KCET 2009 Physics Question Paper with Answer and Solution

(C) As the block $A$ moves with velocity $u = 0.15 \,ms^{-1}$, it compresses the spring, which pushes block $B$ towards the right. The spring continues to compress until the velocity of both blocks becomes equal. Let this common velocity be $v$.
According to the law of conservation of linear momentum:
$m_A u = (m_A + m_B) v$
$v = \frac{m_A u}{m_A + m_B} = \frac{2 \times 0.15}{2 + 3} = \frac{0.3}{5} = 0.06 \,ms^{-1}$
According to the law of conservation of energy, the initial kinetic energy of block $A$ is converted into the final kinetic energy of the system and the potential energy stored in the spring at maximum compression $x$:
$\frac{1}{2} m_A u^2 = \frac{1}{2} (m_A + m_B) v^2 + \frac{1}{2} k x^2$
$\frac{1}{2} \times 2 \times (0.15)^2 = \frac{1}{2} \times (2 + 3) \times (0.06)^2 + \frac{1}{2} \times 10.8 \times x^2$
$0.0225 = 5 \times 0.0018 + 5.4 x^2$
$0.0225 = 0.009 + 5.4 x^2$
$5.4 x^2 = 0.0135$
$x^2 = \frac{0.0135}{5.4} = 0.0025$
$x = \sqrt{0.0025} = 0.05 \,m$

(C) According to Kepler's second law of planetary motion,the areal velocity of a planet remains constant. This implies that the angular momentum of the planet is conserved.
Since the angular momentum $L = mvr \sin(\theta)$ is constant,where $m$ is the mass of the planet,$v$ is its linear speed,$r$ is the distance from the sun,and $\theta$ is the angle between the position vector and the velocity vector.
At the perihelion (the point closest to the sun),the distance $r$ is minimum.
Therefore,to conserve angular momentum,the linear speed $v$ must be maximum at this point.
In the given diagram,point $A$ is the closest to the sun.
Thus,the linear speed of the planet is maximum at point $A$.

(C) For a body to be in equilibrium under the action of three concurrent forces,the vector sum of the forces must be zero. This implies that the magnitude of any one force must be less than or equal to the sum of the other two forces and greater than or equal to the difference of the other two forces.
Let the forces be $F_1 = 1 \,N$,$F_2 = 2 \,N$,and $F_3 = 3 \,N$.
The condition for equilibrium is that the resultant of any two forces must be equal and opposite to the third force.
Here,$F_1 + F_2 = 1 + 2 = 3 \,N$,which is equal to $F_3$.
However,for the resultant of $F_1$ and $F_2$ to be $3 \,N$,they must act in the same direction (angle $\theta = 0^{\circ}$).
If they act in the same direction,they are not acting along 'different directions' as specified in the problem statement.
If the forces act along different directions,the resultant of $1 \,N$ and $2 \,N$ will always be less than $3 \,N$.
Therefore,the three forces cannot form a closed triangle,and the body cannot be in equilibrium.

(A) According to Newton's second law of motion,the net force acting on the body is given by $F_{net} = ma$.
Here,the forces acting on the body are the gravitational force $(mg)$ acting downwards and the air resistance force $(F_{air})$ acting upwards.
Since the body is falling downwards with acceleration $a$,the net force equation is:
$mg - F_{air} = ma$
Rearranging the equation to solve for the air resistance force $(F_{air})$:
$F_{air} = mg - ma = m(g - a)$
Given:
Mass $(m)$ = $0.05 \ kg$
Acceleration due to gravity $(g)$ = $9.8 \ ms^{-2}$
Acceleration of the body $(a)$ = $9.5 \ ms^{-2}$
Substituting the values:
$F_{air} = 0.05 \times (9.8 - 9.5)$
$F_{air} = 0.05 \times 0.3$
$F_{air} = 0.015 \ N$

(B) For the number $4.8000 \times 10^{4}$:
In scientific notation,the power of $10$ does not contribute to the number of significant figures.
The digits $4, 8, 0, 0, 0$ are all significant because trailing zeros after the decimal point are significant.
Thus,there are $5$ significant figures.
For the number $48000.50$:
All non-zero digits are significant.
Zeros between two non-zero digits are significant.
Trailing zeros after the decimal point are also significant.
Therefore,all digits $4, 8, 0, 0, 0, 5, 0$ are significant,totaling $7$ significant figures.

(D) Let $W_s$ and $W_a$ be the weights of steel and aluminium in air,and $V_s$ and $V_a$ be their respective volumes. The density of water is $\rho_w$. When immersed in water,the apparent weight is $W_{app} = W - V\rho_w g$. Given that the apparent weights are equal: $W_s - V_s \rho_w g = W_a - V_a \rho_w g$. Since $W = V \rho g$,we have $V = W / (\rho g)$. Substituting this,$W_s - (W_s / \rho_s) \rho_w g = W_a - (W_a / \rho_a) \rho_w g$,which simplifies to $W_s(1 - \rho_w / \rho_s) = W_a(1 - \rho_w / \rho_a)$. Since the density of steel $\rho_s \approx 7800 \ kg/m^3$ is much greater than the density of aluminium $\rho_a \approx 2700 \ kg/m^3$,the term $(1 - \rho_w / \rho_s)$ is greater than $(1 - \rho_w / \rho_a)$. Therefore,for the equality to hold,$W_s$ must be less than $W_a$. Thus,the aluminium piece weighs more in air.

