AIPMT 1989 Physics Question Paper with Answer and Solution

(C) According to the principle of homogeneity of dimensions,the dimensions of each term in a physical equation must be the same.
Given the equation $x = at + bt^2$,where $x$ is distance and $t$ is time.
Therefore,the dimensions of $bt^2$ must be equal to the dimensions of $x$.
$[bt^2] = [x]$
$[b] = [x] / [t^2]$
Since $x$ is in kilometres $(km)$ and $t$ is in seconds $(s)$,the unit of $b$ is $km/s^2$.

(D) Energy per unit volume = $\frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$.
Force per unit area = $\frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.
Product of voltage and charge per unit volume = $\frac{V \times Q}{\text{Volume}} = \frac{\text{Energy}}{\text{Volume}} = [ML^{-1}T^{-2}]$.
Angular momentum per unit mass = $\frac{[ML^2T^{-1}]}{[M]} = [L^2T^{-1}]$.
Thus,angular momentum per unit mass has dimensions different from the other three.

(A) The energy $E$ stored in an inductor is given by the formula $E = \frac{1}{2} L i^2$.
Rearranging for $L$,we get $L = \frac{2E}{i^2}$.
The dimensional formula for energy $E$ is $[M L^2 T^{-2}]$.
The dimensional formula for current $i$ is $[A]$.
Substituting these into the expression for $L$:
$L = \frac{[M L^2 T^{-2}]}{[A]^2} = [M L^2 T^{-2} A^{-2}]$.
Thus,the correct option is $A$.

(A) Torque $( \tau)$ is defined as the product of force and the perpendicular distance from the axis of rotation.
Mathematically,$\tau = \text{Force} \times \text{Distance}$.
The dimensional formula for force is $[M L T^{-2}]$.
The dimensional formula for distance is $[L]$.
Therefore,the dimensional formula for torque is $[M L T^{-2}] \times [L] = [M L^2 T^{-2}]$.
Thus,the correct option is $A$.

(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 $1:2$.

(D) Given total mass $M = 5 \,kg$. The masses of the fragments are in the ratio $1:1:3$,so the masses are $m_1 = 1 \,kg$,$m_2 = 1 \,kg$,and $m_3 = 3 \,kg$.
The two fragments of mass $1 \,kg$ move in mutually perpendicular directions with speed $v = 21 \,m/s$.
The momentum of the first fragment is $P_x = m_1 v = 1 \times 21 = 21 \,kg \cdot m/s$ along the $x$-axis.
The momentum of the second fragment is $P_y = m_2 v = 1 \times 21 = 21 \,kg \cdot m/s$ along the $y$-axis.
The resultant momentum of these two fragments is $P_{res} = \sqrt{P_x^2 + P_y^2} = \sqrt{21^2 + 21^2} = 21\sqrt{2} \,kg \cdot m/s$.
According to the law of conservation of linear momentum,the initial momentum of the body at rest is zero. Therefore,the momentum of the third (heaviest) fragment must be equal and opposite to the resultant momentum of the first two fragments.
$P_3 = P_{res} = 21\sqrt{2} \,kg \cdot m/s$.
Since $P_3 = m_3 v_3$,we have $3 \times v_3 = 21\sqrt{2}$.
$v_3 = \frac{21\sqrt{2}}{3} = 7\sqrt{2} \,m/s$.

(A) The escape velocity of a body from the surface of a planet is given by the formula $v_e = \sqrt{\frac{2GM}{R}}$.
This expression depends only on the mass of the planet $(M)$ and the radius of the planet $(R)$.
It is independent of the mass of the object being launched and the angle of projection.
Therefore,regardless of the angle at which the satellite is launched,the escape velocity remains constant at $11 \ km/s$.

(B) The gravitational force provides the necessary centripetal force for the particle to move in a circular orbit.
For a circular orbit,the centripetal force is given by $F_c = \frac{mv^2}{R}$.
According to the problem,the gravitational force is $F_g \propto \frac{1}{R}$,which can be written as $F_g = \frac{K}{R}$ for some constant $K$.
Equating the two forces: $\frac{mv^2}{R} = \frac{K}{R}$.
Canceling $R$ from both sides,we get $mv^2 = K$.
Since $m$ and $K$ are constants,$v^2$ is constant,which means $v$ is constant.
Therefore,$v \propto R^0$.

