AIIMS 1995 Physics Question Paper with Answer and Solution

(A) Let the initial position be the origin $(0, 0, 0)$.
First,the aeroplane moves $400 \,m$ North (along $y$-axis) and $300 \,m$ South (along negative $y$-axis).
The net displacement in the horizontal plane is $400 \,m - 300 \,m = 100 \,m$ towards the North.
Then,it flies $1200 \,m$ upwards (along $z$-axis).
The final position vector is $\vec{r} = 100 \hat{j} + 1200 \hat{k}$.
The magnitude of the net displacement is $r = \sqrt{(100)^2 + (1200)^2}$.
$r = \sqrt{10000 + 1440000} = \sqrt{1450000}$.
$r = 100 \sqrt{145} \approx 100 \times 12.04 = 1204 \,m$.

(C) In projectile motion,the horizontal displacement $x$ is given by $x = u_x t$,which is a linear function of time. The vertical displacement $y$ is given by $y = u_y t - \frac{1}{2} g t^2$,which is a downward-opening parabola. At the highest point,the vertical velocity $v_y = \frac{dy}{dt} = 0$. On a displacement-time graph for vertical motion,the slope is $\frac{dy}{dt}$. At the highest point,the slope is $0$. Since the path is a downward-opening parabola ($y = at^2 + bt + c$ with $a < 0$),the second derivative $\frac{d^2y}{dt^2} = -g$ is negative. However,the question asks for the displacement-time graph of the particle's motion. The vertical displacement $y(t)$ reaches a maximum at the peak,where the slope is $0$ and the curvature (concavity) is negative. Thus,the correct description is zero slope and negative curvature.

(D) According to the law of conservation of linear momentum,the total momentum before projection must equal the total momentum after projection.
Initially,both the tank and the body are at rest,so the initial momentum is $0$.
Let $M = 100 \, kg$ be the mass of the tank and $m = 0.25 \, kg$ be the mass of the body.
Let $V$ be the recoil velocity of the tank and $v = 100 \, m/s$ be the velocity of the body.
$M \times V + m \times v = 0$
$100 \times V + 0.25 \times 100 = 0$
$100 \times V = -25$
$V = -25 / 100 = -0.25 \, m/s$.
The magnitude of the recoil velocity is $0.25 \, m/s$.

(D) The potential energy of the particle is given by $U(x) = k(1 - e^{-x^2})$.
The force acting on the particle is $F = -\frac{dU}{dx}$.
$F = -\frac{d}{dx} [k(1 - e^{-x^2})] = -k[0 - e^{-x^2} \cdot (-2x)] = -2kxe^{-x^2}$.
For small displacements from the origin $(x \approx 0)$,we can use the Taylor expansion $e^{-x^2} \approx 1 - x^2 + \dots \approx 1$.
Thus,$F \approx -2kx$.
Since $F \propto -x$,the restoring force is proportional to the displacement,which is the condition for simple harmonic motion $(SHM)$.

(B) The orbital speed of a satellite at a distance $r$ from the center of a planet is given by $v = \sqrt{\frac{GM}{r}}$.
Since $v \propto \frac{1}{\sqrt{r}}$,we can write the ratio of the speeds of satellites $A$ and $B$ as:
$\frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A}}$
Given $r_A = 4R$ and $r_B = R$,we have:
$\frac{v_A}{v_B} = \sqrt{\frac{R}{4R}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$
Given the speed of satellite $A$ is $v_A = 3V$,we substitute this into the ratio:
$\frac{3V}{v_B} = \frac{1}{2}$
Therefore,$v_B = 3V \times 2 = 6V$.

(D) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the orbital radius $(R)$: $T^2 \propto R^3$,which implies $T \propto R^{3/2}$.
Given the initial time period $T_1 = 5$ hours.
Let the initial separation be $R_1$ and the new separation be $R_2 = 4R_1$.
The new time period $T_2$ is given by the ratio: $\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} \right)^{3/2}$.
Substituting the values: $\frac{T_2}{5} = \left( \frac{4R_1}{R_1} \right)^{3/2} = (4)^{3/2}$.
Calculating the exponent: $(4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.
Therefore,$T_2 = 5 \times 8 = 40$ hours.

