KCET 2011 Physics Question Paper with Answer and Solution

(C) According to Kepler's third law of planetary motion,the square of the time period $T$ of a satellite is proportional to the cube of the orbital radius $R$,given by $T^2 \propto R^3$.
This implies $T = 2\pi \sqrt{\frac{R^3}{GM}}$,where $G$ is the gravitational constant and $M$ is the mass of the planet.
Note that the time period $T$ depends only on the radius of the orbit $R$ and the mass of the central planet $M$.
It is independent of the mass of the satellite $m$.
Since both satellites orbit the same planet at the same radius $R$,their time periods will be equal.
Therefore,the ratio of their periods of revolution is $1: 1$.

(A) The heat required to raise the temperature of a body is given by $Q = mc\Delta T$,where $m$ is the mass,$c$ is the specific heat capacity,and $\Delta T$ is the change in temperature.
Since the spheres are made of the same material (copper),$c$ is constant. For a given $\Delta T = 1 \ K$,$Q \propto m$.
The mass $m$ of a sphere is given by $m = \rho V$,where $\rho$ is the density and $V$ is the volume.
$V = \frac{4}{3}\pi r^3$,so $m \propto r^3$.
Therefore,$Q \propto r^3$.
The ratio of heat required is $\frac{Q_1}{Q_2} = \frac{r_1^3}{r_2^3} = \left(\frac{r_1}{r_2}\right)^3$.
Given $r_1 = 1.5 r_2$,we have $\frac{r_1}{r_2} = 1.5 = \frac{3}{2}$.
Thus,$\frac{Q_1}{Q_2} = \left(\frac{3}{2}\right)^3 = \frac{27}{8}$.

(D) According to Charles's Law,for a fixed mass of gas at constant pressure,the volume is directly proportional to its absolute temperature: $V \propto T$ or $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
Given initial temperature $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$.
Let initial volume $V_1 = V$. Then final volume $V_2 = 2V$.
Substituting these values into the formula: $\frac{V}{300} = \frac{2V}{T_2}$.
Solving for $T_2$: $T_2 = 300 \times 2 = 600 \ K$.
Converting the final temperature to Celsius: $T_2 = 600 - 273 = 327^{\circ} C$.
The increase in temperature is $\Delta T = T_2 - T_1 = 327^{\circ} C - 27^{\circ} C = 300^{\circ} C$.

(D) $1$. Total mass of the system $(A + B) = 2 \, kg + 8 \, kg = 10 \, kg$.
$2$. The maximum static friction (limiting friction) between block $B$ and the horizontal floor is given by $f_{max} = \mu_s N$, where $N$ is the normal reaction force from the floor.
$3$. The normal reaction $N = (m_A + m_B)g = 10 \, kg \times 10 \, ms^{-2} = 100 \, N$.
$4$. Therefore, $f_{max} = 0.5 \times 100 \, N = 50 \, N$.
$5$. The applied horizontal force on block $B$ is $F_{app} = 10 \, N$.
$6$. Since $F_{app} < f_{max}$ $(10 \, N < 50 \, N)$, the entire system remains at rest.
$7$. Because the system is at rest and no external force is acting on block $A$ to cause relative motion between $A$ and $B$, the static friction force between blocks $A$ and $B$ is zero.

(D) Let the two forces be $F_1$ and $F_2$. The angle between them is $\theta = 120^{\circ}$.
Let the resultant $R = 10 \,kg$-wt be perpendicular to force $F_1$.
Using the formula for the angle $\alpha$ that the resultant makes with force $F_1$:
$\tan \alpha = \frac{F_2 \sin \theta}{F_1 + F_2 \cos \theta}$
Since the resultant is perpendicular to $F_1$, $\alpha = 90^{\circ}$, so $\tan 90^{\circ} = \infty$.
This implies the denominator must be zero: $F_1 + F_2 \cos 120^{\circ} = 0$.
$F_1 + F_2 (-0.5) = 0 \implies F_1 = 0.5 F_2 \implies F_2 = 2 F_1$.
The magnitude of the resultant is given by $R^2 = F_1^2 + F_2^2 + 2 F_1 F_2 \cos 120^{\circ}$.
$10^2 = F_1^2 + (2 F_1)^2 + 2 F_1 (2 F_1) (-0.5)$.
$100 = F_1^2 + 4 F_1^2 - 2 F_1^2 = 3 F_1^2$.
$F_1^2 = \frac{100}{3} \implies F_1 = \frac{10}{\sqrt{3}} \,kg$-wt.
Thus, the force perpendicular to the resultant is $\frac{10}{\sqrt{3}} \,kg$-wt.

(C) The height of water in a capillary tube is given by the formula:
$h = \frac{2T \cos \theta}{r \rho g}$
where $T$ is the surface tension,$\theta$ is the angle of contact,$r$ is the radius of the capillary tube,$\rho$ is the density of the liquid,and $g$ is the acceleration due to gravity.
From this formula,we can see that $h \propto \frac{1}{r}$.
Since the height $h$ is inversely proportional to the radius $r$ of the tube,the rise of water will be greater in the tube with the smaller diameter (smaller radius).

(A) Let the radius of each small drop be $r$ and the radius of the big drop be $R$. Since the volume remains constant, the volume of $8$ small drops equals the volume of the big drop:
$8 \cdot (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$
$R^3 = 8r^3 \implies R = 2r$.
Terminal velocity $v_t$ is given by the formula $v_t = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta}$, which implies $v_t \propto r^2$.
Let $v_1$ be the terminal velocity of the small drop and $v_2$ be the terminal velocity of the big drop.
$\frac{v_2}{v_1} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4$.
Given $v_1 = 10 \,cm \,s^{-1}$, we have $v_2 = 4 \cdot 10 \,cm \,s^{-1} = 40 \,cm \,s^{-1}$.

