IIT JEE 1982 Physics Question Paper with Answer and Solution

(B) Initial velocity $\vec{v}_i = 5\hat{i}\,m/s$.
Final velocity $\vec{v}_f = 5\hat{j}\,m/s$.
Change in velocity $\Delta \vec{v} = \vec{v}_f - \vec{v}_i = 5\hat{j} - 5\hat{i}$.
Magnitude of change in velocity $|\Delta \vec{v}| = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\,m/s$.
The direction of $\Delta \vec{v}$ is North-West (since it points in the direction of $-\hat{i} + \hat{j}$).
Average acceleration $\vec{a}_{avg} = \frac{\Delta \vec{v}}{\Delta t} = \frac{5\sqrt{2}}{10} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\,m/s^2$.
Thus,the average acceleration is $\frac{1}{\sqrt{2}}\,m/s^2$ in the North-West direction.

(D) The modulus of rigidity $(\eta)$ is defined as the ratio of shear stress to shear strain.
Shear stress is defined as force per unit area,so its dimensional formula is $[F]/[A] = [MLT^{-2}]/[L^2] = [ML^{-1}T^{-2}]$.
Shear strain is a dimensionless quantity because it is the ratio of two lengths.
Therefore,the dimensional formula for the modulus of rigidity is the same as that of stress,which is $[ML^{-1}T^{-2}]$.
Thus,the correct option is $D$.

(D) Let $l$ be the length of the string from the pulley to the mass $M$. From the geometry of the figure,we have the relation $l^2 = b^2 + y^2$,where $b$ is the horizontal distance from the pulley to the vertical axis of the mass,and $y$ is the vertical distance of the mass from the line joining the pulleys.
Since the string is unstretchable and the ends $P$ and $Q$ move downwards with speed $U$,the length $l$ of the string segment decreases at a rate of $U$. Thus,$\frac{dl}{dt} = -U$.
Differentiating the relation $l^2 = b^2 + y^2$ with respect to time $t$,we get:
$2l \frac{dl}{dt} = 2b \frac{db}{dt} + 2y \frac{dy}{dt}$.
Since the pulleys are fixed,the horizontal distance $b$ remains constant,so $\frac{db}{dt} = 0$.
Substituting $\frac{dl}{dt} = -U$ and $\frac{db}{dt} = 0$ into the equation:
$2l(-U) = 2y \frac{dy}{dt}$.
Solving for the vertical speed of the mass $v_y = \frac{dy}{dt}$:
$\frac{dy}{dt} = -\frac{l}{y} U$.
From the triangle formed,$\cos \theta = \frac{y}{l}$,so $\frac{l}{y} = \frac{1}{\cos \theta}$.
Therefore,the speed of the mass $M$ is $v = |\frac{dy}{dt}| = \frac{U}{\cos \theta}$.

(A) The upthrust (buoyant force) on a body submerged in a liquid is given by the formula:
$F_B = V \rho_{\text{liquid}} (g - a)$
where $V$ is the volume of the liquid displaced,$\rho_{\text{liquid}}$ is the density of the liquid,$g$ is the acceleration due to gravity,and $a$ is the downward acceleration of the system.
In the case of free fall,the entire system (beaker and liquid) accelerates downwards with an acceleration equal to the acceleration due to gravity,i.e.,$a = g$.
Substituting $a = g$ into the formula:
$F_B = V \rho_{\text{liquid}} (g - g) = V \rho_{\text{liquid}} (0) = 0$.
Therefore,the upthrust on the body is zero.

(A) From the ideal gas equation,$PV = \mu RT$,we have $V = (\frac{\mu R}{P})T$.
The slope of the $V-T$ graph is given by $m = \tan \theta = \frac{V}{T} = \frac{\mu R}{P}$.
Since the slope is inversely proportional to pressure $(m \propto \frac{1}{P})$,a smaller slope corresponds to a higher pressure.
From the graph,it is clear that $\theta_1 < \theta_2$,which implies $\tan \theta_1 < \tan \theta_2$.
Therefore,$(\frac{V}{T})_1 < (\frac{V}{T})_2$.
Since $(\frac{V}{T}) \propto \frac{1}{P}$,we have $(\frac{1}{P})_1 < (\frac{1}{P})_2$,which leads to $P_1 > P_2$.

(D) Comparing the given equation $y = 10^4 \sin(60t + 2x)$ with the standard wave equation $y = a \sin(\omega t + kx)$:
$1$. Since the sign between $\omega t$ and $kx$ is positive,the wave is travelling in the negative $X$-direction.
$2$. The angular frequency is $\omega = 60 \, rad/s$. The frequency $f$ is given by $f = \frac{\omega}{2\pi} = \frac{60}{2\pi} = \frac{30}{\pi} \, Hz$.
$3$. The wave number is $k = 2 \, m^{-1}$. The wavelength $\lambda$ is given by $k = \frac{2\pi}{\lambda} \implies \lambda = \frac{2\pi}{k} = \frac{2\pi}{2} = \pi \, m$.
$4$. The wave velocity $v$ is given by $v = \frac{\omega}{k} = \frac{60}{2} = 30 \, m/s$.
Since all the statements are correct,the correct option is $(d)$.

(A) The system consists of two particles moving under the influence of their mutual internal attraction.
According to the law of conservation of momentum,if the net external force acting on a system is zero,the velocity of the centre of mass remains constant.
Initially,both particles are at rest,which means the initial velocity of the centre of mass is $v_{CM, initial} = 0$.
Since there is no external force acting on the system,the velocity of the centre of mass remains constant at $0$ at any instant.
Therefore,the speed of the centre of mass of the system is $0$.

