AIIMS 1998 Physics Question Paper with Answer and Solution

(C) The displacement vector can be represented as $\vec{r} = 12\hat{i} + 5\hat{j} + 6\hat{k}$.
The magnitude of the resultant displacement $R$ is given by the formula $R = \sqrt{x^2 + y^2 + z^2}$.
Substituting the values: $R = \sqrt{12^2 + 5^2 + 6^2}$.
$R = \sqrt{144 + 25 + 36}$.
$R = \sqrt{205}$.
$R \approx 14.31 \, m$.

(A) Angular velocity $(\omega)$ is defined as the rate of change of angular displacement $( \theta)$ with respect to time $(t)$.
Mathematically,$\omega = \frac{\theta}{t}$.
The dimensional formula for angular displacement $(\theta)$ is dimensionless,represented as $[M^0L^0T^0]$.
The dimensional formula for time $(t)$ is $[T]$.
Therefore,the dimensional formula for angular velocity is $[\omega] = \frac{[M^0L^0T^0]}{[T]} = [M^0L^0T^{-1}]$.

(B) The formula for centripetal force is given by $F_c = m r \omega^2$.
Given:
Mass $m = 5\, kg$
Radius $r = 1\, m$
Angular velocity $\omega = 2\, rad/s$
Substituting these values into the formula:
$F_c = 5 \times 1 \times (2)^2$
$F_c = 5 \times 1 \times 4$
$F_c = 20\, N$.
Therefore,the centripetal force is $20\, N$.

(D) The horizontal range $R$ of a projectile is given by $R = \frac{u^2 \sin(2\theta)}{g} = \frac{2u^2 \sin \theta \cos \theta}{g}$.
The maximum height $H$ is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
According to the problem,the horizontal range is three times the maximum height,so $R = 3H$.
Substituting the formulas:
$\frac{2u^2 \sin \theta \cos \theta}{g} = 3 \left( \frac{u^2 \sin^2 \theta}{2g} \right)$.
Simplifying the equation:
$2 \cos \theta = \frac{3}{2} \sin \theta$.
Rearranging to find $\tan \theta$:
$\tan \theta = \frac{4}{3}$.
Therefore,$\theta = \tan^{-1}(1.333) \approx 53^\circ 8'$.

(C) The apparent weight of a body in an accelerating frame is given by $W_{app} = m(g + a)$,where $a$ is the upward acceleration.
Initially,as the aeroplane takes off and gains altitude,it experiences upward acceleration,causing the reading of the spring balance to increase.
As the aeroplane reaches a high altitude,the acceleration becomes zero (or negligible) and the value of the acceleration due to gravity $g$ decreases with height $(g' = g(1 - 2h/R))$.
Therefore,the reading of the spring balance will first increase due to the upward acceleration and then decrease due to the reduction in the gravitational force at higher altitudes.

(C) The motion of a rocket is based on the principle of conservation of linear momentum.
As the fuel burns,hot gases are ejected from the nozzle of the rocket at a very high velocity in the downward direction.
According to Newton's third law of motion,these gases exert an equal and opposite reaction force on the rocket in the upward direction.
This upward force provides the necessary thrust to lift the rocket against gravity.

(D) The spring constant $k$ is given by Hooke's Law: $k = \frac{F}{x}$.
Given $F = 10 \, N$ and $x = 1 \, mm = 1 \times 10^{-3} \, m$.
So,$k = \frac{10}{1 \times 10^{-3}} = 10^4 \, N/m$.
The work done $W$ in stretching the spring by a displacement $x = 40 \, mm = 40 \times 10^{-3} \, m$ is given by $W = \frac{1}{2} k x^2$.
Substituting the values: $W = \frac{1}{2} \times 10^4 \times (40 \times 10^{-3})^2$.
$W = \frac{1}{2} \times 10^4 \times (1600 \times 10^{-6}) = \frac{1}{2} \times 10^4 \times 1.6 \times 10^{-3} = 0.5 \times 16 = 8 \, J$.

(A) The relationship between kinetic energy $E$ and momentum $P$ is given by the formula $P = \sqrt{2mE}$.
Since $m$ is constant,we have $P \propto \sqrt{E}$.
Let the initial kinetic energy be $E_1$ and initial momentum be $P_1$. Then $P_1 = \sqrt{2mE_1}$.
Let the new kinetic energy be $E_2 = 4E_1$ and the new momentum be $P_2$.
Then $P_2 = \sqrt{2mE_2} = \sqrt{2m(4E_1)} = 2\sqrt{2mE_1} = 2P_1$.
Therefore,the new momentum becomes twice its initial value.