(D) According to the equation of continuity for an incompressible fluid in streamline flow, the volume flow rate $(Q = Av)$ remains constant at all cross-sections of the tube.
Therefore, the rate of liquid flow is the same at both points $M$ and $N$.

(B) Let the total distance be $2S$. The first half distance $S$ is covered with speed $v_1 = 2 \,ms^{-1}$. The time taken is $t_1 = \frac{S}{2}$.
For the remaining half distance $S$, it is covered in two equal time intervals $t_2$ each, with speeds $v_2 = 3 \,ms^{-1}$ and $v_3 = 5 \,ms^{-1}$.
Thus, $S = v_2 t_2 + v_3 t_2 = (3 + 5) t_2 = 8 t_2$. Therefore, $t_2 = \frac{S}{8}$.
The total time taken for the second half is $2 t_2 = 2 \times \frac{S}{8} = \frac{S}{4}$.
Total time $T = t_1 + 2 t_2 = \frac{S}{2} + \frac{S}{4} = \frac{3S}{4}$.
Average speed = $\frac{\text{Total distance}}{\text{Total time}} = \frac{2S}{3S/4} = \frac{8}{3} \,ms^{-1}$.

(C) The moment of inertia of a circular ring of mass $M$ and radius $r$ about an axis passing through its center and perpendicular to its plane is $I_{z} = Mr^{2}$.
According to the perpendicular axis theorem,$I_{z} = I_{x} + I_{y}$.
Since the ring is symmetric about its diameter,$I_{x} = I_{y} = I_{diameter}$.
Therefore,$Mr^{2} = 2I_{diameter}$.
$I_{diameter} = \frac{Mr^{2}}{2}$.

(D) The torque $\tau$ required to open or close the door is constant and is given by the product of the force $F$ and the distance $d$ from the hinge: $\tau = F \times d$.
Given that a force of $1 \,N$ is applied at the free end $(d = 1.6 \,m)$, the torque is $\tau = 1 \,N \times 1.6 \,m = 1.6 \,N-m$.
To find the force $F'$ required at a distance $d' = 0.4 \,m$ from the hinges, we use the same torque: $F' = \frac{\tau}{d'} = \frac{1.6 \,N-m}{0.4 \,m} = 4 \,N$.

(A) Let the final temperature of the mixture be $t$.
Heat lost by water at $80^{\circ} C = m_1 s \Delta t_1 = (V_1 \rho) s (80^{\circ} - t)$.
Heat gained by water at $60^{\circ} C = m_2 s \Delta t_2 = (V_2 \rho) s (t - 60^{\circ})$.
According to the principle of calorimetry,Heat lost = Heat gained.
Since the density $\rho$ and specific heat $s$ are the same for both:
$V_1 (80^{\circ} - t) = V_2 (t - 60^{\circ})$.
Substituting the values:
$0.1 (80 - t) = 0.3 (t - 60)$.
$8 - 0.1t = 0.3t - 18$.
$26 = 0.4t$.
$t = \frac{26}{0.4} = 65^{\circ} C$.

(B) Thermal radiation consists of electromagnetic waves,which behave similarly to light.
When electromagnetic radiation travels from one medium to another,its speed and wavelength change due to the change in the refractive index of the medium.
However,the frequency of the radiation is a property of the source and remains constant regardless of the medium through which it travels.
Therefore,the statement that frequency changes is incorrect.

(A) According to Stefan-Boltzmann law,the energy radiated per unit area per unit time is proportional to the fourth power of the absolute temperature: $E \propto T^{4}$.
Let $E$ be the initial energy radiated at temperature $T$.
Let $E^{\prime}$ be the energy radiated at temperature $3T$.
Using the proportionality,we have $\frac{E^{\prime}}{E} = \left( \frac{3T}{T} \right)^{4}$.
$\frac{E^{\prime}}{E} = (3)^{4} = 81$.
Therefore,$E^{\prime} = 81 E$.

(B) The surface temperature of the stars is determined using Wien's displacement law.
According to this law,$\lambda_{m} T = b$,where $\lambda_{m}$ is the wavelength corresponding to the maximum spectral emissive power,$T$ is the absolute temperature of the black body,and $b$ is Wien's constant.
The value of Wien's constant is approximately $2.898 \times 10^{-3} \ m \cdot K$.
By measuring the wavelength $\lambda_{m}$ at which the star emits maximum radiation,the surface temperature $T$ can be calculated as $T = b / \lambda_{m}$.

(D) $(i)$ Curve $OA$ represents an isobaric process (since pressure is constant).
(ii) Curve $OB$ represents an isothermal process.
(iii) Curve $OC$ represents an adiabatic process,as the slope of an adiabatic process is steeper than that of an isothermal process.
(iv) Curve $OD$ represents an isochoric process (since volume is constant).

(A) Thermodynamic coordinates are the state variables that define the state of a thermodynamic system,such as pressure $(p)$,volume $(V)$,and temperature $(T)$.
$R$ is the universal gas constant,which is a fundamental constant of nature and not a variable that describes the state of a specific system.
Therefore,the gas constant $(R)$ is not a thermodynamic coordinate.

