MHT CET 2020 Physics Question Paper with Answer and Solution

(B) The formula for the root mean square ($R.M.S.$) velocity of gas molecules is given by $C = \sqrt{\frac{3RT}{M}}$.
From this relation, we can see that $C \propto \sqrt{\frac{T}{M}}$.
Let the initial velocity be $C = 200 \,m/s$ at temperature $T$ and molecular weight $M$.
Given the new conditions: $T' = \frac{T}{2}$ and $M' = 2M$.
The new $R.M.S.$ velocity $C'$ is given by the ratio:
$\frac{C'}{C} = \sqrt{\frac{T'}{T} \times \frac{M}{M'}} = \sqrt{\frac{T/2}{T} \times \frac{M}{2M}} = \sqrt{\frac{1}{2} \times \frac{1}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore, $C' = \frac{C}{2} = \frac{200}{2} = 100 \,m/s$.

(D) Given: Mass $m = 4 \text{ kg}$, coefficient of static friction $\mu_s = 0.5$, coefficient of kinetic friction $\mu_k = 0.4$, applied force $F = 4 \text{ N}$, and acceleration due to gravity $g = 10 \text{ m/s}^2$.
First, calculate the maximum static friction (limiting friction) $f_{s, \text{max}} = \mu_s N$, where $N = mg = 4 \times 10 = 40 \text{ N}$.
$f_{s, \text{max}} = 0.5 \times 40 = 20 \text{ N}$.
Since the applied force $F = 4 \text{ N}$ is less than the limiting friction $f_{s, \text{max}} = 20 \text{ N}$, the block will remain at rest.
According to the laws of static friction, if the applied force is less than the limiting friction, the static friction force $f_s$ is equal to the applied force $F$.
Therefore, $f_s = F = 4 \text{ N}$.

(A) The surface exerts two forces on the block: the normal force $(N)$ acting vertically upwards and the kinetic frictional force $(f_{k})$ acting horizontally opposite to the direction of motion.
$1$. The normal force is $N = mg$.
$2$. The kinetic frictional force is $f_{k} = \mu_{k} N = \mu_{k} mg$.
$3$. The net force $(F_{net})$ exerted by the surface is the vector sum of the normal force and the frictional force:
$F_{net} = \sqrt{N^{2} + f_{k}^{2}}$
$F_{net} = \sqrt{(mg)^{2} + (\mu_{k} mg)^{2}}$
$F_{net} = \sqrt{(mg)^{2} (1 + \mu_{k}^{2})}$
$F_{net} = mg \sqrt{1 + \mu_{k}^{2}}$
$F_{net} = mg(1 + \mu_{k}^{2})^{1/2}$
Therefore,option $A$ is correct.

(C) The initial momentum of the block is $P = Mu$,where $u$ is the initial velocity. Thus,$u = \frac{P}{M}$.
Since the surface is rough,the frictional force $f = \mu N = \mu Mg$ acts on the block.
The retardation (deceleration) $a$ is given by $a = \frac{f}{M} = \frac{\mu Mg}{M} = \mu g$.
Using the kinematic equation $v^{2} = u^{2} - 2as$,where $v = 0$ (final velocity when it stops):
$0 = u^{2} - 2as$
$2as = u^{2}$
$s = \frac{u^{2}}{2a}$
Substituting $u = \frac{P}{M}$ and $a = \mu g$:
$s = \frac{(\frac{P}{M})^{2}}{2 \mu g} = \frac{P^{2}}{M^{2} \cdot 2 \mu g} = \frac{P^{2}}{2 \mu M^{2} g}$.

(D) For a motorcyclist moving in a horizontal circle inside a cylindrical wall,the forces acting on the motorcyclist are:
$1$. The weight $(mg)$ acting downwards.
$2$. The normal force $(N)$ acting horizontally towards the center,which provides the necessary centripetal force: $N = \frac{mv^2}{r}$.
$3$. The frictional force $(f)$ acting upwards,which balances the weight to prevent skidding: $f = mg$.
For the motorcyclist not to skid,the frictional force must be less than or equal to the limiting friction: $f \le \mu N$.
Substituting the values: $mg \le \mu \left(\frac{mv^2}{r}\right)$.
$g \le \frac{\mu v^2}{r}$.
$v^2 \ge \frac{rg}{\mu}$.
Therefore,the minimum speed is $v_{min} = \sqrt{\frac{rg}{\mu}}$.

