AIPMT 2008 Physics Question Paper with Answer and Solution

(A) The dimension of energy density is given by: $\text{Energy density} = \frac{\text{Energy}}{\text{Volume}} = \frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}]$.
Young's modulus $(Y)$ is defined as: $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\text{Force/Area}}{\text{Change in length/Original length}}$.
Since strain is dimensionless,the dimensions of Young's modulus are the same as stress: $[Y] = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.
Refractive index and dielectric constant are dimensionless quantities.
Magnetic field has dimensions $[MT^{-2}A^{-1}]$.
Thus,energy density and Young's modulus have the same dimensions.

(C) The percentage error in the radius is given as $\frac{\Delta r}{r} \times 100 = 2\%$.
The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Using the rules of propagation of errors for powers,the relative error in volume is $\frac{\Delta V}{V} = 3 \times \frac{\Delta r}{r}$.
To find the percentage error,multiply both sides by $100$:
$\frac{\Delta V}{V} \times 100 = 3 \times (\frac{\Delta r}{r} \times 100)$.
Substituting the given value:
$\frac{\Delta V}{V} \times 100 = 3 \times 2\% = 6\%$.

(B) Given: Initial velocity $u = 10 \, m/s$,final velocity $v = 20 \, m/s$,and distance $s = 135 \, m$.
Using the equation of motion $v^2 - u^2 = 2as$:
$(20)^2 - (10)^2 = 2 \times a \times 135$
$400 - 100 = 270a$
$300 = 270a$
$a = \frac{300}{270} = \frac{10}{9} \, m/s^2$.
Now,using the equation $v = u + at$ to find $t$:
$20 = 10 + (\frac{10}{9})t$
$10 = (\frac{10}{9})t$
$t = 9 \, s$.

(A) The distance travelled by a particle in the $n^{th}$ second is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$.
Given,initial velocity $u = 0 \ m/s$,acceleration $a = \frac{4}{3} \ m/s^2$,and time $n = 3 \ s$.
Substituting these values into the formula:
$S_3 = 0 + \frac{4/3}{2}(2(3) - 1)$
$S_3 = \frac{4}{6}(6 - 1)$
$S_3 = \frac{2}{3}(5)$
$S_3 = \frac{10}{3} \ m$.
Therefore,the distance travelled in the third second is $\frac{10}{3} \ m$.

(A) The initial velocity of the particle is $\vec{u} = v \cos \theta \hat{i} + v \sin \theta \hat{j}$.
The final velocity of the particle when it lands on the ground is $\vec{v}_f = v \cos \theta \hat{i} - v \sin \theta \hat{j}$.
The initial momentum is $\vec{p}_i = m(v \cos \theta \hat{i} + v \sin \theta \hat{j})$.
The final momentum is $\vec{p}_f = m(v \cos \theta \hat{i} - v \sin \theta \hat{j})$.
The change in momentum is $\Delta \vec{p} = \vec{p}_f - \vec{p}_i = -2mv \sin \theta \hat{j}$.
The magnitude of the change in momentum is $|\Delta \vec{p}| = 2mv \sin \theta$.
Given $\theta = 45^\circ$,we have $\sin 45^\circ = \frac{1}{\sqrt{2}}$.
Therefore,$|\Delta \vec{p}| = 2mv \times \frac{1}{\sqrt{2}} = \sqrt{2}mv$.

(D) The instantaneous velocity of a particle is given by the slope of the distance-time graph,which is $v = \frac{ds}{dt}$.
In a distance-time graph,the slope is represented by the steepness of the curve at any given point.
By observing the figure,the curve is steepest at point $C$,meaning the slope is maximum at this point.
Therefore,the instantaneous velocity is maximum around point $C$.

(C) The force $F$ required to maintain the motion of the conveyor belt is given by the rate of change of momentum.
Since the velocity $v$ is constant,the force is given by $F = \frac{dp}{dt} = \frac{d}{dt}(mv)$.
Using the product rule,$F = v \frac{dm}{dt} + m \frac{dv}{dt}$.
Since the velocity $v$ is constant,$\frac{dv}{dt} = 0$.
Given that the rate of mass deposition is $\frac{dm}{dt} = M \ kg/s$,we substitute these values into the equation:
$F = v \cdot M + m \cdot 0 = Mv$.
Therefore,the force required to keep the conveyor belt moving at a constant velocity $v$ is $Mv \ N$.