(A) The kinetic energy $E$ of a body is given by the formula $E = \frac{P^2}{2m}$,where $P$ is the momentum and $m$ is the mass of the body.
According to the law of conservation of linear momentum,when a bullet is fired from a rifle,the magnitude of the momentum of the rifle is equal to the magnitude of the momentum of the bullet $(P_{rifle} = P_{bullet} = P)$.
Since $P$ is constant for both,the kinetic energy is inversely proportional to the mass $(E \propto \frac{1}{m})$.
Because the mass of the rifle $(M)$ is much greater than the mass of the bullet $(m)$,the kinetic energy of the rifle will be significantly less than the kinetic energy of the bullet.

(D) The gravitational force $F$ acting on a mass $m$ in a gravitational field $g$ is given by $F = mg$. Here,the gravitational force is directly proportional to the gravitational mass $m$. The gravitational field intensity $g$ is defined as the force per unit mass,$g = F/m$,which implies $F = mg$. Since the gravitational field $g$ is a property of the space around a mass,and the force $F$ is the interaction,the gravitational mass is fundamentally related to all these quantities. Thus,the correct option is $D$.

(B) The correct option is $(B)$.
Due to surface tension,the surface of a liquid tends to minimize its surface area for a given volume.
Among all geometric shapes,a sphere has the minimum surface area for a fixed volume.
Therefore,to minimize the potential energy associated with the surface area,rain drops naturally acquire a spherical shape.

(B) The work done $W$ in splitting a large drop of radius $R$ into $n$ smaller droplets of radius $r$ is given by the change in surface area multiplied by surface tension $T$.
$W = T \times \Delta A = T \times (n \times 4\pi r^2 - 4\pi R^2)$.
Since the volume remains constant,$\frac{4}{3}\pi R^3 = n \times \frac{4}{3}\pi r^3$,which gives $r = R / n^{1/3}$.
Substituting $r$,we get $W = 4\pi R^2 T (n^{1/3} - 1)$.
Given $R = 10^{-3} \, m$,$n = 10^6$,and $T = 72 \times 10^{-3} \, J/m^2$.
$W = 4 \times 3.1416 \times (10^{-3})^2 \times 72 \times 10^{-3} \times ( (10^6)^{1/3} - 1)$.
$W = 4 \times 3.1416 \times 10^{-6} \times 72 \times 10^{-3} \times (100 - 1)$.
$W = 4 \times 3.1416 \times 72 \times 99 \times 10^{-9} \approx 8.95 \times 10^{-5} \, J$.

(C) The 'stem correction' in a platinum resistance thermometer arises due to the resistance of the connecting wires (leads) that connect the platinum coil to the measuring instrument.
As the temperature of the leads changes,their resistance changes,which introduces an error in the temperature measurement.
To eliminate this error,'compensating leads' are used. These leads are identical to the main leads and are placed in the same environment,ensuring that the change in resistance of the main leads is balanced out by the change in resistance of the compensating leads.

(C) Absolute zero is defined as $0 \,K$ or $-273.15 \,^\circ C$.
According to the kinetic theory of gases, the root mean square velocity $(v_{rms})$ of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.
At $T = 0 \,K$, the velocity $v_{rms}$ becomes zero.
This implies that the random thermal motion of molecules ceases at absolute zero.
Therefore, option $(C)$ is correct.

(D) The quantity of heat required to change the unit mass of a solid substance from the solid state to the liquid state at its melting point,while the temperature remains constant,is defined as the latent heat of fusion. Therefore,the correct option is $D$.

(C) The relative humidity is defined as the ratio of the partial pressure of water vapour to the saturated vapour pressure of water at the same temperature.
Given:
Partial pressure of water vapour $(P_W)$ = $0.012 \times 10^5 \, Pa$
Saturated vapour pressure of water $(P_V)$ = $0.016 \times 10^5 \, Pa$
Relative Humidity = $\frac{P_W}{P_V} \times 100\%$
Relative Humidity = $\frac{0.012 \times 10^5}{0.016 \times 10^5} \times 100\%$
Relative Humidity = $\frac{0.012}{0.016} \times 100\% = \frac{12}{16} \times 100\% = 0.75 \times 100\% = 75\%$.

