KCET 2010 Physics Question Paper with Answer and Solution

(B) The kinetic energy $K$ of a body is related to its linear momentum $p$ by the formula $K = \frac{p^2}{2m}$,where $m$ is the mass of the body.
Since the mass $m$ remains constant,we have $K \propto p^2$.
Let the initial momentum be $p_1 = p$ and the final momentum be $p_2 = p + 0.50p = 1.5p$.
The ratio of the final kinetic energy $K_2$ to the initial kinetic energy $K_1$ is given by:
$\frac{K_2}{K_1} = \left(\frac{p_2}{p_1}\right)^2 = \left(\frac{1.5p}{p}\right)^2 = (1.5)^2 = 2.25$.
To find the percentage increase in kinetic energy,we use the formula:
$\text{Percentage increase} = \left(\frac{K_2}{K_1} - 1\right) \times 100 \%$.
Substituting the value: $(2.25 - 1) \times 100 \% = 1.25 \times 100 \% = 125 \%$.

(A) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.
Let $M_e$ and $R_e$ be the mass and radius of the Earth,and $g_e = \frac{GM_e}{R_e^2}$ be the acceleration due to gravity on Earth.
For the strange planet,we are given $g_p = 2g_e$.
Checking option $A$: If $M_p = \frac{M_e}{2}$ and $R_p = \frac{R_e}{2}$,then $g_p = \frac{G(M_e/2)}{(R_e/2)^2} = \frac{GM_e/2}{R_e^2/4} = 2 \frac{GM_e}{R_e^2} = 2g_e$.
Thus,option $A$ is correct.

(A) For a gas in a closed vessel of constant volume,Gay-Lussac's Law states that $P \propto T$,where $P$ is the pressure and $T$ is the absolute temperature in Kelvin.
Taking the derivative,we have $\frac{dP}{P} = \frac{dT}{T}$.
Given that the pressure increases by $1 \%$,we have $\frac{dP}{P} = 0.01$.
Given that the temperature increases by $dT = 1 \ K$ (since a change of $1^{\circ} C$ is equivalent to a change of $1 \ K$).
Substituting these values into the equation: $0.01 = \frac{1}{T}$.
Therefore,$T = \frac{1}{0.01} = 100 \ K$.

(A) Case $(1)$: If there is no friction between the paper and the ball,the paper exerts no horizontal force on the ball. Due to the inertia of rest,the ball remains at its initial position on the table.
Case $(2)$: If there is friction between the paper and the ball,the paper exerts a kinetic frictional force on the ball in the direction of the paper's motion (to the right). This force acts at the point of contact. This creates a torque about the center of mass of the ball,causing it to rotate such that its bottom surface moves to the left relative to the table. Consequently,the ball starts rolling backwards (to the left) while the frictional force also provides a net acceleration to the left relative to the table.
Therefore,both statements $(1)$ and $(2)$ are correct.

(D) When a ball is thrown in the air,two main forces act on it:
$1$. The gravitational force (weight),which always acts vertically downwards,given by $mg$.
$2$. The air resistance (frictional force),which always acts in the direction opposite to the instantaneous velocity of the ball.
At position $X$,the ball is moving along a parabolic path. Its velocity vector is tangent to the path at that point,directed upwards and forwards. Therefore,the air resistance acts downwards and backwards (opposite to the velocity vector).
Combining these,the weight acts vertically downwards,and the air resistance acts at an angle downwards and backwards. This corresponds to the vector diagram in option $D$.

(D) According to the equation of continuity, $A_{1} v_{1} = A_{2} v_{2}$.
Given $A_{1} = 10 \,cm^{2}$, $v_{1} = 1 \,ms^{-1}$, $A_{2} = 5 \,cm^{2}$.
Substituting the values: $10 \times 1 = 5 \times v_{2} \implies v_{2} = 2 \,ms^{-1}$.
Using Bernoulli's equation for a horizontal pipe: $P_{1} + \frac{1}{2} \rho v_{1}^{2} = P_{2} + \frac{1}{2} \rho v_{2}^{2}$.
Here, $\rho = 1000 \,kg/m^{3}$ (density of water).
$2000 + \frac{1}{2} \times 1000 \times (1)^{2} = P_{2} + \frac{1}{2} \times 1000 \times (2)^{2}$.
$2000 + 500 = P_{2} + 2000$.
$2500 = P_{2} + 2000 \implies P_{2} = 500 \,Pa$.

(C) The force exerted by a liquid on the base of a vessel is equal to the weight of the liquid contained in it,provided the vessel has vertical walls (like a cubical vessel).
The force $F$ is given by $F = mg$.
Since the masses of the liquids in all three vessels are equal $(m_{A} = m_{B} = m_{C} = m)$,the force exerted on the base of each vessel is $F_{A} = F_{B} = F_{C} = mg$.
Therefore,the force exerted by the liquid on the base is the same in all the vessels.

(B) Elasticity is defined by the ability of a material to regain its original shape after the removal of a deforming force.
Quantitatively,it is measured by Young's modulus $(Y)$.
$A$ material with a higher Young's modulus requires more stress to produce a given strain,meaning it is more elastic.
Among the given substances,steel has the highest value of Young's modulus $(Y \approx 200 \ GPa)$.
Therefore,steel is the most elastic material among the options provided.

(D) Let the velocity of the train with respect to the ground be $v_T = 2 \,ms^{-1}$ (taking the direction of motion as positive).
The velocity of the passenger with respect to the train is $v_{P/T} = -2 \,ms^{-1}$ (since the passenger is walking towards the back of the train).
The velocity of the passenger with respect to the ground (observer on the platform) is given by the relative velocity formula:
$v_P = v_{P/T} + v_T$
$v_P = -2 \,ms^{-1} + 2 \,ms^{-1} = 0 \,ms^{-1}$.
Therefore, the velocity of the passenger appears to be zero to the observer on the platform.