(C) To have the resultant force only along the $y-$ direction,the net force along the $x-$ direction must be zero.
Let the additional force be $\vec{F}_{add} = F_x \hat{i} + F_y \hat{j}$.
The $x-$ components of the given forces are:
$F_{1x} = 1 \cos(60^{\circ}) = 1 \times 0.5 = 0.5 \ N$
$F_{2x} = 2 \cos(60^{\circ}) = 2 \times 0.5 = 1.0 \ N$
$F_{4x} = -4 \sin(30^{\circ}) = -4 \times 0.5 = -2.0 \ N$
Sum of $x-$ components = $0.5 + 1.0 - 2.0 = -0.5 \ N$.
To make the net $x-$ component zero,we need an additional force $F_x$ such that $-0.5 + F_x = 0$,which gives $F_x = 0.5 \ N$.
The magnitude of the minimum additional force is $0.5 \ N$.

(C) At the top of the hill, the forces acting on the rider are the gravitational force $(mg)$ acting downwards and the normal reaction $(N)$ acting upwards.
The centripetal force required for circular motion is provided by the net force towards the center: $mg - N = \frac{mv^2}{R}$.
For the condition of "weightlessness", the normal reaction $N$ must be zero.
Therefore, $mg = \frac{mv^2}{R}$.
Solving for velocity $v$: $v = \sqrt{Rg}$.
Given $R = 20\, m$ and taking $g = 10\, m/s^2$, we have:
$v = \sqrt{20 \times 10} = \sqrt{200} \approx 14.14\, m/s$.
Thus, the speed is between $14\, m/s$ and $15\, m/s$.

(C) The rate of mass flow of water is $\frac{dm}{dt} = 15\, kg/s$.
The height is $h = 60\, m$ and acceleration due to gravity is $g = 10\, m/s^2$.
The total potential energy available per second (input power) is $P_{in} = \frac{dm}{dt} \cdot g \cdot h = 15 \times 10 \times 60 = 9000\, W$.
The losses due to friction are $10\%$,so the efficiency of the turbine is $90\%$.
The power generated by the turbine is $P_{out} = P_{in} \times 0.90 = 9000 \times 0.90 = 8100\, W$.
Converting to kilowatts,$P_{out} = 8.1\, kW$.

(C) Let $m = 0.2\, kg$ be the mass of the shell and $M = 4\, kg$ be the mass of the gun. Let $v$ be the velocity of the shell and $V$ be the recoil velocity of the gun.
By the law of conservation of momentum,$mv = MV$,so $V = (m/M)v$.
The total energy generated by the explosion is equal to the sum of the kinetic energies of the shell and the gun:
$E = \frac{1}{2}mv^2 + \frac{1}{2}MV^2$
Substituting $V = (m/M)v$:
$E = \frac{1}{2}mv^2 + \frac{1}{2}M(\frac{m}{M}v)^2 = \frac{1}{2}mv^2 (1 + \frac{m}{M})$
Given $E = 1.05\, kJ = 1050\, J$,$m = 0.2\, kg$,and $M = 4\, kg$:
$1050 = \frac{1}{2} \times 0.2 \times v^2 \times (1 + \frac{0.2}{4})$
$1050 = 0.1 \times v^2 \times (1 + 0.05)$
$1050 = 0.1 \times 1.05 \times v^2$
$1050 = 0.105 \times v^2$
$v^2 = \frac{1050}{0.105} = 10000$
$v = 100\, m/s$.

(B) From the ideal gas equation,$PV = nRT = \frac{m}{M}RT$,where $m$ is the mass and $M$ is the molar mass.
Density $\rho = \frac{m}{V} = \frac{PM}{RT}$.
The ratio of density to pressure is $\frac{\rho}{P} = \frac{M}{RT}$.
Since $M$ and $R$ are constants,$\frac{\rho}{P} \propto \frac{1}{T}$.
Let the ratio at $T_1 = 10^{\circ} C = 283 \ K$ be $x_1 = x$.
Let the ratio at $T_2 = 110^{\circ} C = 383 \ K$ be $x_2$.
Since $x \cdot T = \text{constant}$,we have $x_1 T_1 = x_2 T_2$.
$x \cdot 283 = x_2 \cdot 383$.
Therefore,$x_2 = \left( \frac{283}{383} \right)x$.

(D) The rod is bent at its midpoint into two halves,each of length $l = L/2$ and mass $m = M/2$.
Each half acts as a thin rod of length $L/2$ and mass $M/2$ rotating about an axis passing through one of its ends.
The moment of inertia of a thin rod of mass $m$ and length $l$ about an axis passing through one end and perpendicular to its length is given by $I = \frac{1}{3}ml^2$.
Here,$m = M/2$ and $l = L/2$.
So,the moment of inertia of one half about the bending point $O$ is $I_{half} = \frac{1}{3} \times (M/2) \times (L/2)^2 = \frac{1}{3} \times \frac{M}{2} \times \frac{L^2}{4} = \frac{ML^2}{24}$.
Since the two halves are identical and rotate about the same axis passing through $O$,the total moment of inertia is $I_{total} = I_{half} + I_{half} = 2 \times \frac{ML^2}{24} = \frac{ML^2}{12}$.

