IIT JEE 1981 Physics Question Paper with Answer and Solution

(C) When a vessel containing liquid is accelerated horizontally with an acceleration $a$,the effective acceleration experienced by the liquid in the frame of the vessel is the vector sum of the gravitational acceleration $g$ (downwards) and the pseudo-acceleration $-a$ (backwards).
The free surface of the liquid must be perpendicular to the net effective acceleration vector. The angle $\theta$ that the surface makes with the horizontal is given by $\tan \theta = \frac{a}{g}$.
Since the acceleration $a$ is towards the right,the pseudo-force acts towards the left. Consequently,the liquid level rises on the left side and falls on the right side of the vessel. This corresponds to the surface profile shown in diagram $C$.

(C) According to Newton's $3^{rd}$ law of motion,when a person walks,they push the ground backward with their foot.
In response,the ground exerts an equal and opposite reaction force on the person.
The vertical component of this reaction force balances the person's weight,preventing vertical motion.
The horizontal component of this reaction force (which is the static friction force) acts in the forward direction and pushes the person forward.
Therefore,the reaction force exerted by the floor on the person is responsible for their forward motion.
Hence,option $C$ is correct.

(C) The acceleration due to gravity $g$ on the surface of the earth is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the earth,and $R$ is its radius.
Since the mass $M$ remains constant,we have $g \propto \frac{1}{R^2}$.
Taking the logarithmic differentiation,we get $\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$.
Given that the radius shrinks by $1\%$,we have $\frac{\Delta R}{R} = -0.01$.
Substituting this value,$\frac{\Delta g}{g} = -2 \times (-0.01) = 0.02$.
Therefore,the acceleration due to gravity increases by $2\%$.

(A) Let the mass of the bullet be $m \, g$. The total heat required for the bullet to reach its melting point and melt is given by $Q_1 = mc\Delta\theta + mL$.
Substituting the values: $Q_1 = m \times 0.03 \times (327 - 27) + m \times 6 = m \times 0.03 \times 300 + 6m = 9m + 6m = 15m \, cal$.
Converting this to Joules: $Q_1 = 15m \times 4.2 = 63m \, J$.
The kinetic energy of the bullet is $K.E. = \frac{1}{2} M v^2$,where $M$ is the mass in kg $(M = m \times 10^{-3} \, kg)$.
Since $25\%$ of the heat is absorbed by the obstacle,$75\%$ of the kinetic energy is absorbed by the bullet to melt it.
Thus,$0.75 \times (\frac{1}{2} \times m \times 10^{-3} \times v^2) = 63m$.
Simplifying: $\frac{0.75}{2} \times 10^{-3} \times v^2 = 63$.
$0.375 \times 10^{-3} \times v^2 = 63$.
$v^2 = \frac{63}{0.375 \times 10^{-3}} = \frac{63000}{0.375} = 168000$.
$v = \sqrt{168000} \approx 409.87 \, m/s \approx 410 \, m/s$.

(A) The kinetic energy $(K.E.)$ of an ideal gas is given by $K.E. = \frac{3}{2} k_B T$. If $T = 0 \, K$,the kinetic energy becomes zero,but the potential energy of the molecules may not be zero. Thus,absolute zero temperature is not zero energy temperature. Hence,option $A$ is correct.
$(B)$ The root mean square $(RMS)$ velocity is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$. Since it depends on the molar mass $(M)$,two different gases at the same temperature will have different $RMS$ velocities. Thus,option $B$ is incorrect.
$(C)$ Similar to option $B$,the $RMS$ speed depends on the molar mass. Therefore,molecules of different ideal gases at the same temperature will have different $RMS$ speeds. Thus,option $C$ is incorrect.
$(D)$ According to Avogadro's law,at the same temperature and pressure,equal volumes of all gases contain an equal number of molecules. Since both samples have $1 \, cc$ volume at $NTP$,they contain the same number of molecules. Thus,option $D$ is incorrect.

(A) Let the thickness of each layer be $d$. Let $K_c$ be the thermal conductivity of copper and $K_b$ be the thermal conductivity of brass. Given $K_c : K_b = 1 : 4$,so $K_b = 4K_c$.
In a steady state,the rate of heat flow through the copper layer must equal the rate of heat flow through the brass layer.
Let $\theta$ be the temperature of the interface.
Rate of heat flow $H = \frac{KA(\Delta T)}{d}$.
For copper: $H_c = \frac{K_c A(\theta - 0)}{d}$.
For brass: $H_b = \frac{K_b A(100 - \theta)}{d}$.
Since $H_c = H_b$,we have $\frac{K_c A \theta}{d} = \frac{4K_c A(100 - \theta)}{d}$.
Canceling common terms $K_c, A, d$: $\theta = 4(100 - \theta)$.
$\theta = 400 - 4\theta$.
$5\theta = 400$.
$\theta = 80^\circ C$.