(A) Magnetic flux $\phi$ is defined as the product of magnetic field $B$ and area $A$,i.e.,$\phi = B \cdot A$.
From the Lorentz force formula,$F = B I L$,we can write $B = \frac{F}{I L}$.
Substituting the dimensions of force $[F] = [M L T^{-2}]$,current $[I] = [A]$,and length $[L] = [L]$,we get the dimensions of $B$ as $[B] = \frac{[M L T^{-2}]}{[A] [L]} = [M T^{-2} A^{-1}]$.
Now,the dimensions of magnetic flux $\phi$ are $[B] \times [A] = [M T^{-2} A^{-1}] \times [L^2] = [M L^2 T^{-2} A^{-1}]$.

(B) In a steady state,the capacitor acts as an open circuit,so no current flows through the branch containing the capacitor.
The circuit simplifies to a $6\,V$ battery connected in series with a $2.8\,\Omega$ resistor and a parallel combination of $2\,\Omega$ and $3\,\Omega$ resistors.
The equivalent resistance of the parallel combination is $R_p = \frac{2 \times 3}{2 + 3} = \frac{6}{5} = 1.2\,\Omega$.
The total resistance of the circuit is $R_{eq} = 2.8 + 1.2 = 4.0\,\Omega$.
The total current supplied by the battery is $I = \frac{V}{R_{eq}} = \frac{6}{4} = 1.5\,A$.
Using the current divider rule,the current through the $2\,\Omega$ resistor is $I_{2\Omega} = I \times \frac{3}{2 + 3} = 1.5 \times \frac{3}{5} = 0.9\,A$.

(A) magnetic needle acts as a magnetic dipole.
In a non-uniform magnetic field,the magnetic force acting on the two poles of the needle will be different in both magnitude and direction.
Because the magnetic field is non-uniform,the net force on the dipole is non-zero,resulting in a translational force.
Additionally,because the forces on the two poles are not collinear and have different magnitudes,they create a net torque,causing the needle to rotate.
Therefore,the magnetic needle experiences both a force and a torque.

(D) For photoelectric emission to occur, the wavelength of the incident radiation $(\lambda)$ must be less than or equal to the threshold wavelength $(\lambda_0)$ of the material.
Given: $\lambda_0 = 5200 \, \mathring{A}$.
Photoelectric emission occurs if $\lambda \le 5200 \, \mathring{A}$.
Ultraviolet $(UV)$ radiation typically has a wavelength range of $100 \, \mathring{A}$ to $4000 \, \mathring{A}$, which is less than $5200 \, \mathring{A}$.
Infrared $(IR)$ radiation has a wavelength greater than $7000 \, \mathring{A}$, which is greater than $5200 \, \mathring{A}$.
Therefore, both the $1 \, \text{W}$ ultraviolet lamp and the $50 \, \text{W}$ ultraviolet lamp will cause photoelectric emission, regardless of their power, as long as the wavelength is in the $UV$ range.

(B) The energy of an $X$-ray photon is given by $E = h\nu = \frac{hc}{\lambda}$.
For the shortest wavelength (cutoff wavelength),the entire kinetic energy of the electron is converted into a single photon: $eV = \frac{hc}{\lambda_{\min}}$.
Rearranging for $\lambda_{\min}$,we get $\lambda_{\min} = \frac{hc}{eV}$.
Since $h$,$c$,and $e$ are constants,$\lambda_{\min} \propto \frac{1}{V}$.
Therefore,the shortest wavelength depends only on the accelerating voltage $V$ applied to the tube.

(A) The speed of light in free space is given by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
The speed of light in a medium is given by $v = \frac{1}{\sqrt{\mu \varepsilon}}$.
The refractive index $n$ of a medium is defined as the ratio of the speed of light in vacuum to the speed of light in the medium:
$n = \frac{c}{v} = \frac{1/\sqrt{\mu_0 \varepsilon_0}}{1/\sqrt{\mu \varepsilon}} = \sqrt{\frac{\mu \varepsilon}{\mu_0 \varepsilon_0}}$.

(B) The power $P$ of a lens is given by $P = \frac{100}{f(cm)}\, D$.
For the convex lens,the focal length $f_1 = +40\, cm$.
Therefore,the power $P_1 = \frac{100}{40} = +2.5\, D$.
For the concave lens,the focal length $f_2 = -25\, cm$.
Therefore,the power $P_2 = \frac{100}{-25} = -4.0\, D$.
The power of the combination of lenses in contact is given by $P = P_1 + P_2$.
$P = 2.5\, D + (-4.0\, D) = -1.5\, D$.

(D) The ratio of maximum intensity to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = 9$.
We know that $\frac{I_{\max}}{I_{\min}} = \left( \frac{a_1 + a_2}{a_1 - a_2} \right)^2 = 9$.
Taking the square root on both sides,we get $\frac{a_1 + a_2}{a_1 - a_2} = 3$.
Solving for the ratio of amplitudes: $a_1 + a_2 = 3a_1 - 3a_2 \Rightarrow 4a_2 = 2a_1 \Rightarrow \frac{a_1}{a_2} = 2$.
Since intensity $I \propto a^2$,the ratio of intensities of individual sources is $\frac{I_1}{I_2} = \left( \frac{a_1}{a_2} \right)^2 = 2^2 = 4$,which means $I_1 : I_2 = 4 : 1$.
Thus,both statements $(b)$ and $(c)$ are correct.