(A) The moment of inertia $(I)$ of a circular disc about its central axis is $I_{disc} = \frac{1}{2}MR^2$. The radius of gyration $(k)$ is defined by $I = Mk^2$. Thus,$k_{disc} = \sqrt{\frac{I}{M}} = \sqrt{\frac{R^2}{2}} = \frac{R}{\sqrt{2}}$.
The moment of inertia $(I)$ of a circular ring about its central axis is $I_{ring} = MR^2$. Thus,$k_{ring} = \sqrt{\frac{I}{M}} = \sqrt{R^2} = R$.
The ratio of the radius of gyration of the disc to that of the ring is $\frac{k_{disc}}{k_{ring}} = \frac{R/\sqrt{2}}{R} = \frac{1}{\sqrt{2}}$.
Therefore,the ratio is $1 : \sqrt{2}$.

(D) The relationship between two linear temperature scales is given by the formula: $\frac{T_W - \text{Freezing point}_W}{\text{Boiling point}_W - \text{Freezing point}_W} = \frac{T_C - \text{Freezing point}_C}{\text{Boiling point}_C - \text{Freezing point}_C}$.
Given for the $W$ scale: Freezing point = $39\,^{\circ}W$,Boiling point = $239\,^{\circ}W$.
Given for the Celsius scale: Freezing point = $0\,^{\circ}C$,Boiling point = $100\,^{\circ}C$.
Substituting the values:
$\frac{T_W - 39}{239 - 39} = \frac{39 - 0}{100 - 0}$
$\frac{T_W - 39}{200} = \frac{39}{100}$
$T_W - 39 = \frac{39}{100} \times 200$
$T_W - 39 = 39 \times 2 = 78$
$T_W = 78 + 39 = 117\,^{\circ}W$.

(C) According to the first law of thermodynamics,$Q = \Delta U + W$,where $Q$ is the heat added,$\Delta U$ is the change in internal energy,and $W$ is the work done.
Internal energy is a state function,meaning it depends only on the initial and final states of the system.
In a closed cyclic process,the system returns to its initial state after completing the cycle.
Since the initial and final states are identical,the change in internal energy $\Delta U$ (denoted here as $E$) must be zero.
Therefore,$E = 0$.

(D) The angular frequencies are given as $\omega_{1} = 100 \, rad \, s^{-1}$ and $\omega_{2} = 1000 \, rad \, s^{-1}$.
Let the common displacement amplitude be $A$.
The maximum acceleration $a_{max}$ for a simple harmonic motion is given by the formula $a_{max} = \omega^2 A$.
For the first motion,$a_{max1} = \omega_{1}^2 A = (100)^2 A$.
For the second motion,$a_{max2} = \omega_{2}^2 A = (1000)^2 A$.
The ratio of their maximum accelerations is $\frac{a_{max1}}{a_{max2}} = \frac{\omega_{1}^2 A}{\omega_{2}^2 A} = \frac{\omega_{1}^2}{\omega_{2}^2}$.
Substituting the values,we get $\frac{a_{max1}}{a_{max2}} = \frac{(100)^2}{(1000)^2} = \frac{10000}{1000000} = \frac{1}{100}$.
Thus,the ratio is $1:10^2$.

(A) The given wave equation is $y = 0.25 \sin(10\pi x - 2\pi t)$.
Comparing this with the standard wave equation $y = A \sin(kx - \omega t)$:
$1$. The amplitude $A = 0.25 \text{ m}$.
$2$. The wave number $k = 10\pi$. Since $k = \frac{2\pi}{\lambda}$,we have $\frac{2\pi}{\lambda} = 10\pi$,which gives $\lambda = 0.2 \text{ m}$.
$3$. The angular frequency $\omega = 2\pi$. Since $\omega = 2\pi f$,we have $2\pi f = 2\pi$,which gives $f = 1 \text{ Hz}$.
$4$. Since the sign between the $kx$ and $\omega t$ terms is negative,the wave is travelling in the positive $x$-direction.
Therefore,the wave travels in the $+ve$ $x$-direction with frequency $1 \text{ Hz}$ and wavelength $\lambda = 0.2 \text{ m}$.

