AIIMS 1980 Physics Question Paper with Answer and Solution

(A) When a train moves along a curved track,it follows a circular path.
For any circular path,the outer rail is at a larger distance from the center of curvature compared to the inner rail.
Let $R_o$ be the radius of the outer rail and $R_i$ be the radius of the inner rail.
Since the outer rail is further from the center of the turn,$R_o > R_i$.
Therefore,the radius of curvature of the outer rail is greater than that of the inner rail.

(B) The Earth revolves around the Sun due to the gravitational pull of the Sun.
Due to this gravitational attraction between these celestial bodies,a centripetal force is generated,which keeps the Earth in its orbit.
According to Newton's laws of motion,a body moving in a circular path must be acted upon by a centripetal force directed towards the center of the circle.
Therefore,the revolution of the Earth around the Sun is the evidence that there must be a force acting on the Earth and directed towards the Sun.

(A) The formula for the extension $l$ in a wire is given by $l = \frac{FL}{AY}$,where $A = \pi r^2$ is the cross-sectional area and $Y$ is the Young's modulus.
Thus,$l = \frac{FL}{\pi r^2 Y}$.
Since the material is the same,$Y$ is constant. Therefore,$l \propto \frac{FL}{r^2}$.
For the first wire: $l_1 = l$,$F_1 = F$,$L_1 = L$,$r_1 = r$.
For the second wire: $F_2 = 2F$,$L_2 = 2L$,$r_2 = 2r$.
Taking the ratio: $\frac{l_2}{l_1} = \left( \frac{F_2}{F_1} \right) \times \left( \frac{L_2}{L_1} \right) \times \left( \frac{r_1}{r_2} \right)^2$.
Substituting the values: $\frac{l_2}{l} = \left( \frac{2F}{F} \right) \times \left( \frac{2L}{L} \right) \times \left( \frac{r}{2r} \right)^2 = 2 \times 2 \times \left( \frac{1}{4} \right) = 1$.
Therefore,$l_2 = l$.

(C) The twisting couple $C$ required to produce an angle of twist $\theta$ in a wire of length $l$,radius $r$,and modulus of rigidity $\eta$ is given by the formula:
$C = \frac{\pi \eta r^4 \theta}{2l}$
From this expression,we can see that for a constant twisting couple $C$,length $l$,and material property $\eta$:
$\theta = \frac{2lC}{\pi \eta r^4}$
This implies that $\theta \propto \frac{1}{r^4}$.
Therefore,for wires $A$ and $B$ with radii $r_1$ and $r_2$ respectively,the ratio of the angles of twist $\theta_1$ and $\theta_2$ is:
$\frac{\theta_1}{\theta_2} = \frac{r_2^4}{r_1^4}$
Thus,the correct option is $C$.

(A) Surface tension is a property arising from the cohesive forces between molecules at the surface of a liquid.
As the temperature of a liquid increases,the kinetic energy of the molecules increases,which weakens the cohesive forces.
At the critical temperature,the distinction between the liquid phase and the vapor phase disappears,and the density of the liquid becomes equal to the density of the vapor.
Consequently,the cohesive forces become negligible,and the surface tension of the liquid drops to $0$.

(B) The angle of contact $\theta$ is determined by the balance of surface tension forces at the interface of liquid,solid,and gas.
As the temperature increases,the cohesive forces between the liquid molecules decrease more rapidly than the adhesive forces between the liquid and the solid surface.
Since the angle of contact is inversely related to the strength of the adhesive forces relative to the cohesive forces,a reduction in cohesive force leads to a decrease in the angle of contact.
Therefore,the correct option is $(b)$.

(B) When a body is floating in a liquid such that its density is equal to the density of the liquid,it is in a state of neutral equilibrium.
If the body is pushed down slightly,the buoyant force and the weight of the body remain equal because the density of the liquid is uniform.
Since the net force on the body remains zero at any depth,the body will remain submerged at the position where it is left.

(B) The general equation for a wave traveling in the negative $x$-direction is given by $y(x, t) = A \sin(kx + \omega t + \phi)$ or $A \cos(kx + \omega t + \phi)$.
Given: Amplitude $A = 0.5\, m$,Wavelength $\lambda = 1\, m$,Frequency $f = 2\, Hz$.
Calculate the angular wave number $k = \frac{2\pi}{\lambda} = \frac{2\pi}{1} = 2\pi\, rad/m$.
Calculate the angular frequency $\omega = 2\pi f = 2\pi(2) = 4\pi\, rad/s$.
Substituting these values into the wave equation: $y(x, t) = 0.5 \cos(2\pi x + 4\pi t)$.
Thus,option $B$ is correct.

(A) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} C V^2$.
Given:
Capacitance $C = 4 \times 10^{-6} \, F$
Potential $V = 100 \, V$
Substituting the values into the formula:
$U = \frac{1}{2} \times (4 \times 10^{-6}) \times (100)^2$
$U = \frac{1}{2} \times 4 \times 10^{-6} \times 10000$
$U = 2 \times 10^{-6} \times 10^4$
$U = 2 \times 10^{-2} \, J$
$U = 0.02 \, J$
Therefore,the energy released is $0.02 \, J$.

(C) The magnetic field $B$ at the centre of a circular coil of $N$ turns and radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 N I}{2r}$.
Since the length of the wire $L$ is constant,$L = 2\pi r_1 N_1 = 2\pi r_2 N_2$.
For the first case,$N_1 = 1$,so $L = 2\pi r_1$,which implies $r_1 = \frac{L}{2\pi}$.
For the second case,$N_2 = 2$,so $L = 2\pi r_2 \times 2$,which implies $r_2 = \frac{L}{4\pi} = \frac{r_1}{2}$.
The magnetic field in the first case is $B_1 = \frac{\mu_0 (1) I}{2r_1}$.
The magnetic field in the second case is $B_2 = \frac{\mu_0 (2) I}{2r_2} = \frac{\mu_0 (2) I}{2(r_1/2)} = \frac{4 \mu_0 I}{2r_1} = 4 B_1$.
Thus,the magnetic field becomes four times its first value.

(A) Soft iron is highly ferromagnetic and has high magnetic permeability and low retentivity.
This means it can be easily magnetized and demagnetized,which is an essential property for the core of an electromagnet.
Therefore,soft iron is the most suitable material for the core of electromagnets.

(C) The self-inductance $L$ of a solenoid is given by the formula $L = \mu_0 N^2 A / l$.
Since the number of turns per unit length $n = N / l$ is constant,we can write $N = nl$.
Substituting this into the formula,we get $L = \mu_0 (nl)^2 A / l = \mu_0 n^2 l A$.
Here,$l$ represents the length of the coil and $A$ represents the cross-sectional area.
When all linear dimensions are increased by a factor of $2$,the length $l$ becomes $2l$ and the area $A$ (which is proportional to the square of linear dimensions) becomes $2^2 A = 4A$.
Substituting these new values into the expression for $L$,the new inductance $L'$ is $L' = \mu_0 n^2 (2l) (4A) = 8 (\mu_0 n^2 l A) = 8L$.
Therefore,the self-inductance increases by a factor of $8$.