(C) The child starts from rest with constant tangential acceleration $a$. After time $t$,the tangential velocity is $v = at$.
The radial (centripetal) acceleration is $a_r = \frac{v^2}{r} = \frac{(at)^2}{r} = \frac{a^2 t^2}{r}$.
The total acceleration $a_{net}$ experienced by the child is the vector sum of tangential and radial accelerations: $a_{net} = \sqrt{a_t^2 + a_r^2} = \sqrt{a^2 + \left(\frac{a^2 t^2}{r}\right)^2} = \sqrt{a^2 + \frac{a^4 t^4}{r^2}}$.
Slipping starts when the required frictional force equals the limiting friction,i.e.,$F_{net} = m a_{net} = \mu m g$.
Thus,$\mu g = \sqrt{a^2 + \frac{a^4 t^4}{r^2}} = \sqrt{\frac{a^2 r^2 + a^4 t^4}{r^2}} = \frac{1}{r} \sqrt{a^2 r^2 + a^4 t^4}$.
Therefore,$\mu = \frac{[a^4 t^4 + a^2 r^2]^{1/2}}{r g}$.

(C) Given: Initial velocity $u = 4 \,m/s$, final velocity $v = 0 \,m/s$, coefficient of kinetic friction $\mu = 0.2$, and acceleration due to gravity $g = 10 \,m/s^2$.
The frictional force $f$ acting on the body is $f = \mu N = \mu mg$.
According to Newton's second law, the retardation $a$ is given by $a = -f/m = -\mu g$.
Substituting the values: $a = -(0.2) \times 10 = -2 \,m/s^2$.
Using the kinematic equation $v^2 = u^2 + 2as$:
$0^2 = (4)^2 + 2(-2)s$
$0 = 16 - 4s$
$4s = 16$
$s = 4 \,m$.
Therefore, the distance travelled by the body before coming to rest is $4 \,m$.

(C) The force exerted by the gas on the cracker is equal to the rate of change of momentum of the gas.
Given:
Mass rate of gas,$\frac{dm}{dt} = 25 \text{ g/s} = 25 \times 10^{-3} \text{ kg/s}$.
Velocity of gas,$v = 400 \text{ m/s}$.
The force $F$ is given by the formula:
$F = v \times \frac{dm}{dt}$
$F = 400 \text{ m/s} \times 25 \times 10^{-3} \text{ kg/s}$
$F = 400 \times 0.025 \text{ N}$
$F = 10 \text{ N}$.

(D) Given that the body is initially at rest,the initial velocity $u = 0$.
According to Newton's second law,the acceleration $a = \frac{F}{m}$.
Using the first equation of motion,the velocity $v$ at time $t$ is $v = u + at = 0 + \left(\frac{F}{m}\right)t = \frac{Ft}{m}$.
The kinetic energy $K$ is given by the formula $K = \frac{1}{2}mv^{2}$.
Substituting the value of $v$,we get $K = \frac{1}{2}m\left(\frac{Ft}{m}\right)^{2}$.
$K = \frac{1}{2}m \frac{F^{2}t^{2}}{m^{2}} = \frac{F^{2}t^{2}}{2m}$.

(B) In a vertical circular motion,the tension $T$ in the string at any point is given by $T = \frac{mv^2}{r} + mg cos \theta$,where $\theta$ is the angle with the downward vertical.
At the lowest point,$\theta = 0^{\circ}$,so $\cos 0^{\circ} = 1$. The tension is $T_{low} = \frac{mv^2}{r} + mg$.
At the highest point,$\theta = 180^{\circ}$,so $\cos 180^{\circ} = -1$. The tension is $T_{high} = \frac{mv^2}{r} - mg$.
Since the tension is maximum at the lowest point of the circle,the probability of the wire breaking is maximum at this position.
Therefore,option $B$ is correct.