(A) According to the principle of calorimetry,the heat lost by the hot water is equal to the heat gained by the ice to melt.
Let $m$ be the mass of ice that melts in grams.
The specific heat capacity of water is $c_w = 1\, cal/g^{\circ}C$.
The latent heat of fusion of ice is $L_f = 80\, cal/g$.
Heat lost by water = $m_w \times c_w \times \Delta T = 80\, g \times 1\, cal/g^{\circ}C \times (30^{\circ}C - 0^{\circ}C) = 2400\, cal$.
Heat gained by ice = $m \times L_f = m \times 80\, cal/g$.
Equating the two: $2400 = m \times 80$.
Therefore,$m = \frac{2400}{80} = 30\, g$.

(B) According to the kinetic theory of gases,the average kinetic energy of gas molecules is directly proportional to the absolute temperature $(KE_{avg} \propto T)$.
As the temperature increases,the average velocity $(v_{rms} \propto \sqrt{T})$ of the gas molecules increases.
Since the molecules move faster,they strike the walls of the container more frequently per unit time.
Therefore,the number of collisions per unit time increases.

(C) According to the law of equipartition of energy,each degree of freedom contributes $\frac{1}{2}kT$ to the average kinetic energy of a molecule.
For a gas molecule with $n$ degrees of freedom,the total mean energy per molecule is the sum of the energies associated with each degree of freedom.
Therefore,the mean energy per molecule $E = n \times (\frac{1}{2}kT) = \frac{nkT}{2}$.
Thus,the correct option is $C$.

(C) The speed of a transverse wave on a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.
Linear mass density $\mu = \frac{M}{L} = \frac{0.035 \; kg}{7 \; m} = 0.005 \; kg/m$.
Given tension $T = 60.5 \; N$.
Substituting the values into the formula:
$v = \sqrt{\frac{60.5}{0.005}}$
$v = \sqrt{12100}$
$v = 110 \; m/s$.

(C) The intensity of a sound wave is given by the formula $I = 2{\pi ^2}{a^2}{n^2}\rho v$,where $a$ is the amplitude and $n$ is the frequency.
From this,we see that $I \propto {a^2}{n^2}$.
Let the initial amplitude be $a_1$ and frequency be $n_1$. Let the final amplitude be $a_2 = 2a_1$ and frequency be $n_2 = n_1/4$.
The ratio of intensities is $\frac{I_2}{I_1} = \left( \frac{a_2}{a_1} \right)^2 \times \left( \frac{n_2}{n_1} \right)^2$.
Substituting the values: $\frac{I_2}{I_1} = (2)^2 \times (1/4)^2 = 4 \times (1/16) = 1/4$.
Therefore,$I_2 = I_1/4$,which means the intensity is decreased by a factor of $4$.

(D) According to the work-energy theorem,the work done by all forces is equal to the change in kinetic energy of the body.
Here,the work done against air resistance $(W_{air})$ is equal to the loss in kinetic energy.
Mass of the bullet,$m = 10 \; g = 0.01 \; kg$.
Initial velocity,$u = 1000 \; m/s$.
Final velocity,$v = 500 \; m/s$.
Work done against air resistance $W = \Delta K = K_i - K_f = \frac{1}{2} m (u^2 - v^2)$.
$W = \frac{1}{2} \times 0.01 \times [(1000)^2 - (500)^2]$.
$W = 0.005 \times [1,000,000 - 250,000]$.
$W = 0.005 \times 750,000$.
$W = 3750 \; J$.

(C) Let the initial velocity be $\vec{v}_1 = 50 \; \text{km/h}$ due north.
After turning left by $90^{\circ}$,the final velocity is $\vec{v}_2 = 50 \; \text{km/h}$ due west.
The change in velocity is given by $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1 = \vec{v}_2 + (-\vec{v}_1)$.
Here,$-\vec{v}_1 = 50 \; \text{km/h}$ due south.
The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{v_2^2 + v_1^2} = \sqrt{50^2 + 50^2} = \sqrt{2500 + 2500} = \sqrt{5000} \approx 70.7 \; \text{km/h}$.
The direction of $\Delta \vec{v}$ is the resultant of a vector pointing west and a vector pointing south,which is in the south-west direction.