(C) Let the depth of the lake be $h$. The pressure at the bottom of the lake is $P_1 = P_0 + h\rho g$,where $P_0$ is the atmospheric pressure and $\rho$ is the density of water.
Given that atmospheric pressure $P_0 = H\rho g$.
So,$P_1 = H\rho g + h\rho g = (H + h)\rho g$.
The volume of the bubble at the bottom is $V_1 = \frac{4}{3}\pi r^3$.
At the surface,the pressure is $P_2 = P_0 = H\rho g$ and the radius becomes $2r$,so the volume is $V_2 = \frac{4}{3}\pi (2r)^3 = 8 \times \frac{4}{3}\pi r^3$.
Assuming the temperature remains constant,we use Boyle's Law: $P_1V_1 = P_2V_2$.
$(H + h)\rho g \times \frac{4}{3}\pi r^3 = H\rho g \times 8 \times \frac{4}{3}\pi r^3$.
$(H + h) = 8H$.
$h = 7H$.

(B) Let the specific gravities of concrete and sawdust be $\rho_1 = 2.4$ and $\rho_2 = 0.3$ respectively.
According to the principle of floatation,the weight of the whole sphere equals the upthrust (buoyant force) on the sphere.
Weight of sphere = Weight of water displaced
$\frac{4}{3}\pi (R^3 - r^3)\rho_1 g + \frac{4}{3}\pi r^3 \rho_2 g = \frac{4}{3}\pi R^3 \times 1 \times g$
Dividing by $\frac{4}{3}\pi g$,we get:
$(R^3 - r^3)\rho_1 + r^3 \rho_2 = R^3$
$R^3 \rho_1 - r^3 \rho_1 + r^3 \rho_2 = R^3$
$R^3(\rho_1 - 1) = r^3(\rho_1 - \rho_2)$
$\frac{R^3}{r^3} = \frac{\rho_1 - \rho_2}{\rho_1 - 1} = \frac{2.4 - 0.3}{2.4 - 1} = \frac{2.1}{1.4} = 1.5$
Now,the ratio of the mass of concrete $(M_c)$ to the mass of sawdust $(M_s)$ is:
$\frac{M_c}{M_s} = \frac{\frac{4}{3}\pi (R^3 - r^3)\rho_1}{\frac{4}{3}\pi r^3 \rho_2} = \left( \frac{R^3}{r^3} - 1 \right) \frac{\rho_1}{\rho_2}$
Substituting the values:
$\frac{M_c}{M_s} = (1.5 - 1) \times \frac{2.4}{0.3} = 0.5 \times 8 = 4$.

(D) The change in entropy $(dS)$ for a reversible process is defined as $dS = \frac{dQ_{rev}}{T}$.
For an adiabatic reversible process,the heat exchange $dQ_{rev} = 0$,which implies $dS = 0$.
Among the given options,the conversion of work into heat isochorically (if done reversibly) or any process that is internally reversible and adiabatic results in no change in entropy.
However,in the context of standard thermodynamic problems,a reversible process is defined as an isentropic process $(dS = 0)$.
Option $(d)$ represents a process where work is converted to heat; if this is performed reversibly,the entropy of the system remains constant.

(B) For an ideal monoatomic gas,the molar heat capacity at constant pressure is $C_P = \frac{5}{2}R$ and at constant volume is $C_V = \frac{3}{2}R$.
The heat supplied at constant pressure is given by $\Delta Q = n C_P \Delta T$.
The change in internal energy is given by $\Delta U = n C_V \Delta T$.
The fraction of heat energy that increases the internal energy is $\frac{\Delta U}{\Delta Q} = \frac{n C_V \Delta T}{n C_P \Delta T} = \frac{C_V}{C_P}$.
Substituting the values,we get $\frac{\Delta U}{\Delta Q} = \frac{3/2 R}{5/2 R} = \frac{3}{5} = 0.6$.

(C) For a cyclic process,the change in internal energy is $\Delta U = 0$.
From the First Law of Thermodynamics $(FLOT)$,$\Delta Q = \Delta U + \Delta W = 0 + \Delta W = \Delta W$.
The work done $\Delta W$ is equal to the area enclosed by the cyclic process in the $P-V$ diagram.
The area of the circle is $\pi r_P r_V$,where $r_P$ is the radius along the pressure axis and $r_V$ is the radius along the volume axis.
$r_P = \frac{30 \text{ kPa} - 10 \text{ kPa}}{2} = 10 \text{ kPa} = 10 \times 10^{3} \text{ Pa}$.
$r_V = \frac{30 \text{ litre} - 10 \text{ litre}}{2} = 10 \text{ litre} = 10 \times 10^{-3} \text{ m}^{3}$.
Therefore,$\Delta Q = \pi \times (10 \times 10^{3} \text{ Pa}) \times (10 \times 10^{-3} \text{ m}^{3}) = \pi \times 100 \text{ J} = 100 \pi \text{ J} = 10^{2} \pi \text{ J}$.