(B) Let $v_{b}$ be the velocity of the motorboat in still water and $v_{w}$ be the velocity of the river water.
When moving downstream,the effective velocity is $(v_{b} + v_{w})$. The distance $x$ covered in $6 \,h$ is:
$x = (v_{b} + v_{w}) \times 6$ ---$(i)$
When moving upstream,the effective velocity is $(v_{b} - v_{w})$. The distance $x$ covered in $10 \,h$ is:
$x = (v_{b} - v_{w}) \times 10$ ---(ii)
Equating the two expressions for distance $x$:
$(v_{b} + v_{w}) \times 6 = (v_{b} - v_{w}) \times 10$
$6v_{b} + 6v_{w} = 10v_{b} - 10v_{w}$
$16v_{w} = 4v_{b}$
$v_{w} = \frac{v_{b}}{4}$
Substituting $v_{w}$ back into equation $(i)$ to find $x$ in terms of $v_{b}$:
$x = (v_{b} + \frac{v_{b}}{4}) \times 6 = (\frac{5v_{b}}{4}) \times 6 = 7.5v_{b}$
The time $t$ taken to cover distance $x$ in still water (where velocity is $v_{b}$) is:
$t = \frac{x}{v_{b}} = \frac{7.5v_{b}}{v_{b}} = 7.5 \,h$

(D) According to the parallel axis theorem,the moment of inertia $I$ about an axis parallel to the axis passing through the center of mass is given by:
$I = I_{cm} + Md^{2}$
Here,$I_{cm} = 2 \,kg \,m^{2}$ is the moment of inertia about the axis passing through the center of mass.
$M = 1 \,kg$ is the mass of the disc.
$d = R = 2 \,m$ is the distance between the two parallel axes (which is equal to the radius of the disc).
Substituting these values into the formula:
$I = 2 + (1)(2)^{2}$
$I = 2 + (1)(4)$
$I = 2 + 4 = 6 \,kg \,m^{2}$
Therefore,the moment of inertia about the axis passing through the edge of the disc is $6 \,kg \,m^{2}$.

(C) For the first slab,the heat current is $H_{1} = \frac{K_{1} A (\theta_{1} - \theta)}{d_{1}}$.
For the second slab,the heat current is $H_{2} = \frac{K_{2} A (\theta - \theta_{2})}{d_{2}}$.
Since the slabs are in series,the heat current through both must be equal in steady state,so $H_{1} = H_{2}$.
$\frac{K_{1} A (\theta_{1} - \theta)}{d_{1}} = \frac{K_{2} A (\theta - \theta_{2})}{d_{2}}$
$\frac{K_{1} (\theta_{1} - \theta)}{d_{1}} = \frac{K_{2} (\theta - \theta_{2})}{d_{2}}$
$K_{1} d_{2} (\theta_{1} - \theta) = K_{2} d_{1} (\theta - \theta_{2})$
$K_{1} d_{2} \theta_{1} - K_{1} d_{2} \theta = K_{2} d_{1} \theta - K_{2} d_{1} \theta_{2}$
$K_{1} d_{2} \theta_{1} + K_{2} d_{1} \theta_{2} = \theta (K_{1} d_{2} + K_{2} d_{1})$
$\theta = \frac{K_{1} \theta_{1} d_{2} + K_{2} \theta_{2} d_{1}}{K_{1} d_{2} + K_{2} d_{1}}$

(D) According to Newton's law of cooling,the rate of cooling is given by $\frac{\theta_{1}-\theta_{2}}{t} = K \left[ \frac{\theta_{1}+\theta_{2}}{2} - \theta_{s} \right]$,where $\theta_{s}$ is the temperature of the surroundings.
For the first $10 \ min$ interval: $\frac{60-50}{10} = K \left[ \frac{60+50}{2} - \theta_{s} \right] \Rightarrow 1 = K(55 - \theta_{s}) \dots (i)$
For the next $10 \ min$ interval: $\frac{50-42}{10} = K \left[ \frac{50+42}{2} - \theta_{s} \right] \Rightarrow 0.8 = K(46 - \theta_{s}) \dots (ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{1}{0.8} = \frac{55 - \theta_{s}}{46 - \theta_{s}}$
$1.25 = \frac{55 - \theta_{s}}{46 - \theta_{s}}$
$1.25(46 - \theta_{s}) = 55 - \theta_{s}$
$57.5 - 1.25\theta_{s} = 55 - \theta_{s}$
$2.5 = 0.25\theta_{s}$
$\theta_{s} = 10^{\circ} C$.

(A) The efficiency $\eta$ of a Carnot heat engine is given by the formula: $\eta = 1 - \frac{T_{2}}{T_{1}}$.
Given that $\eta = 0.5$, we have $0.5 = 1 - \frac{T_{2}}{T_{1}}$, which implies $\frac{T_{2}}{T_{1}} = 0.5$, or $T_{1} = 2T_{2}$.
For another Carnot engine to have the same efficiency of $0.5$, the ratio of the sink temperature to the source temperature must remain the same, i.e., $\frac{T_{2}'}{T_{1}'} = 0.5$.
If we multiply both the source and sink temperatures by the same factor $k$, the new efficiency is $\eta' = 1 - \frac{k T_{2}}{k T_{1}} = 1 - \frac{T_{2}}{T_{1}} = 0.5$.
Therefore, if the source temperature is $2T_{1}$ and the sink temperature is $2T_{2}$, the efficiency remains $0.5$.

(C) The dimensional formula for resistance $R$ is derived from $R = \frac{V}{I} = \frac{W}{qI}$.
Substituting the dimensions: $R = \frac{[ML^2 T^{-2}]}{[AT][A]} = [ML^2 T^{-3} A^{-2}]$.
Now,let us check the dimensions of $\frac{h}{e^2}$.
The dimensions of Planck's constant $h$ are $[ML^2 T^{-1}]$.
The dimensions of charge $e$ are $[AT]$.
Therefore,the dimensions of $\frac{h}{e^2} = \frac{[ML^2 T^{-1}]}{[AT]^2} = \frac{[ML^2 T^{-1}]}{[A^2 T^2]} = [ML^2 T^{-3} A^{-2}]$.
Since the dimensions of $R$ and $\frac{h}{e^2}$ are identical,the correct option is $C$.

