IIT JEE 1991 Physics Question Paper with Answer and Solution

(B) According to the Stefan-Boltzmann law,the rate of cooling is given by: $\frac{dQ}{dt} = \sigma A (T^4 - T_0^4)$.
Since $dQ = mc \, dT$ and $m = \rho V = \rho (\frac{4}{3} \pi r^3)$,we have: $mc \frac{dT}{dt} = -\sigma (4 \pi r^2) T^4$ (as $T_0 = 0 \, K$).
Substituting $m$: $(\rho \frac{4}{3} \pi r^3) c \frac{dT}{dt} = -4 \pi r^2 \sigma T^4$.
Simplifying: $\frac{dT}{T^4} = -\frac{3 \sigma}{\rho c r} dt$.
Integrating from $T_i = 200 \, K$ to $T_f = 100 \, K$: $\int_{200}^{100} T^{-4} dT = -\frac{3 \sigma}{\rho c r} \int_{0}^{t} dt$.
$[-\frac{1}{3} T^{-3}]_{200}^{100} = -\frac{3 \sigma}{\rho c r} t$.
$\frac{1}{3} [\frac{1}{100^3} - \frac{1}{200^3}] = \frac{3 \sigma}{\rho c r} t$.
$\frac{1}{3} [\frac{8 - 1}{8 \times 10^6}] = \frac{3 \sigma}{\rho c r} t$.
$\frac{7}{24 \times 10^6} = \frac{3 \sigma}{\rho c r} t$.
$t = \frac{7}{72} \frac{r \rho c}{\sigma} \times 10^{-6} \, s$.
Since the time is requested in $\mu s$ $(10^{-6} \, s)$,the value is $\frac{7}{72} \frac{r \rho c}{\sigma} \, \mu s$.

(B) The frequency of a vibrating wire is given by $n = \frac{1}{2l}\sqrt{\frac{T}{m}}$.
Since the wires are identical,$l$ and $m$ are constant,so $n \propto \sqrt{T}$.
Let the initial frequencies be $n_1$ and $n_2$ corresponding to tensions $T_1$ and $T_2$ respectively,where $T_1 > T_2$,which implies $n_1 > n_2$.
The beat frequency is $n_1 - n_2 = 6$.
If the beat frequency remains $6$ after changing the tension,the new frequencies $n_1'$ and $n_2'$ must satisfy $|n_1' - n_2'| = 6$.
Case $(i)$: If $n_1$ remains constant,$n_2$ must increase to $n_2'$ such that $n_1 - n_2' = 6$ is not possible unless $n_2$ increases past $n_1$. However,if $n_2$ increases to a value $n_2'$ such that $n_2' - n_1 = 6$,the beat frequency is still $6$. This requires increasing $T_2$.
Case $(ii)$: If $n_2$ remains constant,$n_1$ must decrease to $n_1'$ such that $n_1' - n_2 = 6$ (not possible as $n_1 > n_1'$) or $n_2 - n_1' = 6$ (not possible as $n_2 < n_1$).
Thus,the only valid physical change to maintain $6$ beats per second while changing one tension is increasing the lower tension $T_2$ such that the new frequency $n_2'$ exceeds $n_1$ by $6$ Hz.

(B) Initial capacitance $C = \frac{{\varepsilon _0}A}{d}$. Initial charge $Q = CV = \frac{{\varepsilon _0}AV}{d}$. Since the battery is disconnected,the charge $Q$ remains constant.
After inserting the dielectric slab of constant $k$,the new capacitance is $C' = kC = \frac{k{\varepsilon _0}A}{d}$.
The new potential difference is $V' = \frac{Q}{C'} = \frac{CV}{kC} = \frac{V}{k}$.
The new electric field is $E = \frac{V'}{d} = \frac{V}{kd}$.
The work done $W$ on the system is the change in stored energy: $W = U_{final} - U_{initial}$.
$U_{initial} = \frac{1}{2}CV^2$.
$U_{final} = \frac{Q^2}{2C'} = \frac{(CV)^2}{2kC} = \frac{CV^2}{2k}$.
$W = \frac{CV^2}{2k} - \frac{1}{2}CV^2 = \frac{1}{2}CV^2 \left( \frac{1}{k} - 1 \right) = -\frac{{\varepsilon _0}AV^2}{2d} \left( 1 - \frac{1}{k} \right)$.
Comparing the options,option $B$ is incorrect as it does not match the derived expression for work done.

(D) The kinetic energy of the particle at point $P$ is $K_P = \frac{1}{2}mv^2$.
The kinetic energy of the particle at point $Q$ is $K_Q = \frac{1}{2}m(2v)^2 = 2mv^2$.
The increase in kinetic energy is $\Delta K = K_Q - K_P = 2mv^2 - \frac{1}{2}mv^2 = \frac{3}{2}mv^2$.
This increase is due to the work done by the electric force $F_e = qE$ as the particle moves a distance $2a$ along the $x$-axis.
Thus,$W = qE(2a) = \frac{3}{2}mv^2$,which gives $E = \frac{3}{4}\frac{mv^2}{qa}$. So,option $A$ is correct.
The rate of work done by the electric field is $P_e = \vec{F_e} \cdot \vec{v}$. At point $P$,$\vec{F_e} = qE\hat{i}$ and $\vec{v} = v\hat{i}$,so $P_e = (qE)(v) = q(\frac{3}{4}\frac{mv^2}{qa})v = \frac{3}{4}\frac{mv^3}{a}$. So,option $B$ is correct.
At point $Q$,the velocity is $\vec{v} = -2v\hat{j}$. The electric force is $\vec{F_e} = qE\hat{i}$. Since $\vec{F_e} \perp \vec{v}$,the rate of work done by the electric field is $P_e = \vec{F_e} \cdot \vec{v} = 0$. The magnetic force $\vec{F_m} = q(\vec{v} \times \vec{B})$ is always perpendicular to the velocity $\vec{v}$,so the rate of work done by the magnetic field is $P_m = \vec{F_m} \cdot \vec{v} = 0$. Thus,the total rate of work done at $Q$ is $0+0=0$. So,option $C$ is correct.
Therefore,all statements are correct.