(B) The time taken by the particle to move from the mean position $(x=0)$ to the extreme position $(x=4 \, cm)$ is $T/4 = 1.2/4 = 0.3 \, s$.
Let $t_1$ be the time taken to move from $x=0$ to $x=2 \, cm$. The equation of motion is $x = A \sin(\omega t)$.
Substituting the values: $2 = 4 \sin(\frac{2\pi}{T} t_1) \Rightarrow 1/2 = \sin(\frac{2\pi}{1.2} t_1)$.
This gives $\frac{\pi}{6} = \frac{2\pi}{1.2} t_1$,which simplifies to $t_1 = 0.1 \, s$.
The time taken to move from $x=2 \, cm$ to $x=4 \, cm$ is $t_2 = (T/4) - t_1 = 0.3 - 0.1 = 0.2 \, s$.
The total time to move from $x=2 \, cm$ to $x=4 \, cm$ and back again is $2 \times t_2 = 2 \times 0.2 = 0.4 \, s$.

(B) The amplitudes of the two waves are given by ${a_1} = 5$ and ${a_2} = 10$.
The intensity of a wave is proportional to the square of its amplitude,$I \propto a^2$.
The ratio of maximum intensity to minimum intensity is given by the formula:
$\frac{I_{\max}}{I_{\min}} = \frac{(a_1 + a_2)^2}{(a_1 - a_2)^2}$.
Substituting the values of the amplitudes:
$\frac{I_{\max}}{I_{\min}} = \left( \frac{5 + 10}{5 - 10} \right)^2 = \left( \frac{15}{-5} \right)^2 = (-3)^2 = 9$.
Therefore,the ratio is $9:1$ or simply $9$.

(D) In a stationary wave,the total energy oscillates between kinetic and potential energy.
When the kinetic energy of the string is maximum,the potential energy is zero.
Potential energy is associated with the deformation (displacement) of the string.
Since potential energy is zero,the displacement of every particle on the string must be zero at that instant.
Therefore,all particles are passing through their mean positions simultaneously.
This results in the string appearing as a straight line along the equilibrium axis.

(C) Let the amplitudes be $A_1 = 10 \,\mu m$,$A_2 = 4 \,\mu m$,and $A_3 = 7 \,\mu m$.
The waves arrive with successive phase differences of $\frac{\pi}{2}$.
Let the phase of the first wave be $0$,the second be $\frac{\pi}{2}$,and the third be $\pi$.
Using the phasor method,the resultant amplitude $A_R$ is given by the vector sum of the amplitudes:
$A_R = \sqrt{(\sum A_i \cos \phi_i)^2 + (\sum A_i \sin \phi_i)^2}$
$A_x = A_1 \cos(0) + A_2 \cos(\frac{\pi}{2}) + A_3 \cos(\pi) = 10(1) + 4(0) + 7(-1) = 10 - 7 = 3 \,\mu m$
$A_y = A_1 \sin(0) + A_2 \sin(\frac{\pi}{2}) + A_3 \sin(\pi) = 10(0) + 4(1) + 7(0) = 4 \,\mu m$
Resultant amplitude $A_R = \sqrt{A_x^2 + A_y^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \,\mu m$.

(C) Let the vertices of the equilateral triangle be $A, B,$ and $C$. The masses are placed at $A, B,$ and $C$. The axis of rotation passes through side $AB$.
$1$. The perpendicular distance of mass at $A$ from the axis $AB$ is $r_A = 0$.
$2$. The perpendicular distance of mass at $B$ from the axis $AB$ is $r_B = 0$.
$3$. The perpendicular distance of mass at $C$ from the axis $AB$ is the altitude $x$ of the equilateral triangle.
Using the Pythagorean theorem in the triangle formed by the altitude:
$x^2 + (a/2)^2 = a^2$
$x^2 = a^2 - a^2/4 = 3a^2/4$
$x = \frac{\sqrt{3}}{2}a$
The moment of inertia $I$ of the system about the axis $AB$ is given by:
$I = \sum m_i r_i^2 = m(r_A^2) + m(r_B^2) + m(r_C^2)$
$I = m(0)^2 + m(0)^2 + m(x)^2$
$I = m \left( \frac{\sqrt{3}}{2}a \right)^2 = m \left( \frac{3}{4}a^2 \right) = \frac{3}{4}m a^2$

