IIT JEE 1989 Physics Question Paper with Answer and Solution

(B) The centripetal force required for the circular motion of the planet is provided by the gravitational force of attraction.
The gravitational force is given as $F_g \propto R^{-5/2}$.
The centripetal force is given by $F_c = m \omega^2 R = m \left( \frac{2\pi}{T} \right)^2 R = \frac{4\pi^2 m R}{T^2}$.
Equating the two forces: $\frac{m R}{T^2} \propto R^{-5/2}$.
Rearranging for $T^2$: $\frac{1}{T^2} \propto \frac{R^{-5/2}}{R} = R^{-7/2}$.
Therefore,$T^2 \propto R^{7/2}$.

(C) The thermal stress $\sigma$ developed in a rod fixed between two rigid walls when heated by a temperature change $\Delta \theta$ is given by the formula:
$\sigma = Y \alpha \Delta \theta$
where $Y$ is the Young's modulus and $\alpha$ is the coefficient of linear expansion.
Given that the thermal stresses are equal $(\sigma_1 = \sigma_2)$ and the increase in temperature $\Delta \theta$ is the same for both rods,we have:
$Y_1 \alpha_1 \Delta \theta = Y_2 \alpha_2 \Delta \theta$
$Y_1 \alpha_1 = Y_2 \alpha_2$
Rearranging the terms to find the ratio $Y_1 : Y_2$:
$\frac{Y_1}{Y_2} = \frac{\alpha_2}{\alpha_1}$
Given $\alpha_1 : \alpha_2 = 2 : 3$,we have $\frac{\alpha_1}{\alpha_2} = \frac{2}{3}$,which implies $\frac{\alpha_2}{\alpha_1} = \frac{3}{2}$.
Therefore,$\frac{Y_1}{Y_2} = \frac{3}{2}$ or $Y_1 : Y_2 = 3 : 2$.

(D) The total mechanical energy of the oscillator is given as $E = 160 \, J$.
For a linear harmonic oscillator,the energy associated with the harmonic oscillation (the kinetic energy part) is given by $E_{osc} = \frac{1}{2} k A^2$.
Substituting the given values: $E_{osc} = \frac{1}{2} \times (2 \times 10^6) \times (0.01)^2 = 10^6 \times 10^{-4} = 100 \, J$.
This $100 \, J$ represents the maximum kinetic energy $(K_{max})$ of the oscillator.
Since the total energy is $160 \, J$ and the oscillatory part is $100 \, J$,there must be an additional constant potential energy $U_0 = 160 - 100 = 60 \, J$.
The potential energy varies as $U(x) = U_0 + \frac{1}{2} k x^2$.
The maximum potential energy occurs at the extreme positions $(x = \pm A)$,which is $U_{max} = U_0 + \frac{1}{2} k A^2 = 60 + 100 = 160 \, J$.
Thus,both statements $(b)$ and $(c)$ are correct.

(A) Let the radius of the solid sphere be $a$ and the radius of the hollow shell be $b$.
Initially,the solid sphere has charge $Q$ and the shell has charge $0$.
The potential at the surface of the solid sphere is $V_{\text{sphere}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{a} + \frac{1}{4\pi\varepsilon_0} \frac{0}{b} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{a}$.
The potential at the surface of the hollow shell is $V_{\text{shell}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{b} + \frac{1}{4\pi\varepsilon_0} \frac{0}{b} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{b}$.
The initial potential difference is $V = V_{\text{sphere}} - V_{\text{shell}} = \frac{Q}{4\pi\varepsilon_0} \left( \frac{1}{a} - \frac{1}{b} \right)$.
Now,the shell is given a charge of $-3Q$.
The new potential at the surface of the solid sphere is $V'_{\text{sphere}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{a} + \frac{1}{4\pi\varepsilon_0} \frac{-3Q}{b}$.
The new potential at the surface of the hollow shell is $V'_{\text{shell}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{b} + \frac{1}{4\pi\varepsilon_0} \frac{-3Q}{b} = \frac{1}{4\pi\varepsilon_0} \left( -\frac{2Q}{b} \right)$.
The new potential difference is $V' = V'_{\text{sphere}} - V'_{\text{shell}} = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{a} - \frac{3Q}{b} - (-\frac{2Q}{b}) \right) = \frac{1}{4\pi\varepsilon_0} \left( \frac{Q}{a} - \frac{Q}{b} \right) = V$.

(B) The electric field is a conservative field,so the work done by the electric field depends only on the initial and final positions,not on the path taken.
Work done $W = \vec{F} \cdot \vec{d} = q\vec{E} \cdot \vec{d}$,where $\vec{d}$ is the displacement vector.
The initial position is $P(a, b, 0)$ and the final position is $S(0, 0, 0)$.
The displacement vector $\vec{d} = \vec{S} - \vec{P} = (0 - a)\hat{i} + (0 - b)\hat{j} + (0 - 0)\hat{k} = -a\hat{i} - b\hat{j}$.
The electric field is $\vec{E} = E\hat{i}$.
Work done $W = q(E\hat{i}) \cdot (-a\hat{i} - b\hat{j}) = qE(-a) = -qEa$.

