IIT JEE 1980 Physics Question Paper with Answer and Solution

(C) Given: Mass $m = 3 \times 10^7 \, kg$,Force $F = 5 \times 10^4 \, N$,Distance $s = 3 \, m$,Initial velocity $u = 0$.
According to Newton's second law,acceleration $a = \frac{F}{m} = \frac{5 \times 10^4}{3 \times 10^7} = \frac{5}{3} \times 10^{-3} \, m/s^2$.
Using the kinematic equation $v^2 - u^2 = 2as$:
$v^2 = 0^2 + 2 \times \left( \frac{5 \times 10^4}{3 \times 10^7} \right) \times 3$.
$v^2 = 2 \times \left( \frac{5}{3} \times 10^{-3} \right) \times 3 = 10 \times 10^{-3} = 10^{-2}$.
$v = \sqrt{10^{-2}} = 0.1 \, m/s$.

(A) The component of the weight acting down the inclined plane is $F_{applied} = mg \sin \theta = 2 \times 9.8 \times \sin 30^\circ = 2 \times 9.8 \times 0.5 = 9.8 \, N$.
The limiting friction is given by $F_l = \mu_s N = \mu_s mg \cos \theta$.
$F_l = 0.7 \times 2 \times 9.8 \times \cos 30^\circ = 0.7 \times 19.6 \times \frac{\sqrt{3}}{2} \approx 11.88 \, N$.
Since the applied force $(9.8 \, N)$ is less than the limiting friction $(11.88 \, N)$,the block remains at rest.
Therefore,the static frictional force is equal to the component of the weight acting down the plane,which is $9.8 \, N$.

(C) The relationship between linear momentum $P$,mass $m$,and kinetic energy $E$ is given by $P = \sqrt{2mE}$.
Since the kinetic energies $E$ are equal for both masses,we have $P \propto \sqrt{m}$.
Therefore,the ratio of their linear momenta is $\frac{P_1}{P_2} = \sqrt{\frac{m_1}{m_2}}$.
Substituting the given values $m_1 = 1 \,g$ and $m_2 = 4 \,g$,we get $\frac{P_1}{P_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(B) Let the thickness of each layer be $x$. Let the thermal conductivity of $B$ be $K$,then the thermal conductivity of $A$ is $2K$.
In steady state,the rate of heat flow through both layers is the same.
Rate of heat flow $H = \frac{KA(\Delta T)}{x}$.
For layer $A$: $H_A = \frac{(2K)A(\theta_1 - \theta)}{x}$.
For layer $B$: $H_B = \frac{KA(\theta - \theta_2)}{x}$.
Since $H_A = H_B$,we have:
$\frac{2KA(\theta_1 - \theta)}{x} = \frac{KA(\theta - \theta_2)}{x}$
$2(\theta_1 - \theta) = (\theta - \theta_2)$
Let $\Delta T_A = (\theta_1 - \theta)$ and $\Delta T_B = (\theta - \theta_2)$.
Then $2\Delta T_A = \Delta T_B$.
We are given the total temperature difference $\Delta T_A + \Delta T_B = 36^{\circ}C$.
Substituting $\Delta T_B = 2\Delta T_A$ into the equation:
$\Delta T_A + 2\Delta T_A = 36^{\circ}C$
$3\Delta T_A = 36^{\circ}C$
$\Delta T_A = 12^{\circ}C$.
Thus,the temperature difference across layer $A$ is $12^{\circ}C$.

(B) The pitch of a sound is primarily determined by its frequency. Higher frequency results in higher pitch.
The quality (or timbre) of a sound is determined by the waveform,which depends on the presence of overtones and harmonics.
The loudness of a sound is primarily determined by its intensity,which is related to the amplitude of the sound wave.
Therefore,the correct matching is: Pitch-frequency,Quality-waveform,Loudness-intensity.
Correct Option: $B$

(B) The resistance of a conductor at temperature $t$ is given by $R_t = R_0(1 + \alpha \Delta t)$,where $R_0$ is the resistance at $0\,^{\circ}C$ and $\Delta t$ is the temperature change from $0\,^{\circ}C$.
Given: $\alpha = 0.00125\,^{\circ}C^{-1}$,$R_1 = 1\,\Omega$ at $T_1 = 300\,K$ $(t_1 = 27\,^{\circ}C)$,and $R_2 = 2\,\Omega$ at $T_2 = ?$.
Using the formula $R_2 = R_1[1 + \alpha(t_2 - t_1)]$:
$2 = 1[1 + 0.00125(t_2 - 27)]$
$2 - 1 = 0.00125(t_2 - 27)$
$1 = 0.00125(t_2 - 27)$
$t_2 - 27 = \frac{1}{0.00125} = 800$
$t_2 = 800 + 27 = 827\,^{\circ}C$
Converting to Kelvin: $T_2 = 827 + 273 = 1100\,K$.

(C) When light travels from one medium to another,its frequency $(f)$ remains constant because it depends on the source of the light.
The speed of light $(v)$ in a medium is given by $v = f \lambda$,where $\lambda$ is the wavelength.
Since the refractive index of glass $(\mu_g)$ is greater than that of air $(\mu_a)$,the speed of light decreases when it enters the glass $(v_g < v_a)$.
Because $v = f \lambda$ and $f$ is constant,the decrease in speed $(v)$ must result in a decrease in wavelength $(\lambda)$.
Therefore,the wavelength decreases while the frequency remains unchanged.

(B) The intensity of illumination $I$ on a surface is given by the formula: $I = \frac{L \cos \theta}{r^2}$,where $L$ is the luminous intensity of the source in candela,$r$ is the distance,and $\theta$ is the angle with the normal.
Given:
$I = 5 \times 10^{-4} \ phot = 5 \times 10^{-4} \ lumen/cm^2 = 5 \times 10^{-4} \times 10^4 \ lumen/m^2 = 5 \ lumen/m^2$.
$r = 2 \ m$.
$\theta = 60^{\circ}$.
Rearranging the formula to find $L$:
$L = \frac{I \times r^2}{\cos \theta}$.
Substituting the values:
$L = \frac{5 \times 2^2}{\cos 60^{\circ}} = \frac{5 \times 4}{0.5} = \frac{20}{0.5} = 40 \ candela$.
Thus,the correct option is $B$.