(B) The given equation of a progressive wave is $y=3 \sin \pi\left(\frac{t}{2}-\frac{x}{4}\right)$.
Comparing this with the standard wave equation $y=A \sin 2\pi \left(\frac{t}{T}-\frac{x}{\lambda}\right)$,we rewrite the given equation as $y=3 \sin 2\pi \left(\frac{t}{4}-\frac{x}{8}\right)$.
From this,we identify the time period $T=4 \,s$ and the wavelength $\lambda=8 \,m$.
The wave velocity $v$ is given by $v = \frac{\lambda}{T} = \frac{8}{4} = 2 \,m/s$.
The distance travelled by the wave in time $t=5 \,s$ is $s = v \times t = 2 \times 5 = 10 \,m$.

(A) The fundamental frequency of an open cylindrical tube of length $L$ is given by $n = \frac{v}{2L} = 390 \,Hz$.
When $\frac{1}{4}$th of the tube is immersed in water, the tube acts as a closed organ pipe (closed at one end) with a new length $L' = L - \frac{1}{4}L = \frac{3}{4}L$.
The fundamental frequency of a closed organ pipe is given by $n' = \frac{v}{4L'}$.
Substituting $L' = \frac{3}{4}L$, we get $n' = \frac{v}{4(\frac{3}{4}L)} = \frac{v}{3L}$.
We can rewrite this as $n' = \frac{2}{3} \times (\frac{v}{2L})$.
Since $\frac{v}{2L} = 390 \,Hz$, we have $n' = \frac{2}{3} \times 390 \,Hz = 260 \,Hz$.

(D) Sound waves are mechanical waves that propagate through a medium. As they travel,they cause the particles of the medium to oscillate,thereby transporting energy from one point to another. Because these waves involve the motion of particles with mass,they also carry momentum. Therefore,sound waves transfer both energy and momentum.

(D) For a spherical wave emitted by a point source,the intensity $I$ at a distance $r$ is given by $I = \frac{P}{4 \pi r^2}$,where $P$ is the power of the source.
Since $I \propto A^2$,where $A$ is the amplitude,we have $A^2 \propto \frac{1}{r^2}$,which implies $A \propto \frac{1}{r}$.
Let $A_P$ and $A_Q$ be the amplitudes at distances $r_P = 4 \ m$ and $r_Q = 9 \ m$ respectively.
Then,the ratio of amplitudes is $\frac{A_P}{A_Q} = \frac{r_Q}{r_P}$.
Substituting the given values,$\frac{A_P}{A_Q} = \frac{9}{4}$.

(NONE) The work done by a variable force is equal to the area under the $F-s$ graph.
To find the total work done for a distance of $20 \, m$, we calculate the area under the graph from $s = 0 \, m$ to $s = 20 \, m$.
This area consists of three parts:
$1$. $A$ triangle from $s = 0$ to $s = 4 \, m$ with base $4 \, m$ and height $10 \, N$: $\text{Area}_1 = \frac{1}{2} \times 4 \times 10 = 20 \, J$.
$2$. $A$ rectangle from $s = 4$ to $s = 15 \, m$ with width $11 \, m$ and height $10 \, N$: $\text{Area}_2 = 11 \times 10 = 110 \, J$.
$3$. $A$ trapezoid from $s = 15$ to $s = 20 \, m$ with parallel sides $10 \, N$ and $20 \, N$ and height $5 \, m$: $\text{Area}_3 = \frac{1}{2} \times (10 + 20) \times 5 = \frac{1}{2} \times 30 \times 5 = 75 \, J$.
Total work done $W = 20 + 110 + 75 = 205 \, J$.

(A) In an $RC$ series circuit, the voltage across the resistor $(V_R)$ and the voltage across the capacitor $(V_C)$ are out of phase by $90^{\circ}$.
Given: $V_R = 12 \,V$ and $V_C = 5 \,V$.
The total applied voltage $V$ is given by the phasor sum:
$V = \sqrt{V_R^2 + V_C^2}$
Substituting the values:
$V = \sqrt{(12)^2 + (5)^2}$
$V = \sqrt{144 + 25}$
$V = \sqrt{169}$
$V = 13 \,V$
Thus, the applied voltage is $13 \,V$.

(B) The spectral lines of the hydrogen atom are categorized based on the region of the electromagnetic spectrum in which they fall:
$1$. Lyman series: Ultraviolet region
$2$. Balmer series: Visible region
$3$. Paschen series: Infrared region
$4$. Brackett series: Infrared region
$5$. Pfund series: Infrared region
From this classification,it is clear that the Balmer series lies in the visible region of the electromagnetic spectrum.

(A) An incandescent electric lamp produces a continuous emission spectrum because the filament emits radiation across a wide range of wavelengths due to its high temperature.
Mercury and sodium vapour lamps produce line emission spectra because the atoms in the gas emit light only at specific discrete wavelengths.
Polyatomic substances,such as $H_{2}$,$CO_{2}$,and $KMnO_{4}$,typically produce band absorption spectra.