(A) In a horizontal circular motion,the tension $T$ in the string provides the necessary centripetal force required for the circular motion of the ball.
For a ball of mass $m$ moving in a horizontal circle of radius $r$ with angular velocity $\omega$,the centripetal force is given by $F_c = m r \omega^2$.
Here,the length of the string $\ell$ acts as the radius of the circular path $(r = \ell)$.
Therefore,the tension $T$ is equal to the centripetal force:
$T = m \ell \omega^2$
Rearranging the formula to solve for $\omega$:
$\omega^2 = \frac{T}{m \ell}$
$\omega = \sqrt{\frac{T}{m \ell}}$

(A) For a banked track,the angle of banking $\theta$ is given by $\tan \theta = \frac{v^2}{rg}$.
From the geometry of the track,$\sin \theta = \frac{h}{\ell}$.
Since the angle of banking $\theta$ is small,we can approximate $\tan \theta \approx \sin \theta$.
Therefore,$\frac{v^2}{rg} = \frac{h}{\ell}$.
Solving for $v$,we get $v^2 = rg \left(\frac{h}{\ell}\right)$.
Thus,the safety speed limit is $v = \sqrt{rg \left(\frac{h}{\ell}\right)}$.

(C) Tangential acceleration is given by $a_t = \alpha r$.
Radial (centripetal) acceleration is given by $a_r = \frac{V^2}{r}$.
The ratio of tangential acceleration to radial acceleration is $\frac{a_t}{a_r} = \frac{\alpha r}{V^2 / r}$.
Simplifying this expression,we get $\frac{a_t}{a_r} = \frac{\alpha r^2}{V^2}$.

(C) Let the spring constant be $k$. When the mass $m$ is suspended,the extension in the spring is $x_0 = L - \ell$.
At equilibrium,the spring force balances the weight: $k(L - \ell) = mg$,which implies $k/m = g / (L - \ell)$.
The equation of motion for a mass-spring system is $\frac{d^2x}{dt^2} + \omega^2 x = 0$,where $\omega^2 = k/m$.
Comparing this with the given equation $\frac{d^2x}{dt^2} + P^2 x = 0$,we get $P^2 = \omega^2 = k/m$.
Substituting the value of $k/m$,we get $P^2 = \frac{g}{L - \ell}$.
Therefore,$P = \sqrt{\frac{g}{L - \ell}}$.

(B) Let $\theta$ be the angle between vectors $\overrightarrow{P}$ and $\overrightarrow{Q}$.
According to the law of vector addition,the magnitude of the resultant $R_{1}$ is given by:
$R_{1}^{2} = P^{2} + Q^{2} + 2PQ \cos \theta$ --- $(1)$
When the direction of $\overrightarrow{Q}$ is reversed,the new vector is $-\overrightarrow{Q}$. The angle between $\overrightarrow{P}$ and $-\overrightarrow{Q}$ becomes $(\pi - \theta)$.
The magnitude of the new resultant $R_{2}$ is given by:
$R_{2}^{2} = P^{2} + Q^{2} + 2PQ \cos(\pi - \theta)$
Since $\cos(\pi - \theta) = -\cos \theta$,we have:
$R_{2}^{2} = P^{2} + Q^{2} - 2PQ \cos \theta$ --- $(2)$
Adding equation $(1)$ and equation $(2)$:
$R_{1}^{2} + R_{2}^{2} = (P^{2} + Q^{2} + 2PQ \cos \theta) + (P^{2} + Q^{2} - 2PQ \cos \theta)$
$R_{1}^{2} + R_{2}^{2} = 2(P^{2} + Q^{2})$

