AIIMS 1999 Physics Question Paper with Answer and Solution

(A) The dot product of two vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$ is defined as $\overrightarrow{P} \cdot \overrightarrow{Q} = PQ \cos \theta$,where $\theta$ is the angle between the two vectors.
Given that $\overrightarrow{P} \cdot \overrightarrow{Q} = PQ$.
Substituting the definition,we get $PQ \cos \theta = PQ$.
Dividing both sides by $PQ$ (assuming $P, Q \neq 0$),we get $\cos \theta = 1$.
Since $\cos 0^\circ = 1$,the angle $\theta$ must be $0^\circ$.

(D) When a body is projected upwards with an initial velocity $u$,it undergoes constant deceleration due to gravity $(g)$.
At the maximum height,its velocity becomes $0$.
As it falls back down to the point of projection,it undergoes constant acceleration due to gravity $(g)$ over the same displacement.
Using the kinematic equation $v^2 = u^2 + 2as$,where $a = -g$ for the upward journey and $a = g$ for the downward journey,the final velocity $v$ at the starting point will have the same magnitude as the initial velocity $u$,but in the opposite direction.
Therefore,the speed of the body upon returning to the point of projection is $v = u$.

(D) The equation of motion for a body projected vertically upwards is given by $v = u - gt$,where $v$ is the final velocity,$u$ is the initial velocity,$g$ is the acceleration due to gravity,and $t$ is time.
Since $g$ is constant,this equation is of the form $y = mx + c$,which represents a straight line.
Therefore,the velocity-time graph for a body projected vertically upwards is a straight line with a negative slope equal to $-g$.

(C) For a cyclist taking a circular turn on a level road,the necessary centripetal force is provided by the static friction between the tyres and the road.
The condition for safe turning is given by the formula: $v^2 \leq \mu rg$,where $v$ is the speed,$\mu$ is the coefficient of friction,$r$ is the radius,and $g$ is the acceleration due to gravity.
To find the minimum coefficient of friction,we use: $\mu = \frac{v^2}{rg}$.
Given: $v = 4.9 \, m/s$,$r = 4 \, m$,and $g = 9.8 \, m/s^2$.
Substituting the values: $\mu = \frac{(4.9)^2}{4 \times 9.8} = \frac{24.01}{39.2} = 0.6125$.
Rounding to two decimal places,we get $\mu = 0.61$.

(C) The relationship between linear momentum $(p)$,mass $(m)$,and kinetic energy $(K.E.)$ is given by the formula: $p = \sqrt{2m(K.E.)}$.
Since the kinetic energy $(K.E.)$ is equal for both bodies,we have $p \propto \sqrt{m}$.
Therefore,the ratio of their linear momentums is $\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}$.
Given $m_1 = m$ and $m_2 = 4m$,we substitute these values:
$\frac{p_1}{p_2} = \sqrt{\frac{m}{4m}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(D) The elasticity of a material is influenced by several external and internal factors:
$1$. Temperature: Generally,the elasticity of most materials decreases as the temperature increases.
$2$. Impurities: The addition of impurities can alter the interatomic forces within the material,thereby changing its elastic properties.
$3$. Hammering and Annealing: These mechanical and thermal treatments change the internal crystalline structure of the material. Hammering breaks down crystal grains into smaller units,while annealing promotes the growth of larger,more uniform grains,both of which significantly affect the material's elasticity.
Therefore,all the given factors affect the elasticity of a substance.

(C) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$: $E \propto T$.
Therefore,the ratio is given by: $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.
Given:
$E_1 = 6.21 \times 10^{-21} \, J$
$T_1 = 27^oC = 27 + 273 = 300 \, K$
$T_2 = 227^oC = 227 + 273 = 500 \, K$
Substituting the values:
$\frac{6.21 \times 10^{-21}}{E_2} = \frac{300}{500} = \frac{3}{5}$.
Solving for $E_2$:
$E_2 = \frac{6.21 \times 10^{-21} \times 5}{3} = 2.07 \times 5 \times 10^{-21} \, J = 10.35 \times 10^{-21} \, J$.