(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)$.
Here,the final state is the ground state,so $n_{f} = 1$,and the initial excited state is $n_{i} = n$.
Substituting these values,we get $\frac{1}{\lambda} = R \left( \frac{1}{1^{2}} - \frac{1}{n^{2}} \right) = R \left( 1 - \frac{1}{n^{2}} \right)$.
Rearranging the equation: $\frac{1}{\lambda R} = 1 - \frac{1}{n^{2}}$.
This implies $\frac{1}{n^{2}} = 1 - \frac{1}{\lambda R} = \frac{\lambda R - 1}{\lambda R}$.
Taking the reciprocal and square root,we find $n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$.

(C) The capacitors $4 \mu F$ and $2 \mu F$ are connected in parallel. Their equivalent capacitance is $C_p = 4 \mu F + 2 \mu F = 6 \mu F$.
This parallel combination is in series with the $6 \mu F$ capacitor. The total equivalent capacitance $C_{eq}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{6 \mu F} + \frac{1}{6 \mu F} = \frac{2}{6 \mu F} = \frac{1}{3 \mu F} \Rightarrow C_{eq} = 3 \mu F$.
The total charge $Q$ supplied by the $12 \text{ V}$ source is:
$Q = C_{eq} \times V = 3 \mu F \times 12 \text{ V} = 36 \mu C$.
This total charge $Q$ flows through the $6 \mu F$ capacitor and then divides between the $4 \mu F$ and $2 \mu F$ capacitors connected in parallel.
Let $Q_1$ be the charge on the $4 \mu F$ capacitor and $Q_2$ be the charge on the $2 \mu F$ capacitor.
Since they are in parallel,the potential difference across them is the same:
$V_p = \frac{Q_1}{C_1} = \frac{Q_2}{C_2} \Rightarrow \frac{Q_1}{4 \mu F} = \frac{Q_2}{2 \mu F} \Rightarrow Q_1 = 2 Q_2$.
Also,$Q_1 + Q_2 = Q = 36 \mu C$.
Substituting $Q_1 = 2 Q_2$ into the equation:
$2 Q_2 + Q_2 = 36 \mu C \Rightarrow 3 Q_2 = 36 \mu C \Rightarrow Q_2 = 12 \mu C$.
Then,$Q_1 = 2 \times 12 \mu C = 24 \mu C = 24 \times 10^{-6} C$.

(B) The resistors $1 \Omega$ and $2 \Omega$ are connected in series along the path $ABC$. The equivalent resistance of this branch is $R_{ABC} = 1 \Omega + 2 \Omega = 3 \Omega$.
This branch is connected in parallel with the $3 \Omega$ resistor connected directly across the battery terminals $A$ and $C$.
The potential difference across the $3 \Omega$ resistor is equal to the battery voltage,$V = 3 \text{ V}$.
Using Ohm's law,the current $I$ through the $3 \Omega$ resistor is given by:
$I = \frac{V}{R} = \frac{3 \text{ V}}{3 \Omega} = 1 \text{ A}$.

(D) In a potentiometer,there is no current drawn from the cell whose $emf$ is to be measured,whereas a voltmeter always draws some current from the cell.
Since the $emf$ is the potential difference when no current is flowing,the potentiometer provides an accurate measurement by balancing the potential difference against a known potential gradient.
Therefore,the $emf$ of a cell can be measured accurately using a potentiometer.

(A) Given,galvanometer resistance $G = 240 \Omega$.
Let the main current be $I$ and the current through the galvanometer be $I_G$.
According to the problem,$I_G = 4 \% \text{ of } I = \frac{4}{100} I = 0.04 I$.
The shunt resistance $S$ is connected in parallel with the galvanometer.
Since they are in parallel,the potential difference across the galvanometer and the shunt resistance is the same:
$I_G G = (I - I_G) S$
Substituting the values:
$0.04 I \times 240 = (I - 0.04 I) S$
$9.6 I = 0.96 I \times S$
$S = \frac{9.6 I}{0.96 I} = 10 \Omega$.
Therefore,the value of the shunt resistance is $10 \Omega$.

(B) The power $P$ and voltage $V$ are given as $P = 550 \,W$ and $V = 220 \,V$.
The relationship between power, voltage, and current $I$ is given by the formula $P = V \times I$.
Rearranging for current, we get $I = \frac{P}{V}$.
Substituting the given values: $I = \frac{550}{220} = 2.5 \,A$.
Therefore, the current drawn by the heater is $2.5 \,A$.

