AIIMS 2017 Physics Question Paper with Answer and Solution

(A) The condition for water not to spill from a bucket at the highest point of a vertical circle is that the centripetal force must be at least equal to the gravitational force.
At the highest point,the minimum velocity $v$ is given by the formula $v = \sqrt{gR}$.
Given,radius $R = 1.6\, m$ and acceleration due to gravity $g = 10\, m/s^2$.
Substituting the values: $v = \sqrt{10 \times 1.6} = \sqrt{16} = 4\, m/s$.
Therefore,the minimum speed required is $4\, m/s$.

(D) In a steady state,the rate of heat flow $H$ through a composite slab of two materials in series is given by $H = \frac{A(T_2 - T_1)}{R_1 + R_2}$,where $R_1 = \frac{x}{KA}$ and $R_2 = \frac{4x}{(2K)A} = \frac{2x}{KA}$.
The total thermal resistance is $R_{eq} = R_1 + R_2 = \frac{x}{KA} + \frac{2x}{KA} = \frac{3x}{KA}$.
Therefore,the rate of heat transfer is $H = \frac{A(T_2 - T_1)}{\frac{3x}{KA}} = \frac{1}{3} \frac{AK(T_2 - T_1)}{x}$.
Comparing this with the given expression $\left( \frac{A(T_2 - T_1)K}{x} \right)f$,we find that $f = \frac{1}{3}$.

(A) For an organ pipe open at one end,the frequency of the first overtone is given by $n_c = \frac{3v}{4l_c}$,where $v$ is the speed of sound and $l_c$ is the length of the closed pipe.
For an organ pipe open at both ends,the frequency of the third harmonic is given by $n_o = \frac{3v}{2l_o}$,where $l_o$ is the length of the open pipe.
Since the pipes are in resonance,$n_c = n_o$.
Equating the two expressions: $\frac{3v}{4l_c} = \frac{3v}{2l_o}$.
Simplifying this,we get $\frac{1}{4l_c} = \frac{1}{2l_o}$,which implies $\frac{l_c}{l_o} = \frac{2}{4} = \frac{1}{2}$.
Thus,the ratio of the length of the closed pipe to the open pipe is $1:2$.

(D) Initial angular speed,$\omega_1 = 1200 \ rpm = \frac{1200 \times 2\pi}{60} \ rad/s = 40\pi \ rad/s$.
Final angular speed,$\omega_2 = 4500 \ rpm = \frac{4500 \times 2\pi}{60} \ rad/s = 150\pi \ rad/s$.
Time interval,$t = 10 \ s$.
Angular acceleration,$\alpha = \frac{\omega_2 - \omega_1}{t} = \frac{150\pi - 40\pi}{10} = \frac{110\pi}{10} = 11\pi \ rad/s^2$.
To convert $rad/s^2$ to $deg/s^2$,multiply by $\frac{180}{\pi}$:
$\alpha = 11\pi \times \frac{180}{\pi} = 11 \times 180 = 1980 \ deg/s^2$.

(A) The observer is on the inclined plane. Since the block does not slide on the wedge,its displacement relative to the inclined plane is zero.
Work done $W = \vec{F} \cdot \vec{S}$.
Since the displacement $\vec{S} = 0$ relative to the observer on the inclined plane,the work done by the force of friction on the block as seen by this observer is zero.

(B) To move block $B$,the applied force $P$ must overcome the frictional forces acting on it.
$1$. The friction between block $A$ and block $B$ $(F_{AB})$ acts on the top surface of block $B$ in the direction opposite to the motion. Since block $A$ is stationary,$F_{AB} = \mu_{AB} \cdot N_A = \mu_{AB} \cdot m_A \cdot g = 0.25 \times 100 \times 10 = 250 \, N$.
$2$. The friction between block $B$ and the ground $(F_{BS})$ acts on the bottom surface of block $B$. The normal force on the ground is the sum of the weights of both blocks: $N_{ground} = (m_A + m_B)g = (100 + 200) \times 10 = 3000 \, N$. Thus,$F_{BS} = \mu_{BS} \cdot N_{ground} = (1/3) \times 3000 = 1000 \, N$.
$3$. The total force $P$ required is the sum of these two frictional forces: $P = F_{AB} + F_{BS} = 250 + 1000 = 1250 \, N$.

(A) Let $u$ be the velocity of the hail stone before impact and $v$ be the velocity after impact.
Since the surface is smooth,there is no impulsive force parallel to the surface. Thus,the component of velocity parallel to the surface remains unchanged:
$u \sin 30^{\circ} = v \sin 60^{\circ}$ $(1)$
For the component normal to the surface,the coefficient of restitution $e$ is defined as the ratio of the velocity of separation to the velocity of approach:
$v \cos 60^{\circ} = e (u \cos 30^{\circ})$ $(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{v \cos 60^{\circ}}{v \sin 60^{\circ}} = \frac{e u \cos 30^{\circ}}{u \sin 30^{\circ}}$
$\cot 60^{\circ} = e \cot 30^{\circ}$
Since $\cot 60^{\circ} = \frac{1}{\sqrt{3}}$ and $\cot 30^{\circ} = \sqrt{3}$:
$\frac{1}{\sqrt{3}} = e (\sqrt{3})$
$e = \frac{1}{\sqrt{3} \cdot \sqrt{3}} = \frac{1}{3}$

(D) The potential energy $V$ has dimensions of work or energy,which is $[M L^2 T^{-2}]$.
In the denominator $(x + B)$,since $x$ is a distance,the dimension of $B$ must be the same as the dimension of $x$. Therefore,$[B] = [L]$.
The expression is $V = \frac{A\sqrt{x}}{x + B}$. Rearranging for $A$,we get $A = \frac{V(x + B)}{\sqrt{x}}$.
Substituting the dimensions: $[A] = \frac{[M L^2 T^{-2}] [L]}{[L^{1/2}]} = [M L^2 T^{-2}] [L^{1/2}] = [M L^{5/2} T^{-2}]$.
Now,the dimension of $AB$ is $[A][B] = [M L^{5/2} T^{-2}] [L] = [M L^{7/2} T^{-2}]$.