(A) The extension $e$ of a wire is given by the formula $e = \frac{FL}{AY}$,where $F$ is the applied force,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.
Since $A = \pi r^2$,we have $e = \frac{FL}{\pi r^2 Y}$.
Given that $F$ and $Y$ are the same for all wires,the extension is proportional to the ratio $\frac{L}{r^2}$,i.e.,$e \propto \frac{L}{r^2}$.
Let us calculate the value of $\frac{L}{r^2}$ for each case:
For option $A$: $\frac{100}{(0.2)^2} = \frac{100}{0.04} = 2500$.
For option $B$: $\frac{200}{(0.4)^2} = \frac{200}{0.16} = 1250$.
For option $C$: $\frac{300}{(0.6)^2} = \frac{300}{0.36} \approx 833.3$.
For option $D$: $\frac{400}{(0.8)^2} = \frac{400}{0.64} = 625$.
Comparing these values,the wire in option $A$ has the largest value of $\frac{L}{r^2}$,and therefore,it will have the largest extension.

(D) By definition,the slope of a displacement-time graph represents the velocity of the particle.
$v = \tan \theta$
Given the angles $\theta_1 = 30^{\circ}$ and $\theta_2 = 45^{\circ}$.
The ratio of their velocities is given by:
$\frac{v_1}{v_2} = \frac{\tan \theta_1}{\tan \theta_2}$
Substituting the values:
$\frac{v_1}{v_2} = \frac{\tan 30^{\circ}}{\tan 45^{\circ}}$
Since $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$ and $\tan 45^{\circ} = 1$,we get:
$\frac{v_1}{v_2} = \frac{1/\sqrt{3}}{1} = \frac{1}{\sqrt{3}}$
Thus,the ratio is $1: \sqrt{3}$.

(C) The equations of motion for the projectile are given as:
$x = 6t$ $(i)$
$y = 8t - 5t^2$ (ii)
Comparing these with the standard equations of motion for a projectile:
$x = (u \cos \theta)t$
$y = (u \sin \theta)t - \frac{1}{2}gt^2$
From equation $(i)$,the horizontal component of velocity is $u_x = u \cos \theta = 6 \text{ m/s}$.
From equation (ii),the vertical component of velocity is $u_y = u \sin \theta = 8 \text{ m/s}$.
The magnitude of the initial projection velocity $u$ is given by:
$u = \sqrt{u_x^2 + u_y^2}$
$u = \sqrt{6^2 + 8^2}$
$u = \sqrt{36 + 64}$
$u = \sqrt{100}$
$u = 10 \text{ m/s}$.

(B) The displacement equation for a particle executing simple harmonic motion starting from the mean position is given by $y = a \sin(\omega t)$, where $\omega = \frac{2\pi}{T}$.
Given that the particle moves from the mean position $(y = 0)$ to half the amplitude $(y = a/2)$, we substitute these values into the equation:
$\frac{a}{2} = a \sin\left(\frac{2\pi}{T} t\right)$
$\frac{1}{2} = \sin\left(\frac{2\pi}{T} t\right)$
Since $\sin(\pi/6) = 1/2$, we have:
$\frac{2\pi}{T} t = \frac{\pi}{6}$
Solving for $t$:
$t = \frac{T}{12}$
Given the period $T = 6 \,s$:
$t = \frac{6}{12} = 0.5 \,s = \frac{1}{2} \,s$.

(D) The total kinetic energy $(KE)$ of a body rolling without slipping is the sum of its translational kinetic energy and rotational kinetic energy.
$KE = KE_{\text{trans}} + KE_{\text{rot}}$
$KE = \frac{1}{2} m v^{2} + \frac{1}{2} I \omega^{2}$
For a solid sphere,the moment of inertia $I = \frac{2}{5} m R^{2}$ and the condition for rolling without slipping is $v = R \omega$,which implies $\omega = \frac{v}{R}$.
Substituting these into the equation:
$KE = \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{5} m R^{2}) (\frac{v}{R})^{2}$
$KE = \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{5} m v^{2})$
$KE = \frac{1}{2} m v^{2} (1 + \frac{2}{5})$
$KE = \frac{1}{2} m v^{2} (\frac{7}{5})$
$KE = \frac{7}{10} m v^{2}$

(A) Given that the rods are identical,let the length of each rod be $l$ and the cross-sectional area be $A_{area}$.
Let $K_{A}, K_{B}, K_{C}$ be the thermal conductivities of rods $A, B, C$ respectively.
According to the problem:
$K_{B} = \frac{1}{2} K_{A}$
$K_{B} = 3 K_{C} \implies K_{C} = \frac{K_{B}}{3} = \frac{K_{A}}{6}$
Since the rods are connected in series,the equivalent thermal resistance $R_{eq}$ is the sum of individual resistances:
$R_{eq} = R_{A} + R_{B} + R_{C}$
$\frac{3l}{K_{eq} A_{area}} = \frac{l}{K_{A} A_{area}} + \frac{l}{K_{B} A_{area}} + \frac{l}{K_{C} A_{area}}$
$\frac{3}{K_{eq}} = \frac{1}{K_{A}} + \frac{1}{K_{A}/2} + \frac{1}{K_{A}/6}$
$\frac{3}{K_{eq}} = \frac{1}{K_{A}} + \frac{2}{K_{A}} + \frac{6}{K_{A}}$
$\frac{3}{K_{eq}} = \frac{9}{K_{A}}$
$K_{eq} = \frac{3 K_{A}}{9} = \frac{1}{3} K_{A}$

(C) According to Wien's displacement law, the wavelength corresponding to maximum intensity $(\lambda_{m})$ is inversely proportional to the absolute temperature $(T)$, i.e., $\lambda_{m} T = \text{constant}$.
Since the frequency $\nu_{m}$ is related to wavelength by $\nu_{m} = c / \lambda_{m}$, we can substitute $\lambda_{m} = c / \nu_{m}$ into the displacement law.
This gives $(c / \nu_{m}) T = \text{constant}$, which simplifies to $\nu_{m} / T = \text{constant}'$ or $\nu_{m} \propto T$.
Therefore, the graph of $\nu_{m}$ versus $T$ for a perfectly black body is a straight line passing through the origin, which corresponds to line $C$ in the given figure.