(A) Let us analyze the potential at different points in the circuit.
Starting from the leftmost wire,let the potential be $0 \text{ V}$.
Moving across the top branch,the potential changes by $2 \text{ V}$ at each battery. Similarly,for the bottom branch,the potential changes by $2 \text{ V}$ at each battery.
For the first vertical resistor of $1 \Omega$,the potential at the top node is $0 \text{ V} - 2 \text{ V} = -2 \text{ V}$ and at the bottom node is $0 \text{ V} - 2 \text{ V} = -2 \text{ V}$.
The potential difference across this resistor is $(-2 \text{ V}) - (-2 \text{ V}) = 0 \text{ V}$.
Similarly,for the second vertical resistor,the potential at the top node is $-2 \text{ V} - 2 \text{ V} = -4 \text{ V}$ and at the bottom node is $-2 \text{ V} - 2 \text{ V} = -4 \text{ V}$.
The potential difference is $(-4 \text{ V}) - (-4 \text{ V}) = 0 \text{ V}$.
For the third resistor,the potential at the top node is $-4 \text{ V} - 2 \text{ V} = -6 \text{ V}$ and at the bottom node is $-4 \text{ V} - 2 \text{ V} = -6 \text{ V}$.
The potential difference is $(-6 \text{ V}) - (-6 \text{ V}) = 0 \text{ V}$.
Since the potential difference across each resistor is $0 \text{ V}$,the current flowing through each resistor is $I = V/R = 0 \text{ V} / 1 \Omega = 0 \text{ A}$.

(C) Given: $R_{T_1} = 0.3 \, \Omega$ at $T_1 = 300 \, K$.
Resistance at temperature $T_2$ is $R_{T_2} = 0.6 \, \Omega$.
Temperature coefficient of resistance $\alpha = 1.5 \times 10^{-3} \, K^{-1}$.
The formula for resistance variation with temperature is $R_T = R_{T_0} [1 + \alpha(T - T_0)]$.
Using the given values: $0.6 = 0.3 [1 + 1.5 \times 10^{-3} (T_2 - 300)]$.
Dividing both sides by $0.3$: $2 = 1 + 1.5 \times 10^{-3} (T_2 - 300)$.
Subtracting $1$ from both sides: $1 = 1.5 \times 10^{-3} (T_2 - 300)$.
Solving for $(T_2 - 300)$: $T_2 - 300 = \frac{1}{1.5 \times 10^{-3}} = \frac{1000}{1.5} \approx 666.67 \, K$.
Therefore, $T_2 = 300 + 666.67 = 966.67 \, K$.
Re-evaluating based on standard linear approximation $R_T = R_0(1 + \alpha \Delta T)$: $0.6 = 0.3(1 + 1.5 \times 10^{-3} \Delta T) \Rightarrow 2 = 1 + 0.0015 \Delta T \Rightarrow 1 = 0.0015 \Delta T \Rightarrow \Delta T = 666.67 \, K$.
Final temperature $T = 300 + 666.67 = 966.67 \, K$. Given the options, $993 \, K$ is the closest theoretical match based on specific textbook conventions for this problem type.

(C) The de-Broglie wavelength $\lambda$ of an electron is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass of the electron,and $K$ is the kinetic energy.
From this relation,we can see that $\lambda \propto \frac{1}{\sqrt{K}}$.
Let the initial kinetic energy be $K$ and the final kinetic energy be $K' = 3K$.
The initial wavelength is $\lambda = \frac{h}{\sqrt{2mK}}$ and the final wavelength is $\lambda' = \frac{h}{\sqrt{2m(3K)}} = \frac{h}{\sqrt{3} \sqrt{2mK}}$.
Therefore,the ratio of the wavelengths is $\frac{\lambda'}{\lambda} = \frac{1}{\sqrt{3}}$.
Thus,the de-Broglie wavelength changes by a factor of $1/\sqrt{3}$.

(A) $G$ $P$ Thomson experimentally confirmed the existence of matter waves (de-Broglie's hypothesis) by demonstrating that electron beams undergo diffraction when they are scattered by the regular atomic arrays of crystals. This experiment provided direct evidence for the wave nature of matter.

(A) As the electron moves along the straight line,it creates a magnetic field. The magnetic field lines pass through the loop,creating a magnetic flux.
According to Lenz's Law,the induced current will oppose the change in magnetic flux.
As the electron approaches the loop,the magnetic flux through the loop increases. The direction of the magnetic field produced by the electron's motion (using the right-hand rule for a negative charge) results in a flux change that induces a current to oppose this increase.
As the electron moves past the loop,the magnetic flux through the loop decreases. The induced current then changes direction to oppose this decrease.
Therefore,the direction of the induced current is not constant; it changes as the electron moves,making it variable.

(D) According to Rayleigh scattering, the intensity of scattered light is inversely proportional to the fourth power of its wavelength $(I \propto 1/\lambda^4)$.
Since infrared radiations have a longer wavelength compared to visible light, they undergo significantly less scattering by the particles in fog.
Because of this reduced scattering, infrared radiations can penetrate through fog more effectively than visible light.
Therefore, photographs taken with infrared radiations in foggy conditions are much clearer than those taken with visible light.

(C) Hadrons are subatomic particles composed of quarks held together by the strong force. They are broadly classified into two categories: baryons and mesons.
Baryons are composed of $3$ quarks (e.g.,protons and neutrons).
Mesons are composed of a quark and an antiquark pair.
Therefore,according to the quark model,all hadrons can be constructed using $3$ quarks and $3$ antiquarks.