(A) The distance covered by an object in the $n^{th}$ second is given by the formula:
$s_{n} = u + \frac{a}{2}(2n - 1)$
Given that the body is falling freely from rest,the initial velocity $u = 0$ and the acceleration $a = g$.
For the $4^{th}$ second $(n = 4)$:
$s_{4} = 0 + \frac{g}{2}(2 \times 4 - 1) = \frac{g}{2}(7) = \frac{7g}{2}$
For the $5^{th}$ second $(n = 5)$:
$s_{5} = 0 + \frac{g}{2}(2 \times 5 - 1) = \frac{g}{2}(9) = \frac{9g}{2}$
The ratio of the distance moved in the $4^{th}$ and $5^{th}$ seconds is:
$\frac{s_{4}}{s_{5}} = \frac{7g/2}{9g/2} = \frac{7}{9}$

(C) The Curie $(Ci)$ is a non-$SI$ unit of radioactivity.
It is defined as the quantity of any radioactive nuclide in which the number of disintegrations per second is $3.7 \times 10^{10}$.
Therefore,Curie is a unit of radioactivity.

(C) Let the resistance of each resistor be $R$.
When $n$ resistors are connected in series,the equivalent resistance is $R_{\max} = nR$.
When $n$ resistors are connected in parallel,the equivalent resistance is $R_{\min} = R/n$.
The ratio of the maximum resistance to the minimum resistance is given by:
$\frac{R_{\max}}{R_{\min}} = \frac{nR}{R/n} = n \times n = n^2$.

(B) Let the resistance of each bulb be $R$.
When $40$ bulbs are connected in series,the total resistance is $R_{eq1} = 40R$.
The current in the circuit is $I_1 = V / (40R)$.
The power consumed by each bulb is $P_1 = I_1^2 R = (V / 40R)^2 R = V^2 / (1600R)$.
When $39$ bulbs are connected in series,the total resistance is $R_{eq2} = 39R$.
The current in the circuit is $I_2 = V / (39R)$.
The power consumed by each bulb is $P_2 = I_2^2 R = (V / 39R)^2 R = V^2 / (1521R)$.
Since $1521R < 1600R$,it follows that $P_2 > P_1$.
Therefore,the illumination is more with $39$ bulbs than with $40$ bulbs.

(B) The heat produced in a conductor is given by Joule's law of heating: $H = I^2Rt$.
Given:
$I = 2\, A$
$H = 80\, J$
$t = 10\, s$
Rearranging the formula to solve for resistance $R$:
$R = \frac{H}{I^2t}$
Substituting the values:
$R = \frac{80}{(2)^2 \times 10} = \frac{80}{4 \times 10} = \frac{80}{40} = 2\,\,\Omega$.
Therefore,the resistance of the conductor is $2\,\,\Omega$.

(C) When the contact is broken,the current in the circuit decreases rapidly.
According to Lenz's Law,the inductor opposes this change in current by inducing an electromotive force (emf) given by $\varepsilon = -L \frac{di}{dt}$.
Since the current drops to zero very quickly when the switch is opened,the rate of change of current $\frac{di}{dt}$ is very high.
This results in a large induced emf across the inductor,which causes a momentary surge of current through the bulb,making it glow suddenly bright.

(A) According to the de-Broglie hypothesis,the wavelength $\lambda$ associated with a particle of momentum $p$ is given by the relation: $\lambda = \frac{h}{p}$.
Since the momentum $p$ of a particle of mass $m$ moving with velocity $v$ is defined as $p = mv$,we substitute this into the equation.
Therefore,the de-Broglie wavelength is $\lambda = \frac{h}{mv}$.

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ of the emitted electron is given by:
$K_{\max} = E - W_0$
Where $E$ is the energy of the incident photon and $W_0$ is the work function of the metal.
Given: $E = 6.2 \, eV$ and $W_0 = 4.2 \, eV$.
$K_{\max} = 6.2 \, eV - 4.2 \, eV = 2.0 \, eV$.
To convert this energy into Joules,we multiply by the charge of an electron $(1.6 \times 10^{-19} \, C)$:
$K_{\max} = 2.0 \times 1.6 \times 10^{-19} \, J = 3.2 \times 10^{-19} \, J$.

(B) Bohr's model of the atom is based on several postulates. One of the fundamental postulates is the quantization of angular momentum. Bohr proposed that an electron can revolve only in those orbits for which its angular momentum is an integral multiple of $h / (2\pi)$,where $h$ is Planck's constant. This is expressed as $L = mvr = n(h / 2\pi)$,where $n = 1, 2, 3, ...$. Thus,Bohr used the principle of quantization of angular momentum to explain the stability of atoms and the observed spectral lines.

(C) The binding energy per nucleon of a nucleus is defined as the total binding energy of the nucleus divided by its mass number $(A)$,which represents the total number of nucleons.
Experimental observations of the binding energy per nucleon versus mass number graph show that for most nuclei (excluding very light nuclei),the value remains relatively constant.
The average binding energy per nucleon for stable nuclei is approximately $8 \, MeV$ per nucleon.