(C) The bat acts as both the source and the observer moving towards a stationary wall.
First, the frequency $f'$ received by the wall is given by the Doppler effect formula for a moving source and stationary observer:
$f' = f \left( \frac{v}{v - v_b} \right)$
where $v = 330 \,ms^{-1}$ is the speed of sound and $v_b = 4 \,ms^{-1}$ is the speed of the bat.
Next, the wall reflects this sound, and the bat acts as a moving observer receiving the reflected frequency $f''$ from a stationary source (the wall):
$f'' = f' \left( \frac{v + v_b}{v} \right)$
Substituting $f'$ into the equation for $f''$:
$f'' = f \left( \frac{v}{v - v_b} \right) \left( \frac{v + v_b}{v} \right) = f \left( \frac{v + v_b}{v - v_b} \right)$
$f'' = 90 \times 10^{3} \left( \frac{330 + 4}{330 - 4} \right) = 90 \times 10^{3} \left( \frac{334}{326} \right)$
$f'' \approx 90 \times 10^{3} \times 1.0245 = 92.21 \times 10^{3} \,Hz$.
Rounding to the nearest provided option, the frequency is $92.1 \times 10^{3} \,Hz$.

(D) Let the length of both pipes be $L$. The fundamental frequency of an open organ pipe is $f_{o} = \frac{v}{2L}$ and that of a closed organ pipe is $f_{c} = \frac{v}{4L}$.
Given that $f_{o} - f_{c} = 2 \text{ Hz}$.
Substituting the expressions,$\frac{v}{2L} - \frac{v}{4L} = 2 \Rightarrow \frac{v}{4L} = 2 \text{ Hz}$.
Thus,$f_{c} = 2 \text{ Hz}$ and $f_{o} = 2f_{c} = 4 \text{ Hz}$.
Now,the length of the open pipe is halved $(L_{o}' = L/2)$,so its new frequency is $f_{o}' = \frac{v}{2(L/2)} = \frac{v}{L} = 2f_{o} = 2 \times 4 = 8 \text{ Hz}$.
The length of the closed pipe is doubled $(L_{c}' = 2L)$,so its new frequency is $f_{c}' = \frac{v}{4(2L)} = \frac{1}{2} \left(\frac{v}{4L}\right) = \frac{1}{2} f_{c} = \frac{1}{2} \times 2 = 1 \text{ Hz}$.
The number of beats produced per second is $|f_{o}' - f_{c}'| = |8 - 1| = 7 \text{ Hz}$.

(A) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu$ is the linear mass density.
Linear mass density $\mu = \text{mass per unit length} = \text{density} \times \text{cross-sectional area} = \rho \times (\pi \frac{D^2}{4})$.
Substituting $\mu$ into the frequency formula:
$f = \frac{1}{2L} \sqrt{\frac{T}{\rho \pi \frac{D^2}{4}}} = \frac{1}{2L} \sqrt{\frac{4T}{\pi \rho D^2}} = \frac{1}{2L} \cdot \frac{2}{D} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{LD} \sqrt{\frac{T}{\pi \rho}}$.
Since $T$,$\rho$,and $\pi$ are constants,we have $f \propto \frac{1}{LD}$.

(B) The first equation is $y_1 = 5[\sin 2 \pi t + \sqrt{3} \cos 2 \pi t]$.
To find the amplitude $A_1$,we rewrite the expression in the form $A_1 \sin(2 \pi t + \phi)$.
Multiply and divide by $2$: $y_1 = 5 \times 2 [\frac{1}{2} \sin 2 \pi t + \frac{\sqrt{3}}{2} \cos 2 \pi t] = 10 [\sin 2 \pi t \cos \frac{\pi}{3} + \cos 2 \pi t \sin \frac{\pi}{3}] = 10 \sin(2 \pi t + \frac{\pi}{3})$.
Thus,the amplitude $A_1 = 10$.
The second equation is $y_2 = 5 \sin [2 \pi t + \frac{\pi}{4}]$.
Comparing this with $y_2 = A_2 \sin(2 \pi t + \phi_2)$,we get the amplitude $A_2 = 5$.
The ratio of their amplitudes is $\frac{A_1}{A_2} = \frac{10}{5} = 2: 1$.

(A) Resistance of the resistor: $R = \frac{V}{I} = \frac{100}{10} = 10 \, \Omega$.
Inductive reactance of the choke at $50 \, Hz$: $X_L = \frac{V}{I} = \frac{100}{8} = 12.5 \, \Omega$.
Since $X_L = 2 \pi f L$, we have $12.5 = 2 \pi \times 50 \times L$, which gives $L = \frac{12.5}{100 \pi} = \frac{1}{8 \pi} \, H$.
Now, for the new source of $40 \, Hz$, the new inductive reactance is $X_L' = 2 \pi f' L = 2 \pi \times 40 \times \frac{1}{8 \pi} = 10 \, \Omega$.
The impedance of the series $RL$ circuit is $Z = \sqrt{R^2 + X_L'^2} = \sqrt{10^2 + 10^2} = 10\sqrt{2} \, \Omega$.
The current in the circuit is $I = \frac{V'}{Z} = \frac{150}{10\sqrt{2}} = \frac{15}{\sqrt{2}} \, A$.

(C) The current $i$ in the series $R-L-C$ circuit is given by $i = \frac{V_R}{R}$.
Substituting the given values: $i = \frac{100 \, V}{1000 \, \Omega} = 0.1 \, A$.
At resonance, the inductive reactance $X_L$ is equal to the capacitive reactance $X_C$, and the potential difference across the inductor $V_L$ is equal to the potential difference across the capacitor $V_C$.
The potential difference across the capacitor is $V_C = i X_C = i \left( \frac{1}{\omega C} \right)$.
Substituting the values: $V_C = 0.1 \times \left( \frac{1}{200 \times 2 \times 10^{-6}} \right)$.
$V_C = 0.1 \times \left( \frac{1}{400 \times 10^{-6}} \right) = 0.1 \times \left( \frac{10^6}{400} \right) = 0.1 \times 2500 = 250 \, V$.
Since $V_L = V_C$ at resonance, the potential difference across the inductance coil is $250 \, V$.