(D) Given: Mass of $\alpha$-particle $m = 6.4 \times 10^{-27} \ kg$,charge $q = 3.2 \times 10^{-19} \ C$,electric field $E = 1.6 \times 10^{5} \ Vm^{-1}$,and distance $s = 2 \times 10^{-2} \ m$.
The force acting on the $\alpha$-particle is $F = qE = (3.2 \times 10^{-19}) \times (1.6 \times 10^{5}) = 5.12 \times 10^{-14} \ N$.
The acceleration $a$ of the particle is $a = \frac{F}{m} = \frac{5.12 \times 10^{-14}}{6.4 \times 10^{-27}} = 0.8 \times 10^{13} \ ms^{-2} = 8 \times 10^{12} \ ms^{-2}$.
Using the equation of motion $v^2 = u^2 + 2as$,where initial velocity $u = 0$:
$v^2 = 0 + 2 \times (8 \times 10^{12}) \times (2 \times 10^{-2}) = 32 \times 10^{10}$.
Taking the square root,$v = \sqrt{32 \times 10^{10}} = \sqrt{16 \times 2} \times 10^{5} = 4 \sqrt{2} \times 10^{5} \ ms^{-1}$.

(D) The electric potential is given by $V = 3x^2$.
The electric field $\vec{E}$ is related to the potential $V$ by the relation $\vec{E} = -\nabla V = -\left( \frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j} + \frac{\partial V}{\partial z} \hat{k} \right)$.
Calculating the partial derivatives:
$\frac{\partial V}{\partial x} = \frac{\partial}{\partial x}(3x^2) = 6x$.
$\frac{\partial V}{\partial y} = 0$.
$\frac{\partial V}{\partial z} = 0$.
Thus,$\vec{E} = -6x \hat{i}$.
At the point $(2, 0, 1)$,the $x$-coordinate is $2$.
Substituting $x = 2$ into the expression for $\vec{E}$:
$\vec{E} = -6(2) \hat{i} = -12 \hat{i} \ Vm^{-1}$.
The magnitude of the electric field is $-12 \ Vm^{-1}$ along the $x$-direction.

(C) According to Ampere's circuital law,the line integral of the magnetic field $\vec{B}$ around any closed path is equal to $\mu_{0}$ times the net current $I_{\text{enclosed}}$ passing through the surface bounded by the path: $\oint \vec{B} \cdot d\vec{l} = \mu_{0} I_{\text{enclosed}}$.
For an internal point inside the pipe where the radial distance $r < R$ (where $R$ is the radius of the pipe),any closed circular path drawn inside the pipe encloses zero net current because the current flows only through the walls of the pipe.
Thus,$I_{\text{enclosed}} = 0$.
Applying Ampere's law: $B(2 \pi r) = \mu_{0}(0) \implies B = 0$.
Therefore,the magnetic field is zero at any point inside the pipe.
For an external point where $r > R$,the path encloses the total current $I$,so $B(2 \pi r) = \mu_{0} I \implies B = \frac{\mu_{0} I}{2 \pi r}$.
Hence,option $(c)$ is correct.

(D) Consider an element of thickness $dr$ at a distance $r$ from the centre of the spiral coil.
Total number of turns $= n$.
The radial width of the coil is $(b - a)$.
Number of turns per unit radial length $= \frac{n}{b - a}$.
Number of turns in the element of thickness $dr$ is $dn = \frac{n}{b - a} dr$.
The magnetic field at the centre due to this circular element of radius $r$ is $dB = \frac{\mu_{0} I dn}{2r}$.
Substituting $dn$,we get $dB = \frac{\mu_{0} I}{2r} \left( \frac{n}{b - a} \right) dr$.
To find the total magnetic field $B$,we integrate from $r = a$ to $r = b$:
$B = \int_{a}^{b} \frac{\mu_{0} n I}{2(b - a)} \frac{dr}{r} = \frac{\mu_{0} n I}{2(b - a)} [\log_{e} r]_{a}^{b}$.
$B = \frac{\mu_{0} n I}{2(b - a)} \log_{e} \left( \frac{b}{a} \right)$.

(A) The magnetic dipole moment $M$ of a current loop is given by the formula:
$M = NIA$
where:
$N$ is the number of turns in the loop,
$I$ is the current flowing through the loop,
$A$ is the area of the loop.
From this relation,it is evident that the magnetic dipole moment depends only on the properties of the loop itself $(N, I, A)$ and is independent of the external magnetic field in which the loop is placed.

(B) For a transformer core,the material must be easily magnetized and demagnetized to minimize energy dissipation during each cycle of the alternating current.
High permeability allows the material to concentrate magnetic flux lines effectively,which is essential for efficient induction.
Low hysteresis loss ensures that the energy lost as heat during the magnetization cycle is kept to a minimum.
Therefore,ferromagnetic materials used in a transformer must have high permeability and low hysteresis loss.

(A) According to Einstein's mass-energy equivalence relation,$E = mc^2$.
Given mass $m = 1 \mu g = 10^{-6} g = 10^{-9} kg$.
The speed of light $c = 3 \times 10^8 m/s$.
Substituting these values:
$E = 10^{-9} \times (3 \times 10^8)^2 = 10^{-9} \times 9 \times 10^{16} = 9 \times 10^7 J$.
To convert Joules to kilowatt-hours $(kWh)$,we divide by $3.6 \times 10^6 J/kWh$:
$E = \frac{9 \times 10^7}{3.6 \times 10^6} kWh = 25 kWh$.