(B) The intensity $I$ of a wave is proportional to the square of its amplitude $A$,i.e.,$I = kA^2$,where $k$ is a constant. Thus,$A = \sqrt{I/k}$.
When two waves superpose,the maximum and minimum amplitudes are given by $A_{\max} = A_1 + A_2$ and $A_{\min} = |A_1 - A_2|$.
The maximum intensity is $I_{\max} = k(A_1 + A_2)^2 = k(A_1^2 + A_2^2 + 2A_1A_2) = I_1 + I_2 + 2\sqrt{I_1I_2}$.
The minimum intensity is $I_{\min} = k(A_1 - A_2)^2 = k(A_1^2 + A_2^2 - 2A_1A_2) = I_1 + I_2 - 2\sqrt{I_1I_2}$.
Adding the maximum and minimum intensities:
$I_{\max} + I_{\min} = (I_1 + I_2 + 2\sqrt{I_1I_2}) + (I_1 + I_2 - 2\sqrt{I_1I_2}) = 2(I_1 + I_2)$.

(B) The equation of motion is $x = A \sin(\omega t + \frac{\pi}{6})$.
The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = A\omega \cos(\omega t + \frac{\pi}{6})$.
The maximum velocity $v_{max}$ is $A\omega$.
We are given that the velocity is half of its maximum velocity: $v = \frac{v_{max}}{2}$.
Substituting the expressions: $A\omega \cos(\omega t + \frac{\pi}{6}) = \frac{A\omega}{2}$.
This simplifies to $\cos(\omega t + \frac{\pi}{6}) = \frac{1}{2}$.
Since $\cos(60^{\circ}) = \frac{1}{2}$,we have $\omega t + \frac{\pi}{6} = \frac{\pi}{3}$.
Using $\omega = \frac{2\pi}{T}$,we get $\frac{2\pi}{T} t + \frac{\pi}{6} = \frac{\pi}{3}$.
Solving for $t$: $\frac{2\pi}{T} t = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.
Therefore,$t = \frac{\pi}{6} \times \frac{T}{2\pi} = \frac{T}{12}$.

(D) Given:
Distance of first point,$x_1 = 10\;m$
Distance of second point,$x_2 = 15\;m$
Period of oscillation,$T = 0.05\;s$
Velocity of wave,$v = 300\;m/s$
Path difference between the two points is:
$\Delta x = x_2 - x_1 = 15 - 10 = 5\;m$
First,calculate the wavelength $\lambda$ using the formula $v = f \lambda$,where $f = \frac{1}{T}$:
$f = \frac{1}{0.05} = 20\;Hz$
$\lambda = \frac{v}{f} = \frac{300}{20} = 15\;m$
The phase difference $\Delta \phi$ is related to the path difference $\Delta x$ by the formula:
$\Delta \phi = \frac{2\pi}{\lambda} \times \Delta x$
Substituting the values:
$\Delta \phi = \frac{2\pi}{15} \times 5 = \frac{2\pi}{3}$

(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 $(B)$.

(C) The binding energy $B$ of a nucleus is defined as the energy equivalent of the mass defect $\Delta m$.
The mass defect is given by $\Delta m = (Z M_p + N M_n) - M(N, Z)$.
According to Einstein's mass-energy equivalence relation,$B = \Delta m c^2$.
Substituting the expression for $\Delta m$,we get $B = [Z M_p + N M_n - M(N, Z)] c^2$.
Rearranging the terms to solve for $M(N, Z)$:
$B / c^2 = Z M_p + N M_n - M(N, Z)$
$M(N, Z) = Z M_p + N M_n - B / c^2$.

(D) When two thin lenses of focal lengths $f_1$ and $f_2$ are placed in contact,the equivalent focal length $F$ of the combination is given by the formula:
$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$
Taking the common denominator,we get:
$\frac{1}{F} = \frac{f_2 + f_1}{f_1 f_2}$
The power $P$ of a lens is defined as the reciprocal of its focal length in meters,i.e.,$P = \frac{1}{F}$.
Therefore,the power of the combination is $P = \frac{f_1 + f_2}{f_1 f_2}$.

(C) The potential difference $V$ between the plates of a parallel plate capacitor in a uniform electric field $E$ is given by $V = Ed$.
The capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
The energy $U$ stored in a capacitor is given by the formula $U = \frac{1}{2} CV^2$.
Substituting the expressions for $C$ and $V$ into the energy formula:
$U = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \right) (Ed)^2$
$U = \frac{1}{2} \left( \frac{\varepsilon_0 A}{d} \right) E^2 d^2$
$U = \frac{1}{2} \varepsilon_0 E^2 Ad$.

(C) The electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by $V = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r}$.
Given $V = Q \times 10^{11} \, V$,we have $\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r} = Q \times 10^{11}$.
This simplifies to $\frac{1}{r} = 4 \pi \varepsilon_0 \times 10^{11}$.
The electric field $E$ at that point is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2} = \left( \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r} \right) \cdot \frac{1}{r}$.
Substituting the values,$E = (Q \times 10^{11}) \times (4 \pi \varepsilon_0 \times 10^{11})$.
Therefore,$E = 4 \pi \varepsilon_0 Q \times 10^{22} \, V/m$.