(B) In a displacement $(X)$ versus time $(t)$ graph,the slope represents the velocity of the particle. For a graph to be physically possible,a particle must have only one unique position at any given instant of time $(t)$.
Option $(A)$ is possible as it represents a particle moving away and then returning.
Option $(C)$ is possible as it shows motion in both directions.
Option $(D)$ is possible as it represents a particle moving with constant velocity,resetting,and repeating.
Option $(B)$ is not possible because at a specific instant of time $(t)$,the graph shows two different values of displacement $(X)$. Since a particle cannot be at two places at the same time,this graph is physically impossible.

(B) The pressure at any point in a static fluid is given by $P = h'\rho g$,where $h'$ is the depth of the point below the free surface of the liquid.
In the given figure,the total height of the liquid column is $H$ and the height of point $P$ from the bottom is $h$. Therefore,the depth of point $P$ below the free surface is $H - h$.
The normal thrust per unit area is equivalent to the hydrostatic pressure at that point.
Thus,the normal thrust per unit area at point $P$ is $(H - h)\rho g$.

(B) The $Assertion$ is correct because physical quantities can be measured directly (e.g.,using a meter scale) or indirectly (e.g.,using parallax method for distance of stars).
The $Reason$ is also correct because the reliability of any measurement depends on the precision and accuracy of the instrument used,and the final result must account for the associated uncertainties or errors.
However,the $Reason$ does not explain why we use direct or indirect methods; it explains the criteria for reporting a measurement. Therefore,the $Reason$ is not the correct explanation of the $Assertion$.

(D) The $Assertion$ states that the man cannot move on the horizontal surface while holding a rope connected to a block. This is incorrect because the man can exert a force on the block by pulling the rope. According to Newton's third law,the block will exert an equal and opposite force on the man,causing him to move towards the block.
The $Reason$ states that a man standing at rest on a smooth horizontal surface cannot start walking due to the absence of friction. This is correct. Walking requires friction to push the ground backward,which in turn provides a forward reaction force. Without friction,the man's feet would simply slip.
Since the $Assertion$ is incorrect and the $Reason$ is correct,the correct option is $D$ (based on standard assertion-reason logic where the assertion is false).

(D) The potential energy $U$ of a spring is given by the formula $U = \frac{1}{2} K x^2$,where $K$ is the spring constant and $x$ is the extension or compression.
This equation shows that $U \propto x^2$.
The graph of $U$ versus $x$ represents a parabola,not a straight line.
Therefore,the $Assertion$ is incorrect because the relationship is quadratic,not linear.
The $Reason$ is correct because it correctly states that the potential energy is proportional to the square of the extension or compression $(U \propto x^2)$.

(D) The radius of gyration $K$ of a body is defined as $K = \sqrt{\frac{\sum m_i r_i^2}{M}}$,where $M$ is the total mass and $r_i$ is the distance of the $i$-th particle from the axis of rotation.
It is not a constant quantity because its value depends on the position and orientation of the axis of rotation.
Therefore,the $Assertion$ is incorrect.
The $Reason$ correctly defines the radius of gyration as the root mean square distance of the particles from the axis of rotation.
Thus,the $Assertion$ is incorrect,but the $Reason$ is correct.

(A) The $Assertion$ is correct: Rockets are launched from west to east to take advantage of the Earth's rotational velocity,which helps in achieving orbital velocity with less fuel.
The $Reason$ is also correct: The effective acceleration due to gravity $g'$ is given by $g' = g - \omega^2 R \cos^2 \lambda$. At the equator,the latitude $\lambda = 0$,so $\cos \lambda = 1$,making $g' = g - \omega^2 R$,which is the minimum value.
However,the $Reason$ does not explain why rockets are launched from west to east. The launch direction is based on the Earth's rotation,not the value of $g$ at the equator. Therefore,both are correct,but the $Reason$ is not the correct explanation for the $Assertion$.

(B) The compressibility of a substance depends on the intermolecular forces and the distance between its molecules.
In solids,the atoms or molecules are held together by strong intermolecular forces and are closely packed,making them least compressible.
In gases,the intermolecular forces are negligible and the molecules are far apart,allowing them to be easily compressed.
While the $Reason$ correctly states the properties of solids and gases regarding shape and volume,it does not directly explain the physical mechanism (intermolecular forces) behind their compressibility. Therefore,both statements are correct,but the $Reason$ is not the correct explanation for the $Assertion$.

(A) According to the equation of continuity,$A_1v_1 = A_2v_2$. When water flows from a narrow pipe (small area $A_1$) to a wider pipe (large area $A_2$),the velocity $v$ decreases because $v \propto 1/A$.
According to Bernoulli's principle,$P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant}$.
For a horizontal pipe $(h = \text{constant})$,as the velocity $v$ decreases,the pressure $P$ must increase to keep the sum constant.
Therefore,the pressure increases when water flows from a narrow pipe to a wider pipe.
Both the Assertion and the Reason are correct,and the Reason correctly explains the Assertion.