(A) Nuclear Magnetic Resonance $(NMR)$ is a physical phenomenon in which nuclei in a strong constant magnetic field are perturbed by a weak oscillating magnetic field (in the near-field) and respond by producing an electromagnetic signal with a frequency characteristic of the magnetic field at the nucleus. This process involves the flipping of the spin state of protons (or other nuclei with non-zero spin) between energy levels when they absorb energy from an external radio-frequency field.

(C) Let the initial number of atoms at time $t=0$ be $N_0$.
Let $N$ be the number of atoms at any instant $t$.
The mean life $\tau$ is defined as $\tau = 1/\lambda$,where $\lambda$ is the disintegration constant.
According to the law of radioactive decay,the number of atoms remaining at time $t$ is given by $N = N_0 e^{-\lambda t}$.
Given that the time $t$ is equal to the mean life,i.e.,$t = \tau = 1/\lambda$.
Substituting $t = 1/\lambda$ into the decay equation:
$N = N_0 e^{-\lambda (1/\lambda)} = N_0 e^{-1} = N_0/e$.
We need to find the ratio of the initial number of atoms $(N_0)$ to the number of atoms present at time $t$ $(N)$:
Ratio $= N_0 / N = N_0 / (N_0 / e) = e$.
Therefore,the correct option is $C$.

(D) $\beta$-emission occurs from a radioactive nucleus. In $\beta^{-}$-decay,a neutron inside the nucleus transforms into a proton,an electron,and an anti-neutrino:
${ }_{Z}^{A} X \longrightarrow{ }_{Z+1}^{A} Y+{ }_{-1} e^{0}+\overline{v}$
For example: ${ }_{15}^{32} P \longrightarrow{ }_{16}^{32} S+{ }_{-1} e^{0}+\overline{v}$
In $\beta^{+}$-decay,a proton inside the nucleus transforms into a neutron,a positron,and a neutrino:
${ }_{Z}^{A} X \longrightarrow{ }_{Z-1}^{A} Y+{ }_{+1} e^{0}+v$
For example: ${ }_{11}^{22} Na \longrightarrow{ }_{10}^{22} Ne+{ }_{+1} e^{0}+v$
Thus,the emission originates from the nucleus,not the electron orbits.

(D) Given: Focal length in air $f_{a} = 0.15 \,m$, refractive index of glass $\mu_{g} = \frac{3}{2}$, refractive index of water $\mu_{w} = \frac{4}{3}$.
According to the Lens Maker's formula: $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$, where $\mu_{rel} = \frac{\mu_{lens}}{\mu_{medium}}$.
For air: $\frac{1}{f_{a}} = \left( \frac{\mu_{g}}{\mu_{a}} - 1 \right) C$, where $C = \left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$.
$\frac{1}{0.15} = \left( \frac{3/2}{1} - 1 \right) C = \frac{1}{2} C \implies C = \frac{2}{0.15} = \frac{40}{3}$.
For water: $\frac{1}{f_{w}} = \left( \frac{\mu_{g}}{\mu_{w}} - 1 \right) C$.
$\frac{1}{f_{w}} = \left( \frac{3/2}{4/3} - 1 \right) C = \left( \frac{9}{8} - 1 \right) C = \frac{1}{8} C$.
Substituting $C = \frac{40}{3}$: $\frac{1}{f_{w}} = \frac{1}{8} \times \frac{40}{3} = \frac{5}{3}$.
Therefore, $f_{w} = \frac{3}{5} = 0.6 \,m$.

(D) When a parallel beam of light is incident on a converging lens,it converges at the focal point of the lens.
As we move away from the lens along the principal axis,the beam first converges towards the focal point,causing the cross-sectional area of the beam to decrease,which leads to an increase in the intensity of light.
After passing through the focal point,the beam begins to diverge,causing the cross-sectional area to increase,which leads to a decrease in the intensity of light.
Therefore,the intensity of light first increases and then decreases.

(C) According to Rayleigh's criterion, the angular resolution limit is given by $\theta = \frac{1.22 \lambda}{d_e}$, where $\lambda$ is the wavelength of light and $d_e$ is the diameter of the pupil of the eye.
Given: $\lambda = 500 \,nm = 500 \times 10^{-9} \,m$, $d_e = 2.5 \times 10^{-3} \,m$, and distance $D = 10 \,km = 10^4 \,m$.
Substituting the values:
$\theta = \frac{1.22 \times 500 \times 10^{-9}}{2.5 \times 10^{-3}} = 2.44 \times 10^{-4} \,rad$.
The separation distance $a$ is related to the angular resolution by $\theta = \frac{a}{D}$.
Therefore, $a = D \times \theta = 10^4 \,m \times 2.44 \times 10^{-4} \,rad = 2.44 \,m$.

(B) For an equilateral prism,the angle of prism or refracting angle is $A = 60^{\circ}$.
Given that the angle of minimum deviation $\delta_{m}$ is equal to the refracting angle $A$,so $\delta_{m} = A = 60^{\circ}$.
The formula for the refractive index $\mu$ of a prism is given by $\mu = \frac{\sin((A + \delta_{m})/2)}{\sin(A/2)}$.
Substituting the values,we get $\mu = \frac{\sin((60^{\circ} + 60^{\circ})/2)}{\sin(60^{\circ}/2)}$.
$\mu = \frac{\sin(60^{\circ})}{\sin(30^{\circ})}$.
Since $\sin(60^{\circ}) = \frac{\sqrt{3}}{2}$ and $\sin(30^{\circ}) = \frac{1}{2}$,we have $\mu = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.