(C) In an adiabatic process,the system is thermally insulated from its surroundings.
By definition,there is no exchange of heat between the system and the surroundings,which means $dQ = 0$.
Therefore,the heat content of the system remains constant during the process.

(A) When the temperature of an object is equal to that of the human body,no heat is transferred between the object and the body.
Since the human body is at a constant temperature,if the objects are also at that same temperature,there is no net flow of heat.
Therefore,both the block of wood and the block of metal will feel neither cold nor hot,or they will feel equally neutral,which is the condition described as feeling 'equally cold or hot' (i.e.,no temperature difference is perceived).
Thus,the correct answer is that their temperatures are equal to the temperature of the human body.

(D) In simple harmonic motion $(SHM)$,the acceleration $a$ is given by $a = -\omega^2 x$,where $\omega$ is the angular frequency and $x$ is the displacement from the equilibrium position.
At the equilibrium position,$x = 0$,so the acceleration $a = 0$,which is the minimum value.
At the extreme positions,$x = \pm A$ (where $A$ is the amplitude),so the acceleration $a = \mp \omega^2 A$,which is the maximum value.
Therefore,the statement that the acceleration is maximum at the equilibrium position is incorrect.
The correct option is $D$.

(D) The maximum acceleration $(a_{\max})$ of a particle executing simple harmonic motion is given by the formula: $a_{\max} = A\omega^2$,where $A$ is the amplitude and $\omega$ is the angular velocity.
Given:
Angular velocity $\omega = 3.5\, rad/s$
Maximum acceleration $a_{\max} = 7.5\, m/s^2$
Rearranging the formula to solve for amplitude $A$:
$A = \frac{a_{\max}}{\omega^2}$
Substituting the given values:
$A = \frac{7.5}{(3.5)^2} = \frac{7.5}{12.25} \approx 0.61\, m$
Therefore,the amplitude of oscillation is $0.61\, m$.

(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.

(A) For the object not to be detached from the platform,the normal reaction $R$ must be greater than or equal to zero. The object is most likely to detach at the highest point of its oscillation where the acceleration is directed downwards.
Applying Newton's second law for the object of mass $m$ at the extreme top position:
$mg - R = ma\omega^2$
For the critical condition where the object is just about to detach,the normal reaction $R = 0$.
Thus,$mg = ma\omega^2$
$g = a\omega^2$
$\omega = \sqrt{\frac{g}{a}}$
Given $g = 9.8 \, m/s^2$ and amplitude $a = 3.92 \times 10^{-3} \, m$:
$\omega = \sqrt{\frac{9.8}{3.92 \times 10^{-3}}} = \sqrt{\frac{9800}{3.92}} = \sqrt{2500} = 50 \, rad/s$
The time period $T$ is given by:
$T = \frac{2\pi}{\omega} = \frac{2 \times 3.14159}{50} = 0.1256 \, s$
Therefore,the least period of oscillation is $0.1256 \, s$.

(B) $SONAR$ stands for Sound Navigation and Ranging.
It works on the principle of reflection of sound waves.
$SONAR$ emits ultrasonic waves (sound waves with frequencies higher than $20,000 \ Hz$) into the water.
These waves travel through the water,strike an object,and reflect back to the receiver,allowing the system to detect the object's distance and location.

(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 frequency of vibration of a stretched string is given by the formula $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $n$ is the frequency,$L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.
From this relation,we can see that $n \propto \sqrt{T}$.
If the frequency is increased by a factor of two,the new frequency $n' = 2n$.
Therefore,$\frac{n'}{n} = \frac{\sqrt{T'}}{\sqrt{T}} = 2$.
Squaring both sides,we get $\frac{T'}{T} = 4$,which implies $T' = 4T$.
Thus,the tension must be made four times the original tension.