(A) Let the atomic number of $A$ be $Z$ and the mass number be $A_{mass}$.
$1$. In the first step,$A \to B + {\;_2}He^4$ ($\alpha$ decay):
The atomic number of $B$ becomes $Z - 2$ and the mass number becomes $A_{mass} - 4$.
$2$. In the second step,$B \to C + 2e^-$ ($\beta$ decay):
Since one $\beta$ decay increases the atomic number by $1$,two $\beta$ decays increase the atomic number by $2$.
Therefore,the atomic number of $C$ becomes $(Z - 2) + 2 = Z$.
Since $A$ and $C$ have the same atomic number $Z$ but different mass numbers ($A_{mass}$ and $A_{mass} - 4$),they are isotopes.

(D) In an $N-$type semiconductor,the majority charge carriers are electrons and the minority charge carriers are holes.
When the semiconductor is heated,thermal energy provides enough energy to break covalent bonds in the crystal lattice.
Each broken bond creates an electron-hole pair.
Therefore,the number of electrons and the number of holes both increase by the same amount due to thermal generation.
Thus,the correct option is $(d)$.

(D) Ray optics is based on the assumption that light travels in straight lines. This approximation holds true when the characteristic dimensions of the obstacles or apertures (such as the size of a slit or an object) are much larger than the wavelength of light $(\lambda)$. If the dimensions are comparable to or smaller than $\lambda$, wave effects like diffraction become significant, and ray optics is no longer valid. Therefore, the correct condition is that the dimensions must be much larger than the wavelength of light.

(D) Diffraction is a general characteristic of all types of waves, including mechanical waves (like sound) and electromagnetic waves (like light).
For diffraction to be significant, the size of the obstacle or aperture must be comparable to the wavelength of the wave.
Since sound waves have wavelengths in the range of centimeters to meters, they easily diffract around common obstacles.
Light waves have very small wavelengths (in the range of $400 \, nm$ to $700 \, nm$), so they require very small apertures or obstacles to exhibit observable diffraction.
Therefore, the diffraction effect can be observed in both sound and light waves.

(D) When a current flows through a coil,it generates a magnetic field around it.
According to the principles of electromagnetism,the energy associated with an inductor or a current-carrying coil is stored in the magnetic field created by the current.
Unlike a capacitor,which stores energy in an electric field,an inductor stores energy in the form of a magnetic field.
Therefore,the energy in a current-carrying coil is stored in the form of a magnetic field only.

(C) The atomic number of silicon is $Z = 14$,which means it has $14$ electrons.
According to the Aufbau principle,electrons fill orbitals in order of increasing energy: $1s, 2s, 2p, 3s, 3p, 4s, \dots$
Filling the orbitals with $14$ electrons:
$1s$ orbital takes $2$ electrons $(1s^2)$.
$2s$ orbital takes $2$ electrons $(2s^2)$.
$2p$ orbital takes $6$ electrons $(2p^6)$.
$3s$ orbital takes $2$ electrons $(3s^2)$.
Remaining electrons: $14 - (2 + 2 + 6 + 2) = 2$ electrons.
These $2$ electrons go into the $3p$ orbital $(3p^2)$.
Thus,the configuration is $1s^2 2s^2 2p^6 3s^2 3p^2$.

(D) The formula for the remaining fraction of a radioactive substance is given by $\frac{N}{N_{0}} = \left(\frac{1}{2}\right)^{n}$,where $n$ is the number of half-lives.
First,calculate the number of half-lives $n = \frac{t}{T_{1/2}} = \frac{6400}{800} = 8$.
Now,substitute the value of $n$ into the formula:
$\frac{N}{N_{0}} = \left(\frac{1}{2}\right)^{8} = \frac{1}{256}$.
Therefore,the fraction remaining after $6400$ years is $\frac{1}{256}$.

(D) For a prism,the angle of the prism is given by $A = r_1 + r_2$.
Since the ray emerges normally from the opposite surface,the angle of emergence $e = 0$,which implies the angle of refraction at the second surface $r_2 = 0$.
Substituting $r_2 = 0$ into the prism equation,we get $r_1 = A$.
Applying Snell's law at the first surface: $\sin i = \mu \sin r_1$.
For small angles,$\sin i \approx i$ and $\sin r_1 \approx r_1$.
Therefore,$i = \mu r_1$.
Substituting $r_1 = A$,we get $i = \mu A$.