AIIMS 1997 Physics Question Paper with Answer and Solution

(B) The potential energy $(U)$ of an object is given by the formula $U = mgh$,where $m$ is the mass,$g$ is the acceleration due to gravity,and $h$ is the height.
Given:
Mass $(m)$ = $20 \,kg$
Height $(h)$ = $50 \,cm = 0.5 \,m$
Acceleration due to gravity $(g)$ = $9.8 \,m/s^2$
Decrease in potential energy $(\Delta U)$ = $mgh$
$\Delta U = 20 \times 9.8 \times 0.5$
$\Delta U = 10 \times 9.8 = 98 \,J$.
Therefore,the correct option is $B$.

(C) The mass of a body is defined as the quantity of matter contained in it.
Mass is an intrinsic property of an object and remains constant regardless of its location in the universe.
Unlike weight,which depends on the acceleration due to gravity $(g)$ at a specific location,mass does not change.
Therefore,if the mass of the body is $M$ on the Earth,it will remain $M$ on the Moon.

(B) The potential energy $U$ stored in a stretched spring is given by the formula $U = \frac{1}{2} k x^2$.
According to Hooke's Law,the tension $T$ in the spring is related to the extension $x$ by the equation $T = kx$.
From this,we can express the extension as $x = \frac{T}{k}$.
Substituting this value of $x$ into the energy formula:
$U = \frac{1}{2} k \left( \frac{T}{k} \right)^2$
$U = \frac{1}{2} k \left( \frac{T^2}{k^2} \right)$
$U = \frac{T^2}{2k}$.
Therefore,the correct option is $B$.

(B) The elastic potential energy stored per unit volume $(u)$ in a material is given by the formula:
$u = \frac{1}{2} \times \text{stress} \times \text{strain}$
We know that Young's modulus $(Y)$ is defined as the ratio of stress $(S)$ to strain $(\epsilon)$:
$Y = \frac{S}{\epsilon} \implies \epsilon = \frac{S}{Y}$
Substituting the expression for strain into the energy formula:
$u = \frac{1}{2} \times S \times \left( \frac{S}{Y} \right)$
$u = \frac{S^2}{2Y}$
Thus,the correct option is $B$.

(C) The formula for heat exchange is given by $Q = m \cdot c \cdot \Delta \theta$,where $Q$ is the heat,$m$ is the mass,$c$ is the specific heat,and $\Delta \theta$ is the change in temperature.
Rearranging for specific heat,we get $c = \frac{Q}{m \cdot \Delta \theta}$.
If the specific heat $c$ is infinite $(c = \infty)$,then the denominator $m \cdot \Delta \theta$ must be zero.
Since mass $m$ cannot be zero,it implies that $\Delta \theta = 0$.
This means that there is no change in temperature regardless of whether heat is absorbed or released by the substance,which is characteristic of a phase change process.

(A) The relationship between temperature in Celsius $(C)$ and Fahrenheit $(F)$ is given by the formula: $\frac{C}{100} = \frac{F - 32}{180}$.
Rearranging this to express $F$ in terms of $C$,we get: $F = \frac{9}{5}C + 32$.
Comparing this with the equation of a straight line,$y = mx + c$,where $y = F$ and $x = C$:
Here,the slope $m = \frac{9}{5}$ (which is positive) and the $Y-$intercept $c = 32$ (which is positive).
Since the $Y-$intercept is positive,the line intersects the $Y-$axis at a point above the origin. Thus,the graph has a positive intercept on the $Y-$axis.

(A) When a molecule of mass $m$ collides elastically with a wall of a vessel with velocity $V$,it rebounds with the same speed in the opposite direction.
Initial momentum of the molecule,$p_i = mV$.
Final momentum of the molecule,$p_f = -mV$.
The change in linear momentum,$\Delta p = p_f - p_i = -mV - (mV) = -2mV$.
Since the question asks for the magnitude of the change in linear momentum,the change is $2mV$.

(B) According to the Stefan-Boltzmann Law,the rate of energy radiation $P$ from a black body is proportional to the fourth power of its absolute temperature $T$ (in Kelvin).
$P \propto T^4$
Initial temperature $T_1 = 7^oC = 7 + 273 = 280 \ K$.
Final temperature $T_2 = 287^oC = 287 + 273 = 560 \ K$.
The ratio of the rates of energy radiation is:
$\frac{P_2}{P_1} = (\frac{T_2}{T_1})^4$
Substituting the values:
$\frac{P_2}{P_1} = (\frac{560}{280})^4 = (2)^4 = 16$.
Therefore,the rate of energy radiation increases by a factor of $16$.