(A) According to the Doppler effect,when the source is moving away from a stationary observer,the observed frequency $f'$ is given by the formula:
$f' = f \left( \frac{v}{v + v_s} \right)$
Where:
$f = 500 \; Hz$ (source frequency)
$v = 330 \; m/s$ (speed of sound)
$v_s = 50 \; m/s$ (speed of the source)
Substituting the values:
$f' = 500 \times \left( \frac{330}{330 + 50} \right)$
$f' = 500 \times \left( \frac{330}{380} \right)$
$f' = 500 \times 0.8684$
$f' \approx 434.2 \; Hz$

(A) On a rainy day,the roads become wet.
Water acts as a lubricant between the tires and the road surface,which significantly lowers the coefficient of friction $(\mu)$.
Since the frictional force $f = \mu N$ (where $N$ is the normal force),a decrease in $\mu$ leads to a decrease in the available frictional force.
This reduced friction makes it difficult to maintain control,increases the braking distance,and increases the chances of skidding at high speeds.
Therefore,both the $Assertion$ and $Reason$ are correct,and the $Reason$ is the correct explanation for the $Assertion$.

(A) The moment of inertia $(I)$ of a circular disc about an axis perpendicular to its plane and passing through its center is given by the formula:
$I = \frac{1}{2} M R^2$
Given:
Mass $(M)$ = $0.4\, kg$
Radius $(R)$ = $100\, cm = 1\, m$
Substituting the values into the formula:
$I = \frac{1}{2} \times 0.4\, kg \times (1\, m)^2$
$I = 0.2 \times 1 = 0.2\, kg\, m^2$
Therefore,the correct option is $A$.

(D) The Assertion is incorrect because the size of a degree in a temperature scale is determined by the number of divisions between the freezing and boiling points of water.
In the Celsius scale $(^oC)$,there are $100$ divisions.
In the Fahrenheit scale $(^oF)$,there are $180$ divisions.
In the Reaumur scale $(^oR)$,there are $80$ divisions.
In the Rankine scale $(^oRa)$,there are $180$ divisions.
Since the Rankine scale has the largest number of divisions for the same temperature interval,its unit size is actually smaller than that of the Fahrenheit scale.
Therefore,the statement that Fahrenheit is the smallest unit is false.
The Reason is correct because Daniel Gabriel Fahrenheit developed the first standardized mercury-in-glass thermometer and the associated Fahrenheit scale in $1724$.

(A) According to the kinetic theory of matter,all bodies at temperatures above absolute zero $(0 \ K)$ possess thermal energy and emit electromagnetic radiation. Thus,the assertion is correct.
According to the Stefan-Boltzmann law,the total energy radiated per unit surface area of a black body per unit time is directly proportional to the fourth power of its absolute temperature,given by $E = \sigma T^4$. This law explains why the rate of radiation depends on the temperature of the body. Therefore,the reason is correct and provides a valid explanation for the assertion.

(C) According to the law of conservation of linear momentum,the total initial momentum of the system must be equal to the total final momentum.
Initial momentum of the system: $P_i = m \cdot v + 2m \cdot 0 = mv$
After the collision,the two particles stick together,forming a combined mass of $(m + 2m) = 3m$.
Let the final velocity of the combined mass be $v'$.
Final momentum of the system: $P_f = (3m) \cdot v'$
Equating initial and final momentum: $mv = 3m \cdot v'$
Solving for $v'$: $v' = \frac{mv}{3m} = \frac{v}{3}$

(C) The dimension of $CR$ is given by the product of capacitance $(C)$ and resistance $(R)$.
$C = \frac{Q}{V}$,where $Q$ is charge and $V$ is potential difference.
$R = \frac{V}{I}$,where $I$ is current.
Therefore,$CR = \left( \frac{Q}{V} \right) \times \left( \frac{V}{I} \right) = \frac{Q}{I}$.
Since $I = \frac{Q}{t}$,where $t$ is time,we have $\frac{Q}{I} = t$.
Thus,the dimensions of $CR$ are the same as those of time,which corresponds to the time period.

(B) According to the quantization of charge,the total charge $q$ is given by $q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge.
Given,$q = 1 \ C$ and $e = 1.6 \times 10^{-19} \ C$.
Therefore,$n = \frac{q}{e} = \frac{1}{1.6 \times 10^{-19}}$.
$n = \frac{1}{1.6} \times 10^{19} = 0.625 \times 10^{19} = 6.25 \times 10^{18}$.
Thus,the number of electrons in one coulomb of charge is $6.25 \times 10^{18}$.