(D) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W = \Delta U + P\Delta V$.
If heat is supplied in such a manner that the volume does not change $(\Delta V = 0)$,i.e.,an isochoric process,then the whole of the heat energy supplied to the system will increase the internal energy only.
However,in any other process where work is done,the heat supplied is distributed between the change in internal energy and the work done by the system.
Therefore,the $Assertion$ is incorrect.
The $Reason$ is also incorrect because when a system changes from one thermal equilibrium to another,heat may be absorbed,evolved,or remain zero depending on the process.

(B) During driving,the friction between the tyre and the road,along with the continuous compression and expansion of the tyre,generates heat. This increases the temperature of the air trapped inside the tyre.
According to Gay-Lussac's Law,for a fixed mass of gas at constant volume,the pressure $P$ is directly proportional to its absolute temperature $T$ $(P \propto T)$. Therefore,as the temperature increases,the air pressure inside the tyre increases. This confirms the Assertion is correct.
Regarding the Reason,absolute zero temperature $(0 \ K)$ is defined as the temperature at which the classical kinetic energy of gas molecules becomes zero. However,in quantum mechanics,the zero-point energy exists even at absolute zero. Thus,the statement that 'Absolute zero temperature is not zero energy temperature' is scientifically correct.
However,the Reason does not explain why the pressure increases in a car tyre; the pressure increase is due to the temperature rise caused by friction,not the nature of absolute zero energy. Therefore,both are correct,but the Reason is not the correct explanation of the Assertion.

(A) The displacement of a particle in $S.H.M.$ is given by $y = a \sin(\omega t + \phi)$.
The velocity is given by $v = \frac{dy}{dt} = a\omega \cos(\omega t + \phi)$.
The velocity is maximum at the mean position,where $v_{max} = a\omega$.
The kinetic energy at the mean position is $K.E._{max} = \frac{1}{2} m v_{max}^2 = \frac{1}{2} m (a\omega)^2$.
Given $m = 0.1 \, kg$,$a = 0.1 \, m$,and $K.E._{max} = 8 \times 10^{-3} \, J$.
Substituting these values: $\frac{1}{2} \times 0.1 \times \omega^2 \times (0.1)^2 = 8 \times 10^{-3}$.
$0.05 \times 0.01 \times \omega^2 = 8 \times 10^{-3} \implies 0.0005 \times \omega^2 = 0.008$.
$\omega^2 = \frac{0.008}{0.0005} = 16 \implies \omega = 4 \, rad/s$.
The initial phase $\phi = 45^o = \pi/4 \, rad$.
Thus,the equation of motion is $y = 0.1 \sin(\pm 4t + \pi/4)$.

(A) The fundamental frequency $f$ of an open organ pipe of length $l$ is given by $f = \frac{v}{2l}$,where $v$ is the velocity of sound in the air column.
As the temperature $T$ increases,the velocity of sound $v$ increases according to the relation $v = \sqrt{\frac{\gamma RT}{M}}$.
While the length $l$ of the pipe may increase slightly due to thermal expansion,the increase in the velocity of sound $v$ is much more significant.
Since $f \propto v$ and $f \propto \frac{1}{l}$,the increase in $v$ dominates the effect of the slight increase in $l$,leading to an overall increase in the fundamental frequency $f$.
Therefore,both the Assertion and the Reason are correct,and the Reason provides a correct explanation for the Assertion.

(A) Given the formula $X = 3 Y Z^{2}$.
We know the dimensions of capacitance $C$ in $MKSQ$ system are $[X] = [C] = [M^{-1} L^{-2} T^{2} Q^{2}]$.
We know the dimensions of magnetic induction $B$ in $MKSQ$ system are $[Z] = [B] = [M T^{-1} Q^{-1}]$.
Rearranging the formula for $Y$,we get $Y = \frac{X}{3 Z^{2}}$.
Ignoring the dimensionless constant $3$,the dimensional formula is $[Y] = \frac{[X]}{[Z^{2}]}$.
Substituting the dimensions: $[Y] = \frac{[M^{-1} L^{-2} T^{2} Q^{2}]}{[M T^{-1} Q^{-1}]^{2}}$.
$[Y] = \frac{[M^{-1} L^{-2} T^{2} Q^{2}]}{[M^{2} T^{-2} Q^{-2}]}$.
$[Y] = [M^{-1-2} L^{-2} T^{2-(-2)} Q^{2-(-2)}] = [M^{-3} L^{-2} T^{4} Q^{4}]$.

(B) The density $\rho$ is given by $\rho = \frac{M}{V}$,where $M$ is the mass and $V$ is the volume.
Taking the logarithmic derivative,we get $\frac{d\rho}{\rho} = -\frac{dV}{V}$.
The bulk modulus $K$ is defined as $K = -\frac{P}{dV/V}$,which implies $-\frac{dV}{V} = \frac{P}{K}$.
Substituting this into the density relation,we get $\frac{d\rho}{\rho} = \frac{P}{K}$.
Therefore,the magnitude of the increase in density is $d\rho = \frac{\rho P}{K}$.