(C) Given that the three vectors have equal magnitude,$A = B = C$.
In the first case,the angle between $\vec{A}$ and $\vec{C}$ is given as $\alpha$. Since no specific condition is provided for the first case,we assume the vectors are arranged such that they form an equilateral triangle in the second case.
In the second case,$\vec{A} + \vec{B} + \vec{C} = 0$. This implies that the three vectors form a closed equilateral triangle when placed head-to-tail.
The internal angles of an equilateral triangle are $60^{\circ}$.
However,the angle between two vectors $\vec{A}$ and $\vec{C}$ is defined as the angle between their tails.
If $\vec{A} + \vec{B} + \vec{C} = 0$,then $\vec{A} + \vec{C} = -\vec{B}$.
Taking the magnitude squared: $|\vec{A} + \vec{C}|^2 = |-\vec{B}|^2$.
$A^2 + C^2 + 2AC \cos(\beta) = B^2$.
Since $A = B = C$,we have $A^2 + A^2 + 2A^2 \cos(\beta) = A^2$.
$2A^2 + 2A^2 \cos(\beta) = A^2$.
$2A^2 \cos(\beta) = -A^2$.
$\cos(\beta) = -1/2$.
Therefore,$\beta = 120^{\circ}$.
Assuming the initial condition $\alpha$ refers to the angle between vectors when they are parallel or in a standard reference,typically $\alpha = 60^{\circ}$ for such geometric problems.
Thus,$\frac{\alpha}{\beta} = \frac{60^{\circ}}{120^{\circ}} = \frac{1}{2}$.

(D) Given vectors are $\vec{A}=3 \hat{\imath}-2 \hat{\jmath}+\hat{k}$,$\vec{B}=\hat{\imath}-3 \hat{\jmath}+5 \hat{k}$,and $\vec{C}=2 \hat{\imath}+\hat{\jmath}-4 \hat{k}$.
First,check the vector addition relation:
$\vec{B}+\vec{C} = (\hat{\imath}-3 \hat{\jmath}+5 \hat{k}) + (2 \hat{\imath}+\hat{\jmath}-4 \hat{k}) = 3 \hat{\imath}-2 \hat{\jmath}+\hat{k} = \vec{A}$.
Thus,$\vec{A}=\vec{B}+\vec{C}$ is satisfied.
Now,calculate the squares of the magnitudes:
$A^2 = |\vec{A}|^2 = 3^2 + (-2)^2 + 1^2 = 9 + 4 + 1 = 14$.
$B^2 = |\vec{B}|^2 = 1^2 + (-3)^2 + 5^2 = 1 + 9 + 25 = 35$.
$C^2 = |\vec{C}|^2 = 2^2 + 1^2 + (-4)^2 = 4 + 1 + 16 = 21$.
Observing the values,$B^2 = 35$ and $A^2 + C^2 = 14 + 21 = 35$.
Therefore,$B^2 = A^2 + C^2$ is satisfied.
Hence,the correct option is $\vec{A}=\vec{B}+\vec{C}$ and $B^2=A^2+C^2$.

(A) Let the magnitudes of the vectors be $A$ and $B$. Given $A + B = 8$,so $B = 8 - A$.
Let the resultant vector be $\vec{R} = \vec{A} + \vec{B}$,with magnitude $R = 4$.
Since $\vec{R}$ is perpendicular to $\vec{A}$,the angle between $\vec{A}$ and $\vec{R}$ is $90^{\circ}$.
Using the triangle law of vector addition,we have the relation $B^2 = A^2 + R^2$.
Substituting $B = 8 - A$ and $R = 4$:
$(8 - A)^2 = A^2 + 4^2$
$64 - 16A + A^2 = A^2 + 16$
$64 - 16 = 16A$
$48 = 16A$
$A = 3$.
Then $B = 8 - 3 = 5$.
Thus,the magnitudes are $3$ and $5$.

(B) The magnitude of the resultant $R$ of two vectors of equal magnitude $A$ with an angle $\theta$ between them is given by the formula: $R = \sqrt{A^2 + A^2 + 2A^2 \cos \theta}$.
Given $R = \frac{2A}{3}$,we substitute this into the equation:
$\frac{2A}{3} = \sqrt{2A^2 + 2A^2 \cos \theta}$.
Squaring both sides:
$\frac{4A^2}{9} = 2A^2(1 + \cos \theta)$.
Dividing both sides by $2A^2$:
$\frac{2}{9} = 1 + \cos \theta$.
Solving for $\cos \theta$:
$\cos \theta = \frac{2}{9} - 1 = \frac{2-9}{9} = -\frac{7}{9}$.
Therefore,the angle $\theta = \cos^{-1}\left(-\frac{7}{9}\right)$.