(C) For a conductor placed in an external electric field $E$, the induced charge $q_{ind}$ is given by the relation $q_{ind} = -q(1 - 1/K)$, where $K$ is the dielectric constant.
For any perfect conductor, the dielectric constant $K$ is considered to be infinity $(\infty)$.
Therefore, the induced charge $q_{ind} = -q(1 - 1/\infty) = -q(1 - 0) = -q$.
Since both copper and aluminium are metals (conductors), they both have an infinite dielectric constant.
Thus, the magnitude of the induced charge on both conductors will be equal.

(B) From the circuit diagram,we can observe that resistors $R_2$ and $R_3$ are connected in parallel.
The equivalent resistance of this parallel combination $(R_p)$ is given by:
$\frac{1}{R_p} = \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$
So,$R_p = 2 \,\Omega$.
Now,this parallel combination is in series with resistors $R_1$ and $R_4$.
The total equivalent resistance $(R_{AB})$ between points $A$ and $B$ is:
$R_{AB} = R_1 + R_p + R_4$
$R_{AB} = 2 + 2 + 2 = 6 \,\Omega$.

(D) In the given circuit,the terminals $A$ and $B$ are open-circuited.
Since no current flows through the circuit $(I = 0)$,there is no voltage drop across any internal resistance of the cell (assuming an ideal cell or negligible internal resistance).
Therefore,the potential difference across the open terminals $A$ and $B$ is equal to the electromotive force $(EMF)$ of the cell.
Thus,the potential difference is $20 \ V$.

(B) To convert a galvanometer into an ammeter,a shunt resistance $S$ must be connected in parallel with the galvanometer.
Given:
Galvanometer resistance $G = 50\,\Omega$
Full-scale deflection current $i_g = 100\,\mu A = 100 \times 10^{-6}\,A = 10^{-4}\,A$
Desired range of ammeter $i = 10\,A$
The formula for shunt resistance is $S = \frac{G \cdot i_g}{i - i_g}$.
Substituting the values:
$S = \frac{50 \times 10^{-4}}{10 - 10^{-4}}$
Since $10^{-4}$ is very small compared to $10$,we can approximate the denominator as $10$.
$S \approx \frac{50 \times 10^{-4}}{10} = 5 \times 10^{-4}\,\Omega$.
Thus,a resistance of $5 \times 10^{-4}\,\Omega$ must be connected in parallel.

(D) The $SI$ unit of magnetic field induction $(B)$ is Tesla $(T)$.
One Tesla is defined as the magnetic field that exerts a force of $1 \ N$ on a charge of $1 \ C$ moving with a velocity of $1 \ m/s$ perpendicular to the field.

(D) When a charged particle enters a magnetic field $B$ at an angle $\theta$ (where $\theta \neq 0^\circ, 90^\circ, 180^\circ$),its velocity $v$ can be resolved into two components:
$1$. The component $v \cos \theta$ parallel to the magnetic field,which causes the particle to move in a straight line along the direction of the field.
$2$. The component $v \sin \theta$ perpendicular to the magnetic field,which causes the particle to move in a circular path.
Since the particle has both a parallel and a perpendicular velocity component,the resultant path is a helix. Given $\theta = 45^\circ$,the path is a helix.

(A) The angle of dip (or magnetic inclination) is defined as the angle that the total magnetic field of the Earth makes with the surface of the Earth.
At the magnetic poles,the Earth's magnetic field lines are perpendicular to the surface of the Earth.
Therefore,the angle of dip at the poles is $90^{\circ}$.

(B) When a magnetic substance is heated,it loses its magnetic property.
This happens because the thermal energy causes the atomic magnets (magnetic dipoles) to become randomly oriented,thereby destroying the net alignment that creates the magnetic field.

(B) According to Lenz's law,the direction of the induced current is such that it opposes the cause that produces it.
When the north pole of a magnet is brought near a metallic ring,the magnetic flux linked with the ring increases.
To oppose this increase in magnetic flux,the ring will develop a north pole on the face towards the magnet.
$A$ face with a north pole polarity corresponds to an anticlockwise direction of current when viewed from the side of the magnet.
Therefore,the induced current in the ring will be anticlockwise.