(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$,where $\Delta Q$ is the heat exchanged,$\Delta U$ is the change in internal energy,and $\Delta W$ is the work done by the system.
In an adiabatic process,there is no exchange of heat with the surroundings,so $\Delta Q = 0$.
Substituting this into the first law equation: $0 = \Delta U + \Delta W$.
This simplifies to $\Delta W = -\Delta U$ or $\Delta U = -\Delta W$.
If we consider the magnitude of work done on the system $(W_{on} = -\Delta W)$,then $W_{on} = \Delta U$.
Thus,in an adiabatic process,the work done on the system is equal to the change in internal energy of the system.

(C) From the figure,the coordinates $(x, y)$ and masses $m$ are:
$m_1 = 1 \text{ kg}$ at $(0, 0)$
$m_2 = 2 \text{ kg}$ at $(2, 0)$
$m_3 = 3 \text{ kg}$ at $(0, 2)$
$m_4 = 4 \text{ kg}$ at $(2, 2)$
$m_5 = 5 \text{ kg}$ at $(1, 1)$
The total mass $M = 1 + 2 + 3 + 4 + 5 = 15 \text{ kg}$.
The $x$-coordinate of the centre of mass is:
$x_{cm} = \frac{\sum m_i x_i}{M} = \frac{(1 \times 0) + (2 \times 2) + (3 \times 0) + (4 \times 2) + (5 \times 1)}{15} = \frac{0 + 4 + 0 + 8 + 5}{15} = \frac{17}{15} \approx 1.13$
The $y$-coordinate of the centre of mass is:
$y_{cm} = \frac{\sum m_i y_i}{M} = \frac{(1 \times 0) + (2 \times 0) + (3 \times 2) + (4 \times 2) + (5 \times 1)}{15} = \frac{0 + 0 + 6 + 8 + 5}{15} = \frac{19}{15} \approx 1.27$
Thus,the coordinates are approximately $(1.1, 1.3)$. The correct option is $C$.

(C) The molar heat capacity at constant volume for a mixture is given by $C_{V, \text{mix}} = \frac{n_1 C_{V_1} + n_2 C_{V_2}}{n_1 + n_2} = \frac{13R}{6}$.
Given the ratio $n_1 : n_2 = 1 : 2$,let $n_1 = n$ and $n_2 = 2n$.
Substituting these values: $\frac{n C_{V_1} + 2n C_{V_2}}{n + 2n} = \frac{13R}{6}$.
$\frac{C_{V_1} + 2C_{V_2}}{3} = \frac{13R}{6} \implies C_{V_1} + 2C_{V_2} = \frac{13R}{2}$.
For monatomic gases,$C_V = 3R/2$. For diatomic gases,$C_V = 5R/2$.
If gas $1$ is monatomic $(C_{V_1} = 3R/2)$ and gas $2$ is diatomic $(C_{V_2} = 5R/2)$:
$3R/2 + 2(5R/2) = 3R/2 + 5R = 13R/2$.
This matches the condition. Thus,the first gas is monatomic $(He)$ and the second is diatomic $(N_2)$.

(B) For the system of three charges to be in equilibrium,the net force on each charge must be zero.
Let the two charges $Q$ be placed at points $A$ and $B$ separated by a distance $x$. The charge $q$ is placed at the midpoint $C$ (distance $x/2$ from both $A$ and $B$).
Consider the equilibrium of charge $Q$ at point $B$. The force exerted by charge $Q$ at $A$ on charge $Q$ at $B$ must be balanced by the force exerted by charge $q$ at $C$ on charge $Q$ at $B$.
$F_{AB} + F_{CB} = 0$
$\frac{1}{4\pi\varepsilon_0} \frac{Q^2}{x^2} + \frac{1}{4\pi\varepsilon_0} \frac{qQ}{(x/2)^2} = 0$
$\frac{Q^2}{x^2} + \frac{4qQ}{x^2} = 0$
$Q^2 + 4qQ = 0$
$4qQ = -Q^2$
$q = -\frac{Q}{4}$

(A) The strength of an electric field is represented by the density of electric field lines.
In the given figure,the electric field lines are closer together (denser) at point $A$ compared to point $B$.
Since the density of field lines is higher at $A$,the magnitude of the electric field at $A$ $(E_A)$ must be greater than the magnitude of the electric field at $B$ $(E_B)$.
Therefore,$E_A > E_B$.

(C) Let the nodes be labeled. The circuit consists of a central node connected to three outer nodes,each with a resistance $R$. The outer nodes form a triangle with resistances $R$ between them. Terminals $A$ and $B$ are connected to two of these outer nodes.
By analyzing the symmetry and redrawing the circuit,we can see that the path from $A$ to $B$ consists of two parallel branches.
One branch is the direct resistance $R$ between $A$ and $B$.
The other branch consists of the remaining resistances in series and parallel combinations.
Specifically,the equivalent resistance of this network between terminals $A$ and $B$ simplifies to $R$.

(B) In a Wheatstone bridge circuit,the condition for balance is given by the ratio of resistances $\frac{P}{Q} = \frac{R}{S}$.
This balance condition is independent of the positions of the cell and the galvanometer.
According to the principle of reciprocity in electrical networks,if the positions of the source (cell) and the detector (galvanometer) are interchanged,the current through the galvanometer remains the same for a given potential difference.
Therefore,the balance point of the Wheatstone bridge remains unchanged.

(B) The Curie temperature $(T_C)$ is a characteristic property of ferromagnetic materials.
At temperatures below $T_C$,the material exhibits ferromagnetic properties due to the alignment of magnetic domains.
As the temperature increases and reaches $T_C$,the thermal agitation becomes strong enough to overcome the exchange coupling that maintains the alignment of these domains.
Consequently,above the Curie temperature,the material loses its spontaneous magnetization and transitions into a paramagnetic state.
Therefore,the correct option is $(B)$.