(A) Let the piston be displaced through a small distance $x$ towards the left. The volume of the gas decreases,and its pressure increases. Let $\Delta P$ be the increase in pressure and $\Delta V$ be the decrease in volume. Assuming the process is isothermal,we have $P_1V_1 = P_2V_2$.
$PV = (P + \Delta P)(V - \Delta V)$
$PV = PV - P\Delta V + \Delta P V - \Delta P \Delta V$
Neglecting the small term $\Delta P \Delta V$,we get $P \Delta V = \Delta P V$.
Since $\Delta V = A x$ and $V = A h$,we have $P(Ax) = \Delta P(Ah)$.
$\Delta P = \frac{Px}{h}$.
This excess pressure creates a restoring force $F$ on the piston of mass $M$:
$F = -(\Delta P)A = -\left(\frac{PA}{h}\right)x$.
Comparing this with the equation for simple harmonic motion $F = -kx$,we identify the force constant $k = \frac{PA}{h}$.
The angular frequency is $\omega = \sqrt{\frac{k}{M}} = \sqrt{\frac{PA}{Mh}}$.
The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{Mh}{PA}}$.

(B) The circuit consists of two parallel branches. The upper branch has a total resistance of $R_2 = 4\, \Omega + 6\, \Omega = 10\, \Omega$. The lower branch has a resistance of $R_1 = 5\, \Omega$.
Since the branches are in parallel, the potential difference $V$ across both branches is the same.
The current in the branches is inversely proportional to their resistance: $\frac{i_1}{i_2} = \frac{R_2}{R_1} = \frac{10}{5} = \frac{2}{1}$.
The heat produced per second (power) is given by $P = i^2 R$.
For the $5\, \Omega$ resistor, $P_5 = i_1^2 \times 5 = 10\, \text{cal/s}$.
For the $4\, \Omega$ resistor, the current is $i_2$. The power dissipated in the upper branch is $P_{\text{upper}} = i_2^2 \times R_2 = i_2^2 \times 10$.
Using the ratio $\frac{i_1}{i_2} = 2$, we have $i_1 = 2i_2$. Substituting this into the power equation for the $5\, \Omega$ resistor: $(2i_2)^2 \times 5 = 10 \Rightarrow 20i_2^2 = 10 \Rightarrow i_2^2 = 0.5$.
The power dissipated in the $4\, \Omega$ resistor is $P_4 = i_2^2 \times 4 = 0.5 \times 4 = 2\, \text{cal/s}$.

(C) The induced electromotive force $(e)$ produced in the axle of the train as it moves through the Earth's magnetic field is given by the formula: $e = B_v \cdot v \cdot l$
Here,$B_v = 0.2 \times 10^{-4} \ Wb/m^2$ is the vertical component of the Earth's magnetic field.
The speed of the train $v = 180 \ km/hr = 180 \times \frac{5}{18} \ m/s = 50 \ m/s$.
The distance between the rails (length of the axle) $l = 1 \ m$.
Substituting these values into the formula:
$e = (0.2 \times 10^{-4}) \times 50 \times 1$
$e = 10 \times 10^{-4} \ V$
$e = 10^{-3} \ V$.

(C) At the distance of closest approach,the initial kinetic energy of the $\alpha$-particle is entirely converted into electrostatic potential energy.
$K.E. = P.E.$
$5 \; MeV = \frac{1}{4\pi\epsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$
Given:
$K.E. = 5 \times 10^6 \times 1.6 \times 10^{-19} \; J = 8 \times 10^{-13} \; J$
$Z = 92$ (for Uranium)
$e = 1.6 \times 10^{-19} \; C$
$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \; N \cdot m^2/C^2$
Substituting the values:
$8 \times 10^{-13} = \frac{9 \times 10^9 \times 92 \times 2 \times (1.6 \times 10^{-19})^2}{r_0}$
Solving for $r_0$:
$r_0 = \frac{9 \times 10^9 \times 184 \times 2.56 \times 10^{-38}}{8 \times 10^{-13}}$
$r_0 \approx 5.3 \times 10^{-14} \; m = 5.3 \times 10^{-12} \; cm$
Thus,the order of magnitude is $10^{-12} \; cm$.