(C) The first wave equation is ${y_1} = 10\sin \left( {3\pi t + \frac{\pi }{3}} \right)$. The amplitude of this wave is $A_1 = 10$.
The second wave equation is ${y_2} = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)$.
We can rewrite this by multiplying and dividing by $2$:
${y_2} = 5 \times 2 \left( \frac{1}{2} \sin 3\pi t + \frac{\sqrt 3}{2} \cos 3\pi t \right)$.
Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,where $\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt 3}{2}$:
${y_2} = 10 \left( \sin 3\pi t \cos \frac{\pi}{3} + \cos 3\pi t \sin \frac{\pi}{3} \right) = 10 \sin \left( 3\pi t + \frac{\pi}{3} \right)$.
The amplitude of the second wave is $A_2 = 10$.
The ratio of their amplitudes is $\frac{A_1}{A_2} = \frac{10}{10} = 1:1$.

(B) The torque $\tau$ is defined as the rate of change of angular momentum $L$ with respect to time $t$,given by the formula: $\tau = \frac{dL}{dt} = \frac{\Delta L}{\Delta t}$.
Given:
Initial angular momentum $L_i = 1\,J\cdot s$
Final angular momentum $L_f = 5\,J\cdot s$
Time interval $\Delta t = 5\,s$
Change in angular momentum $\Delta L = L_f - L_i = 5\,J\cdot s - 1\,J\cdot s = 4\,J\cdot s$.
Substituting the values into the formula:
$\tau = \frac{4\,J\cdot s}{5\,s} = 0.8\,N\cdot m$.
Thus,the torque is $0.8\,N\cdot m$,which is equivalent to $\frac{4}{5}\,N\cdot m$.

(B) The moment of inertia $I$ of a system of particles is defined as $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle of mass $m_i$ from the axis of rotation.
From this definition,it is clear that $I$ depends on the distribution of mass relative to the axis of rotation and the position/orientation of the axis itself. Thus,the $Assertion$ is correct.
$Moment$ of inertia is defined as the rotational analogue of mass in linear motion. Just as mass represents the inertia of a body in linear motion (resistance to change in linear velocity),the moment of inertia represents the rotational inertia of a body (resistance to change in angular velocity). Thus,the $Reason$ is also correct.
However,the $Reason$ explains what the moment of inertia represents (rotational inertia),but it does not explain *why* it depends on the axis of rotation and mass distribution. Therefore,the $Reason$ is not the correct explanation of the $Assertion$. The correct option is $B$.

(A) The volume of the big drop is equal to the sum of the volumes of the $729$ small drops.
Let $V_{big}$ be the volume of the big drop and $V_{small}$ be the volume of one small drop.
$V_{big} = 729 \times V_{small}$
Using the formula for the volume of a sphere,$\frac{4}{3}\pi R^3 = 729 \times \frac{4}{3}\pi r^3$.
Canceling $\frac{4}{3}\pi$ from both sides,we get $R^3 = 729 \times r^3$.
Taking the cube root of both sides,$R = \sqrt[3]{729} \times r$.
Since $9^3 = 729$,we have $R = 9r$.
Therefore,the radius of each small drop is $r = \frac{R}{9}$.

(C) For two plates of equal thickness $d$ and thermal conductivities $K_1$ and $K_2$ placed in series,the total thermal resistance is $R_{eq} = R_1 + R_2$.
Since $R = \frac{d}{KA}$,we have $\frac{2d}{K_{eq}A} = \frac{d}{K_1A} + \frac{d}{K_2A}$.
Simplifying this,we get $\frac{2}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2}$,which implies $K_{eq} = \frac{2K_1K_2}{K_1 + K_2}$.
This is the harmonic mean of $K_1$ and $K_2$. The harmonic mean of two numbers is always less than the larger number and greater than the smaller number,but specifically,the equivalent conductivity $K_{eq}$ satisfies $K_{min} < K_{eq} < K_{max}$.
Wait,let's re-evaluate: If $K_1 = 10$ and $K_2 = 2$,then $K_{eq} = \frac{2(10)(2)}{10+2} = \frac{40}{12} \approx 3.33$.
Here $3.33 > 2$ (the smaller value). Thus,the Assertion is incorrect.