(D) Diffusion is the process of movement of particles from a region of higher concentration to a region of lower concentration.
In gases,the intermolecular forces are very weak and the particles have high kinetic energy,allowing them to move freely and rapidly.
In liquids,particles are more closely packed than in gases,and in solids,they are tightly packed with very little space to move.
Therefore,the rate of diffusion is highest in gases,followed by liquids,and lowest in solids due to the differences in intermolecular space and kinetic energy.

(C) According to the kinetic theory of gases,the root mean square velocity of gas molecules is given by the formula $v_{rms} = \sqrt{\frac{3RT}{M}}$.
At absolute temperature,$T = 0 \ K$.
Substituting this value into the formula,we get $v_{rms} = \sqrt{\frac{3R(0)}{M}} = 0$.
Since the root mean square velocity represents the average motion of the molecules,a velocity of $0$ implies that all molecular motion ceases at absolute zero.

(B) In a thermodynamic process,the work done by a gas is equal to the area under the $PV$ curve with respect to the volume axis.
For a given expansion from volume ${V_1}$ to ${V_2}$,the pressure $P$ remains constant in an isobaric process,whereas it decreases in both isothermal and adiabatic processes.
Since the isobaric process maintains the highest pressure throughout the expansion,the area under the $PV$ curve is the largest.
Therefore,the work done follows the order: ${W_{adiabatic}} < {W_{isothermal}} < {W_{isobaric}}$.
Thus,the work done is greatest for an isobaric expansion.

(B) When you touch a surface,heat flows from your body to the surface if the surface is colder than your body.
Metals are good conductors of heat,meaning they have high thermal conductivity.
Because of this high thermal conductivity,heat is transferred away from your skin into the metal very rapidly.
In contrast,wood is a poor conductor of heat (an insulator),so it transfers heat away from your skin much more slowly.
Therefore,the metal feels colder because it extracts heat from your body at a much faster rate.

(C) The correct option is $C$.
In winter,the ambient temperature is significantly lower than the human body temperature (approximately $37.4 ^\circ C$).
Woolen clothes contain a large amount of trapped air between their fibers.
Since air is a very poor conductor of heat and wool itself is a bad conductor,these clothes act as an insulator.
This prevents the heat generated by the human body from escaping into the colder surroundings,thereby keeping the body warm.

(C) Heat transfer occurs through three main processes: conduction,convection,and radiation.
Conduction requires a material medium and involves the transfer of energy through molecular collisions.
Convection also requires a material medium (liquid or gas) and involves the actual movement of matter.
Radiation is the process of heat transfer in the form of electromagnetic waves,which do not require any material medium to propagate.
Therefore,heat can travel through a vacuum only by radiation.

(B) The maximum velocity $(v_{\max})$ of a simple harmonic oscillator is given by the formula: $v_{\max} = A\omega$,where $A$ is the amplitude and $\omega$ is the angular frequency.
Given: Amplitude $A = 50\, mm = 50 \times 10^{-3}\, m = 0.05\, m$.
Time period $T = 2\, s$.
The angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi\, rad/s$.
Substituting the values: $v_{\max} = 0.05 \times \pi$.
Using $\pi \approx 3.14159$,we get $v_{\max} = 0.05 \times 3.14159 \approx 0.157\, m/s$.
Rounding to two decimal places,we get $0.16\, m/s$.
Thus,the correct option is $B$.

(C) The time period $T$ of a simple pendulum is given by the formula $T = 2\pi \sqrt{\frac{L}{g}}$, where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity.
As seen from the formula, the time period $T$ depends only on the length of the pendulum and the acceleration due to gravity.
It is independent of the mass, material, or density of the bob.
Therefore, replacing the metal bob with a wooden bob will not change the time period of the pendulum.

(D) The time period $T$ of a mass $m$ suspended from a spring with spring constant $k$ is given by the formula: $T = 2\pi \sqrt{\frac{m}{k}}$.
Given that for mass $m$,the period $T_1 = 2 \, s$.
For mass $m_2 = 4m$,the new period $T_2$ is given by: $T_2 = 2\pi \sqrt{\frac{4m}{k}}$.
Dividing $T_2$ by $T_1$,we get: $\frac{T_2}{T_1} = \sqrt{\frac{4m}{m}} = \sqrt{4} = 2$.
Therefore,$T_2 = 2 \times T_1 = 2 \times 2 \, s = 4 \, s$.