(A) The transformation ratio of a transformer is given by the formula: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$.
Given:
Primary turns $(N_p)$ = $500$
Secondary turns $(N_s)$ = $5000$
Primary voltage $(V_p)$ = $20\, V$
Substituting the values:
$\frac{V_s}{20} = \frac{5000}{500}$
$\frac{V_s}{20} = 10$
$V_s = 200\, V$.
In a transformer,the frequency of the output voltage remains the same as the input frequency because the magnetic flux changes at the same rate as the input current. Therefore,the frequency is $50\, Hz$.

(C) In a transformer,the primary purpose is to change the voltage and current levels through electromagnetic induction. However,the frequency of the alternating current $(AC)$ remains constant because the magnetic flux in the core oscillates at the same rate as the input supply frequency. Therefore,the frequency at the input is equal to the frequency at the output.

(A) choke coil is designed to have high inductance and low resistance.
Since an inductor is an ideal non-resistive device,it does not consume power in the form of heat,unlike a resistor.
By using a coil with high inductance $(L)$ and very low resistance $(R)$,we can control the current in an $AC$ circuit while minimizing power loss $(P = I^2 R)$.
Therefore,the correct option is $A$.

(A) Cathode rays consist of high-speed electrons. When these high-energy electrons strike a metal target with a high melting point (such as tungsten or molybdenum),they are suddenly decelerated by the strong electric fields of the target nuclei. This rapid deceleration causes the emission of high-energy electromagnetic radiation known as $X$-rays. This is the fundamental principle behind the operation of an $X$-ray tube.

(D) According to the Bohr model of the hydrogen atom,the radius of the $n^{th}$ stationary orbit is given by the formula:
$r_n = \frac{\varepsilon_0 n^2 h^2}{\pi Z m e^2}$
In this expression,$\varepsilon_0$ is the permittivity of free space,$h$ is Planck's constant,$Z$ is the atomic number,$m$ is the mass of the electron,and $e$ is the elementary charge.
Since all these parameters are constants for a given atom,the radius $r$ is directly proportional to the square of the principal quantum number $n$.
Therefore,$r \propto n^2$.

(B) The correct answer is $(b)$. $A$ neutron moderator is a medium that reduces the speed of fast neutrons,thereby turning them into thermal neutrons capable of sustaining a nuclear chain reaction involving uranium$-235$.
Heavy water $(D_2O)$ serves as an effective neutron moderator in a nuclear reactor. It slows down fast-moving neutrons through elastic collisions,which increases the probability of these neutrons causing further fission in uranium nuclei.
Explanation:
In a fission reaction,neutrons are released with high kinetic energy. These fast neutrons are less likely to cause further fission. By using a moderator like heavy water,the neutrons lose kinetic energy and become 'thermal' or 'slow' neutrons,which are much more efficient at sustaining the chain reaction.

(D) The energy in the sun is generated primarily through the process of nuclear fusion.
In the core of the sun,hydrogen nuclei (protons) undergo fusion to form helium nuclei.
This process releases a tremendous amount of energy due to the mass defect between the reactants and the products,as described by Einstein's mass-energy equivalence principle,$E = \Delta mc^2$.

(D) In an $N$-type semiconductor,pentavalent impurity atoms (like Phosphorus or Arsenic) are added to the intrinsic semiconductor (like Silicon or Germanium).
These impurity atoms provide extra electrons to the conduction band.
Therefore,in $N$-type semiconductors,electrons are the majority charge carriers,while holes are the minority charge carriers.

(B) Boolean algebra is a branch of algebra in which the values of the variables are the truth values $true$ and $false$,usually denoted by $1$ and $0$ respectively.
It is essentially based on logic,as it deals with logical operations such as $AND$,$OR$,and $NOT$ to manipulate these truth values.
Therefore,the correct option is $B$.

(C) diode is a semiconductor device that allows current to flow in only one direction. Due to this unidirectional property,it is primarily used as a rectifier to convert alternating current $(AC)$ into direct current $(DC)$.