(D) The total electric field at the centre $O$ of a uniformly charged ring is zero.
Let the ring be divided into four equal parts: $AK$,$KB$,$BD$,and $DC$. Each part carries a charge of $+Q/4$.
The electric field at the centre $O$ due to the part $AKB$ is the vector sum of the fields due to $AK$ and $KB$. Let this resultant be $E$ directed along $KO$.
The part $ACDB$ consists of the segments $AC$,$CD$,and $DB$.
Due to symmetry,the electric field at $O$ due to the charge on part $AC$ is equal and opposite to the field due to the charge on part $BD$. Thus,they cancel each other out.
Therefore,the net electric field at $O$ due to the part $ACDB$ is equal to the electric field due to the part $CD$ alone.
Since the total field of the entire ring is zero,the field due to $AKB$ must be equal and opposite to the field due to $ACDB$.
If the field due to $AKB$ is $E$ directed along $KO$,then the field due to $ACDB$ must be $E$ directed along $OK$ to ensure the vector sum is zero.

(C) Let the initial length be $l$ and area be $A$. The initial resistance is $R = \rho \frac{l}{A}$.
When the wire is stretched by $10\%$,the new length $l' = l + 0.1l = 1.1l$.
Since the volume $V = Al$ remains constant,$A'l' = Al \Rightarrow A' = \frac{Al}{1.1l} = \frac{A}{1.1}$.
The new resistance $R' = \rho \frac{l'}{A'} = \rho \frac{1.1l}{A/1.1} = (1.1)^2 \rho \frac{l}{A} = 1.21 R$.
Specific resistance (resistivity) $\rho$ is a property of the material and depends only on the temperature,not on the dimensions of the wire.
Therefore,the new resistance becomes $1.21$ times the original,and the specific resistance remains the same.

(C) Power $P = V \times I = 220\, V \times 4\, A = 880\, W = 880\, J/s$.
Heat required to raise the temperature of $1\, kg$ $(1000\, g)$ of water from $20\, ^oC$ to $100\, ^oC$:
$Q = mc\Delta T$
$Q = 1000\, g \times 4.2\, J/g\, ^oC \times (100\, ^oC - 20\, ^oC)$
$Q = 1000 \times 4.2 \times 80 = 336,000\, J$.
Time taken $t = \frac{Q}{P} = \frac{336,000\, J}{880\, J/s} \approx 381.8\, s$.
Converting to minutes: $t = \frac{381.8}{60} \approx 6.36\, min \approx 6.3\, min$.

(C) Let $\varepsilon$ be the $EMF$ of the cell and $r$ be its internal resistance. Let $V/L$ be the potential gradient of the potentiometer wire.
When the cell is not short-circuited,the balancing length is $l_1 = 110 \, cm$. The $EMF$ is balanced as:
$\varepsilon = (V/L) \times 110$ ....$(i)$
When the cell is short-circuited through an external resistance $R = 10 \, \Omega$,the terminal voltage $V_t$ is balanced at $l_2 = 100 \, cm$. The terminal voltage is given by $V_t = \frac{\varepsilon R}{R + r}$.
$V_t = (V/L) \times 100$ ....$(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{\varepsilon}{V_t} = \frac{110}{100}$
$\frac{\varepsilon}{\varepsilon R / (R + r)} = \frac{11}{10}$
$\frac{R + r}{R} = 1.1$
$1 + \frac{r}{R} = 1.1$
$\frac{r}{R} = 0.1$
$r = 0.1 \times R = 0.1 \times 10 \, \Omega = 1 \, \Omega$.

(B) The initial total resistance is $R_{total,1} = R_G + R_1 = 50\, \Omega + 2950\, \Omega = 3000\, \Omega$.
The current for full-scale deflection ($30$ divisions) is $I_1 = \frac{V}{R_{total,1}} = \frac{3\, V}{3000\, \Omega} = 1 \times 10^{-3}\, A = 1\, mA$.
To reduce the deflection to $20$ divisions,the new current $I_2$ must be proportional to the divisions: $I_2 = I_1 \times \frac{20}{30} = 1\, mA \times \frac{2}{3} = \frac{2}{3}\, mA$.
Using Ohm's law for the new circuit,$V = I_2 \times R_{total,2}$,where $R_{total,2} = R_G + R_2$.
$3\, V = (\frac{2}{3} \times 10^{-3}\, A) \times R_{total,2}$.
$R_{total,2} = \frac{3 \times 3}{2} \times 10^3\, \Omega = 4500\, \Omega$.
Since $R_{total,2} = R_G + R_2$,the required series resistance is $R_2 = 4500\, \Omega - 50\, \Omega = 4450\, \Omega$.