(B) The given wave equation is $y = 10 \sin \left( \frac{2 \pi}{45} t + \alpha \right)$.
At $t = 0$,the displacement $y = 5 \text{ cm}$.
Substituting these values: $5 = 10 \sin \left( \frac{2 \pi}{45} \times 0 + \alpha \right) = 10 \sin \alpha$.
Thus,$\sin \alpha = \frac{5}{10} = \frac{1}{2}$,which gives $\alpha = \frac{\pi}{6}$.
The total phase of the wave is given by $\phi = \frac{2 \pi}{45} t + \alpha$.
At $t = 7.5 \text{ s} = \frac{15}{2} \text{ s}$,the total phase is:
$\phi = \frac{2 \pi}{45} \times \frac{15}{2} + \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{6}$.
$\phi = \frac{2 \pi + \pi}{6} = \frac{3 \pi}{6} = \frac{\pi}{2}$.

(C) Let the unknown frequency be $n \,Hz$.
The beat frequency with $A$ $(n_A = 258 \,Hz)$ is $x = |n - 258|$.
The beat frequency with $B$ $(n_B = 262 \,Hz)$ is $2x = |n - 262|$.
If $n < 258$, then $x = 258 - n$.
Then $2x = |n - 262| = 262 - n$.
Substituting $x$: $2(258 - n) = 262 - n$.
$516 - 2n = 262 - n$.
$n = 516 - 262 = 254 \,Hz$.
Checking the condition: If $n = 254 \,Hz$, beats with $A$ is $|254 - 258| = 4 \,Hz$.
Beats with $B$ is $|254 - 262| = 8 \,Hz$.
Since $8 = 2 \times 4$, the condition is satisfied.

$(A)$ The fundamental frequency of a stretched wire is given by the formula: $f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$, where $\mu = \pi r^2 \rho$ is the linear mass density.
Substituting $\mu$, we get: $f = \frac{1}{2lr} \sqrt{\frac{T}{\pi \rho}}$.
Given the initial conditions: $f_1 = 600 \,Hz$, length $l_1 = l$, radius $r_1 = r$, and tension $T_1 = T$.
New conditions: $l_2 = 2l$, $r_2 = r/2$, and $T_2 = T/9$.
The ratio of frequencies is: $\frac{f_2}{f_1} = \frac{l_1}{l_2} \times \frac{r_1}{r_2} \times \sqrt{\frac{T_2}{T_1}}$.
Substituting the values: $\frac{f_2}{600} = \frac{l}{2l} \times \frac{r}{r/2} \times \sqrt{\frac{T/9}{T}}$.
$\frac{f_2}{600} = \frac{1}{2} \times 2 \times \sqrt{\frac{1}{9}} = 1 \times \frac{1}{3} = \frac{1}{3}$.
Therefore, $f_2 = \frac{600}{3} = 200 \,Hz$.

(A) Given: Mass $m = 5 \,kg$,Initial Kinetic Energy $KE_i = 490 \,J$,Acceleration due to gravity $g = 9.8 \,ms^{-2}$.
At the required height $h$,the kinetic energy $KE_f$ becomes half of the initial value:
$KE_f = \frac{1}{2} KE_i = \frac{490}{2} = 245 \,J$.
According to the law of conservation of energy,the total mechanical energy remains constant:
$KE_i = KE_f + PE_f$
$KE_i = KE_f + mgh$
Substituting the values:
$490 = 245 + 5 \times 9.8 \times h$
$490 - 245 = 49 \times h$
$245 = 49h$
$h = \frac{245}{49} = 5 \,m$.
Therefore,the height is $5 \,m$.

(D) transformer is an electrical device that transfers electrical energy between two or more circuits through electromagnetic induction.
It consists of two coils,the primary coil and the secondary coil,which are magnetically coupled.
When an alternating current flows through the primary coil,it produces a changing magnetic flux.
This changing magnetic flux is linked to the secondary coil,inducing an electromotive force $(EMF)$ in it.
This phenomenon,where a change in current in one coil induces an $EMF$ in a nearby coil,is known as mutual induction.
Therefore,a transformer works on the principle of mutual induction.

(A) The given equations are $V = V_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t + \phi)$.
Comparing with the given values,$V_0 = 150 \text{ V}$,$I_0 = 150 \text{ A}$,and the phase difference $\phi = \pi/3 = 60^\circ$.
The average power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.
Since $V_{rms} = V_0 / \sqrt{2}$ and $I_{rms} = I_0 / \sqrt{2}$,the formula becomes $P = \frac{1}{2} V_0 I_0 \cos \phi$.
Substituting the values: $P = \frac{1}{2} \times 150 \times 150 \times \cos(60^\circ)$.
Since $\cos(60^\circ) = 0.5$,we get $P = 0.5 \times 150 \times 150 \times 0.5$.
$P = 0.25 \times 22500 = 5625 \text{ W}$.

