.
The photoelectric effect is a phenomenon that demonstrates the particle nature of light (photons).
Therefore,the combination of diffraction and the photoelectric effect exhibits the dual nature of light.
Thus,the correct option is $D$.

(D) The correct answer is $D$.
According to the Doppler effect,when a source of light moves away from an observer,the frequency of the emitted light decreases,which corresponds to an increase in wavelength.
This shift towards longer wavelengths (the red end of the visible spectrum) is known as the red shift.
Since light observed from distant galaxies consistently shows a red shift,it indicates that these galaxies are moving away from us.
This observation provides strong evidence that the universe is expanding.

(D) The correct answer is $D$. Fraunhofer lines are produced by the absorption of light rays from the Sun by the gases and vapors present in the solar atmosphere. When white light from the photosphere passes through the cooler chromosphere,the atoms and molecules present in the chromosphere absorb specific wavelengths of light. This process results in the formation of dark lines in the otherwise continuous spectrum of the Sun,which are known as Fraunhofer lines. Therefore,they represent a line absorption spectrum.

(A) According to the principle of electrostatic shielding,the electric field inside a closed metallic conductor is always zero,regardless of any external electric field.
This phenomenon occurs because the free electrons in the metal redistribute themselves on the surface to cancel out the effect of the external field.
Since the electric field inside the hollow shell is zero,it effectively blocks external electric fields from entering the interior region.
Therefore,the Assertion is correct,and the Reason provides the correct physical explanation for why such a shield works.

(A) All electromagnetic waves travel at the speed of light $(c \approx 3 \times 10^8 \ m/s)$ in a vacuum.
Since $X-$rays are a part of the electromagnetic spectrum,they are electromagnetic waves.
Therefore,the Assertion is correct because the Reason correctly identifies their nature as electromagnetic waves,which dictates their speed.

(C) The phenomenon where an image persists on the retina after the stimulus is removed is known as persistence of vision.
This persistence of vision lasts for approximately $\frac{1}{16} \, s$.
To perceive motion in a movie, the frame rate must be higher than the reciprocal of the persistence time, which is $16 \, \text{frames per second}$.
Since $24 \, \text{frames per second}$ is greater than $16 \, \text{frames per second}$, the motion appears smooth.
The Reason provided states the persistence time is $1/10 \, s$, which is factually incorrect as it is approximately $1/16 \, s$.
Therefore, the Assertion is correct, but the Reason is incorrect.

(A) According to Rayleigh's law of scattering, the intensity of scattered light is inversely proportional to the fourth power of its wavelength $(I \propto 1/\lambda^4)$.
Since blue light has a shorter wavelength compared to other colours in the visible spectrum, it is scattered to the maximum extent by the molecules and fine particles in the atmosphere.
Therefore, the blue colour appears to be coming from the sky. Both the Assertion and the Reason are correct, and the Reason is the correct explanation of the Assertion.

(D) In Young's double slit experiment,the fringe width is given by $\beta = \frac{\lambda D}{d}$. Since the fringe width depends only on the wavelength $\lambda$,the distance between slits $d$,and the distance to the screen $D$,the fringe width for dark and bright fringes is the same. Thus,the Assertion is incorrect.
When white light is used as a source,the central fringe is white,and the subsequent fringes are coloured because different wavelengths have different fringe widths. Black fringes are not observed in the same way as monochromatic light. Thus,the Reason is also incorrect.

(A) Let us analyze the decay process for the parent nucleus ${}_{Z}{X^{A}}$:
$1$. After $2\alpha$-decays: Each $\alpha$-decay reduces the mass number by $4$ and atomic number by $2$. Thus,$2\alpha$-decays result in a change of $\Delta A = -8$ and $\Delta Z = -4$. The product is ${}_{Z-4}{X^{A-8}}$.
$2$. After $2\beta$-decays: Each $\beta^-$-decay increases the atomic number by $1$ while the mass number remains unchanged. Thus,$2\beta$-decays result in $\Delta Z = +2$. The product is ${}_{(Z-4)+2}{X^{A-8}} = {}_{Z-2}{X^{A-8}}$.
$3$. $\gamma$-decays do not change the mass number or atomic number.
Therefore,the final product is indeed ${}_{Z-2}{X^{A-8}}$. The Assertion is correct.
The Reason correctly describes the individual effects of $\alpha$ and $\beta$ decays on the mass and atomic numbers. Thus,the Reason is the correct explanation for the Assertion.

(D) In an alternating current circuit,the phase relationship between voltage and current depends on the components present.
For a purely inductive circuit (containing only an inductor $L$),the voltage leads the current by a phase angle of $\pi / 2$ $(90^{\circ})$,which is equivalent to saying the current lags behind the voltage by $\pi / 2$.
For a purely capacitive circuit (containing only a capacitor $C$),the current leads the voltage by a phase angle of $\pi / 2$ $(90^{\circ})$.
For a purely resistive circuit (containing only a resistor $R$),the current and voltage are in the same phase (phase difference is $0$).
Therefore,since the current lags behind the voltage by $\pi / 2$,the circuit must contain only an inductor $L$.

(A) The intensity of light $I$ is directly proportional to the slit width $w$ $(I \propto w)$.
Given the ratio of slit widths is $w_1 / w_2 = 1 / 9$,the ratio of intensities of the two sources is $I_1 / I_2 = 1 / 9$.
Since intensity $I \propto a^2$ (where $a$ is the amplitude),the ratio of amplitudes is $a_1 / a_2 = \sqrt{I_1 / I_2} = \sqrt{1 / 9} = 1 / 3$.
Let $a_1 = a$ and $a_2 = 3a$.
The ratio of minimum intensity to maximum intensity is given by the formula:
$\frac{I_{\text{min}}}{I_{\text{max}}} = \left( \frac{a_1 - a_2}{a_1 + a_2} \right)^2$
Substituting the values:
$\frac{I_{\text{min}}}{I_{\text{max}}} = \left( \frac{a - 3a}{a + 3a} \right)^2 = \left( \frac{-2a}{4a} \right)^2 = \left( -\frac{1}{2} \right)^2 = \frac{1}{4}$.