IIT JEE 1986 Physics Question Paper with Answer and Solution

(A) The dimensional formula for Torque is $[ML^2T^{-2}]$.
The dimensional formula for Work is $[ML^2T^{-2}]$.
Since both have the same dimensional formula,the correct pair is Torque and Work.
Therefore,the correct option is $A$.

(C) In an inelastic collision,the kinetic energy of the ball is not conserved as some of it is converted into other forms of energy like heat or sound.
However,for the system consisting of the ball and the Earth,there is no external force acting on the system during the collision.
According to the law of conservation of momentum,in the absence of an external force,the total momentum of the system remains constant.
Therefore,the total momentum of the ball and the Earth is conserved.

(B) The atmosphere consists of various gas molecules that possess mass.
According to Newton's law of universal gravitation,a gravitational force acts between the earth and these gas molecules.
This gravitational force pulls the gas molecules toward the center of the earth,preventing them from escaping into space.
Therefore,the atmosphere is held to the earth primarily due to gravity.
Thus,the correct option is $B.$

(A) For a mixture of gases,the adiabatic index $\gamma_{\text{mix}}$ is given by the formula:
$\gamma_{\text{mix}} = \frac{n_1 C_{p,1} + n_2 C_{p,2}}{n_1 C_{v,1} + n_2 C_{v,2}}$
Alternatively,using the relation $C_v = \frac{R}{\gamma - 1}$,we have:
$\gamma_{\text{mix}} = \frac{n_1 + n_2}{\frac{n_1}{\gamma_1 - 1} + \frac{n_2}{\gamma_2 - 1}}$
Given $n_1 = 1, \gamma_1 = 5/3$ and $n_2 = 1, \gamma_2 = 7/5$:
$\gamma_{\text{mix}} = \frac{1 + 1}{\frac{1}{5/3 - 1} + \frac{1}{7/5 - 1}} = \frac{2}{\frac{1}{2/3} + \frac{1}{2/5}} = \frac{2}{3/2 + 5/2} = \frac{2}{8/2} = \frac{2}{4} = 1/2$ (Wait,re-calculating).
Correct calculation:
$C_{v,1} = \frac{R}{5/3 - 1} = \frac{3R}{2}$,$C_{v,2} = \frac{R}{7/5 - 1} = \frac{5R}{2}$.
$C_{v,\text{mix}} = \frac{n_1 C_{v,1} + n_2 C_{v,2}}{n_1 + n_2} = \frac{1(1.5R) + 1(2.5R)}{2} = 2R$.
$C_{p,\text{mix}} = C_{v,\text{mix}} + R = 2R + R = 3R$.
$\gamma_{\text{mix}} = \frac{C_{p,\text{mix}}}{C_{v,\text{mix}}} = \frac{3R}{2R} = 1.5 = 3/2$.

(D) For a simple pendulum oscillating in a vertical plane,the forces acting on the bob are the tension $T$ in the string (directed towards the pivot) and the gravitational force $Mg$ (directed vertically downwards).
$1$. Radial direction: The net force towards the center of the circular path is the centripetal force,which is given by $\frac{Mv^2}{L}$. The radial component of the gravitational force is $Mg \cos \theta$ (acting away from the pivot). Thus,the net radial force is $T - Mg \cos \theta$. Equating these,we get $T - Mg \cos \theta = \frac{Mv^2}{L}$. This matches option $(b)$.
$2$. Tangential direction: The component of the gravitational force acting tangentially to the path is $Mg \sin \theta$. According to Newton's second law,$F_T = Ma_T$,so $Mg \sin \theta = Ma_T$. Therefore,the magnitude of the tangential acceleration is $|a_T| = g \sin \theta$. This matches option $(c)$.
Since both $(b)$ and $(c)$ are correct,the correct option is $(d)$.