(B) The number of waves per unit length is defined as the wave number.
It is the reciprocal of the wavelength $(\lambda)$.
Mathematically,it is expressed as $\overline{n} = \frac{1}{\lambda}$.
Therefore,the correct option is $B$.

(D) The speed of sound in an ideal gas is given by the formula $v = \sqrt{\frac{\gamma RT}{M}}$.
Here,$\gamma$ is the adiabatic index,$R$ is the universal gas constant,$T$ is the absolute temperature,and $M$ is the molar mass of the gas.
Since $v$ depends only on the temperature $T$ and the nature of the gas (represented by $\gamma$ and $M$),it is independent of the pressure $P$ and density $\rho$ of the gas at a constant temperature.
Therefore,the correct option is $D$.

(A) In transverse waves,the particles of the medium vibrate in a direction perpendicular to the direction of wave propagation.
Since the question describes this specific motion,the correct answer is $A$.

(B) stationary wave is formed by the superposition of two waves of the same frequency and amplitude traveling in opposite directions.
Given the incident wave is $y_1 = a \cos (kx - \omega t)$.
For a point $x = 0$ to be a node,the resultant displacement at $x = 0$ must be zero for all time $t$.
Let the second wave be $y_2 = a \cos (kx + \omega t + \phi)$.
The resultant wave is $y = y_1 + y_2 = a [\cos (kx - \omega t) + \cos (kx + \omega t + \phi)]$.
Using the identity $\cos A + \cos B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$,we get:
$y = 2a \cos (kx + \phi/2) \cos (\omega t + \phi/2)$.
At $x = 0$,$y = 2a \cos (\phi/2) \cos (\omega t + \phi/2)$.
For this to be a node (zero displacement) for all $t$,we must have $\cos (\phi/2) = 0$,which implies $\phi/2 = \pi/2$,or $\phi = \pi$.
Substituting $\phi = \pi$ into the equation for the second wave:
$y_2 = a \cos (kx + \omega t + \pi) = -a \cos (kx + \omega t)$.

(A) stationary wave (also known as a standing wave) is formed by the superposition of two identical waves traveling in opposite directions.
In a stationary wave,the energy is trapped between the nodes and does not propagate through the medium.
While the energy oscillates between kinetic and potential forms at different points,there is no net transport of energy across the medium.
In contrast,progressive waves,transverse waves,and electromagnetic waves are all characterized by the transport of energy from one point to another.
Therefore,the correct answer is $A$.

(C) The length of the string is $l = 10 \; m$. The number of segments is $p = 5$. The wave velocity is $v = 20 \; m/s$.
For a string vibrating in $p$ segments,the length $l$ is related to the wavelength $\lambda$ by the formula $l = p \cdot \frac{\lambda}{2}$.
Substituting the values,we get $10 = 5 \cdot \frac{\lambda}{2}$,which simplifies to $\lambda = \frac{10 \cdot 2}{5} = 4 \; m$.
The frequency $f$ is given by the relation $f = \frac{v}{\lambda}$.
Substituting the values,$f = \frac{20 \; m/s}{4 \; m} = 5 \; Hz$.

(C) The excess pressure inside a soap bubble of radius $r$ is given by $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension.
This implies that $\Delta P \propto \frac{1}{r}$.
Given the ratio of excess pressures is $\frac{\Delta P_1}{\Delta P_2} = 3$,we have $\frac{r_2}{r_1} = 3$,which means $\frac{r_1}{r_2} = \frac{1}{3}$.
The volume of a spherical bubble is $V = \frac{4}{3}\pi r^3$,so the ratio of their volumes is $\frac{V_1}{V_2} = \left( \frac{r_1}{r_2} \right)^3$.
Substituting the ratio of radii: $\frac{V_1}{V_2} = \left( \frac{1}{3} \right)^3 = \frac{1}{27}$.

(C) The angular displacement is given by $\theta(t) = 2t^3 + 0.5$.
Angular velocity $\omega$ is defined as the rate of change of angular displacement with respect to time: $\omega = \frac{d\theta}{dt}$.
Differentiating $\theta$ with respect to $t$: $\omega = \frac{d}{dt}(2t^3 + 0.5) = 6t^2$.
To find the angular velocity at $t = 2 \ s$,substitute $t = 2$ into the expression for $\omega$:
$\omega = 6(2)^2 = 6 \times 4 = 24 \ rad/s$.