(D) Given: Inductance $L = 1 \text{ H}$,Capacitance $C = 20 \mu\text{F} = 20 \times 10^{-6} \text{ F}$,Resistance $R = 300 \Omega$,Frequency $f = \frac{50}{\pi} \text{ Hz}$.
First,calculate the inductive reactance $X_{L}$:
$X_{L} = 2 \pi f L = 2 \pi \left(\frac{50}{\pi}\right) \times 1 = 100 \Omega$.
Next,calculate the capacitive reactance $X_{C}$:
$X_{C} = \frac{1}{2 \pi f C} = \frac{1}{2 \pi \left(\frac{50}{\pi}\right) \times 20 \times 10^{-6}} = \frac{1}{100 \times 20 \times 10^{-6}} = \frac{1}{2 \times 10^{-3}} = 500 \Omega$.
The impedance $Z$ of a series $L-C-R$ circuit is given by:
$Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}$.
Substituting the values:
$Z = \sqrt{(300)^{2} + (500 - 100)^{2}} = \sqrt{300^{2} + 400^{2}} = \sqrt{90000 + 160000} = \sqrt{250000} = 500 \Omega$.

(A) The flash spectrum is a series of bright emission lines observed in the solar spectrum for a few seconds during a total solar eclipse. When the moon completely covers the bright photosphere of the sun,the thin layer of the chromosphere becomes visible,producing this unique emission spectrum.

(A) The wavelength $\lambda$ of the radiation absorbed when an electron transitions from orbit $n_1$ to $n_2$ is given by the Rydberg formula:
$\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$
Given $n_1 = 2$ and $n_2 = 4$:
$\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} \right]$
$\frac{1}{\lambda} = R \left[ \frac{1}{4} - \frac{1}{16} \right]$
$\frac{1}{\lambda} = R \left[ \frac{4 - 1}{16} \right]$
$\frac{1}{\lambda} = \frac{3 R}{16}$
Therefore,the wavelength is $\lambda = \frac{16}{3 R}$.

(C) Rutherford's alpha-particle scattering experiment led to the discovery of the nucleus. His atomic model proposed that the entire positive charge and most of the mass of an atom are concentrated in a very small central region called the nucleus. Therefore,it could successfully account for the positively charged central core of an atom. It failed to explain the stability of atoms and the origin of line spectra,which were later addressed by Bohr's model.

(C) First, simplify the network to the of capacitor $C$. The $6 \mu F$ and $12 \mu F$ capacitors are in series, giving $C_{6,12} = \frac{6 \times 12}{6 + 12} = 4 \mu F$. This $4 \mu F$ is in parallel with the $2 \mu F$ capacitor, giving $C_{p1} = 4 + 2 = 6 \mu F$. This $6 \mu F$ is in series with the $4 \mu F$ capacitor, giving $C_{s1} = \frac{6 \times 4}{6 + 4} = 2.4 \mu F$.
Next, the $1 \mu F$ and $2 \mu F$ capacitors are in series, giving $C_{1,2} = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \mu F$. This is in parallel with the $2 \mu F$ capacitor, giving $C_{p2} = \frac{2}{3} + 2 = \frac{8}{3} \mu F$.
These two branches are in parallel, so $C_{eq_rest} = 2.4 + \frac{8}{3} = \frac{12}{5} + \frac{8}{3} = \frac{36 + 40}{15} = \frac{76}{15} \mu F$.
Finally, this is in series with the $8 \mu F$ capacitor, giving $C_{total_rest} = \frac{(\frac{76}{15}) \times 8}{(\frac{76}{15}) + 8} = \frac{608}{76 + 120} = \frac{608}{196} = \frac{152}{49} \mu F$.
Given the total equivalent capacitance is $3 \mu F$, we have $\frac{C \times (152/49)}{C + (152/49)} = 3$. Solving for $C$ gives $C = 48 \mu F$.

(B) The color code for resistors is determined by the sequence: Green,Blue,Brown,Silver.
$1$. The first color (Green) represents the first digit: $5$.
$2$. The second color (Blue) represents the second digit: $6$.
$3$. The third color (Brown) represents the multiplier: $10^1$.
$4$. The fourth color (Silver) represents the tolerance: $\pm 10 \%$.
Combining these,the resistance $R = 56 \times 10^1 \Omega \pm 10 \% = 560 \Omega \pm 10 \%$.

(B) First,calculate the equivalent resistance of the parallel combination of $12 \Omega$ and $4 \Omega$ resistors:
$R_p = \frac{12 \times 4}{12 + 4} = \frac{48}{16} = 3 \Omega$
Now,calculate the total resistance of the circuit including the series resistor and internal resistance:
$R_{total} = R_p + 2 \Omega + 1 \Omega = 3 \Omega + 2 \Omega + 1 \Omega = 6 \Omega$
The total current $I$ flowing from the battery is:
$I = \frac{V}{R_{total}} = \frac{12 \text{ V}}{6 \Omega} = 2 \text{ A}$
Using the current divider rule for parallel resistors:
$i_1 = I \times \frac{R_2}{R_1 + R_2} = 2 \times \frac{4}{12 + 4} = 2 \times \frac{4}{16} = 0.5 \text{ A}$
$i_2 = I \times \frac{R_1}{R_1 + R_2} = 2 \times \frac{12}{12 + 4} = 2 \times \frac{12}{16} = 1.5 \text{ A}$
Thus,$i_1 = 0.5 \text{ A}$ and $i_2 = 1.5 \text{ A}$.

(B) The relationship between current $I$ and drift velocity $v_{d}$ is given by the formula: $I = n A e v_{d}$.
Here, $n = 10^{29} \,m^{-3}$ is the electron density, $A = 1 \,mm^{2} = 10^{-6} \,m^{2}$ is the cross-sectional area, $e = 1.6 \times 10^{-19} \,C$ is the charge of an electron, and $I = 20 \,A$.
Rearranging the formula for drift velocity: $v_{d} = \frac{I}{n A e}$.
Substituting the values: $v_{d} = \frac{20}{10^{29} \times 10^{-6} \times 1.6 \times 10^{-19}}$.
$v_{d} = \frac{20}{1.6 \times 10^{4}} = \frac{20}{16000} = 1.25 \times 10^{-3} \,ms^{-1}$.