(D) The current in an $AC$ circuit containing a capacitor is given by $i = \frac{V}{\sqrt{R^2 + X_C^2}}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
When a dielectric is introduced into the gap between the plates of the capacitor,the capacitance $C$ increases $(C = K C_0)$.
As $C$ increases,the capacitive reactance $X_C = \frac{1}{\omega C}$ decreases.
Since the impedance $Z = \sqrt{R^2 + X_C^2}$ decreases,the current $i$ in the circuit increases.
Consequently,the brightness of the bulb,which depends on the power dissipated $(P = i^2 R)$,increases.

(B) An oil flame produces light due to the incandescence of solid carbon particles present in the flame. Since the spectrum of an oil flame consists of a continuous range of wavelengths without any gaps,it is an example of a continuous emission spectrum.

(A) The frequency of a spectral line is given by $v = RC \left[ \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right]$.
For the series limit of the Lyman series $(n_{1}=1, n_{2}=\infty)$: $v_{1} = RC \left[ 1 - \frac{1}{\infty} \right] = RC$.
For the first line of the Lyman series $(n_{1}=1, n_{2}=2)$: $v_{2} = RC \left[ 1 - \frac{1}{4} \right] = \frac{3}{4} RC$.
For the series limit of the Balmer series $(n_{1}=2, n_{2}=\infty)$: $v_{3} = RC \left[ \frac{1}{4} - \frac{1}{\infty} \right] = \frac{RC}{4}$.
Comparing these values,we see that $v_{1} - v_{2} = RC - \frac{3}{4} RC = \frac{1}{4} RC = v_{3}$.
Therefore,$v_{1} - v_{2} = v_{3}$.

(D) The number of spectral lines $N$ obtained due to transitions of electrons from the $n$th orbit to lower orbits is given by the formula: $N = \frac{n(n-1)}{2}$.
For the first case,$N = 6$:
$6 = \frac{n_1(n_1-1)}{2} \Rightarrow n_1^2 - n_1 - 12 = 0 \Rightarrow (n_1-4)(n_1+3) = 0$. Since $n > 0$,we have $n_1 = 4$.
For the second case,$N = 3$:
$3 = \frac{n_2(n_2-1)}{2} \Rightarrow n_2^2 - n_2 - 6 = 0 \Rightarrow (n_2-3)(n_2+2) = 0$. Since $n > 0$,we have $n_2 = 3$.
The velocity of an electron in the $n$th orbit of a hydrogen atom is given by $v_n \propto \frac{1}{n}$.
Therefore,the ratio of the velocities is:
$\frac{v_1}{v_2} = \frac{n_2}{n_1} = \frac{3}{4}$.

(A) By analyzing the circuit,we can see that the capacitors are arranged in two groups.
In the first group,three capacitors are connected in parallel between the input and the intermediate node.
Therefore,the equivalent capacitance of this group is $C_{123} = C + C + C = 3 C$.
Similarly,in the second group,three capacitors are connected in parallel between the intermediate node and point $B$.
Therefore,the equivalent capacitance of this group is $C_{456} = C + C + C = 3 C$.
These two groups are connected in series with each other.
Therefore,the total effective capacitance $C_{\text{eq}}$ between points $A$ and $B$ is given by:
$C_{\text{eq}} = \frac{C_{123} \times C_{456}}{C_{123} + C_{456}} = \frac{(3 C)(3 C)}{3 C + 3 C} = \frac{9 C^2}{6 C} = 1.5 C$.

(B) The current $i$ drawn from the battery is given by the formula $i = \frac{E}{R + r}$.
As the variable resistance $R$ is gradually reduced to $0$,the current $i$ approaches the value $i = \frac{E}{0 + r} = \frac{E}{r}$.
The terminal potential difference $V$ is given by $V = E - ir$ or $V = iR$.
Using $V = iR$,as $R$ approaches $0$,the potential difference $V$ approaches $0 \times \frac{E}{r} = 0$.
Therefore,as $R \to 0$,$i \to \frac{E}{r}$ and $V \to 0$.

(C) Let the resistance of each identical bulb be $R$ and the $EMF$ of the battery be $E$ with internal resistance $r$.
When key $K$ is closed, bulbs $B_{2}$ and $B_{3}$ are in parallel, and this combination is in series with $B_{1}$. The equivalent resistance is $R_{eq} = R + (R/2) = 1.5R$. The total current is $I = E / (1.5R + r)$. The voltage across $B_{1}$ is $V_{1} = I \cdot R = E \cdot R / (1.5R + r)$.
When key $K$ is opened, bulb $B_{3}$ is disconnected. The circuit now consists of $B_{1}$ and $B_{2}$ in series. The new equivalent resistance is $R'_{eq} = 2R$. The new total current is $I' = E / (2R + r)$.
Comparing the currents: Since $2R > 1.5R$, the total current $I'$ is less than $I$. Thus, the current through $B_{1}$ decreases, so its brightness decreases.
The voltage across $B_{2}$ is $V'_{2} = I' \cdot R = E \cdot R / (2R + r)$. Comparing $V'_{2}$ with the previous voltage across $B_{2}$ (which was $V_{2} = I \cdot (R/2) = E \cdot (R/2) / (1.5R + r) = E \cdot R / (3R + 2r)$), we find that $V'_{2} > V_{2}$. Therefore, the brightness of $B_{2}$ increases.

(A) Let the total current flowing through the circuit be $I$.
Voltmeter $A$ has resistance $R$,voltmeter $B$ has resistance $1.5 R$,and voltmeter $C$ has resistance $3 R$.
Voltmeters $B$ and $C$ are connected in parallel. Let the potential difference across them be $V_p$.
Since they are in parallel,$V_2 = V_3 = V_p$.
Using the formula for parallel resistance,the equivalent resistance of $B$ and $C$ is $R_p = \frac{(1.5 R)(3 R)}{1.5 R + 3 R} = \frac{4.5 R^2}{4.5 R} = R$.
The total resistance of the circuit is $R_{eq} = R_A + R_p = R + R = 2 R$.
The reading of voltmeter $A$ is $V_1 = I \times R$.
The potential difference across the parallel combination is $V_p = I \times R_p = I \times R$.
Thus,$V_1 = V_2 = V_3 = IR$.