(A) When a charged particle enters a uniform magnetic field perpendicular to its velocity,it experiences a magnetic Lorentz force given by $\vec{F} = Q(\vec{v} \times \vec{B})$.
Since the force is always perpendicular to the velocity of the particle,the magnetic field does no work on the particle $(W = \int \vec{F} \cdot d\vec{r} = 0)$.
According to the work-energy theorem,the change in kinetic energy is equal to the work done by the net force.
Since the work done is zero,the kinetic energy of the particle remains constant throughout its motion.
Therefore,after $3$ $seconds$,the kinetic energy remains $K$.

(A) For a closed loop carrying a current in a uniform magnetic field,the net magnetic force on the loop is zero.
Let the force on segment $QP$ be $F_4$.
Since the loop is in equilibrium under the influence of these forces,the vector sum of all forces must be zero:
$\vec{F_1} + \vec{F_2} + \vec{F_3} + \vec{F_4} = 0$
$\vec{F_4} = -(\vec{F_1} + \vec{F_2} + \vec{F_3})$
Resolving the forces into horizontal and vertical components:
Let the horizontal direction be along the $x$-axis (positive towards the right) and the vertical direction be along the $y$-axis (positive upwards).
$F_{4x} = -(F_{3x} + F_{2x} + F_{1x}) = -(F_3 + 0 - F_1) = F_1 - F_3$
$F_{4y} = -(F_{3y} + F_{2y} + F_{1y}) = -(0 - F_2 + 0) = F_2$
The magnitude of the force $F_4$ is given by:
$F_4 = \sqrt{F_{4x}^2 + F_{4y}^2} = \sqrt{(F_1 - F_3)^2 + F_2^2} = \sqrt{(F_3 - F_1)^2 + F_2^2}$

(C) The angular diameter of the sun as seen from the earth is $\alpha = \frac{\text{Diameter of Sun}}{\text{Distance of Sun}} = \frac{1.39 \times 10^9}{1.5 \times 10^{11}} \approx 9.26 \times 10^{-3}\, \text{radians}$.
When the sunlight is focused by a lens of focal length $f = 10\, cm = 0.1\, m$,the image of the sun is formed at the focus.
The diameter of the image $d$ is given by $d = f \times \alpha$.
Substituting the values: $d = 0.1\, m \times (\frac{1.39 \times 10^9}{1.5 \times 10^{11}})$.
$d = 0.1 \times 0.926 \times 10^{-2} = 0.0926 \times 10^{-2} = 9.26 \times 10^{-4}\, m$.
Rounding to the nearest provided option,the diameter is $9.2 \times 10^{-4}\, m$.

(C) The work function is given by $\Phi = 6.2\, eV$.
The stopping potential is $V_s = 5\, V$,so the maximum kinetic energy is $K_{\max} = e V_s = 5\, eV$.
According to Einstein's photoelectric equation,the energy of the incident photon is $E = \Phi + K_{\max} = 6.2\, eV + 5\, eV = 11.2\, eV$.
The wavelength $\lambda$ is given by $\lambda = \frac{hc}{E}$. Using $hc \approx 12400\, eV\cdot\mathring{A}$,we get $\lambda = \frac{12400}{11.2} \approx 1107\, \mathring{A}$.
Since the wavelength $1107\, \mathring{A}$ falls in the range of $100\, \mathring{A}$ to $4000\, \mathring{A}$,it lies in the ultraviolet region.

(C) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Since the wavelengths are equal,we have $\frac{h}{m_p v_p} = \frac{h}{m_e v_e}$.
This simplifies to $m_p v_p = m_e v_e$.
Given: mass of particle $m_p = 1 \, mg = 10^{-6} \, kg$,mass of electron $m_e = 9.1 \times 10^{-31} \, kg$,and velocity of electron $v_e = 3 \times 10^6 \, m/s$.
Substituting the values: $10^{-6} \, kg \times v_p = 9.1 \times 10^{-31} \, kg \times 3 \times 10^6 \, m/s$.
$v_p = \frac{27.3 \times 10^{-25}}{10^{-6}} \, m/s = 27.3 \times 10^{-19} \, m/s = 2.73 \times 10^{-18} \, m/s$.
Rounding to the nearest option,the velocity of the particle is $2.7 \times 10^{-18} \, m/s$.