(A) For a conductor,the resistance $R$ is given by the slope of the $V-I$ graph,where $R = \frac{V}{I}$.
From the figure,the slope of the line corresponding to $T_{1}$ is greater than the slope of the line corresponding to $T_{2}$.
Therefore,$R_{1} > R_{2}$.
For a metallic conductor,the resistance increases with an increase in temperature $(R \propto T)$.
Since $R_{1} > R_{2}$,it follows that $T_{1} > T_{2}$.

(D) Let the potential at point $A$ be $V_A$ and at point $B$ be $V_B$. Let the potential at the top junction be $V_1$ and at the bottom junction be $V_2$.
By applying Kirchhoff's Current Law $(KCL)$ at the top junction:
$\frac{V_1 - V_A}{R} + \frac{V_1 - V_B}{R} + \frac{V_1 - V_2}{G} = 0$
By applying $KCL$ at the bottom junction:
$\frac{V_2 - V_A}{2R} + \frac{V_2 - V_B}{2R} + \frac{V_2 - V_1}{G} = 0$
Adding these two equations,the terms involving $G$ cancel out:
$\frac{V_1 - V_A}{R} + \frac{V_1 - V_B}{R} + \frac{V_2 - V_A}{2R} + \frac{V_2 - V_B}{2R} = 0$
$\frac{2(V_1 - V_A) + 2(V_1 - V_B) + (V_2 - V_A) + (V_2 - V_B)}{2R} = 0$
$2V_1 + 2V_1 + V_2 + V_2 = 3V_A + 3V_B$
$4V_1 + 2V_2 = 3(V_A + V_B)$
Since the ratio of resistances in the two branches is $\frac{R}{2R} = \frac{1}{2}$,the bridge is balanced if the potentials at the nodes connected by $G$ are equal. However,here the ratio is $\frac{R}{2R} = \frac{1}{2}$ on both sides,so the bridge is balanced. Thus,$V_1 = V_2$,and the current through $G$ is zero. Statement $(3)$ is true.
Since the current through $G$ is zero,we can remove $G$. The circuit becomes two parallel branches: one with $R+R = 2R$ and one with $2R+2R = 4R$.
$R_{\text{eff}} = \frac{(2R)(4R)}{2R + 4R} = \frac{8R^2}{6R} = \frac{4}{3} R$. Statement $(2)$ is true.
Since the equivalent resistance is independent of $G$ when the bridge is balanced,statement $(1)$ is also true.

(B) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the speed.
For an electron,the wavelength is $\lambda_{e} = \frac{h}{m_{e}v}$.
For a proton,the wavelength is $\lambda_{p} = \frac{h}{m_{p}v}$.
Taking the ratio of the two wavelengths:
$\frac{\lambda_{e}}{\lambda_{p}} = \frac{h / m_{e}v}{h / m_{p}v} = \frac{m_{p}}{m_{e}}$.
Given that the mass of a proton $m_{p} \approx 1.67 \times 10^{-27} \ kg$ and the mass of an electron $m_{e} \approx 9.11 \times 10^{-31} \ kg$,the ratio is:
$\frac{\lambda_{e}}{\lambda_{p}} = \frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}} \approx 1833 \approx 1836$ (using standard mass ratios).
Thus,the ratio $\lambda_{e} / \lambda_{p} = 1836$.

(D) According to Einstein's photoelectric equation,the energy of the incident photon is equal to the sum of the work function $(W)$ and the maximum kinetic energy $(KE)$ of the ejected electron.
$E = W + KE$
$KE = E - W$
Since $E = \frac{hc}{\lambda}$ and $W = \frac{hc}{\lambda_{0}}$,we have:
$KE = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}}$
$KE = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$
$KE = hc \left( \frac{\lambda_{0} - \lambda}{\lambda \lambda_{0}} \right)$

(B) The magnetic potential energy $U$ stored in an inductor with self-inductance $L$ carrying a current $I$ is given by the formula:
$U = \frac{1}{2} L I^{2}$
This energy is stored in the magnetic field generated by the current flowing through the inductor.

(A) Let $m$ be the mass of each sphere and $V$ be its volume. The density of the sphere is $\rho = m/V$,so $m = V\rho$.
In air,the forces acting on the sphere are tension $T$,weight $mg$,and electrostatic force $F_e = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2}$.
For equilibrium,$\tan(\theta/2) = \frac{F_e}{mg} = \frac{q^2}{4\pi\epsilon_0 r^2 mg}$.
In a liquid of density $\sigma$,the sphere experiences an upthrust $F_b = V\sigma g$. The effective weight is $mg' = mg - V\sigma g = Vg(\rho - \sigma)$.
The electrostatic force in the liquid is $F_e' = \frac{F_e}{K}$.
Since the angle $\theta$ remains the same,$\tan(\theta/2) = \frac{F_e'}{mg'} = \frac{F_e}{K Vg(\rho - \sigma)}$.
Equating the two expressions for $\tan(\theta/2)$:
$\frac{F_e}{V\rho g} = \frac{F_e}{K Vg(\rho - \sigma)}$.
Solving for $K$,we get $K = \frac{\rho}{\rho - \sigma}$.

(C) Let the electric dipole be along the $x$-axis. The electric field $\vec{E}$ at a point $(r, \theta)$ makes an angle $\alpha$ with the radial vector $\vec{r}$,given by $\tan \alpha = \frac{1}{2} \tan \theta$.
Here,$\theta$ is the angle the position vector makes with the dipole axis.
The angle that the electric field vector makes with the dipole axis is $\phi = \theta + \alpha$.
For the electric field to be perpendicular to the dipole axis,the angle $\phi$ must be $90^{\circ}$.
Therefore,$\theta + \alpha = 90^{\circ}$,which implies $\alpha = 90^{\circ} - \theta$.
Substituting this into the relation $\tan \alpha = \frac{1}{2} \tan \theta$,we get:
$\tan(90^{\circ} - \theta) = \frac{1}{2} \tan \theta$
$\cot \theta = \frac{1}{2} \tan \theta$
$\tan^2 \theta = 2$
$\tan \theta = \sqrt{2}$
$\theta = \tan^{-1}(\sqrt{2})$.