(C) When an object of mass $m$ slides down an inclined plane without friction,the potential energy is converted entirely into translational kinetic energy: $mgh = \frac{1}{2}mv^2$,which gives $v = \sqrt{2gh}$.
When the object is a ring rolling down the inclined plane,the potential energy is converted into both translational and rotational kinetic energy: $mgh = \frac{1}{2}mv_{ring}^2 + \frac{1}{2}I\omega^2$.
For a ring,the moment of inertia $I = mR^2$ and $\omega = \frac{v_{ring}}{R}$.
Substituting these,we get: $mgh = \frac{1}{2}mv_{ring}^2 + \frac{1}{2}(mR^2)(\frac{v_{ring}}{R})^2 = \frac{1}{2}mv_{ring}^2 + \frac{1}{2}mv_{ring}^2 = mv_{ring}^2$.
Thus,$v_{ring} = \sqrt{gh}$.
Since $v = \sqrt{2gh}$,we can write $\sqrt{gh} = \frac{v}{\sqrt{2}}$.
Therefore,the velocity of the ring is $\frac{v}{\sqrt{2}}$.

(A) For the mass to remain in contact with the surface,the downward acceleration of the surface must not exceed the acceleration due to gravity $g$.
The maximum acceleration of a particle in $S.H.M.$ is given by $a_{max} = \omega^2 A$.
To maintain contact,we require $a_{max} \leq g$.
Substituting $\omega = 2 \pi f$,we get $(2 \pi f)^2 A \leq g$.
Given $A = 1 \, cm = 0.01 \, m$ and taking $g \approx 10 \, m/s^2$:
$4 \pi^2 f^2 (0.01) = 10$
$f^2 = \frac{10}{4 \pi^2 \times 0.01} = \frac{10}{0.04 \pi^2} = \frac{250}{\pi^2}$
$f = \sqrt{\frac{250}{\pi^2}} = \frac{\sqrt{250}}{\pi} \approx \frac{15.81}{3.14} \approx 5.03 \, Hz$.
Thus,the maximum frequency is approximately $5 \, Hz$.

(A) For point charges,the force is given by Coulomb's law: $F_e = \frac{k Q^2}{d^2}$.
When the charges are placed on spheres of radius $R = 0.3 \, d$,the distance between the centers is $d$.
Since the spheres are close to each other ($R$ is a significant fraction of $d$),the electrostatic induction effect causes the charges to redistribute on the surfaces of the spheres.
Specifically,the positive charge on one sphere is attracted towards the negative charge on the other sphere,causing the charges to shift closer to each other on the inner faces of the spheres.
Because the effective distance between the centers of charge becomes less than $d$,the force of attraction increases.
Therefore,the new force is greater than $F_e$.

(C) The distance of each corner from the center $O$ of the square is $r = \frac{a}{\sqrt{2}}$.
The electric potential $V_O$ at the center $O$ due to the four charges $Q$ at the corners is:
$V_O = 4 \times \left( \frac{1}{4\pi \varepsilon_0} \cdot \frac{Q}{r} \right) = 4 \times \left( \frac{Q}{4\pi \varepsilon_0 (a/\sqrt{2})} \right) = \frac{4\sqrt{2} Q}{4\pi \varepsilon_0 a} = \frac{\sqrt{2} Q}{\pi \varepsilon_0 a}$.
The work done $W$ in moving a charge $q = -Q$ from the center to infinity is given by $W = q(V_{\infty} - V_O)$.
Since the potential at infinity $V_{\infty} = 0$,we have:
$W = (-Q)(0 - V_O) = Q V_O$.
Substituting the value of $V_O$:
$W = Q \times \left( \frac{\sqrt{2} Q}{\pi \varepsilon_0 a} \right) = \frac{\sqrt{2} Q^2}{\pi \varepsilon_0 a}$.

(C) Let the equivalent resistance of the infinite network be $R$. Since the network is infinite,adding or removing one repeating unit does not change the equivalent resistance.
The circuit can be viewed as a $2\,\Omega$ resistor in series with the top branch,a $2\,\Omega$ resistor in series with the bottom branch,and the equivalent resistance $R$ in parallel with the vertical $2\,\Omega$ resistor.
Thus,the equivalent resistance $R$ is given by:
$R = 2 + 2 + \frac{2 \times R}{2 + R}$
$R = 4 + \frac{2R}{2 + R}$
Multiplying by $(2 + R)$:
$R(2 + R) = 4(2 + R) + 2R$
$2R + R^2 = 8 + 4R + 2R$
$R^2 - 4R - 8 = 0$
Using the quadratic formula $R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$R = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(-8)}}{2(1)}$
$R = \frac{4 \pm \sqrt{16 + 32}}{2} = \frac{4 \pm \sqrt{48}}{2} = \frac{4 \pm 4\sqrt{3}}{2} = 2 \pm 2\sqrt{3}$
Since resistance cannot be negative,we take the positive root:
$R = 2 + 2\sqrt{3} \approx 2 + 2(1.732) = 5.464\,\Omega$.
This value is greater than $4\,\Omega$ and less than $12\,\Omega$.