(C) Let the potential at point $D$ be $V_D$. Applying Kirchhoff's Current Law $(KCL)$ at node $D$,the sum of currents leaving the node must be zero:
$I_{AD} + I_{DB} + I_{DC} = 0$
$\frac{V_D - 70}{10} + \frac{V_D - 0}{20} + \frac{V_D - 10}{30} = 0$
Multiplying by $60$ to clear the denominators:
$6(V_D - 70) + 3(V_D) + 2(V_D - 10) = 0$
$6V_D - 420 + 3V_D + 2V_D - 20 = 0$
$11V_D = 440$
$V_D = 40 \,V$
Now,calculate the currents:
$I_{AD} = \frac{70 - 40}{10} = 3 \,A$ (current flows from $A$ to $D$)
$I_{DB} = \frac{40 - 0}{20} = 2 \,A$ (current flows from $D$ to $B$)
$I_{DC} = \frac{40 - 10}{30} = 1 \,A$ (current flows from $D$ to $C$)
Thus,the ratio of currents in paths $AD$,$DB$,and $DC$ is $3: 2: 1$.

(B) According to Bohr's quantization of angular momentum,the condition is given by:
$mvr = \frac{nh}{2\pi}$
Rearranging this,we get:
$\frac{h}{mv} = \frac{2\pi r}{n} \quad \dots(i)$
By definition,the de-Broglie wavelength $\lambda$ is given by:
$\lambda = \frac{h}{mv} \quad \dots(ii)$
Comparing equations $(i)$ and $(ii)$,we obtain:
$\lambda = \frac{2\pi r}{n}$
For the ground state of the hydrogen atom,$n = 1$ and the radius $r = 0.53 \ \text{Å}$.
Substituting these values:
$\lambda = \frac{2 \times \pi \times 0.53 \ \text{Å}}{1}$
$\lambda = 2 \times 3.14159 \times 0.53 \ \text{Å} \approx 3.33 \ \text{Å}$.

(C) According to Einstein's photoelectric equation:
$KE_{max} = h\nu - \phi_{0}$
Comparing this with the equation of a straight line $y = mx + c$,where $y = KE_{max}$,$x = \nu$,$m$ is the slope,and $c$ is the y-intercept:
$KE_{max} = h\nu + (-\phi_{0})$
Here,the slope $m = h$ (Planck's constant).
Since $h$ is a universal constant,the slope is the same for all metals and is independent of the intensity of the incident radiation.

(A) For a given perimeter,the area of a circle is the maximum among all plane shapes. As the irregular loop changes into a circular loop,its area $A$ increases.
Since the magnetic field $B$ is uniform and directed into the plane,the magnetic flux $\phi = B \cdot A$ increases.
According to Lenz's law,the induced current will create a magnetic field to oppose this increase in flux.
Therefore,the induced magnetic field must be directed out of the plane.
Using the right-hand thumb rule,a magnetic field directed out of the plane corresponds to an anticlockwise induced current.

(C) The speed of all electromagnetic waves in a vacuum is given by the formula $c = \frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}$.
Here,$\mu_{0}$ is the permeability of free space and $\varepsilon_{0}$ is the permittivity of free space.
Since both $\mu_{0}$ and $\varepsilon_{0}$ are universal constants,the speed of electromagnetic waves in a vacuum is a constant value,approximately $3 \times 10^{8} \ m/s$,regardless of the frequency,wavelength,or the source of the radiation.
Therefore,the speed is the same for all electromagnetic waves in a vacuum.

(B) Let the original charges on spheres $A$ and $B$ be $q_{1}$ and $q_{2}$ respectively.
Let the distance between the two spheres be $r$.
The initial force between them is $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$.
Since both spheres are identical,when they are brought into contact,the total charge is redistributed equally between them.
Thus,the new charge on each sphere is $q^{\prime} = \frac{q_{1} + q_{2}}{2}$.
The new force of repulsion between spheres $A$ and $B$ is $F^{\prime} = \frac{1}{4 \pi \varepsilon_{0}} \frac{q^{\prime} q^{\prime}}{r^{2}} = \frac{1}{4 \pi \varepsilon_{0}} \frac{(\frac{q_{1} + q_{2}}{2})^{2}}{r^{2}}$.
According to the Arithmetic Mean-Geometric Mean inequality,for any two positive numbers $q_{1}$ and $q_{2}$ where $q_{1} \neq q_{2}$,we have $(\frac{q_{1} + q_{2}}{2})^{2} > q_{1} q_{2}$.
Therefore,$F^{\prime} > F$.

(B) For sphere $1$,in equilibrium:
$T_{1} \cos \theta_{1} = M_{1} g$
$T_{1} \sin \theta_{1} = F$
Dividing the two equations,we get:
$\tan \theta_{1} = \frac{F}{M_{1} g}$
Similarly,for sphere $2$,in equilibrium:
$T_{2} \cos \theta_{2} = M_{2} g$
$T_{2} \sin \theta_{2} = F$
Dividing the two equations,we get:
$\tan \theta_{2} = \frac{F}{M_{2} g}$
Here,$F$ is the electrostatic force of repulsion between the two spheres,which is the same for both spheres according to Newton's third law.
If $\theta_{1} = \theta_{2}$,then $\tan \theta_{1} = \tan \theta_{2}$.
Therefore,$\frac{F}{M_{1} g} = \frac{F}{M_{2} g}$,which implies $M_{1} = M_{2}$.