(C) Since there is no external torque acting on the system,the angular momentum is conserved.
$L_i = L_f$
$I_1 \omega_1 = I_2 \omega_2$
For the first disc,the moment of inertia is $I_1 = \frac{1}{2} M R^2$ and angular velocity is $\omega_1 = \omega$.
When the second disc of mass $M/4$ is placed coaxially,the new moment of inertia of the system is $I_2 = I_1 + I_{disc2} = \frac{1}{2} M R^2 + \frac{1}{2} (M/4) R^2$.
$I_2 = \frac{1}{2} M R^2 + \frac{1}{8} M R^2 = \frac{4+1}{8} M R^2 = \frac{5}{8} M R^2$.
Using the conservation of angular momentum:
$\frac{1}{2} M R^2 \cdot \omega = \frac{5}{8} M R^2 \cdot \omega_2$
$\omega_2 = \frac{1/2}{5/8} \omega = \frac{1}{2} \cdot \frac{8}{5} \omega = \frac{4}{5} \omega$.

(A) In the absence of gravity,the only force acting on the spheres is the electrostatic repulsive force between them.
Since both spheres have the same charge $+Q$,they repel each other.
To maximize the distance between them and reach an equilibrium position,the threads will stretch out in opposite directions,forming a straight line.
Thus,the angle between the two threads is $180^\circ$.
The distance between the two charges is $r = L + L = 2L$.
According to Coulomb's law,the electrostatic force $F$ between them is $F = \frac{1}{4\pi \varepsilon_0} \frac{Q \cdot Q}{(2L)^2} = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{(2L)^2}$.
This force is equal to the tension $T$ in each thread.

(B) The magnetic field $B$ produced by one wire at a distance $b$ is given by the formula: $B = \frac{\mu_0 i}{2\pi b}$.
According to the Lorentz force law,the force $F$ on a length $L$ of the second wire carrying current $i$ in this magnetic field is $F = i L B \sin(\theta)$.
Since the wires are parallel,the angle $\theta = 90^\circ$,so $\sin(90^\circ) = 1$.
Therefore,the force per unit length $f = \frac{F}{L} = i B$.
Substituting the value of $B$,we get $f = i \left( \frac{\mu_0 i}{2\pi b} \right) = \frac{\mu_0 i^2}{2\pi b}$.

(D) The mass number $(A)$ is the sum of the number of protons $(Z)$ and neutrons $(N)$ in a nucleus,so $A = Z + N$.
For the hydrogen nucleus $(_{1}^{1}H)$,the number of protons is $1$ and the number of neutrons is $0$,so the mass number is $1$,which is equal to the atomic number $(A = Z)$.
For all other nuclei,the number of neutrons is at least $1$ or more,making the mass number greater than the atomic number $(A > Z)$.
Therefore,the mass number is sometimes equal to and sometimes greater than the atomic number.

(A) radioactive nucleus undergoes decay by emitting either an $\alpha$-particle or a $\beta$-particle at a given time.
It is physically impossible for a single nucleus to emit both $\alpha$ and $\beta$ particles simultaneously.
However,$\gamma$-rays are often emitted as a result of the nucleus transitioning from an excited state to a lower energy state following the emission of an $\alpha$ or $\beta$ particle.
Therefore,at any single instant,a nucleus emits only one type of particle ($\alpha$ or $\beta$),which may be accompanied by $\gamma$-radiation.

(D) When the upper half of the lens is covered,the light rays from the object still pass through the remaining lower half of the lens.
Since every point on the object sends light rays to all parts of the lens,the lower half of the lens is sufficient to form the complete image of the object at the same position.
However,because the total amount of light passing through the lens is reduced (as half of the aperture is blocked),the intensity of the image decreases.
Therefore,the complete image is formed,but with decreased intensity.

(B) For coherent sources,the resultant intensity is given by $I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
At the central point,the path difference is zero,so $\phi = 0$. Given $I_1 = I_2 = I_0$,the intensity for coherent sources is $I_{coh} = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(0) = 4I_0$.
For incoherent sources,the phase difference $\phi$ varies randomly with time,so the average value of $\cos \phi$ is $0$.
Thus,the resultant intensity for incoherent sources is $I_{incoh} = I_1 + I_2 = I_0 + I_0 = 2I_0$.
The ratio of the intensities is $\frac{I_{coh}}{I_{incoh}} = \frac{4I_0}{2I_0} = \frac{2}{1}$.