(A) Let the real dip be $\phi$. The vertical component of the Earth's magnetic field is $B_V$ and the horizontal component is $B_H$. The real dip is given by $\tan \phi = \frac{B_V}{B_H}$.
When the magnet is suspended at an angle $\beta = 30^o$ to the magnetic meridian,the effective horizontal component becomes $B_H' = B_H \cos \beta$.
The apparent dip $\phi'$ is given by $\tan \phi' = \frac{B_V}{B_H \cos \beta}$.
Given $\phi' = 45^o$ and $\beta = 30^o$,we have $\tan 45^o = \frac{B_V}{B_H \cos 30^o}$.
Since $\tan 45^o = 1$ and $\cos 30^o = \frac{\sqrt{3}}{2}$,we get $1 = \frac{B_V}{B_H (\sqrt{3}/2)}$.
This simplifies to $\frac{B_V}{B_H} = \frac{\sqrt{3}}{2}$.
Therefore,$\tan \phi = \frac{\sqrt{3}}{2}$,which implies $\phi = \tan^{-1}\left(\frac{\sqrt{3}}{2}\right)$.

(D) The given current is $i = 5 \sin \left( 100 t - \frac{\pi}{2} \right)$ and the potential is $V = 200 \sin (100 t)$.
Comparing these with the standard forms $i = I_0 \sin(\omega t + \phi_1)$ and $V = V_0 \sin(\omega t + \phi_2)$,we get the phase of current $\phi_1 = -\frac{\pi}{2}$ and the phase of voltage $\phi_2 = 0$.
The phase difference $\phi$ between voltage and current is $\phi = \phi_2 - \phi_1 = 0 - \left( -\frac{\pi}{2} \right) = \frac{\pi}{2}$.
The average power consumption in an $ac$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.
Since $\phi = \frac{\pi}{2}$,the power factor $\cos \phi = \cos \left( \frac{\pi}{2} \right) = 0$.
Therefore,the power consumption $P = V_{rms} I_{rms} \times 0 = 0 \text{ watts}$.

(A) For the Brackett series,the wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$,where $n_1 = 4$ and $n_2 = 5, 6, 7, \dots$
The longest wavelength $(\lambda_{\max})$ corresponds to the transition from $n_2 = 5$ to $n_1 = 4$:
$\frac{1}{\lambda_{\max}} = R \left[ \frac{1}{4^2} - \frac{1}{5^2} \right] = R \left[ \frac{1}{16} - \frac{1}{25} \right] = R \left[ \frac{25 - 16}{400} \right] = \frac{9R}{400}$
$\lambda_{\max} = \frac{400}{9R}$
The shortest wavelength $(\lambda_{\min})$ corresponds to the transition from $n_2 = \infty$ to $n_1 = 4$:
$\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{4^2} - \frac{1}{\infty^2} \right] = R \left[ \frac{1}{16} - 0 \right] = \frac{R}{16}$
$\lambda_{\min} = \frac{16}{R}$
The ratio of the longest to shortest wavelength is:
$\frac{\lambda_{\max}}{\lambda_{\min}} = \frac{400/9R}{16/R} = \frac{400}{9 \times 16} = \frac{25}{9}$

(B) The nuclear fusion reaction is given by: ${}_1^2H + {}_1^2H \to {}_2^4He + Q$.
Total binding energy of the reactants: $2 \times (2 \times 1.1 \; MeV) = 4.4 \; MeV$.
Total binding energy of the product: $4 \times 7.1 \; MeV = 28.4 \; MeV$.
The energy released $Q$ is the difference between the total binding energy of the product and the reactants:
$Q = 28.4 \; MeV - 4.4 \; MeV = 24 \; MeV$.

(B) The output of the $OR$ gate $(G_1)$ is $(A + B)$.
The output of the $NAND$ gate $(G_2)$ is $\overline{A \cdot B}$.
These two outputs are fed into the $AND$ gate $(G_3)$,so the final output $Y$ is:
$Y = (A + B) \cdot (\overline{A \cdot B})$
Using De Morgan's theorem,$\overline{A \cdot B} = \bar{A} + \bar{B}$.
Substituting this into the equation:
$Y = (A + B) \cdot (\bar{A} + \bar{B})$
$Y = A \cdot \bar{A} + A \cdot \bar{B} + B \cdot \bar{A} + B \cdot \bar{B}$
Since $A \cdot \bar{A} = 0$ and $B \cdot \bar{B} = 0$,the expression simplifies to:
$Y = 0 + A \cdot \bar{B} + \bar{A} \cdot B + 0$
$Y = A \cdot \bar{B} + \bar{A} \cdot B$
This is the Boolean expression for an $XOR$ gate.

(A) In vacuum, all electromagnetic $(EM)$ waves travel with the same speed, which is the speed of light, $c \approx 3 \times 10^8 \ m/s$.
Since the frequency $(f)$ and wavelength $(\lambda)$ are related by the equation $c = f \lambda$, and different types of electromagnetic waves have different frequencies, they must have different wavelengths.
Therefore, radio waves and visible light have the same velocity but different wavelengths.