(D) The first law of thermodynamics is given by $\Delta Q = \Delta U + P\Delta V$.
For an ideal gas,the internal energy $U$ is a function of temperature $T$ only $(U = f(T))$.
In an isothermal process,the temperature remains constant,so $\Delta T = 0$.
Since $\Delta U = nC_v\Delta T$,it follows that $\Delta U = 0$ for an isothermal process.
Substituting this into the first law: $\Delta Q = 0 + P\Delta V$,which means $\Delta Q = P\Delta V$.
Therefore,the heat supplied is converted into work done by the system,not internal energy.
Thus,the Assertion is incorrect,and the Reason is a correct statement of the first law of thermodynamics.

(A) According to Coulomb's Law,the force between two charges in a vacuum or air is given by $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d^2}$.
When the charges are immersed in a medium with dielectric constant $K$,the force becomes $F_m = \frac{1}{4\pi\epsilon_0 K} \frac{q_1 q_2}{d^2}$.
Therefore,the relationship between the force in the medium and the force in air is $F_m = \frac{F}{K}$.
Given that $K = 2$,the new force is $F_m = \frac{F}{2}$.

(C) The electric potential $V$ at a distance $r$ from a point charge $Q$ is given by $V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r}$.
Since the charge $q$ is moved in a circle of radius $r$ around the charge $Q$,every point on the circular path is at the same distance $r$ from $Q$.
Therefore,the entire circular path acts as an equipotential surface.
The work done $W$ in moving a charge $q$ between two points on an equipotential surface is given by $W = q(V_f - V_i)$.
Since $V_f = V_i$,the work done $W = 0$.

(C) Let the sphere have a uniform charge density $\rho = \frac{3Q}{4\pi R^3}$,and let $E$ be the electric field at a distance $x$ from the centre of the sphere,where $x < R$.
Applying Gauss's law to a Gaussian surface of radius $x$:
$E \cdot 4\pi x^2 = \frac{q_{enclosed}}{\varepsilon_0} = \frac{\rho V'}{\varepsilon_0} = \frac{\rho}{\varepsilon_0} \cdot \frac{4}{3}\pi x^3$
Here,$V' = \frac{4}{3}\pi x^3$ is the volume of the sphere of radius $x$.
Simplifying the equation:
$E = \frac{\rho}{3\varepsilon_0} x$
Since $\rho$,$3$,and $\varepsilon_0$ are constants,we have $E \propto x$.
Therefore,the electric field is directly proportional to $x$.

(D) The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.
Thus,$R = \rho \frac{L}{\pi d^2 / 4} = \frac{4 \rho L}{\pi d^2}$.
When a wire is drawn to reduce its diameter,its volume $V = A \times L$ remains constant.
Since $V = (\pi d^2 / 4) \times L$ is constant,$L \propto \frac{1}{d^2}$.
Substituting this into the resistance formula: $R \propto \frac{L}{d^2} \propto \frac{1/d^2}{d^2} = \frac{1}{d^4}$.
If the diameter is reduced to half $(d' = d/2)$,the new resistance $R'$ is:
$R' = R \times (d/d')^4 = R \times (d / (d/2))^4 = R \times (2)^4 = 16R$.
Therefore,the resistance becomes $16$ times the original value.

(B) The induced $e.m.f.$ is given by Faraday's law: $e = -N \frac{d\phi}{dt} = -N \frac{\phi_2 - \phi_1}{\Delta t}$.
Here,$\phi = BA \cos \theta$. Initially,the field is perpendicular to the plane,so $\theta_1 = 0^o$. After rotating by $180^o$,the angle becomes $\theta_2 = 180^o$.
Given: $N = 200$,$A = 70 \ cm^2 = 70 \times 10^{-4} \ m^2$,$B = 0.3 \ Wb/m^2$,$\Delta t = 0.1 \ s$.
$e = -N \frac{BA(\cos 180^o - \cos 0^o)}{\Delta t}$
$e = -200 \times \frac{0.3 \times 70 \times 10^{-4} \times (-1 - 1)}{0.1}$
$e = -200 \times \frac{0.3 \times 70 \times 10^{-4} \times (-2)}{0.1}$
$e = 200 \times 0.3 \times 70 \times 10^{-4} \times 20 = 8.4 \ V$.

(A) The induced $e.m.f.$ $(e)$ in a coil due to self-induction is given by the formula: $e = -L \frac{di}{dt}$.
Here,the inductance $L = 5 \, H$.
The rate of change of current $\frac{di}{dt} = -2 \, A/s$ (since the current is decreasing).
Substituting these values into the formula:
$e = -5 \times (-2) = 10 \, V$.
Therefore,the $e.m.f.$ developed across the coil is $10 \, V$.