(A) In a uniform electric field,the electric potential $V$ at a point $(x, y, z)$ is given by $V = -E \cdot r + C$,where $E$ is the electric field vector and $r$ is the position vector.
For a uniform electric field directed along the $x$-axis,the potential depends only on the $x$-coordinate $(V = -Ex + C)$.
Therefore,all points lying on a plane perpendicular to the electric field lines have the same electric potential.
In the given figure,there are $4$ rows of $3$ points each,forming a grid of $12$ points.
The shaded point is in the middle column.
The points that have the same potential as the shaded point are those that lie on the same vertical line (perpendicular to the electric field lines).
There are $3$ points in that vertical column,including the shaded point itself.
Thus,there are $2$ other points that have the same electric potential as the shaded point.

(C) The magnetic field at the centre of a circular coil of radius $R$ carrying current $i$ is given by $B = \frac{\mu_{0} i}{2 R}$.
The magnetic field at a point on the axis of the coil at a distance $x$ from the centre is given by $B_{axis} = \frac{\mu_{0} i R^{2}}{2(R^{2} + x^{2})^{3/2}}$.
Given $x = R$,we substitute this into the formula:
$B_{axis} = \frac{\mu_{0} i R^{2}}{2(R^{2} + R^{2})^{3/2}}$
$B_{axis} = \frac{\mu_{0} i R^{2}}{2(2R^{2})^{3/2}}$
$B_{axis} = \frac{\mu_{0} i R^{2}}{2(2^{3/2} R^{3})}$
$B_{axis} = \frac{\mu_{0} i}{2 R \cdot 2^{3/2}}$
Since $2^{3/2} = \sqrt{8}$,we have:
$B_{axis} = \frac{B}{\sqrt{8}}$.

(D) The magnetic field at the centre of a loop is zero if the current splits into two paths such that the magnetic fields produced by the two paths at the centre are equal in magnitude and opposite in direction.
In configuration $P$,the two paths consist of one thick and one thin wire each. Since the total resistance of each path is the same,the current divides equally. The magnetic fields produced by these two symmetric paths cancel each other out.
In configuration $R$,the two paths are also symmetric,each consisting of one thick and one thin wire connected in series. Thus,the current divides equally,and the magnetic field at the centre is zero.
In configuration $Q$,the two paths consist of two thick wires and two thin wires respectively. Since the resistances are different,the currents are unequal,and the magnetic fields do not cancel out.
Therefore,the magnetic field at the centre is zero in cases $P$ and $R$.

(D) The loop consists of four parts: two radial segments ($AB$ and $CD$) and two circular arcs ($DA$ and $BC$).
$(i)$ For the radial segments $AB$ and $CD$,the magnetic field at the center $M$ is zero because the point $M$ lies on the line of the current elements.
(ii) For the circular arc $DA$ with radius $R$ and angle $\theta_1 = 270^{\circ} = \frac{3\pi}{2}$ radians,the magnetic field at $M$ is $B_1 = \frac{\mu_0 i \theta_1}{4\pi R} = \frac{\mu_0 i (3\pi/2)}{4\pi R} = \frac{3\mu_0 i}{8R}$. By the right-hand rule,the direction is into the plane of the paper.
(iii) For the circular arc $BC$ with radius $2R$ and angle $\theta_2 = 90^{\circ} = \frac{\pi}{2}$ radians,the magnetic field at $M$ is $B_2 = \frac{\mu_0 i \theta_2}{4\pi (2R)} = \frac{\mu_0 i (\pi/2)}{8\pi R} = \frac{\mu_0 i}{16R}$. By the right-hand rule,the direction is into the plane of the paper.
(iv) The total magnetic field at $M$ is $B = B_1 + B_2 = \frac{3\mu_0 i}{8R} + \frac{\mu_0 i}{16R} = \frac{6\mu_0 i + \mu_0 i}{16R} = \frac{7\mu_0 i}{16R}$.
Since both fields are directed into the plane of the paper,the resultant field is $\frac{7}{16} \frac{\mu_0 i}{R}$ into the plane of the paper.

(B) Let the distance of $M$ from each wire be $r$. The magnetic field due to a long straight wire carrying current $I$ at a distance $r$ is given by $B = \frac{\mu_0 I}{2 \pi r}$.
In the first case,the currents are $I_1 = 2 \text{ A}$ and $I_2 = 1 \text{ A}$ in the same direction. Using the right-hand rule,the magnetic fields at $M$ due to these wires are in opposite directions.
$B_{PQ} = \frac{\mu_0 (2)}{2 \pi r} = \frac{\mu_0}{\pi r}$
$B_{RS} = \frac{\mu_0 (1)}{2 \pi r} = \frac{\mu_0}{2 \pi r}$
The net magnetic field at $M$ is $B = B_{PQ} - B_{RS} = \frac{\mu_0}{\pi r} - \frac{\mu_0}{2 \pi r} = \frac{\mu_0}{2 \pi r}$.
When the current $2 \text{ A}$ is switched off,only the current $I_2 = 1 \text{ A}$ remains.
The new magnetic field at $M$ is $B' = B_{RS} = \frac{\mu_0 (1)}{2 \pi r} = \frac{\mu_0}{2 \pi r}$.
Comparing the two,we get $B' = B$.

(C) Since the electron moves without any change in direction,the net force on it must be zero. This means the electrostatic force is balanced by the magnetic force.
$F_e = F_m$
$qE = qvB$
$v = \frac{E}{B}$
For a parallel plate capacitor,the electric field is given by $E = \frac{\sigma}{\varepsilon_{0}}$.
Substituting this into the velocity equation:
$v = \frac{\sigma}{\varepsilon_{0} B}$
The time $t$ taken to travel a distance $l$ is given by:
$t = \frac{l}{v} = \frac{l}{\frac{\sigma}{\varepsilon_{0} B}} = \frac{\varepsilon_{0} l B}{\sigma}$

(D) The force on a moving charge in a magnetic field is given by the Lorentz force formula: $\vec{F} = q(\vec{v} \times \vec{B})$.
Here,the velocity vector $\vec{v}$ is directed upwards (let this be the $+z$ direction).
The magnetic field $\vec{B}$ is directed towards the north (let this be the $+y$ direction).
Using the right-hand rule for the cross product $\vec{v} \times \vec{B}$:
Point your fingers in the direction of $\vec{v}$ (upwards) and curl them towards $\vec{B}$ (north).
The thumb points in the direction of the force,which is towards the west ($-x$ direction).
Therefore,the correct option is $D$.