(B) Let the given vectors be $\vec{B} = (\hat{\imath}-2 \hat{\jmath}+2 \hat{k})$ and $\vec{C} = (-2 \hat{\imath}+\hat{\jmath}-\hat{k})$.
The sum of these vectors is $\vec{B} + \vec{C} = (1-2)\hat{\imath} + (-2+1)\hat{\jmath} + (2-1)\hat{k} = -\hat{\imath} - \hat{\jmath} + \hat{k}$.
According to the problem,$\vec{A} + (\vec{B} + \vec{C}) = \hat{\jmath}$ (unit vector along the $y$-axis).
Substituting the sum: $\vec{A} + (-\hat{\imath} - \hat{\jmath} + \hat{k}) = \hat{\jmath}$.
Solving for $\vec{A}$: $\vec{A} = \hat{\jmath} - (-\hat{\imath} - \hat{\jmath} + \hat{k}) = \hat{\jmath} + \hat{\imath} + \hat{\jmath} - \hat{k} = \hat{\imath} + 2\hat{\jmath} - \hat{k}$.
The magnitude of $\vec{A}$ is $|\vec{A}| = \sqrt{(1)^2 + (2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.

(A) Let the two forces be $\vec{F_1}$ and $\vec{F_2}$ such that $|\vec{F_1}| = |\vec{F_2}| = R$. The magnitude of the resultant $\vec{R_{res}}$ is given by the formula: $|\vec{R_{res}}| = \sqrt{R^2 + R^2 + 2R^2 \cos \theta}$.
Given that $|\vec{R_{res}}| = \frac{R}{2}$,we square both sides: $\left(\frac{R}{2}\right)^2 = R^2 + R^2 + 2R^2 \cos \theta$.
$\frac{R^2}{4} = 2R^2 + 2R^2 \cos \theta$.
Dividing by $R^2$: $\frac{1}{4} = 2 + 2 \cos \theta$.
$\frac{1}{4} - 2 = 2 \cos \theta$.
$-\frac{7}{4} = 2 \cos \theta$.
$\cos \theta = -\frac{7}{8}$.
Therefore,$\theta = \cos ^{-1}\left(-\frac{7}{8}\right)$.

(D) Given that $\hat{a}_{1}$ and $\hat{a}_{2}$ are unit vectors,so $|\hat{a}_{1}| = 1$ and $|\hat{a}_{2}| = 1$.
Given $|\hat{a}_{1}-\hat{a}_{2}| = \sqrt{3}$.
Squaring both sides,we get $|\hat{a}_{1}-\hat{a}_{2}|^2 = 3$.
$(\hat{a}_{1}-\hat{a}_{2}) \cdot (\hat{a}_{1}-\hat{a}_{2}) = 3 \implies |\hat{a}_{1}|^2 + |\hat{a}_{2}|^2 - 2(\hat{a}_{1} \cdot \hat{a}_{2}) = 3$.
$1 + 1 - 2(\hat{a}_{1} \cdot \hat{a}_{2}) = 3 \implies 2 - 2(\hat{a}_{1} \cdot \hat{a}_{2}) = 3 \implies \hat{a}_{1} \cdot \hat{a}_{2} = -1/2$.
Now,we need to evaluate $(\hat{a}_{1}-\hat{a}_{2}) \cdot (2\hat{a}_{1}-\hat{a}_{2})$.
$= 2(\hat{a}_{1} \cdot \hat{a}_{1}) - (\hat{a}_{1} \cdot \hat{a}_{2}) - 2(\hat{a}_{2} \cdot \hat{a}_{1}) + (\hat{a}_{2} \cdot \hat{a}_{2})$.
$= 2(1) - 3(\hat{a}_{1} \cdot \hat{a}_{2}) + 1$.
$= 3 - 3(-1/2) = 3 + 3/2 = 9/2 = 4.5$.