(A) First,simplify the right side of the circuit. The resistors $7\,\Omega$,$1\,\Omega$,and $10\,\Omega$ are in series,so their equivalent resistance is $R_s = 7 + 1 + 10 = 18\,\Omega$.
This $18\,\Omega$ resistor is in parallel with the $6\,\Omega$ resistor. Their equivalent resistance $R_p$ is given by $\frac{1}{R_p} = \frac{1}{6} + \frac{1}{18} = \frac{3+1}{18} = \frac{4}{18}$,so $R_p = \frac{18}{4} = 4.5\,\Omega$.
Now,the total resistance of the circuit is $R_{total} = 2\,\Omega + 0.5\,\Omega + 4.5\,\Omega + 8\,\Omega = 15\,\Omega$.
The current from the battery is $i = \frac{V}{R_{total}} = \frac{15\,V}{15\,\Omega} = 1\,A$.

(D) The magnetic flux $\phi$ linked with the loop is given by $\phi = B \cdot A$,where $A = L^2$ is the area of the square loop.
Since the magnetic field $B$ is uniform in time and space,and the loop moves within this field,the area $A$ enclosed by the loop remains constant.
According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is given by $\varepsilon = -\frac{d\phi}{dt}$.
Since $\phi = B \cdot L^2$ is constant,$\frac{d\phi}{dt} = 0$.
Therefore,the induced $EMF$ $\varepsilon = 0$.
Consequently,the induced current $I = \frac{\varepsilon}{R} = 0$.

(B) The half-life $(T_{1/2})$ of a radioactive substance is defined as the time required for half of the radioactive nuclei to decay.
It is given by the formula: $T_{1/2} = \frac{ln 2}{\lambda} = \frac{log_e 2}{\lambda}$.
The mean life or average life $(\tau)$ is defined as the reciprocal of the decay constant.
It is given by the formula: $\tau = \frac{1}{\lambda}$.
Therefore,the half-life and mean life are $\frac{log_e 2}{\lambda}$ and $\frac{1}{\lambda}$ respectively.

(B) In circuit $(1)$,the $N$-region of the first diode is connected to the $N$-region of the second diode. This configuration does not allow for a proper series biasing of the junctions.
In circuit $(2)$,the $P$-region of the first diode is connected to the $N$-region of the second diode. The external battery is connected such that both diodes are forward-biased (or both reverse-biased depending on polarity),allowing for an equal potential drop across both junctions.
In circuit $(3)$,the $P$-region of the first diode is connected to the $N$-region of the second diode. Similar to circuit $(2)$,this configuration allows for a symmetric series connection where the potential difference is distributed equally across the junctions.
Therefore,circuits $(2)$ and $(3)$ are the correct connections.

(D) For an astronomical telescope with the final image at infinity (normal adjustment),the angular magnification is given by $|m| = \frac{f_o}{f_e} = 5$.
This implies $f_o = 5f_e$ ... $(i)$.
The length of the telescope tube is the sum of the focal lengths of the objective and the eyepiece: $L = f_o + f_e = 36 \, cm$ ... $(ii)$.
Substituting $(i)$ into $(ii)$,we get $5f_e + f_e = 36 \, cm$,which simplifies to $6f_e = 36 \, cm$.
Thus,$f_e = 6 \, cm$.
Substituting $f_e$ back into $(i)$,we get $f_o = 5 \times 6 \, cm = 30 \, cm$.
Therefore,the focal lengths are $f_o = 30 \, cm$ and $f_e = 6 \, cm$.

(A) As the beam of light is incident normally on the face $AB$ of the right-angled prism $ABC$,no refraction occurs at face $AB$. The light passes straight and strikes the face $AC$ at an angle of incidence $i = 45^{\circ}$.
For total internal reflection to take place at face $AC$,the condition is $i > i_c$,where $i_c$ is the critical angle.
We know that $\sin i_c = \frac{1}{\mu}$. Therefore,the condition for total internal reflection is $\sin i > \frac{1}{\mu}$,or $\mu > \frac{1}{\sin i}$.
Given $i = 45^{\circ}$,we have $\sin 45^{\circ} = \frac{1}{\sqrt{2}} \approx 0.707$. Thus,the condition becomes $\mu > \sqrt{2} \approx 1.414$.
Comparing the refractive indices:
For red: $\mu_{\text{red}} = 1.39 < 1.414$.
For green: $\mu_{\text{green}} = 1.44 > 1.414$.
For blue: $\mu_{\text{blue}} = 1.47 > 1.414$.
Since $\mu_{\text{red}} < 1.414$,the red light will be refracted out of the prism through face $AC$. Since $\mu_{\text{green}}$ and $\mu_{\text{blue}}$ are both greater than $1.414$,both green and blue light will undergo total internal reflection at face $AC$.
Therefore,the prism will separate the red colour from the green and blue colours.