(B) Energy is released in a nuclear reaction if the total binding energy of the products is greater than the total binding energy of the reactants. This corresponds to an increase in the specific binding energy (binding energy per nucleon).
From the graph:
$1$. For $1 < A < 100$,the specific binding energy is $2 \text{ MeV/nucleon}$.
$2$. For $100 < A < 200$,the specific binding energy is $8 \text{ MeV/nucleon}$.
$3$. For $200 < A < 250$,the specific binding energy is $4 \text{ MeV/nucleon}$.
If two nuclei with mass numbers in the range $51 < A < 100$ (each with specific binding energy $2 \text{ MeV/nucleon}$) fuse to form a nucleus with mass number in the range $100 < A < 200$ (with specific binding energy $8 \text{ MeV/nucleon}$),the final state has a higher specific binding energy. Thus,energy is released.

(B) The packing fraction $(f)$ is defined as the mass defect per nucleon,given by $f = \frac{m - A}{A}$,where $m$ is the mass of the nucleus and $A$ is the mass number.
Packing fraction is a measure of the stability of a nucleus.
$A$ smaller (or more negative) value of the packing fraction indicates higher stability of the nucleus.
The packing fraction can be positive,negative,or zero depending on the mass number of the nucleus.
Therefore,the statement that the packing fraction may be positive or negative is correct.

(A) The activity $A$ of a radioactive sample is given by $A = \lambda N = \frac{0.693}{T_{1/2}} N$,where $N$ is the number of nuclei and $T_{1/2}$ is the half-life.
From the given information,$N_{1} = 2 N_{2}$ and $A_{2} = 2 A_{1}$.
We can write the ratio of activities as:
$\frac{A_{1}}{A_{2}} = \frac{N_{1} / T_{1}}{N_{2} / T_{2}} = \frac{N_{1}}{N_{2}} \times \frac{T_{2}}{T_{1}}$
Rearranging for the ratio of half-lives $\frac{T_{1}}{T_{2}}$:
$\frac{T_{1}}{T_{2}} = \frac{A_{2}}{A_{1}} \times \frac{N_{1}}{N_{2}}$
Substituting the given values:
$\frac{T_{1}}{T_{2}} = \frac{2 A_{1}}{A_{1}} \times \frac{2 N_{2}}{N_{2}} = 2 \times 2 = 4$
Thus,the ratio of the half-life of $S_{1}$ to the half-life of $S_{2}$ is $4$.

(D) $\beta^{-}$-particle is an electron. The emission of a $\beta^{-}$-particle involves the transformation of a neutron into a proton,an electron,and a third particle called an antineutrino $(\bar{\nu})$.
The nuclear reaction is given by:
${ }_{0} n^{1} \rightarrow { }_{1} p^{1} + { }_{-1} e^{0} + \bar{\nu}$

(A) Given: Object distance $u = -15 \,cm$ (as it is placed to the left of the pole).
Radius of curvature $R = +30 \,cm$ (as the center of curvature is to the right of the pole).
Refractive index of air $\mu_1 = 1$.
Refractive index of glass $\mu_2 = 1.5$.
Using the refraction formula at a spherical surface:
$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$
Substituting the values:
$\frac{1.5}{v} - \frac{1}{-15} = \frac{1.5 - 1}{30}$
$\frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{30}$
$\frac{1.5}{v} = \frac{1}{60} - \frac{1}{15}$
$\frac{1.5}{v} = \frac{1 - 4}{60} = \frac{-3}{60} = -\frac{1}{20}$
$v = 1.5 \times (-20) = -30 \,cm$.
The negative sign indicates that the image is formed $30 \,cm$ to the left of the pole,on the same side as the object.

(C) The lens maker's formula is given by $\frac{1}{f} = (\mu - 1) \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]$.
For a plano-convex lens, one surface is plane $(R_1 = 60 \,cm)$ and the other is plane $(R_2 = \infty)$.
Given refractive index $\mu = 1.6$ and radius of curvature $R_1 = 60 \,cm$.
Substituting the values in the formula:
$\frac{1}{f} = (1.6 - 1) \left[ \frac{1}{60} - \frac{1}{\infty} \right]$
$\frac{1}{f} = 0.6 \times \left[ \frac{1}{60} - 0 \right]$
$\frac{1}{f} = \frac{0.6}{60} = \frac{6}{600} = \frac{1}{100}$
Therefore, the focal length $f = 100 \,cm$.

(D) For a prism,the angle of deviation is given by $\delta = (i + e) - A$.
In the position of minimum deviation,the angle of incidence is equal to the angle of emergence,i.e.,$i = e$.
Under this condition,the light ray inside the prism travels parallel to the base,and the angle of refraction $r$ is given by $r_1 = r_2 = r$.
Since $A = r_1 + r_2$,we have $A = 2r$.
Given the prism angle $A = 60^{\circ}$,the angle of refraction is $r = \frac{A}{2} = \frac{60^{\circ}}{2} = 30^{\circ}$.
This condition is independent of the wavelength or colour of the light.
Therefore,the angle of refraction is $30^{\circ}$ for both red and violet colours.

(A) Applying Snell's law at the interface of media $A$ and $B$:
$n_{1} \sin i = n_{2} \sin r_{1} \quad \text{...(i)}$
Applying Snell's law at the interface of media $B$ and $C$:
$n_{2} \sin r_{1} = n_{3} \sin r_{2} \quad \text{...(ii)}$
Since the ray grazes the surface of separation of media $B$ and $C$,the angle of refraction $r_{2} = 90^{\circ}$.
Substituting $r_{2} = 90^{\circ}$ in equation (ii):
$n_{2} \sin r_{1} = n_{3} \sin 90^{\circ} = n_{3}$
Now,substitute $n_{2} \sin r_{1} = n_{3}$ into equation $(i)$:
$n_{1} \sin i = n_{3}$
$\sin i = \frac{n_{3}}{n_{1}}$

(D) When light travels from a rarer medium (air) to a denser medium (water),its frequency remains constant,but its wavelength changes.
The wavelength in water is given by the formula $\lambda_{w} = \frac{\lambda_{a}}{n_{w}}$,where $\lambda_{a}$ is the wavelength in air and $n_{w}$ is the refractive index of water.
Substituting the given values: $\lambda_{w} = \frac{500 \ nm}{4/3} = 500 \times \frac{3}{4} \ nm = 375 \ nm$.
Rounding this value,we get $\lambda_{w} \approx 376 \ nm$.
Since the wavelength of light shifts towards the blue end of the spectrum ($376 \ nm$ falls in the blue region),the diver will observe the light as blue.