(B) Gauss's law states that the total electric flux $\phi_E$ through any closed surface (a Gaussian surface) is equal to $\frac{1}{\varepsilon_0}$ times the net charge $Q_{enclosed}$ enclosed by the surface.
Mathematically,it is expressed as $\phi_E = \oint_S \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\varepsilon_0}$.
Therefore,option $B$ is the correct statement.

(C) The number of electrons entering the emitter is $n_E = 10^{10}$.
Since $4\%$ of the electrons are lost in the base, the number of electrons reaching the collector is $n_C = n_E - (0.04 \times n_E) = 0.96 \times n_E$.
The emitter current is given by $I_E = \frac{n_E \times e}{t}$, where $e$ is the electronic charge and $t$ is the time.
The collector current is given by $I_C = \frac{n_C \times e}{t} = \frac{0.96 \times n_E \times e}{t}$.
The current transfer ratio $\alpha$ is defined as the ratio of collector current to emitter current:
$\alpha = \frac{I_C}{I_E} = \frac{n_C}{n_E} = \frac{0.96 \times n_E}{n_E} = 0.96$.

(B) Given:
Incident wavelength,$\lambda = 200 \, nm$
Work function,$\phi_{0} = 5.01 \, eV$
According to Einstein's photoelectric equation:
$K_{max} = E - \phi_{0}$
$e V_{s} = \frac{hc}{\lambda} - \phi_{0}$
Using the relation $hc \approx 1240 \, eV \cdot nm$:
$e V_{s} = \frac{1240 \, eV \cdot nm}{200 \, nm} - 5.01 \, eV$
$e V_{s} = 6.2 \, eV - 5.01 \, eV = 1.19 \, eV \approx 1.2 \, eV$
Thus,the stopping potential $V_{s} = 1.2 \, V$.
The potential difference that must be applied to stop the photoelectrons is the negative of the stopping potential,which is $-1.2 \, V$.

(D) Given: $\lambda = 600 \, nm = 600 \times 10^{-9} \, m$,slit width $a = 1 \, mm = 10^{-3} \, m$,and distance $D = 2 \, m$.
The distance between the first dark fringes on either side of the central bright fringe is equal to the width of the central maximum.
The formula for the width of the central maximum is given by $w = \frac{2 \lambda D}{a}$.
Substituting the values:
$w = \frac{2 \times (600 \times 10^{-9} \, m) \times 2 \, m}{10^{-3} \, m}$
$w = \frac{2400 \times 10^{-9}}{10^{-3}} \, m$
$w = 2400 \times 10^{-6} \, m = 2.4 \times 10^{-3} \, m = 2.4 \, mm$.

(B) For a system of two identical positive charges,the electric field lines originate from each charge and repel each other.
Equipotential surfaces are surfaces where the electric potential is constant.
Near each individual charge,the equipotential surfaces are approximately spherical.
As we move further away from the charges,the surfaces become distorted and eventually merge to form a single,larger,roughly oval-shaped surface that encloses both charges.
Figure $B$ correctly depicts this behavior,where the inner surfaces are centered around each charge,and the outer surfaces encompass both charges,reflecting the repulsive nature of the two positive charges.

(D) At point $A$,by Snell's law:
$1 \cdot \sin 45^{\circ} = \mu \cdot \sin r$
$\sin r = \frac{1}{\mu \sqrt{2}}$ ..... $(i)$
At point $B$,for total internal reflection to occur at the vertical face,the angle of incidence $i_1$ must be greater than or equal to the critical angle $C$.
$\sin i_1 = \frac{1}{\mu}$
From the geometry of the triangle,$i_1 = 90^{\circ} - r$.
Therefore,$\sin(90^{\circ} - r) = \frac{1}{\mu} \Rightarrow \cos r = \frac{1}{\mu}$ ..... $(ii)$
Using the identity $\sin^2 r + \cos^2 r = 1$:
$\left(\frac{1}{\mu \sqrt{2}}\right)^2 + \left(\frac{1}{\mu}\right)^2 = 1$
$\frac{1}{2\mu^2} + \frac{1}{\mu^2} = 1$
$\frac{1 + 2}{2\mu^2} = 1$
$\frac{3}{2\mu^2} = 1 \Rightarrow \mu^2 = \frac{3}{2}$
$\mu = \sqrt{\frac{3}{2}}$

(D) The torque on a current-carrying coil in a uniform magnetic field is given by $\vec{\tau} = \vec{m} \times \vec{B}$,where $\vec{m}$ is the magnetic moment and $\vec{B}$ is the magnetic field.
The coil experiences zero torque when the magnetic moment $\vec{m}$ is parallel to the magnetic field $\vec{B}$.
The magnetic moment vector $\vec{m}$ is always perpendicular to the plane of the coil.
For $\vec{m}$ to be parallel to $\vec{B}$,the plane of the coil must be perpendicular to the magnetic field $\vec{B}$.

(D) According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is given by $e = -\frac{d\phi}{dt}$.
Given $\phi = 6t^2 - 5t + 1$,we differentiate with respect to $t$:
$\frac{d\phi}{dt} = \frac{d}{dt}(6t^2 - 5t + 1) = 12t - 5$.
The induced current $i$ is given by $i = \frac{|e|}{R} = \frac{1}{R} |\frac{d\phi}{dt}|$.
Given resistance $R = 10 \, \Omega$,at $t = 0.25 \, s$:
$|\frac{d\phi}{dt}| = |12(0.25) - 5| = |3 - 5| = |-2| = 2 \, Wb/s$.
Therefore,$i = \frac{2}{10} = 0.2 \, A$.