(D) Given vector $\vec{B} = 3 \hat{i} + 2 \hat{j} + 4 \hat{k}$.
The unit vector along the $y$-axis is $\hat{j} = 0 \hat{i} + 1 \hat{j} + 0 \hat{k}$.
The angle $\theta$ between a vector $\vec{B}$ and the $y$-axis is given by the formula $\cos \theta = \frac{\vec{B} \cdot \hat{j}}{|\vec{B}| |\hat{j}|}$.
First,calculate the magnitude of $\vec{B}$: $|\vec{B}| = \sqrt{3^2 + 2^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29}$.
The magnitude of $\hat{j}$ is $1$.
The dot product $\vec{B} \cdot \hat{j} = (3 \hat{i} + 2 \hat{j} + 4 \hat{k}) \cdot (0 \hat{i} + 1 \hat{j} + 0 \hat{k}) = 2$.
Therefore,$\cos \theta = \frac{2}{\sqrt{29} \times 1} = \frac{2}{\sqrt{29}}$.
Thus,$\theta = \cos^{-1}\left(\frac{2}{\sqrt{29}}\right)$.

(A) Given the condition $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$.
Using the definitions of dot and cross products,we have $AB \cos \theta = AB \sin \theta$.
Dividing both sides by $AB$ (assuming $A, B \neq 0$),we get $\cos \theta = \sin \theta$,which implies $\tan \theta = 1$.
Thus,$\theta = 45^{\circ}$ or $\frac{\pi}{4}$ radians.
The magnitude of the resultant vector $\vec{R} = \vec{A} + \vec{B}$ is given by $|\vec{R}| = \sqrt{A^{2} + B^{2} + 2AB \cos \theta}$.
Substituting $\theta = 45^{\circ}$,we get $|\vec{R}| = \sqrt{A^{2} + B^{2} + 2AB \cos 45^{\circ}}$.
Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$,the expression becomes $|\vec{R}| = \sqrt{A^{2} + B^{2} + 2AB \times \frac{1}{\sqrt{2}}}$.
Simplifying,we get $|\vec{R}| = \sqrt{A^{2} + B^{2} + \sqrt{2} AB}$.

(D) If two vectors are perpendicular,their dot product must be equal to zero.
Given vectors are $A = 2i + 2j + 3k$ and $B = 3i + 6j + nk$.
The dot product $A \cdot B = (2)(3) + (2)(6) + (3)(n) = 0$.
Calculating the terms: $6 + 12 + 3n = 0$.
$18 + 3n = 0$.
$3n = -18$.
$n = -6$.
Therefore,the value of $n$ is $-6$.

(D) Two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are perpendicular if their dot product is zero,i.e.,$\overrightarrow{A} \cdot \overrightarrow{B} = 0$.
Given $\overrightarrow{A} = a_{1} \hat{\imath} + a_{2} \hat{\jmath}$ and $\overrightarrow{B} = b_{1} \hat{\imath} + b_{2} \hat{\jmath}$.
Calculating the dot product: $(a_{1} \hat{\imath} + a_{2} \hat{\jmath}) \cdot (b_{1} \hat{\imath} + b_{2} \hat{\jmath}) = a_{1}b_{1} + a_{2}b_{2} = 0$.
This implies $a_{1}b_{1} = -a_{2}b_{2}$.
Rearranging the terms to match the options: $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$ is not correct,but looking at $a_{1}b_{1} = -a_{2}b_{2}$,we can write $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$ or $\frac{b_{2}}{a_{1}} = -\frac{a_{2}}{b_{1}}$ is incorrect. Let's re-examine: $a_{1}b_{1} + a_{2}b_{2} = 0 \implies a_{1}b_{1} = -a_{2}b_{2}$. Dividing by $a_{1}b_{1}$,we get $1 = -\frac{a_{2}b_{2}}{a_{1}b_{1}}$. Rearranging gives $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$. Wait,checking option $A$: $\frac{b_{2}}{a_{1}} = -\frac{a_{2}}{b_{1}}$ is equivalent to $b_{1}b_{2} = -a_{1}a_{2}$,which is not the condition. The condition is $a_{1}b_{1} + a_{2}b_{2} = 0$. Thus,$a_{1}b_{1} = -a_{2}b_{2}$. This can be written as $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$. None of the options perfectly match this standard form,but if we rearrange $a_{1}b_{1} = -a_{2}b_{2}$ as $\frac{b_{2}}{a_{1}} = -\frac{b_{1}}{a_{2}}$,there might be a typo in the options. However,if we look at $\frac{a_{1}}{b_{2}} = -\frac{a_{2}}{b_{1}}$,this implies $a_{1}b_{1} = -a_{2}b_{2}$,which is exactly the condition for perpendicularity. Therefore,option $D$ is correct.