(C) In the astronomical magnitude scale,the brightness of stars is classified by their apparent magnitude. The brightest stars have lower (or negative) magnitude values,while the faintest stars visible to the human eye under ideal conditions are classified as sixth magnitude stars.

(B) The magnetic moment of the coil is $M = N I A$.
Given: $N = 10$, $I = \frac{21}{44} \,A$, $A = 1 \,mm^{2} = 10^{-6} \,m^{2}$.
$M = 10 \times \frac{21}{44} \times 10^{-6} \,A-m^{2}$.
The magnetic field inside the solenoid is $B = \mu_{0} n I_{s}$.
Given: $n = 10^{3} \,turns/m$, $I_{s} = 2.5 \,A$, $\mu_{0} = 4\pi \times 10^{-7} \,T-m/A$.
$B = (4 \times \frac{22}{7} \times 10^{-7}) \times 10^{3} \times 2.5 \,T$.
The torque $\tau$ required to hold the coil with its axis perpendicular to the magnetic field is $\tau = M B \sin(90^{\circ}) = M B$.
$\tau = (10 \times \frac{21}{44} \times 10^{-6}) \times (4 \times \frac{22}{7} \times 10^{-7} \times 10^{3} \times 2.5)$.
$\tau = (10 \times \frac{21}{44} \times 10^{-6}) \times (4 \times \frac{22}{7} \times 2.5 \times 10^{-4})$.
$\tau = (10 \times \frac{21}{44} \times 10^{-6}) \times (22 \times 10^{-4}) = 1.5 \times 10^{-8} \,N-m$.

(B) Let $G$ be the resistance of the galvanometer and $I$ be the total current.
When the galvanometer is shunted with a resistance $S = 40 \Omega$,the current flowing through the galvanometer $I_G$ is given by the current divider rule:
$I_G = I \left( \frac{S}{G + S} \right)$
Given that the deflection is reduced to half,the current through the galvanometer becomes half of the total current,i.e.,$I_G = \frac{I}{2}$.
Substituting this into the equation:
$\frac{I}{2} = I \left( \frac{40}{G + 40} \right)$
$\frac{1}{2} = \frac{40}{G + 40}$
$G + 40 = 80$
$G = 40 \Omega$
Therefore,the resistance of the galvanometer is $40 \Omega$.

(A) When a charged particle moves perpendicular to a uniform magnetic field,the magnetic Lorentz force provides the necessary centripetal force for circular motion.
$F_m = F_c$
$q v B = \frac{m v^2}{r}$
Here,the charge $q = e$.
Substituting $q = e$ into the equation:
$e v B = \frac{m v^2}{r}$
Solving for $r$:
$r = \frac{m v^2}{e v B}$
$r = \frac{m v}{e B}$

(D) The radius of the circular path of a charged particle moving in a uniform magnetic field is given by $r = \frac{mv}{qB}$.
Since the particles enter with the same velocity $v$ in the same magnetic field $B$,the radius $r$ is directly proportional to the mass-to-charge ratio,i.e.,$r \propto \frac{m}{q}$.
$1$. $A$ neutron is neutral $(q=0)$,so it experiences no magnetic force and follows a straight path,which is track $C$.
$2$. The electron is negatively charged,while the proton and $\alpha$-particle are positively charged. According to Fleming's Left-Hand Rule,the electron will deflect in the opposite direction to the positively charged particles.
$3$. Among the charged particles,the electron has the smallest mass-to-charge ratio $(m/q)$. Therefore,it will have the smallest radius of curvature.
$4$. Since the electron is negatively charged and has the smallest radius,it follows track $D$.

(B) The principle of a tangent galvanometer is given by the formula $I = K \tan \theta$,where $I$ is the current,$K$ is the reduction factor,and $\theta$ is the deflection angle.
Given current $I = \frac{2}{\sqrt{3}} \text{ A}$ and deflection $\theta = 60^{\circ}$.
Rearranging the formula to solve for the reduction factor $K$:
$K = \frac{I}{\tan \theta}$
Substituting the given values:
$K = \frac{2 / \sqrt{3}}{\tan 60^{\circ}}$
Since $\tan 60^{\circ} = \sqrt{3}$,we have:
$K = \frac{2 / \sqrt{3}}{\sqrt{3}} = \frac{2}{\sqrt{3} \times \sqrt{3}} = \frac{2}{3} \text{ A}$.
Thus,the reduction factor is $\frac{2}{3} \text{ A}$.

(D) In the $CNO$ cycle (Carbon-Nitrogen-Oxygen cycle),hydrogen nuclei fuse to form helium nuclei within stars.
Carbon acts as a catalyst in this process.
It is consumed in the initial steps to form nitrogen and oxygen isotopes,but it is regenerated at the end of the cycle.
Therefore,the net reaction is the fusion of four hydrogen nuclei into one helium nucleus,with carbon remaining unchanged in total quantity.

(D) The law of radioactive decay is given by $N(t) = N_{0} e^{-\lambda t}$.
We know that the half-life $T_{1/2} = \frac{\ln 2}{\lambda}$,which implies $\lambda = \frac{\ln 2}{T_{1/2}}$.
Given time $t = \frac{1}{2} T_{1/2}$.
Substituting these into the decay equation:
$N(t) = N_{0} e^{-(\frac{\ln 2}{T_{1/2}}) (\frac{1}{2} T_{1/2})}$
$N(t) = N_{0} e^{-\frac{1}{2} \ln 2} = N_{0} e^{\ln(2^{-1/2})}$
$N(t) = N_{0} (2^{-1/2}) = \frac{N_{0}}{\sqrt{2}}$.
The fraction remaining is $\frac{N(t)}{N_{0}} = \frac{1}{\sqrt{2}}$.