(A) The self-inductance $L$ of a solenoid is defined by the relation $\Phi_{total} = L \cdot I$,where $\Phi_{total}$ is the total magnetic flux linked with the solenoid.
The total magnetic flux linked with the solenoid is given by $\Phi_{total} = N \cdot \phi$,where $N$ is the number of turns and $\phi$ is the magnetic flux linked with each turn.
Given:
Number of turns $N = 500$
Current $I = 2 \, A$
Flux per turn $\phi = 4 \times 10^{-3} \, Wb$
Total flux $\Phi_{total} = 500 \times 4 \times 10^{-3} \, Wb = 2 \, Wb$.
Using the formula $L = \frac{\Phi_{total}}{I}$:
$L = \frac{2 \, Wb}{2 \, A} = 1 \, H$.
Therefore,the self-inductance of the solenoid is $1 \, H$.

(C) The magnetic flux $\phi$ is given by the formula $\phi = \vec{B} \cdot \vec{A} = BA \cos \theta$,where $\theta$ is the angle between the magnetic field vector $\vec{B}$ and the area vector $\vec{A}$ (which is normal to the surface).
Given,radius $R = 0.2 \; m$,so the area $A = \pi R^2 = \pi (0.2)^2 = 0.04\pi \; m^2$.
The magnetic field $B = \frac{1}{\pi} \; Wb/m^2$.
The axis of the disc makes an angle of $60^{\circ}$ with $\vec{B}$. Since the area vector $\vec{A}$ is along the axis of the disc,the angle $\theta$ between $\vec{B}$ and $\vec{A}$ is $60^{\circ}$.
Therefore,$\phi = B A \cos 60^{\circ} = \left(\frac{1}{\pi}\right) \times (0.04\pi) \times \cos 60^{\circ}$.
$\phi = 0.04 \times \frac{1}{2} = 0.02 \; Wb$.

(B) The instantaneous power $p$ in an $A.C.$ circuit is given by the product of instantaneous $e.m.f.$ and instantaneous current:
$p = E \cdot i = (E_o \sin \omega t) \cdot (I_o \sin(\omega t - \phi))$
Using the trigonometric identity $\sin A \sin B = \frac{1}{2} [\cos(A-B) - \cos(A+B)]$,we get:
$p = \frac{E_o I_o}{2} [\cos \phi - \cos(2\omega t - \phi)]$
The average power $P_{av}$ over one complete cycle $T$ is the average of the instantaneous power:
$P_{av} = \frac{1}{T} \int_0^T p \, dt = \frac{1}{T} \int_0^T \frac{E_o I_o}{2} [\cos \phi - \cos(2\omega t - \phi)] \, dt$
Since the average of $\cos(2\omega t - \phi)$ over a complete cycle is $0$,we have:
$P_{av} = \frac{E_o I_o}{2} \cos \phi$
Thus,the average power is $\frac{E_o I_o}{2} \cos \phi$.

(A) The velocity $v$ of electromagnetic radiation in a medium is given by the relation $v = \frac{1}{\sqrt{\mu \varepsilon}}$,where $\mu$ is the permeability and $\varepsilon$ is the permittivity of the medium.
For vacuum (or free space),the permeability is $\mu_0$ and the permittivity is $\varepsilon_0$.
Therefore,the velocity of electromagnetic radiation in vacuum is $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.

(A) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \, eV$.
For the ground state $(n=1)$,$E_1 = -13.6 \, eV$.
For the first excited state $(n=2)$,$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, eV$.
The excitation energy required to move the electron from the ground state to the first excited state is $\Delta E = E_2 - E_1$.
$\Delta E = -3.4 \, eV - (-13.6 \, eV) = 10.2 \, eV$.

(A) Let the initial number of nuclei for both materials be $N_0$.
The number of nuclei remaining at time $t$ is given by $N(t) = N_0 e^{-\lambda t}$.
For material $X_1$: $N_1(t) = N_0 e^{-(5\lambda)t}$.
For material $X_2$: $N_2(t) = N_0 e^{-\lambda t}$.
The ratio is given as $\frac{N_1(t)}{N_2(t)} = \frac{1}{e} = e^{-1}$.
Substituting the expressions: $\frac{N_0 e^{-5\lambda t}}{N_0 e^{-\lambda t}} = e^{-1}$.
$e^{-5\lambda t + \lambda t} = e^{-1}$.
$e^{-4\lambda t} = e^{-1}$.
Equating the exponents: $-4\lambda t = -1$.
Therefore,$t = \frac{1}{4\lambda}$.