(B) The initial temperature of boiling water is $100\,^{\circ}C$ or $212\,^{\circ}F$.
The final temperature in Fahrenheit is $F = 140\,^{\circ}F$.
Using the conversion formula $\frac{C}{100} = \frac{F-32}{180}$,we find the final temperature in Celsius:
$\frac{C}{100} = \frac{140-32}{180} = \frac{108}{180} = 0.6$
$C = 60\,^{\circ}C$.
The fall in temperature on the Centigrade scale is $\Delta C = 100\,^{\circ}C - 60\,^{\circ}C = 40\,^{\circ}C$.

(A) When a body is thrown vertically upwards,at the highest point,its velocity becomes $0 \ m/s$.
However,the gravitational force continues to act on it,resulting in a constant downward acceleration of $g \approx 9.8 \ m/s^2$.
Thus,a body can have acceleration even when its velocity is zero at a specific instant.
Therefore,the $Assertion$ is correct.
When a body reverses its direction of motion,it must momentarily come to rest (velocity becomes zero) before moving in the opposite direction.
This change in velocity is caused by the acceleration acting on the body.
Thus,the $Reason$ is correct and it provides a valid explanation for why the velocity can be zero while acceleration is non-zero.

(D) The $Assertion$ is incorrect because a body with constant acceleration does not necessarily move in a straight line. For example,in projectile motion,the acceleration due to gravity is constant,but the path is a parabola.
The $Reason$ is correct because a body with constant acceleration may not speed up. For example,in uniform circular motion,the centripetal acceleration is constant in magnitude,but the speed of the body remains constant.
Therefore,the correct option is $D$.

(B) Using the third equation of motion: $v^2 - u^2 = 2as$.
Here,initial velocity $u = 0 \ m/s$ (as water falls from rest),acceleration $a = g = 9.8 \ m/s^2$,and displacement $s = 19.6 \ m$.
Substituting the values:
$v^2 - 0^2 = 2 \times 9.8 \times 19.6$
$v^2 = 2 \times 9.8 \times (2 \times 9.8)$
$v^2 = (2 \times 9.8)^2$
$v = 2 \times 9.8 = 19.6 \ m/s$.
Therefore,the velocity of water at the turbines is $19.6 \ m/s$.

(D) During the melting process,the temperature of the substance remains constant,but the heat supplied (latent heat) is used to overcome the intermolecular forces of attraction between the particles of the solid.
This process increases the potential energy of the molecules,which in turn increases the total internal energy of the system.
Therefore,the Assertion is incorrect.
Latent heat is defined as the amount of heat energy required to change the state of a unit mass of a substance without changing its temperature,which makes the Reason correct.
Thus,the correct option is $D$.

(A) The fundamental unit of charge is the electron,which carries a negative charge.
When a neutral body gains excess electrons,the total number of electrons becomes greater than the number of protons.
Since protons are positively charged and electrons are negatively charged,an excess of electrons results in a net negative charge on the body.
Therefore,giving excess electrons to a body makes it negatively charged.

(B) The Earth is considered a good conductor. Due to its massive size,it can accept or provide an almost unlimited amount of charge without a significant change in its potential. By convention,the electric potential of the Earth is taken to be $0 \ V$.

(A) The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 A}{d}$.
Rearranging the formula to solve for the area $A$,we get $A = \frac{Cd}{\varepsilon_0}$.
Given values are $C = 3\,F$,$d = 5\,mm = 5 \times 10^{-3}\,m$,and $\varepsilon_0 = 8.854 \times 10^{-12}\,F/m$.
Substituting these values into the formula:
$A = \frac{3 \times 5 \times 10^{-3}}{8.854 \times 10^{-12}}$
$A = \frac{15 \times 10^{-3}}{8.854 \times 10^{-12}}$
$A \approx 1.694 \times 10^9\,m^2$.