(D) For any closed surface enclosing a net charge of $0$, the total electric flux through the surface is $0$ according to Gauss's Law $(\Phi = \oint E \cdot dA = q_{enclosed} / \epsilon_0 = 0)$.
This means the number of electric field lines entering the surface must exactly equal the number of lines leaving the surface.
Therefore, the Assertion is incorrect because the number of lines terminating on the sphere cannot be more than those emerging from it.
The Reason is correct because, due to electrostatic induction, the side of the conductor closer to the positive point charge develops a negative induced charge, which has the highest surface charge density in magnitude.
Since the Assertion is incorrect and the Reason is correct, the correct choice is $D$ (based on standard logic for this specific question type).

(C) When two conductors are connected by a conducting wire,they reach the same electric potential.
Since the inner sphere $A$ is connected to the outer sphere $B$ by a copper wire,all the charge given to the system will reside on the outer surface of the outer sphere $B$ because the inner sphere is at a higher potential than the outer sphere if they were isolated,but the wire forces them to be at the same potential.
Effectively,the entire system acts as a single isolated spherical conductor of radius $b$.
The capacitance of an isolated spherical conductor of radius $R$ is given by $C = 4\pi \varepsilon_0 R$.
Therefore,the equivalent capacitance of this system is $C = 4\pi \varepsilon_0 b$.

(C) Electric potential $(V)$ at a point is defined as the work done per unit positive charge in bringing it from infinity to that point,given by $V = W/q$.
Electric potential energy $(U)$ is the total work done in assembling a system of charges,given by $U = qV$.
Since $U$ and $V$ have different physical dimensions ($[ML^2T^{-3}A^{-1}]$ for $V$ and $[ML^2T^{-2}A^{-1}]$ for $U$),they are distinct physical quantities.
The Assertion is correct.
The Reason states that $U = V$,which is dimensionally inconsistent and physically incorrect.
Therefore,the Assertion is correct but the Reason is incorrect.

(C) Kirchhoff's junction rule is based on the principle of conservation of charge,which states that the total charge entering a junction must equal the total charge leaving it.
Kirchhoff's loop rule is based on the principle of conservation of energy,which states that the algebraic sum of potential changes in any closed loop is zero.
Since the Assertion is correct and the Reason is incorrect,the correct option is $C$.

(C) The magnetic field $B$ produced by a solenoid is given by $B = \mu_0 n I$. When the current $I$ is reversed,the direction of the magnetic field $B$ is reversed,but its magnitude remains the same.
The magnetic field energy density $u$ is given by the formula $u = \frac{B^2}{2\mu_0}$.
Since the energy density depends on the square of the magnetic field magnitude $(B^2)$,the sign of $B$ does not affect the energy density.
Therefore,the total magnetic field energy stored in the solenoid,$U = u \times V$ (where $V$ is the volume),remains unchanged.
However,the Reason states that the energy density is proportional to the magnetic field,which is incorrect because it is proportional to the square of the magnetic field $(B^2)$.
Thus,the Assertion is correct,but the Reason is incorrect.

(B) The Assertion is correct because magnetic monopoles do not exist; breaking a magnet simply creates two smaller magnets,each with its own north and south pole.
The Reason is also correct. The magnetic moment $M$ of a bar magnet is given by $M = m \times 2l$,where $m$ is the pole strength and $2l$ is the length. When the magnet is cut into two equal pieces perpendicular to its length,the pole strength $m$ remains the same,but the length of each piece becomes $l$. Therefore,the new magnetic moment $M' = m \times l = M/2$.
Since the Reason explains why the magnetic properties change but does not explain why poles cannot be isolated,the correct option is $B$.

(C) Faraday's law,specifically Lenz's law,is a direct consequence of the law of conservation of energy. If this were not true,we could create energy out of nothing,which violates physical laws.
In a purely resistive $AC$ circuit,the voltage $(emf)$ and current are in the same phase. The phase difference between them is $0$. Therefore,the statement in the Reason is incorrect.
Thus,the Assertion is correct,but the Reason is incorrect.

(D) Large eddy currents are produced in a non-laminated iron core of a transformer by the induced $emf$,as the resistance of a bulk iron core is very small.
By using thin iron sheets as a core,the electrical resistance is increased.
Laminating the core substantially reduces the eddy currents.
Eddy currents heat up the core of the transformer,leading to energy loss in the form of heat.
Therefore,more eddy currents result in greater energy loss,which decreases the efficiency of the transformer.
Since both the Assertion and the Reason are false,the correct option is $D$.

(D) The Assertion is incorrect because optical waves (like those used in optical fibres) have a much higher frequency and bandwidth compared to microwaves,making them superior for signal transmission.
The Reason is also incorrect because all electromagnetic waves,including microwaves and optical waves,travel at the same speed $(c = 3 \times 10^8 \ m/s)$ in a vacuum.
Therefore,both the Assertion and the Reason are incorrect.

(B) plane mirror forms a real image when the object is virtual. $A$ virtual object is formed when converging light rays are incident on the mirror. The reflected rays then converge at a point in front of the mirror,forming a real image.
Conversely,if the object is real (diverging rays from a point source),the plane mirror forms a virtual image behind the mirror.
Therefore,the Assertion is correct because a plane mirror can form a real image if the incident rays are converging (virtual object).
The Reason is also correct because a plane mirror forms a virtual image for a real object.
However,the Reason does not explain why the Assertion is true; it describes a different case (real object vs. virtual object). Thus,both are correct,but the Reason is not the correct explanation of the Assertion.