(A) The critical angle $C$ is given by the formula $\sin C = \frac{1}{\mu}$,where $\mu$ is the refractive index of the medium with respect to air.
Since the refractive index of water $\mu_w \approx 1.33$ and the refractive index of glass $\mu_g \approx 1.50$,we have $\mu_w < \mu_g$.
Because $C = \arcsin(\frac{1}{\mu})$,a smaller refractive index results in a larger critical angle.
Therefore,$C_w > C_g$.

(A) According to Rayleigh's law of scattering, the intensity of scattered light $I$ is inversely proportional to the fourth power of its wavelength, given by $I \propto \frac{1}{\lambda^4}$.
Since the wavelength of blue light $(\lambda_{blue})$ is the shortest among the visible spectrum, it undergoes maximum scattering by the atmospheric particles.
Therefore, the sky appears blue to our eyes.

(D) Interference is a general property of waves. It occurs when two or more waves of the same frequency and constant phase difference superpose each other. This phenomenon is observed in all types of waves,including longitudinal mechanical waves (e.g.,sound waves),transverse mechanical waves (e.g.,waves on a string),and electromagnetic waves (e.g.,light waves). Therefore,the correct option is $D$.

(B) The shift in the interference pattern due to the insertion of a transparent plate is given by the formula: $x = \frac{(\mu - 1)tD}{d}$.
Here,$\mu = 1.5$,$t = 2.5 \times 10^{-5} \, m$,$D = 100 \, cm = 1 \, m$,and $d = 0.5 \, mm = 0.5 \times 10^{-3} \, m$.
Substituting these values into the formula:
$x = \frac{(1.5 - 1) \times 2.5 \times 10^{-5} \times 1}{0.5 \times 10^{-3}}$
$x = \frac{0.5 \times 2.5 \times 10^{-5}}{0.5 \times 10^{-3}}$
$x = 2.5 \times 10^{-2} \, m$
$x = 2.5 \, cm$.

(D) When a thin glass plate of thickness $t$ and refractive index $\mu$ is placed in the path of one of the beams,an additional path difference is introduced.
This path difference is given by $\Delta x = (\mu - 1)t$.
Due to this additional path difference,the entire interference pattern shifts by a distance $y = \frac{D}{d}(\mu - 1)t$ towards the side where the glass plate is placed.
However,the fringe width $\beta = \frac{\lambda D}{d}$ depends only on the wavelength $\lambda$,the distance between the slits $d$,and the distance to the screen $D$.
Since these parameters remain unchanged,the fringe width remains constant.
Therefore,the correct option is $(d)$.

(B) Nuclear fission is a process in which a heavy nucleus splits into two lighter nuclei.
According to the binding energy per nucleon curve,for heavy nuclei (high mass number $A$),the binding energy per nucleon decreases as the mass number increases.
This means that heavy nuclei are less stable compared to intermediate-mass nuclei.
When a heavy nucleus splits into two lighter nuclei,the binding energy per nucleon of the resulting products is higher than that of the parent nucleus.
This increase in binding energy per nucleon results in the release of energy,making the fission process possible.
Therefore,the correct reason is that the binding energy per nucleon decreases with mass number at high mass numbers.

(B) The half-life of the substance is $T_{1/2} = 20 \, min$. The decay constant is $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{20}$.
Let $N_0$ be the initial amount of the substance.
For $20\%$ decay,the amount remaining is $N_1 = N_0 - 0.20 N_0 = 0.80 N_0$. The time taken is $t_1$,where $0.80 N_0 = N_0 e^{-\lambda t_1}$.
For $80\%$ decay,the amount remaining is $N_2 = N_0 - 0.80 N_0 = 0.20 N_0$. The time taken is $t_2$,where $0.20 N_0 = N_0 e^{-\lambda t_2}$.
Dividing the two equations:
$\frac{0.80 N_0}{0.20 N_0} = \frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}$
$4 = e^{\lambda(t_2 - t_1)}$
Taking the natural logarithm on both sides:
$\ln 4 = \lambda(t_2 - t_1)$
$2 \ln 2 = \left( \frac{\ln 2}{20} \right) (t_2 - t_1)$
$2 = \frac{t_2 - t_1}{20}$
$t_2 - t_1 = 40 \, min$.