(C) When a $p-n$ junction is formed,electrons diffuse from the $n$-region to the $p$-region,and holes diffuse from the $p$-region to the $n$-region. This leaves behind immobile ionized donors in the $n$-region (positive charge) and immobile ionized acceptors in the $p$-region (negative charge).
This creates a depletion region at the junction with a built-in potential difference. The $n$-side becomes positive relative to the $p$-side.
Since the electric field lines are directed from positive potential to negative potential,the electric field at the junction is directed from the $n$-type side to the $p$-type side.

(C) The energy required to generate an electron-hole pair is equal to the forbidden energy gap $E_{g}$.
For a photon to create an electron-hole pair,its energy must be at least equal to $E_{g}$.
The relationship between energy and wavelength is given by $E = \frac{hc}{\lambda}$.
To find the maximum wavelength $\lambda_{max}$,we use the minimum energy $E_{g}$:
$\lambda_{max} = \frac{hc}{E_{g}}$
Substituting the given values:
$\lambda_{max} = \frac{12400 eV-Å}{0.72 eV}$
$\lambda_{max} = 17222.22 Å$
Rounding to the nearest integer,we get $\lambda_{max} = 17222 Å$.

(A) The given circuit consists of a $NOR$ gate followed by a $NOR$ gate with its inputs shorted together,which acts as a $NOT$ gate.
Let the inputs to the first $NOR$ gate be $A$ and $B$. The output of the first $NOR$ gate is $Y' = \overline{A+B}$.
This output $Y'$ is fed into both inputs of the second $NOR$ gate. The output of the second $NOR$ gate is $Y = \overline{Y' + Y'} = \overline{Y'} = \overline{\overline{A+B}}$.
Using the law of double negation,$\overline{\overline{X}} = X$,we get $Y = A+B$.
This is the Boolean expression for an $OR$ gate.

(B) The condition for principal maxima in a diffraction grating is given by $d \sin \theta = n \lambda$, where $d = \frac{1}{N}$ is the grating element and $N$ is the number of lines per unit length.
Thus, $\frac{\sin \theta}{N} = n \lambda$, which implies $n = \frac{\sin \theta}{N \lambda}$.
Given $\lambda = 625 \, nm = 6.25 \times 10^{-7} \, m$ and $N = 2 \times 10^{5} \, \text{lines}/m$.
The maximum possible order $n$ is determined by the condition $\sin \theta \leq 1$, so $n < \frac{1}{N \lambda}$.
$n < \frac{1}{(2 \times 10^{5}) \times (6.25 \times 10^{-7})} = \frac{1}{0.125} = 8$.
Since $n$ must be an integer, the maximum order is $n = 8$.
The total number of maxima observed is given by $2n + 1$ (including the central maximum at $n=0$, and $n$ orders on both sides).
Total maxima $= 2(8) + 1 = 17$.

(A) The condition for constructive interference in reflected light from a thin film is given by $2 \mu t = (2n + 1) \frac{\lambda}{2}$, where $n = 0, 1, 2, \dots$ and $\mu$ is the refractive index, $t$ is the thickness, and $\lambda$ is the wavelength.
For the minimum thickness, we take $n = 0$.
Substituting the values: $2 \mu t = \frac{\lambda}{2}$.
Rearranging for $t$: $t = \frac{\lambda}{4 \mu}$.
Given $\lambda = 600 \, nm$ and $\mu = 1.5$, we get $t = \frac{600}{4 \times 1.5} = \frac{600}{6} = 100 \, nm$.

(C) The critical angle $C$ is given by $\sin(C) = 0.6$.
The refractive index $\mu$ of the medium is related to the critical angle by the formula $\mu = \frac{1}{\sin(C)}$.
Substituting the value,$\mu = \frac{1}{0.6} = \frac{10}{6} = 1.6667$.
According to Brewster's Law,the polarizing angle $i_p$ is given by $\tan(i_p) = \mu$.
Therefore,$i_p = \tan^{-1}(\mu) = \tan^{-1}(1.6667)$.

(D) The phase difference $\phi$ is related to the path difference $\Delta x$ by the formula: $\phi = \frac{2\pi}{\lambda} \times \Delta x$.
Given the path difference $\Delta x = \frac{\lambda}{6}$,we calculate the phase difference: $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3} = 60^{\circ}$.
The intensity $I$ at any point on the screen is given by $I = I_{0} \cos^{2}\left(\frac{\phi}{2}\right)$,where $I_{0}$ is the maximum intensity.
Substituting the value of $\phi$: $\frac{I}{I_{0}} = \cos^{2}\left(\frac{60^{\circ}}{2}\right) = \cos^{2}(30^{\circ})$.
Since $\cos(30^{\circ}) = \frac{\sqrt{3}}{2}$,we have $\frac{I}{I_{0}} = \left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{3}{4} = 0.75$.

(D) According to the Rayleigh scattering law,the intensity of scattered light $(I)$ is proportional to the fourth power of its frequency $(f)$,i.e.,$I \propto f^4$.
Given the ratio of intensities $\frac{I_1}{I_2} = \frac{256}{81}$.
Therefore,the ratio of frequencies is $\frac{f_1}{f_2} = (\frac{I_1}{I_2})^{1/4} = (\frac{256}{81})^{1/4} = \frac{4}{3}$.
Since the ratio $4:3$ is not provided in the options,the correct answer is 'None of these'.