(B) The unit of radioactivity,the curie $(Ci)$,is defined as the activity of $1 \ g$ of radium-$226$.
By definition,$1$ curie is equal to $3.7 \times 10^{10}$ disintegrations per second.
This value is based on the activity of $1 \ gram$ of radium-$226$ in equilibrium with its decay products.

(C) The refractive index $\mu$ of a medium is given by the ratio of the wavelength in vacuum (or air) $\lambda_a$ to the wavelength in the medium $\lambda_m$:
$\mu = \frac{\lambda_a}{\lambda_m} = \frac{6000 \ Å}{4000 \ Å} = \frac{6}{4} = \frac{3}{2}$.
The relationship between the critical angle $C$ and the refractive index $\mu$ is given by:
$\sin C = \frac{1}{\mu}$.
Substituting the value of $\mu$:
$\sin C = \frac{1}{3/2} = \frac{2}{3}$.
Therefore,the critical angle is:
$C = \sin ^{-1}\left(\frac{2}{3}\right)$.

(C) Given,the sum of powers of two thin lenses is $P_1 + P_2 = 9 \text{ D}$.
When separated by a distance $d = 20 \text{ cm} = 0.2 \text{ m}$,the equivalent power $P$ is given by the formula:
$P = P_1 + P_2 - d P_1 P_2$
Substituting the given values:
$\frac{27}{5} = 9 - 0.2 \times P_1 P_2$
$5.4 = 9 - 0.2 \times P_1 P_2$
$0.2 \times P_1 P_2 = 9 - 5.4 = 3.6$
$P_1 P_2 = \frac{3.6}{0.2} = 18$
We have $P_1 + P_2 = 9$ and $P_1 P_2 = 18$. The quadratic equation $x^2 - (P_1+P_2)x + P_1 P_2 = 0$ becomes $x^2 - 9x + 18 = 0$.
Solving this,$(x-3)(x-6) = 0$,so $x = 3$ or $x = 6$.
Thus,the individual powers are $3 \text{ D}$ and $6 \text{ D}$.

(D) For a light ray to retrace its path after reflection from a silvered surface,it must strike the silvered surface normally (at an angle of $90^{\circ}$ to the surface).
Since the light ray strikes the silvered surface normally,the angle of refraction at the second surface is $r_{2} = 0^{\circ}$.
In a prism,the sum of the angles of refraction is equal to the prism angle: $r_{1} + r_{2} = A$.
Given $A = 30^{\circ}$ and $r_{2} = 0^{\circ}$,we get $r_{1} = 30^{\circ}$.
Applying Snell's Law at the first surface: $n = \frac{\sin i}{\sin r_{1}}$.
Given $n = 1.414 = \sqrt{2}$,we have $\sqrt{2} = \frac{\sin i}{\sin 30^{\circ}}$.
$\sin i = \sqrt{2} \times \sin 30^{\circ} = \sqrt{2} \times \frac{1}{2} = \frac{1}{\sqrt{2}}$.
Therefore,$i = 45^{\circ}$.

(B) The refractive index $n$ of a medium is defined as the ratio of the speed of light in vacuum $(c)$ to the speed of light in the medium $(v)$: $n = c/v$.
Given, $n = 1.5$ and $c = 3 \times 10^{8} \,m/s$.
Therefore, the speed of light in the glass slab is $v = c/n = (3 \times 10^{8}) / 1.5 = 2 \times 10^{8} \,m/s$.
The thickness of the glass slab is $d = 4 \,mm = 4 \times 10^{-3} \,m$.
The time $t$ taken to pass through the slab is given by $t = d/v$.
Substituting the values: $t = (4 \times 10^{-3} \,m) / (2 \times 10^{8} \,m/s) = 2 \times 10^{-11} \,s$.

(D) Case $1$: Curved surface in contact with the table.
The real depth of the lens is $t = 6 \,cm$. The apparent depth is $d' = 4 \,cm$.
The refractive index $n$ is given by $n = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{6}{4} = 1.5$.
Case $2$: Plane surface in contact with the table.
We use the refraction formula at a spherical surface: $\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$.
Here, the light travels from the lens $(n = 1.5)$ to the air $(n = 1)$.
Real depth $u = 6 \,cm$, apparent depth $v = -\frac{17}{4} \,cm$ (virtual image).
Using the formula $\frac{1}{v} - \frac{1.5}{u} = \frac{1 - 1.5}{R}$:
$\frac{1}{-17/4} - \frac{1.5}{6} = \frac{-0.5}{R}$
$-\frac{4}{17} - 0.25 = -\frac{0.5}{R}$
$-\frac{4}{17} - \frac{1}{4} = -\frac{1}{2R}$
$-\frac{16 + 17}{68} = -\frac{1}{2R}$
$\frac{33}{68} = \frac{1}{2R} \implies R = \frac{68}{66} \times 33 = 34 \,cm$.

(D) In forward biasing,the positive terminal of the battery is connected to the $p$-region and the negative terminal to the $n$-region.
Conventional current flows from the positive terminal to the negative terminal through the diode.
Inside the $p$-region,the majority charge carriers are holes,and the conventional current flows in the same direction as the flow of holes (towards the junction).
Inside the $n$-region,the majority charge carriers are electrons,and the conventional current flows in the direction opposite to the flow of electrons (towards the junction).
Therefore,in both regions,the conventional current is directed towards the depletion layer.
This corresponds to the arrows pointing towards the junction in both the $p$ and $n$ regions,which is shown in figure $D$.