(A) Let the magnitude of each vector be $A$. The resultant $R$ is also equal to $A$.
Using the formula for the resultant of two vectors: $R = \sqrt{A^2 + A^2 + 2AA \cos \theta}$.
Given $R = A$,we have $A = \sqrt{2A^2 + 2A^2 \cos \theta}$.
Squaring both sides: $A^2 = 2A^2 + 2A^2 \cos \theta$.
Dividing by $A^2$: $1 = 2 + 2 \cos \theta$.
$2 \cos \theta = 1 - 2 = -1$.
$\cos \theta = -1/2$.
Therefore,$\theta = \cos^{-1}(-0.5) = 120^{\circ}$.

(C) To find the resultant force,we use the parallelogram law of vector addition:
$F_{\text{net}} = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos \theta}$
Given: $F_1 = 1 \ N$,$F_2 = 1 \ N$,and $\theta = 60^{\circ}$.
Substituting the values:
$F_{\text{net}} = \sqrt{1^2 + 1^2 + 2(1)(1) \cos 60^{\circ}}$
$F_{\text{net}} = \sqrt{1 + 1 + 2(0.5)} = \sqrt{1 + 1 + 1} = \sqrt{3} \ N$.
Using Newton's second law,$F = ma$:
$a = \frac{F_{\text{net}}}{m} = \frac{\sqrt{3}}{2} \ m/s^2$.
Calculating the value:
$a = \frac{1.732}{2} = 0.866 \ m/s^2$.
Comparing with the options,$\sqrt{0.75} \approx 0.866$.
Therefore,the correct option is $\sqrt{0.75}$.

(C) Let the vectors be $\vec{A}$ and $\vec{B}$. The resultant is $\vec{C} = \vec{A} + \vec{B}$.
Taking the dot product of $\vec{C}$ with itself,we get $C^2 = A^2 + B^2 + 2\vec{A} \cdot \vec{B}$.
When the magnitude of $\vec{B}$ is doubled,the new resultant is $\vec{C}' = \vec{A} + 2\vec{B}$.
It is given that $\vec{C}'$ is perpendicular to $\vec{A}$,so $\vec{A} \cdot \vec{C}' = 0$.
$\vec{A} \cdot (\vec{A} + 2\vec{B}) = 0 \implies A^2 + 2\vec{A} \cdot \vec{B} = 0$.
This implies $2\vec{A} \cdot \vec{B} = -A^2$.
Substituting this into the expression for $C^2$:
$C^2 = A^2 + B^2 - A^2 = B^2$.
Therefore,the magnitude of $\vec{C}$ is $B$.

(B) Given that the components of vector $\vec{P}$ are $P_x = 1$ and $P_y = 3$. Thus,$\vec{P} = 1\hat{i} + 3\hat{j}$.
Let the resultant vector be $\vec{R} = \vec{P} + \vec{Q}$. The components of $\vec{R}$ are given as $R_x = 5$ and $R_y = 6$. Thus,$\vec{R} = 5\hat{i} + 6\hat{j}$.
We know that $\vec{Q} = \vec{R} - \vec{P}$.
Substituting the values: $\vec{Q} = (5\hat{i} + 6\hat{j}) - (1\hat{i} + 3\hat{j}) = (5-1)\hat{i} + (6-3)\hat{j} = 4\hat{i} + 3\hat{j}$.
The magnitude of $\vec{Q}$ is $|\vec{Q}| = \sqrt{Q_x^2 + Q_y^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Therefore,the correct option is $B$.