(A) Let the inputs be $A$ and $B$.
$1$. The first gate is a $NOR$ gate,so its output is $X = \overline{A+B}$.
$2$. The second gate is a $NAND$ gate with both inputs tied together. $A$ $NAND$ gate with tied inputs acts as a $NOT$ gate. Thus,the output of the $NAND$ gate is $Z = \overline{X \cdot X} = \overline{X}$.
$3$. Substituting $X = \overline{A+B}$ into $Z$,we get $Z = \overline{\overline{A+B}} = A+B$.
$4$. The third gate is a $NOT$ gate,so its final output is $Y = \overline{Z} = \overline{A+B}$.
$5$. The expression $\overline{A+B}$ represents a $NOR$ gate.
Therefore,the circuit is equivalent to a $NOR$ gate.

(D) The energy of the photon required to excite an electron across the band gap is given by $E = h\nu$.
For the minimum frequency $\nu$,the energy of the photon must be equal to the band gap energy $E_g$.
Given $E_g = 2.0\, eV$.
We know that $1\, eV = 1.6 \times 10^{-19}\, J$,so $E_g = 2.0 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}\, J$.
Using the relation $E = h\nu$,we have $\nu = \frac{E_g}{h}$.
Substituting the values $h = 6.63 \times 10^{-34}\, J\cdot s$:
$\nu = \frac{3.2 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 0.482 \times 10^{15}\, Hz$.
$\nu \approx 4.82 \times 10^{14}\, Hz$.
Rounding this value,we get $\nu \approx 5 \times 10^{14}\, Hz$.

(B) The resistors $2 \, \Omega$,$4 \, \Omega$,and the series combination of $(1 \, \Omega + 5 \, \Omega)$ are connected in parallel. Therefore,the potential difference across each branch is the same.
The potential difference $V$ across the $2 \, \Omega$ resistor is given by $V = I \times R = 3 \, A \times 2 \, \Omega = 6 \, V$.
Since the branches are in parallel,the same potential difference $V = 6 \, V$ is applied across the branch containing the $1 \, \Omega$ and $5 \, \Omega$ resistors in series.
The total resistance of this branch is $R_{branch} = 1 \, \Omega + 5 \, \Omega = 6 \, \Omega$.
The current $i_3$ flowing through this branch is $i_3 = \frac{V}{R_{branch}} = \frac{6 \, V}{6 \, \Omega} = 1 \, A$.
The power dissipated in the $5 \, \Omega$ resistor is $P = i_3^2 \times R = (1 \, A)^2 \times 5 \, \Omega = 5 \, W$.

(B) The resistors $4\,\Omega$ and $3\,\Omega$ are in parallel. Since the potential difference across parallel branches is the same,the current $i_1$ through the $3\,\Omega$ resistor is given by $4\,\Omega \times 1\,A = 3\,\Omega \times i_1$,which gives $i_1 = \frac{4}{3}\,A$.
The total current flowing through the upper branch is $I_{upper} = 1\,A + \frac{4}{3}\,A = \frac{7}{3}\,A$.
The equivalent resistance of the $4\,\Omega$ and $3\,\Omega$ parallel combination is $R_{eq} = \frac{4 \times 3}{4 + 3} = \frac{12}{7}\,\Omega$.
The potential difference between points $P$ and $M$ is $V_{PM} = I_{upper} \times R_{eq} = \frac{7}{3}\,A \times \frac{12}{7}\,\Omega = 4\,V$.
The lower branch consists of two $0.5\,\Omega$ resistors in parallel,which is equivalent to $R_{lower1} = \frac{0.5 \times 0.5}{0.5 + 0.5} = 0.25\,\Omega$. This is in series with a $1\,\Omega$ resistor. The total resistance of the lower branch is $R_{lower} = 0.25\,\Omega + 1\,\Omega = 1.25\,\Omega$.
The current in the lower branch is $I_{lower} = \frac{V_{PM}}{R_{lower}} = \frac{4\,V}{1.25\,\Omega} = 3.2\,A$.
The potential difference between points $M$ and $N$ is the voltage drop across the $1\,\Omega$ resistor: $V_{MN} = I_{lower} \times 1\,\Omega = 3.2\,A \times 1\,\Omega = 3.2\,V$.

(D) The nuclear density $\rho$ is defined as the ratio of the mass of the nucleus to its volume.
Since the mass of a nucleus is approximately $M = A \cdot m_p$ (where $A$ is the mass number and $m_p$ is the mass of a nucleon) and the volume is $V = \frac{4}{3} \pi R^3$,where $R = R_0 A^{1/3}$.
Substituting the expression for $R$,we get $V = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$.
Thus,$\rho = \frac{A \cdot m_p}{\frac{4}{3} \pi R_0^3 A} = \frac{m_p}{\frac{4}{3} \pi R_0^3}$.
This expression shows that the nuclear density is independent of the mass number $A$.
Therefore,the ratio of the nuclear densities of any two nuclei is always $1: 1$.