(B) For a purely capacitive circuit,the applied alternating e.m.f. is given by $e = e_0 \sin(\omega t)$.
In a capacitor,the current $i$ leads the voltage $e$ by a phase angle of $\pi / 2$.
Therefore,the expression for current is $i = i_0 \sin(\omega t + \pi / 2)$.
This indicates that the current is ahead of the e.m.f. by $\pi / 2$.

(D) The minimum wavelength (cutoff wavelength) of $X$-rays produced in an $X$-ray tube is given by the formula:
${\lambda _{\min }} = \frac{{hc}}{{eV}}$
where $h$ is Planck's constant,$c$ is the speed of light,$e$ is the charge of an electron,and $V$ is the accelerating potential difference.
From the relation ${\lambda _{\min }} \propto \frac{1}{V}$,it is clear that the minimum wavelength is inversely proportional to the applied potential difference.
Therefore,when the potential difference $V$ is increased,the minimum wavelength ${\lambda _{\min }}$ decreases.
Thus,the correct option is $(d)$.

(C) The radius of the $n^{th}$ orbit of an electron in a hydrogen-like atom is given by the formula $r_n = r_0 n^2 / Z$.
For a hydrogen atom,the atomic number $Z = 1$,which simplifies the expression to $r_n = r_0 n^2$.
For the third orbit,we substitute $n = 3$ into the formula.
Therefore,the radius of the third orbit is $r_3 = r_0 \times (3)^2 = 9r_0$.

(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For a fixed transition ($n_i = 4$ to $n_f = 2$),the wavelength $\lambda$ is inversely proportional to the square of the atomic number $Z$,i.e.,$\lambda \propto \frac{1}{Z^2}$.
For hydrogen $(H)$,$Z = 1$,so $\lambda_H = 20.397 \, cm$.
For helium ion $(He^+)$,$Z = 2$,so $\lambda_{He^+} = \frac{\lambda_H}{Z^2} = \frac{20.397}{2^2} = \frac{20.397}{4} = 5.099 \, cm$.

(C) The binding energy of a nucleus is given by the product of the number of nucleons and the binding energy per nucleon.
For the reactant,two deuterons $(1^2H)$ are involved. Each deuteron has $2$ nucleons.
Total binding energy of two deuterons $= 2 \times (2 \times 1.112 \, MeV) = 4 \times 1.112 \, MeV = 4.448 \, MeV$.
For the product,one $\alpha$-particle $(2^4He)$ is formed. It has $4$ nucleons.
Total binding energy of one $\alpha$-particle $= 4 \times 7.047 \, MeV = 28.188 \, MeV$.
The energy released $Q$ in the fusion reaction is the difference between the total binding energy of the products and the total binding energy of the reactants.
$Q = 28.188 \, MeV - 4.448 \, MeV = 23.74 \, MeV$.
Rounding to the nearest provided option,$Q \approx 23.8 \, MeV$.

(D) In nuclear reactions,the total energy (including mass-energy equivalence),linear momentum,angular momentum,and charge are conserved. According to the law of conservation of mass-energy,the total mass-energy of the system remains constant. Since no external forces act on the system,the total linear momentum is also conserved. Therefore,mass (in terms of mass-energy),energy,and momentum are all conserved.

(C) An intrinsic semiconductor is a pure semiconductor material in which the electrical conductivity is solely due to the thermal excitation of electrons from the valence band to the conduction band,which involves the breaking of covalent bonds.
In contrast,an extrinsic semiconductor is a semiconductor that has been doped with impurities to increase its conductivity,meaning its conductivity is due to both the breaking of covalent bonds and the presence of excess charge carriers introduced by the impurity atoms.
Therefore,when conductivity is due to the breaking of covalent bonds in a pure material,it is classified as an intrinsic semiconductor.

(A) In a $P-$type semiconductor,the doping is done using trivalent impurities,which create an excess of holes.
Since holes are the majority charge carriers,the current in a $P-$type semiconductor is mainly carried by holes.
Therefore,option $A$ is correct.

(B) The plate resistance $({r_p})$ is defined as the ratio of the change in plate voltage $(\delta V)$ to the change in plate current $(\delta I)$,expressed as $r_p = \frac{\delta V}{\delta I}$.
At the saturation point,the current reaches its maximum value and does not change with further increases in voltage,meaning the change in current $\delta I = 0$.
Substituting this into the formula,we get $r_p = \frac{\delta V}{0} = \infty$.
Therefore,at saturation,the plate resistance is infinite.