(B) The radioactive decay law is given by $N = N_0 e^{-\lambda t}$,where $N/N_0 = 1/20$.
The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.6931}{3.8 \ days}$.
Substituting the values into the decay equation: $\frac{1}{20} = e^{-\lambda t}$,which implies $20 = e^{\lambda t}$.
Taking the natural logarithm on both sides: $\ln 20 = \lambda t$.
Converting to base $10$ logarithm: $2.303 \log_{10} 20 = \lambda t$.
Given $\log_{10} 20 = \log_{10} (2 \times 10) = 0.3010 + 1 = 1.3010$.
Substituting $\lambda = \frac{0.6931}{3.8}$ and $\log_{10} e = 0.4343$,we use the relation $\ln x = 2.303 \log_{10} x$ or $\lambda = \frac{0.6931}{T_{1/2}}$.
$t = \frac{\ln 20}{\lambda} = \frac{2.303 \times 1.3010 \times 3.8}{0.6931} \approx 16.43 \ days$.
Rounding to the nearest provided option,$t \approx 16.5 \ days$.

(A) For total internal reflection $(TIR)$ to occur at the surface $AC$,the angle of incidence $i$ at the surface $AC$ must be greater than the critical angle $C$ between glass and water.
From the geometry of the prism,the light ray incident normally on $AB$ strikes $AC$ at an angle of incidence $i = \theta$.
For $TIR$ at $AC$,we require $i > C$,which implies $\sin i > \sin C$.
Substituting $i = \theta$,we get $\sin \theta > \sin C$.
The critical angle $C$ is given by $\sin C = \frac{\mu_w}{\mu_g}$,where $\mu_w = 4/3$ and $\mu_g = 1.5 = 3/2$.
Thus,$\sin C = \frac{4/3}{3/2} = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}$.
Therefore,for $TIR$ to occur,$\sin \theta > 8/9$. Given the options,the condition is $\sin \theta \ge 8/9$.

(D) The formula for fringe width in Young's double-slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slit and the screen,and $d$ is the distance between the slits.
Given that the new distance between the slits is $d' = \frac{d}{2}$ and the new distance between the slit and the screen is $D' = 2D$.
The new fringe width $\beta'$ will be $\beta' = \frac{\lambda D'}{d'} = \frac{\lambda (2D)}{(d/2)} = 4 \left( \frac{\lambda D}{d} \right) = 4\beta$.
Therefore,the fringe width will become four times the original value.

(C) Let $q_1$ and $q_2$ be the charges on the spheres of radii $r$ and $R$ respectively.
Given $q_1 + q_2 = Q$.
Since surface charge densities are equal,$\sigma_1 = \sigma_2$.
$\frac{q_1}{4\pi r^2} = \frac{q_2}{4\pi R^2} \implies \frac{q_1}{r^2} = \frac{q_2}{R^2}$.
Using componendo and dividendo,$\frac{q_1}{r^2} = \frac{q_2}{R^2} = \frac{q_1 + q_2}{r^2 + R^2} = \frac{Q}{r^2 + R^2}$.
Thus,$q_1 = \frac{Q r^2}{R^2 + r^2}$ and $q_2 = \frac{Q R^2}{R^2 + r^2}$.
The electric potential at the common centre is $V = \frac{1}{4\pi \varepsilon_0} \left( \frac{q_1}{r} + \frac{q_2}{R} \right)$.
Substituting the values of $q_1$ and $q_2$:
$V = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q r^2}{r(R^2 + r^2)} + \frac{Q R^2}{R(R^2 + r^2)} \right) = \frac{1}{4\pi \varepsilon_0} \left( \frac{Q r + Q R}{R^2 + r^2} \right)$.
$V = \frac{1}{4\pi \varepsilon_0} \frac{(R + r)Q}{R^2 + r^2}$.

(C) $1$. The light beam is incident normally on face $AB$,so it enters the prism without deviation and strikes face $AC$.
$2$. Let the angle of incidence at face $AC$ be $i$. From the geometry of the prism,the angle between the normal to face $AC$ and the incident ray is equal to the angle $\theta$ of the prism at $A$. Thus,$i = \theta$.
$3$. For total internal reflection $(TIR)$ to occur at face $AC$,the angle of incidence must be greater than the critical angle $C$ for the glass-water interface.
$4$. The condition for $TIR$ is $i > C$,which implies $\sin i > \sin C$.
$5$. Since $i = \theta$,we have $\sin \theta > \sin C$.
$6$. The critical angle $C$ is given by $\sin C = \frac{\mu_{\text{water}}}{\mu_{\text{glass}}} = \frac{4/3}{3/2} = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9}$.
$7$. Therefore,the condition for total internal reflection is $\sin \theta > \frac{8}{9}$.