(A) Diffraction is a general characteristic of all types of waves,including mechanical (like sound) and non-mechanical (like light),as well as transverse and longitudinal waves. Thus,the Assertion is correct.
For diffraction to be significant or perceptible,the size of the obstacle or aperture must be of the order of the wavelength of the wave. If the wavelength is much smaller than the obstacle,the wave behaves like a ray and diffraction is negligible. Thus,the Reason is also correct and explains why diffraction is observed in specific conditions. Therefore,the correct option is $A$.

(B) According to Einstein's photoelectric equation, $K_{max} = h\nu - \Phi$, where $h\nu$ is the energy of the incident photon and $\Phi$ is the work function of the metal.
While the maximum kinetic energy is fixed for a given frequency, photoelectrons are emitted from different depths within the metal surface.
Electrons emitted from deeper layers lose some energy due to collisions before escaping the surface, resulting in a range of kinetic energies from $0$ to $K_{max}$.
Thus, the Assertion is correct.
The work function $\Phi$ is indeed a characteristic property of the metal, making the Reason correct.
However, the Reason does not explain why there is a spread in kinetic energies; the spread is due to the varying depths of emission and subsequent energy loss.
Therefore, both are correct, but the Reason is not the correct explanation of the Assertion.

(B) According to classical physics,any charged particle undergoing acceleration must radiate electromagnetic energy. Since an electron revolving in an orbit around the nucleus is undergoing centripetal acceleration,classical electrodynamics predicts it should continuously lose energy and spiral into the nucleus.
To resolve this instability,Bohr postulated that electrons in specific 'stationary' orbits do not radiate energy.
Both the Assertion and the Reason are scientifically correct statements. However,the Reason describes the classical conflict,while the Assertion describes Bohr's specific quantum postulate to overcome that conflict. The Reason does not explain *why* Bohr's postulate is true; it only explains why the postulate was *necessary*. Therefore,the Reason is not the correct explanation of the Assertion.

(A) In an absorption transition,an electron in a lower energy state absorbs a photon and jumps to a higher energy state. For a system with energy levels $A, B, C$ (where $A < B < C$),absorption can only occur from the ground state $A$ to $B$ or $A$ to $C$. Thus,there are $2$ possible absorption transitions.
In an emission transition,an electron in a higher energy state drops to a lower energy state by emitting a photon. For the same levels,emission can occur from $C \rightarrow B$,$C \rightarrow A$,and $B \rightarrow A$. Thus,there are $3$ possible emission transitions.
Since $2 < 3$,the number of absorption transitions is less than the number of emission transitions. The Reason correctly explains that absorption is restricted to starting from the lower level,whereas emission can occur between any two levels where the initial state is higher than the final state.

(C) The Assertion is correct because energy is released in nuclear fission of heavy nuclei and nuclear fusion of light nuclei due to the increase in binding energy per nucleon of the product nuclei compared to the reactants.
The Reason is incorrect because the binding energy per nucleon curve shows that for heavy nuclei,the binding energy per nucleon decreases with increasing atomic number $Z$ (making them unstable and prone to fission),while for light nuclei,the binding energy per nucleon increases with increasing $Z$ (making them prone to fusion to reach a more stable state).

(A) Diode lasers are widely used in optical communication because they are compact,have high efficiency,and can be easily modulated at high frequencies.
They also consume significantly less energy compared to other light sources like gas lasers or incandescent bulbs.
Since the high efficiency and low energy consumption are key reasons for their preference in optical communication,the Reason correctly explains the Assertion.

(B) When a radioactive material decays through two different processes simultaneously,the effective decay constant $\lambda_{eff}$ is the sum of the individual decay constants: $\lambda_{eff} = \lambda_{\alpha} + \lambda_{\beta}$.
Given $T_{\alpha} = 16$ years and $T_{\beta} = 48$ years.
The decay constants are $\lambda_{\alpha} = \frac{\ln 2}{16}$ and $\lambda_{\beta} = \frac{\ln 2}{48}$.
Thus,$\lambda_{eff} = \frac{\ln 2}{16} + \frac{\ln 2}{48} = \ln 2 \left( \frac{3+1}{48} \right) = \frac{\ln 2}{12}$.
The effective half-life $T_{eff} = \frac{\ln 2}{\lambda_{eff}} = 12$ years.
We want to find the time $t$ when $3/4$th of the material has decayed,which means $1/4$th of the material remains.
Using the radioactive decay law: $\frac{N}{N_0} = \left( \frac{1}{2} \right)^{t/T_{eff}}$.
$\frac{1}{4} = \left( \frac{1}{2} \right)^{t/12} \implies \left( \frac{1}{2} \right)^2 = \left( \frac{1}{2} \right)^{t/12}$.
Equating the exponents: $2 = \frac{t}{12} \implies t = 24$ years.

(C) The de-Broglie wavelength $(\lambda)$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum.
Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Substituting this into the wavelength formula: $\lambda = \frac{h}{\sqrt{2mE}}$.
Given that the energy $E$ is the same for all particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.
Comparing the masses: $m_e < m_p \approx m_n < m_\alpha$.
Since $\lambda$ is inversely proportional to the square root of mass,the order of wavelengths will be $\lambda_\alpha < \lambda_n \approx \lambda_p < \lambda_e$.
Thus,the increasing order is $\lambda_\alpha, \lambda_n, \lambda_p, \lambda_e$.