(D) For two short bar magnets placed coaxially,the force of interaction $F$ is inversely proportional to the fourth power of the distance $r$ between their centers,i.e.,$F \propto \frac{1}{r^4}$.
Given the initial force $F_1 = 4.8 \ N$ at distance $r_1 = r$.
When the distance is increased to $r_2 = 2r$,the new force $F_2$ is given by:
$\frac{F_2}{F_1} = \left( \frac{r_1}{r_2} \right)^4$
$\frac{F_2}{4.8} = \left( \frac{r}{2r} \right)^4 = \left( \frac{1}{2} \right)^4 = \frac{1}{16}$
$F_2 = \frac{4.8}{16} = 0.3 \ N$.

(B) Let the energies of the states $A, B,$ and $C$ be $E_A, E_B,$ and $E_C$ respectively.
From the principle of conservation of energy,the energy of the transition from $C$ to $A$ is equal to the sum of the energies of the transitions from $C$ to $B$ and $B$ to $A$.
Thus,$(E_C - E_A) = (E_C - E_B) + (E_B - E_A)$.
Using the relation $E = \frac{hc}{\lambda}$,we can write:
$\frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$
Dividing both sides by $hc$,we get:
$\frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$
$\frac{1}{\lambda_3} = \frac{\lambda_1 + \lambda_2}{\lambda_1 \lambda_2}$
Therefore,$\lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$.

(B) When lenses are in contact,the effective focal length $F$ is given by $\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$. Since the object is at infinity,the image is formed at the focus $F = 60 \ cm$. Thus,$\frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{60}$ ... $(i)$.
When the lenses are separated by $d = 10 \ cm$,the equivalent focal length $F'$ is given by $\frac{1}{F'} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2}$. The image shifts by $30 \ cm$ towards the combination,so the new image distance is $60 \ cm - 30 \ cm = 30 \ cm$. Thus,$\frac{1}{30} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{10}{f_1 f_2}$ ... $(ii)$.
Substituting $(i)$ into $(ii)$: $\frac{1}{30} = \frac{1}{60} - \frac{10}{f_1 f_2} \implies \frac{10}{f_1 f_2} = \frac{1}{60} - \frac{1}{30} = -\frac{1}{60} \implies f_1 f_2 = -600$.
From $(i)$,$\frac{f_1 + f_2}{f_1 f_2} = \frac{1}{60} \implies f_1 + f_2 = \frac{-600}{60} = -10$.
Solving $x^2 - (f_1+f_2)x + f_1 f_2 = 0 \implies x^2 + 10x - 600 = 0 \implies (x+30)(x-20) = 0$. Thus,the focal lengths are $20 \ cm$ and $-30 \ cm$.

(A) For the light ray to retrace its path,it must strike the silvered surface normally (at an angle of $90^{\circ}$).
In the triangle formed inside the prism,the angle at the silvered surface is $90^{\circ}$,and the angle at the apex is $A$. Thus,the angle of refraction $r$ at the first surface must be $r = 90^{\circ} - A$.
According to Snell's Law at the first surface: $\mu = \frac{\sin i}{\sin r}$.
Given $i = 2A$ and $r = 90^{\circ} - A$,we have:
$\mu = \frac{\sin(2A)}{\sin(90^{\circ} - A)}$
$\mu = \frac{2 \sin A \cos A}{\cos A}$
$\mu = 2 \sin A$.

(D) Let $P$ be the point on the screen directly in front of slit $S_1$. The path difference between the light rays reaching $P$ from $S_1$ and $S_2$ is given by:
$\Delta x = S_2P - S_1P = \sqrt{b^2 + d^2} - d$
Using binomial expansion for $d >> b$:
$\Delta x = d(1 + \frac{b^2}{d^2})^{1/2} - d \approx d(1 + \frac{b^2}{2d^2}) - d = \frac{b^2}{2d}$
For destructive interference (missing wavelengths),the path difference must be an odd multiple of $\frac{\lambda}{2}$:
$\Delta x = (2n - 1)\frac{\lambda}{2}$,where $n = 1, 2, 3, ...$
Equating the two expressions for path difference:
$\frac{b^2}{2d} = (2n - 1)\frac{\lambda}{2}$
$\lambda = \frac{b^2}{(2n - 1)d}$
For $n = 1$,$\lambda = \frac{b^2}{d}$.
For $n = 2$,$\lambda = \frac{b^2}{3d}$.
Thus,both $(a)$ and $(c)$ are correct.