(A) The circuit consists of a $3 \ V$ battery connected in parallel with a resistor $R_1 = 2 \ \Omega$ and a series combination of resistors $R_2, R_3, R_4$.
First,calculate the equivalent resistance of the series branch containing $R_2, R_3, R_4$:
$R_s = R_2 + R_3 + R_4 = 2 \ \Omega + 2 \ \Omega + 2 \ \Omega = 6 \ \Omega$.
Now,this $R_s$ is in parallel with $R_1 = 2 \ \Omega$. The equivalent resistance $R_{eq}$ of the circuit is:
$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_s} = \frac{1}{2} + \frac{1}{6} = \frac{3+1}{6} = \frac{4}{6} = \frac{2}{3} \ \Omega^{-1}$.
Therefore,$R_{eq} = 1.5 \ \Omega$ or $\frac{3}{2} \ \Omega$.
The total current $i$ drawn from the battery is given by Ohm's law:
$i = \frac{V}{R_{eq}} = \frac{3 \ V}{1.5 \ \Omega} = 2 \ A$.

(A) The Earth's magnetic field is approximately uniform over a small region on its surface. The horizontal component of the Earth's magnetic field $(B_H)$ represents the magnetic field lines running parallel to the Earth's surface in a specific direction. In a localized region,these magnetic field lines are represented as parallel straight lines.

(A) transformer consists of two coils,the primary and the secondary,wound on a common core. When an alternating current flows through the primary coil,it produces a changing magnetic flux. This changing magnetic flux is linked with the secondary coil,which induces an electromotive force (emf) in it. This phenomenon,where a change in current in one coil induces an emf in a nearby coil,is known as mutual induction. Therefore,a transformer is based on the principle of mutual inductance.

(A) In a transformer,energy losses occur due to various factors,including hysteresis loss.
Soft magnetic materials,such as soft iron,possess a narrow hysteresis loop,which results in low hysteresis loss.
Therefore,to minimize these losses and increase the efficiency of the transformer,a soft iron core is used.

(A) The kinetic energy $(K)$ gained by an electron accelerated through a potential difference $(V)$ is given by the formula $K = e \cdot V$.
Here,the charge of an electron $e = 1.602 \times 10^{-19} \, C$ and the potential difference $V = 100 \, V$.
Substituting these values into the formula:
$K = (1.602 \times 10^{-19} \, C) \times (100 \, V)$
$K = 1.602 \times 10^{-17} \, J$.
Therefore,the correct option is $A$.

(B) In a photoelectric experiment,the photoelectric current is directly proportional to the intensity of the incident radiation,provided the frequency is above the threshold frequency.
- The work function depends only on the nature of the metal surface.
- The stopping potential and the maximum kinetic energy of the emitted photoelectrons depend only on the frequency of the incident radiation and the work function of the metal,not on the intensity.
Therefore,the correct option is $B$.

(C) The half-life period $T_{1/2}$ of a radioactive substance is related to the decay constant $\lambda$ by the formula: $T_{1/2} = \frac{0.693}{\lambda}$.
Given the decay constant $\lambda = 1.07 \times 10^{-4} \text{ year}^{-1}$.
Substituting the value into the formula:
$T_{1/2} = \frac{0.693}{1.07 \times 10^{-4}}$
$T_{1/2} = \frac{0.693}{1.07} \times 10^4$
$T_{1/2} \approx 0.6476 \times 10^4 = 6476 \text{ years}$.
Thus,the half-life period is $6476 \text{ years}$.

(C) Mirage is an optical illusion observed in hot deserts or on hot roads.
It occurs due to the total internal reflection of light.
As light travels from a denser medium (cooler air near the ground) to a rarer medium (hotter air above),the refractive index of the air changes continuously.
When the angle of incidence exceeds the critical angle for the interface between the layers of air,the light undergoes total internal reflection,creating an inverted image of distant objects.

(C) simple microscope consists of a convex lens of short focal length.
When an object is placed between the optical centre and the principal focus of the convex lens,a virtual,erect,and magnified image is formed on the same side as the object.
Therefore,the person sees an upright virtual image.

(B) In an astronomical telescope,the objective lens is used to collect light from distant objects,so it requires a large aperture and a long focal length $(f_0)$.
The eye-piece acts as a simple magnifier and requires a short focal length $(f_e)$.
Therefore,the condition for an astronomical telescope is $f_0 > f_e$.