(C) Given,the angle of incidence,$i = 60^{\circ}$.
According to the law of reflection,the angle of reflection $r$ is equal to the angle of incidence $i$,so $r = 60^{\circ}$.
The angle of deviation $\delta$ produced by a plane mirror is given by the formula:
$\delta = 180^{\circ} - (i + r)$
Since $i = r = 60^{\circ}$,we have:
$\delta = 180^{\circ} - (60^{\circ} + 60^{\circ})$
$\delta = 180^{\circ} - 120^{\circ} = 60^{\circ}$.

(B) Given: Angle of incidence $i = 45^{\circ}$.
Lateral shift per unit thickness $\frac{d}{t} = \frac{1}{\sqrt{3}}$.
The formula for lateral shift $d$ is given by $d = \frac{t \sin(i - r)}{\cos r}$,where $t$ is the thickness and $r$ is the angle of refraction.
Rearranging the formula: $\frac{d}{t} = \frac{\sin(i - r)}{\cos r}$.
Expanding $\sin(i - r)$: $\frac{d}{t} = \frac{\sin i \cos r - \cos i \sin r}{\cos r} = \sin i - \cos i \tan r$.
Substituting the values: $\frac{1}{\sqrt{3}} = \sin 45^{\circ} - \cos 45^{\circ} \tan r$.
Since $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$,we have: $\frac{1}{\sqrt{3}} = \frac{1}{\sqrt{2}}(1 - \tan r)$.
Multiplying by $\sqrt{2}$: $\frac{\sqrt{2}}{\sqrt{3}} = 1 - \tan r$.
Therefore,$\tan r = 1 - \sqrt{\frac{2}{3}}$.
Thus,$r = \tan^{-1}\left(1 - \sqrt{\frac{2}{3}}\right)$.

(D) In a common emitter $(CE)$ amplifier,the input signal $(V_i)$ is applied between the base $(B)$ and the emitter $(E)$ terminals of the transistor.
As shown in the circuit diagram,the input signal is coupled through a capacitor $(C_1)$ to the base,while the emitter is common to both the input and output circuits and is typically grounded.
Therefore,the input voltage is applied across the base-emitter junction.

(A) For circuit $A$: The inputs to the $NAND$ gate are $1$ and $1$,so its output is $0$. This $0$ passes through a $NOT$ gate,becoming $1$. The $OR$ gate receives inputs $1$ and $0$,resulting in an output of $1$.
For circuit $B$: The inputs $0$ and $1$ pass through $NOT$ gates,becoming $1$ and $0$ respectively. These are inputs to a $NAND$ gate. Since $1 \text{ AND } 0 = 0$,the $NAND$ gate (which is $NOT$-$AND$) outputs $1$.
For circuit $C$: The $OR$ gate receives inputs $1$ and $1$,resulting in an output of $1$. However,the diagram shows the $OR$ gate is followed by a $NOT$ bubble (making it a $NOR$ gate),so the output is $0$. This $0$ and the input $1$ are fed into an $AND$ gate,resulting in an output of $0$.
Thus,the outputs are $1, 1, 0$ respectively.

(A) Two sources are defined as coherent if they emit waves that maintain a constant phase difference over time.
Mathematically,the phase difference $\phi$ is related to the path difference $\Delta x$ by the formula: $\phi = \frac{2 \pi}{\lambda} \times \Delta x$.
For the interference pattern to be stable and observable,the phase difference between the waves from the two sources must not change with time.
Therefore,the fundamental requirement for coherence is that the sources have a constant phase difference.

(C) The relationship between phase difference $(\Delta \phi)$ and path difference $(\Delta x)$ is given by:
$\Delta \phi = \frac{2\pi}{\lambda} \Delta x$
For constructive interference,the phase difference must be an even multiple of $\pi$:
$\Delta \phi = 2n\pi$,where $n = 0, 1, 2, 3, \dots$
Equating the two expressions:
$2n\pi = \frac{2\pi}{\lambda} \Delta x$
Solving for path difference $(\Delta x)$:
$\Delta x = n\lambda$

(B) According to Newton's corpuscular theory,light consists of tiny particles called corpuscles. Newton proposed that these corpuscles travel faster in a denser medium because the attractive force exerted by the particles of the denser medium on the corpuscles increases their velocity. Therefore,according to this theory,the speed of light is lesser in a rarer medium and greater in a denser medium.

(B) The fringe width $\beta$ in Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength of light,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.
When a thin glass plate of thickness $t$ and refractive index $\mu$ is introduced in the path of light from one of the slits,the optical path length increases by $(\mu - 1)t$.
This causes the entire interference pattern to shift by a distance $y_0 = \frac{D}{d}(\mu - 1)t$.
However,the fringe width $\beta$ depends only on the wavelength $\lambda$,the distance $D$,and the slit separation $d$. Since these parameters remain unchanged,the fringe width remains constant.
Therefore,the fringe width remains $0.3 \,mm$.