(D) The electric field $E$ is directed from a region of higher potential to a region of lower potential.
An electron carries a negative charge $q = -e$.
The force on a charged particle in an electric field is given by $F = qE$.
Since the charge is negative,the force on the electron is in the direction opposite to the electric field.
Therefore,an electron moves from a region of lower potential to a region of higher potential.
Thus,the Assertion is incorrect,and the Reason is correct.

(D) For total internal reflection to occur at the prism-liquid interface,the angle of incidence $i$ must be greater than or equal to the critical angle $c$.
In the given figure,the ray strikes the interface at an angle of incidence $i = 45^o$.
For total internal reflection,we require $i \ge c$,so $45^o \ge c$,or $\sin 45^o \ge \sin c$.
Since $\sin c = \frac{\mu_{liquid}}{\mu_{prism}}$,and assuming the prism is glass with $\mu_{prism} \approx 1.5$,the condition for total internal reflection is $\sin 45^o \ge \frac{\mu_{liquid}}{\mu_{prism}}$.
However,the question asks for the maximum refractive index of the liquid such that total internal reflection still occurs. This occurs when $c = 45^o$.
Using the relation $\sin c = \frac{\mu_{liquid}}{\mu_{prism}}$,we get $\mu_{liquid} = \mu_{prism} \sin 45^o = 1.5 \times \frac{1}{\sqrt 2} \approx 1.06$.
The assertion states the value is $\sqrt 2$,which is incorrect as it assumes the prism refractive index is $2$. Given the standard context of such problems,the assertion and reason are both technically incorrect based on standard glass prisms.

(A) The Lens Maker's Formula is given by $\frac{1}{f} = \left( \frac{\mu_l}{\mu_m} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,which can be written as $\frac{1}{f} = \left( \frac{\mu_l - \mu_m}{\mu_m} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$. Thus,the Reason is correct.
For the lens in air $(\mu_m = 1)$:
$\frac{1}{10} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \frac{1}{10} = 0.5 \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \implies \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \frac{1}{5} \dots (i)$
For the lens in water $(\mu_m = 4/3)$:
$\frac{1}{f'} = \left( \frac{1.5 - 4/3}{4/3} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$
$\frac{1}{f'} = \left( \frac{4.5/3 - 4/3}{4/3} \right) \left( \frac{1}{5} \right) = \left( \frac{0.5/3}{4/3} \right) \left( \frac{1}{5} \right) = \left( \frac{0.5}{4} \right) \left( \frac{1}{5} \right) = \frac{1}{8} \times \frac{1}{5} = \frac{1}{40}$
Therefore,$f' = 40 \, cm$. The Assertion is also correct and the Reason explains it.

(D) The kinetic energy of emitted photoelectrons depends on the frequency of the incident photon, not the intensity. Intensity only determines the number of photoelectrons emitted per unit time.
According to Einstein's photoelectric equation, $K_{max} = h\nu - \Phi_0$, where $\nu$ is the frequency of incident light and $\Phi_0$ is the work function.
Furthermore, the ejection of electrons from a metallic surface is only possible if the frequency of the incident photon is greater than or equal to the threshold frequency $(\nu \ge \nu_0)$.
Therefore, both the Assertion and the Reason are incorrect.

(A) The specific charge $\frac{e}{m}$ of positive rays is defined as the ratio of charge to mass. According to the theory of relativity,the mass $m$ of a particle moving with a velocity $v$ is given by $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$,where $m_0$ is the rest mass and $c$ is the speed of light.
Since the mass $m$ depends on the speed $v$,the specific charge $\frac{e}{m}$ also changes with speed.
Therefore,the specific charge of positive rays is not constant because the mass of ions varies with speed.
Thus,both the Assertion and the Reason are correct,and the Reason is the correct explanation of the Assertion.

(D) Isotopes of an element have the same number of electrons and protons, but they differ in the number of neutrons, which leads to a difference in their atomic masses.
Because of this difference in atomic mass, isotopes can be separated using a mass spectrometer, which separates ions based on their mass-to-charge ratio $(m/q)$.
Since the Assertion claims that separation is due to the difference in electron numbers, the Assertion is incorrect.
However, the Reason stating that isotopes can be separated by a mass spectrometer is correct.
Therefore, the correct option is $D$.