(B) An $n-p-n$ transistor consists of two $p-n$ junctions.
In an $n-p-n$ transistor,the emitter-base junction is a $p-n$ junction and the collector-base junction is also a $p-n$ junction.
The base is the common $p$-type region.
For an $n-p-n$ transistor,the emitter is $n$-type,the base is $p$-type,and the collector is $n$-type.
This means the emitter-base diode has its $n$-side at the emitter and $p$-side at the base.
The collector-base diode has its $n$-side at the collector and $p$-side at the base.
Therefore,the two diodes are connected with their cathodes (n-sides) facing outwards and their anodes (p-sides) connected together at the base.
Looking at the provided figures,option $B$ correctly shows two diodes with their anodes connected at the base $B$ and cathodes pointing towards the emitter $E$ and collector $C$ respectively.

(C) The given circuit consists of two $OR$ gates followed by an $AND$ gate.
$1$. The inputs to the first $OR$ gate are $A$ and $B$. Therefore,its output is $(A+B)$.
$2$. The inputs to the second $OR$ gate are $A$ and $C$. Therefore,its output is $(A+C)$.
$3$. These two outputs $(A+B)$ and $(A+C)$ are fed as inputs to the final $AND$ gate.
$4$. The output $Y$ of the $AND$ gate is the product of its inputs: $Y = (A+B) \cdot (A+C)$.

(A) The energy stored in a capacitor is given by the formula $U = \frac{1}{2} C V^{2}$.
Since $\frac{1}{2}$ is a dimensionless constant,the dimensional formula of $C V^{2}$ is equivalent to the dimensional formula of energy $(U)$.
The dimensional formula for energy is $[Work] = [Force \times Displacement] = [MLT^{-2} \times L] = [ML^{2} T^{-2}]$.
Therefore,the dimensional formula of $C V^{2}$ is $[ML^{2} T^{-2} A^{0}]$.

(A) For a single slit diffraction, the condition for the $n^{th}$ minima is given by $a \sin \theta = n \lambda$, where $a$ is the slit width, $\lambda$ is the wavelength, and $\theta$ is the angular position.
Given: $\lambda = 6000 \text{ Å} = 6000 \times 10^{-10} \text{ m} = 6 \times 10^{-7} \text{ m}$ and $a = 0.3 \text{ mm} = 0.3 \times 10^{-3} \text{ m} = 3 \times 10^{-4} \text{ m}$.
For the first minima, $n = 1$.
Substituting the values: $3 \times 10^{-4} \sin \theta = 1 \times 6 \times 10^{-7}$.
$\sin \theta = \frac{6 \times 10^{-7}}{3 \times 10^{-4}} = 2 \times 10^{-3}$.
Since $\theta$ is very small, $\sin \theta \approx \theta$.
Therefore, $\theta = 2 \times 10^{-3} \text{ rad}$.

(D) The resultant amplitude $R$ of two interfering waves with amplitudes $A_1$ and $A_2$ and phase difference $\phi$ is given by $R = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos \phi}$.
Given $A_1 = 3 A$,$A_2 = 2 A$,and $\phi = 60^{\circ}$.
The intensity $I$ is proportional to the square of the resultant amplitude,$I \propto R^2$.
$R^2 = (3 A)^2 + (2 A)^2 + 2(3 A)(2 A) \cos(60^{\circ})$.
$R^2 = 9 A^2 + 4 A^2 + 12 A^2 \times (0.5)$.
$R^2 = 13 A^2 + 6 A^2 = 19 A^2$.
Therefore,the intensity is proportional to $19 A^2$.

(B) The critical angle $C$ is given by $\sin C = \frac{3}{5}$.
We know that the refractive index $\mu$ of the medium is given by $\mu = \frac{1}{\sin C} = \frac{1}{3/5} = \frac{5}{3}$.
According to Brewster's Law,the polarizing angle $i_p$ is related to the refractive index $\mu$ by the formula $\tan i_p = \mu$.
Substituting the value of $\mu$,we get $\tan i_p = \frac{5}{3}$.
Therefore,the polarizing angle is $i_p = \tan^{-1}\left(\frac{5}{3}\right)$.

(A) By definition,a wavefront is the locus of all points in a medium that are in the same state of vibration,which means they have the same phase.
As the wave propagates,all points on a wavefront reach their maximum displacement at the same time,maintaining a constant phase difference of zero between them.
Therefore,the correct option is $A$.

(D) In Young's double slit experiment,the primary requirement for interference is the presence of coherent sources.
Statement $(1)$ is correct: An ordinary extended source of light is incoherent,so a single slit $S$ is required to act as a point source to ensure spatial coherence.
Statement $(2)$ is incorrect: Even if a beam is well-collimated,it does not guarantee the spatial coherence required for the two slits to act as secondary coherent sources.
Statement $(3)$ is correct: If the source is already spatially coherent,the light waves arriving at the two slits maintain a constant phase difference,making the initial slit $S$ redundant.
Therefore,statements $(1)$ and $(3)$ are correct.

(C) According to Rayleigh's law of scattering, the scattering intensity $I$ is inversely proportional to the fourth power of the wavelength $\lambda$:
$I \propto \frac{1}{\lambda^{4}}$
Given $I_{1} = 8 \text{ units}$ at $\lambda_{1} = 500 \text{ nm}$ and we need to find $I_{2}$ at $\lambda_{2} = 400 \text{ nm}$.
Using the ratio formula:
$\frac{I_{2}}{I_{1}} = \left(\frac{\lambda_{1}}{\lambda_{2}}\right)^{4}$
$\frac{I_{2}}{8} = \left(\frac{500}{400}\right)^{4} = (1.25)^{4}$
$(1.25)^{4} = 2.4414 \approx 2.5$
$I_{2} = 8 \times 2.4414 \approx 19.53 \text{ units}$
Rounding to the nearest provided option, $I_{2} \approx 20 \text{ units}$.