(C) The luminous intensity $I$ is defined as the luminous flux per unit solid angle, given by $I = \frac{d\phi}{d\Omega}$.
For a point source emitting light uniformly in all directions, the total solid angle is $4\pi \, \text{steradians}$.
However, for a unidirectional bulb, the flux is emitted into a specific solid angle. Assuming the standard interpretation for such problems where the source is treated as isotropic over a full sphere unless specified otherwise, the total luminous flux $\phi$ is given by $\phi = I \times \Omega$.
Given $I = 100 \, \text{candela}$ and $\Omega = 4\pi \, \text{steradians}$:
$\phi = 100 \times 4 \times 3.14159$
$\phi = 1256.6 \, \text{lumen}$.
Rounding to the nearest provided option, the correct value is $1256 \, \text{lumen}$.

(C) The phenomenon of light bending around obstacles is known as diffraction. Diffraction is significant only when the size of the obstacle or aperture is comparable to the wavelength of the light. Since the wavelength of visible light is extremely small ($400 \, nm$ to $700 \, nm$), it does not show noticeable diffraction around common objects in our daily life. Consequently, light appears to travel in straight lines, a principle known as rectilinear propagation of light. Therefore, the correct option is $C$.

(C) According to Maxwell's $EM$ theory,a time-varying electric field produces a displacement current,which in turn acts as a source of a magnetic field.
This is represented by the Maxwell-Ampere law: $\oint B \cdot dl = \mu_0 (I_c + I_d)$,where $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$ is the displacement current.
Therefore,a changing electric field gives rise to a magnetic field.

(A) In a series $LCR$ circuit,the impedance $Z$ is given by $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Resonance occurs when the inductive reactance $X_L = \omega L$ and capacitive reactance $X_C = \frac{1}{\omega C}$ are equal and opposite,such that $X_L - X_C = 0$.
At this condition,the impedance $Z$ becomes minimum $(Z = R)$,and the current $I = \frac{V}{Z}$ becomes maximum.
Since the condition for resonance is indeed $X_L = X_C$,the Reason correctly explains the Assertion.

(A) In the standard Newton's rings experiment,the air film is between the lens $(n_l)$ and the glass plate $(n_g)$,where $n_l > n_{air}$ and $n_g > n_{air}$. The reflection at the bottom surface of the air film occurs at the glass-air interface (denser to rarer),while the top surface reflection occurs at the air-glass interface (rarer to denser). This introduces a phase change of $\pi$ for one ray,resulting in a dark central spot.
When the space is filled with a liquid of refractive index $n_{liq} > n_g$,the reflection at the bottom surface occurs at the glass-liquid interface (rarer to denser),and the reflection at the top surface occurs at the liquid-glass interface (denser to rarer).
Since both reflections now occur under similar conditions (either both undergo phase change or neither does),the path difference at the center is effectively zero,leading to constructive interference. Thus,the central spot becomes bright. Both the Assertion and the Reason are correct,and the Reason explains why the phase change conditions lead to a bright center.

(A) Newton's corpuscular theory suggests that light consists of tiny particles called corpuscles.
According to this theory,the velocity of light increases as it enters a denser medium because the particles are attracted by the medium.
However,experimental evidence shows that the velocity of light is lower in a denser medium compared to a rarer medium.
Therefore,the corpuscular theory fails to explain the observed velocities of light in different media.
Since the reason correctly explains why the theory fails,both statements are correct and the reason is the correct explanation of the assertion.
Thus,option $(A)$ is correct.

(D) In vacuum,all electromagnetic waves (including different colours of light) travel with the same speed,$c \approx 3 \times 10^8 \ m/s$. Thus,the Assertion is incorrect.
The speed of light in a medium is given by $v = c/n$,where $n$ is the refractive index. Since $n$ depends on the wavelength of light $(\lambda)$ in the medium,the speed $v$ also depends on the wavelength. Therefore,the Reason is correct.

(D) The half-life $(T_{1/2})$ of the radioactive substance is $40 \ days$. The amount of substance remaining after time $t$ is given by $N = N_0 \left( \frac{1}{2} \right)^{t/T_{1/2}}$.
For $t = 20 \ days$,the number of half-lives is $n = \frac{20}{40} = 0.5$.
The amount remaining is $N = N_0 \left( \frac{1}{2} \right)^{0.5} = N_0 \left( \frac{1}{\sqrt{2}} \right) \approx 0.707 N_0$.
The amount decayed is $N_0 - N = N_0 - 0.707 N_0 = 0.293 N_0$,which is $29.3\%$.
Since $29.3\% \neq 25\%$,the Assertion is incorrect.
The formula provided in the Reason is the standard law of radioactive decay,which is correct.
Therefore,the Assertion is incorrect but the Reason is correct.