(C) The number of images $n$ formed by two plane mirrors inclined at an angle $\theta$ is given by the formula $n = \frac{360^{\circ}}{\theta} - 1$,provided that $\frac{360^{\circ}}{\theta}$ is an even integer.
Given $\theta = 60^{\circ}$,we calculate the ratio: $\frac{360^{\circ}}{60^{\circ}} = 6$.
Since $6$ is an even integer,the number of images is $n = 6 - 1 = 5$.

(C) The intensity of the resultant wave in interference is given by $I_{res} = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi$.
For maximum intensity, $\cos \phi = 1$, so $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2$.
Given $I_1 = I$ and $I_2 = 4I$, we have $I_{max} = (\sqrt{I} + \sqrt{4I})^2 = (\sqrt{I} + 2\sqrt{I})^2 = (3\sqrt{I})^2 = 9I$.
For minimum intensity, $\cos \phi = -1$, so $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2$.
Thus, $I_{min} = (\sqrt{I} - \sqrt{4I})^2 = (\sqrt{I} - 2\sqrt{I})^2 = (-\sqrt{I})^2 = I$.
Therefore, the maximum and minimum intensities are $9I$ and $I$ respectively.

(C) According to the Doppler effect for light,when a source of light moves towards an observer,the observed frequency increases,which corresponds to a decrease in wavelength.
This shift towards shorter wavelengths (higher frequencies) in the visible spectrum is known as a 'blue shift'.
Therefore,the spectral lines are shifted towards the blue end of the spectrum.

(C) The capacitance of a parallel plate capacitor is given by $C = \frac{K \epsilon_0 A}{d}$,where $K$ is the dielectric constant,$A$ is the area,and $d$ is the distance between the plates.
Given that the new distance $d' = \frac{d}{2}$ and the new dielectric constant $K' = 3K$.
The new capacitance $C'$ becomes $C' = \frac{(3K) \epsilon_0 A}{(d/2)} = 6 \left( \frac{K \epsilon_0 A}{d} \right) = 6C$.
Thus,the Assertion is correct.
The capacity of a capacitor depends on the dielectric constant $(K)$ of the material placed between the plates. Therefore,the Reason is incorrect.

(C) cyclotron is a particle accelerator used to accelerate charged particles,such as positive ions,to high energies. Thus,the Assertion is correct.
The cyclotron frequency $f$ is given by the formula:
$f = \frac{Bq}{2 \pi m}$
where $B$ is the magnetic field,$q$ is the charge,and $m$ is the mass of the particle.
From this formula,it is clear that the cyclotron frequency depends only on the magnetic field,the charge,and the mass of the particle. It is independent of the velocity $v$ and the radius $r$ of the orbit.
Therefore,the Reason is incorrect.

(A) green flower reflects only green light and absorbs all other wavelengths of light.
Red glass acts as a filter that transmits only red light and absorbs or reflects all other colours.
When light passes through red glass,only red light reaches the green flower.
Since the green flower absorbs red light and reflects only green light (which is absent in the incident light),it reflects almost no light to the observer's eye.
Therefore,the flower appears dark or black when viewed through red glass.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) green flower reflects light in the green region of the visible spectrum and absorbs other wavelengths.
When this reflected green light is incident on red glass,the red glass acts as a filter that transmits only red light and absorbs all other wavelengths,including green.
Since the green light is absorbed by the red glass,no light from the flower reaches the observer's eye.
Therefore,the flower appears dark when viewed through red glass.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(A) The photosensitivity of a metal refers to its ability to emit photoelectrons when light of a suitable frequency is incident on it. A metal with a low work function $(\Phi_0)$ requires less energy to eject an electron, making it more photosensitive. Thus, the Assertion is correct.
The work function is defined as $\Phi_0 = hf_0$, where $h$ is Planck's constant and $f_0$ is the threshold frequency. This is a standard physical relationship for the photoelectric effect. Thus, the Reason is also correct.
Since the work function is defined by the threshold frequency $(f_0)$, and a smaller $f_0$ (and thus smaller $\Phi_0$) makes it easier to eject electrons, the Reason correctly explains why a small work function leads to high photosensitivity. Therefore, the correct option is $A$.

(B) By definition,isobars are atoms of different elements that have the same mass number $(A)$ but different atomic numbers $(Z)$.
Therefore,the Assertion is correct.
Protons and neutrons are indeed present inside the nucleus,which is a correct statement regarding nuclear composition.
However,the fact that protons and neutrons exist in the nucleus does not explain why certain atoms are isobars.
Thus,the Reason is a correct statement but is not the correct explanation for the Assertion.