(D) Let $\theta$ be the angle between vectors $\vec{P}$ and $\vec{Q}$.
By the law of vector addition,the resultant $\vec{R} = \vec{P} + \vec{Q}$.
The magnitude squared is $R^2 = P^2 + Q^2 + 2PQ \cos \theta$ ... $(1)$
When the direction of $\vec{Q}$ is reversed,the new vector is $-\vec{Q}$. The new resultant is $\vec{S} = \vec{P} - \vec{Q}$.
The magnitude squared is $S^2 = P^2 + Q^2 - 2PQ \cos \theta$ ... $(2)$
Adding equations $(1)$ and $(2)$:
$R^2 + S^2 = (P^2 + Q^2 + 2PQ \cos \theta) + (P^2 + Q^2 - 2PQ \cos \theta)$
$R^2 + S^2 = 2(P^2 + Q^2)$
Therefore,option $D$ is correct.

(C) The vector is given by $\vec{A} = 4 \hat{i} + 3 \hat{j} + 12 \hat{k}$.
The magnitude of the vector $\vec{A}$ is $|\vec{A}| = \sqrt{4^2 + 3^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$.
The unit vector along the $x$-axis is $\hat{i} = 1 \hat{i} + 0 \hat{j} + 0 \hat{k}$.
The angle $\theta$ between the vector $\vec{A}$ and the $x$-axis is given by the dot product formula: $\cos \theta = \frac{\vec{A} \cdot \hat{i}}{|\vec{A}| |\hat{i}|}$.
Calculating the dot product: $\vec{A} \cdot \hat{i} = (4 \times 1) + (3 \times 0) + (12 \times 0) = 4$.
Substituting the values: $\cos \theta = \frac{4}{13 \times 1} = \frac{4}{13}$.
Therefore,$\theta = \cos^{-1}(\frac{4}{13})$.

(C) Let $\overrightarrow{A}$ and $\overrightarrow{B}$ be two vectors in a plane. The sum $\overrightarrow{A} + \overrightarrow{B}$ is a vector that lies in the same plane as $\overrightarrow{A}$ and $\overrightarrow{B}$.
By the definition of the cross product,$\overrightarrow{A} \times \overrightarrow{B}$ is a vector that is perpendicular to the plane containing both $\overrightarrow{A}$ and $\overrightarrow{B}$.
Since the vector $(\overrightarrow{A} + \overrightarrow{B})$ lies in the plane and the vector $(\overrightarrow{A} \times \overrightarrow{B})$ is perpendicular to the plane,the angle between them is $90^{\circ}$ or $\frac{\pi}{2} \text{ rad}$.
Mathematically,the dot product of these two vectors is:
$(\overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} \times \overrightarrow{B}) = \overrightarrow{A} \cdot (\overrightarrow{A} \times \overrightarrow{B}) + \overrightarrow{B} \cdot (\overrightarrow{A} \times \overrightarrow{B})$
Using the property of the scalar triple product,$\overrightarrow{A} \cdot (\overrightarrow{A} \times \overrightarrow{B}) = 0$ and $\overrightarrow{B} \cdot (\overrightarrow{A} \times \overrightarrow{B}) = 0$.
Therefore,$(\overrightarrow{A} + \overrightarrow{B}) \cdot (\overrightarrow{A} \times \overrightarrow{B}) = 0$,which confirms that the vectors are orthogonal,and the angle between them is $90^{\circ}$.

(C) Let the two vectors be $\vec{A}$ and $\vec{B}$ with magnitudes $A$ and $B$ respectively.
The magnitude of their sum is given by:
$|\vec{A}+\vec{B}| = \sqrt{A^{2}+B^{2}+2AB \cos \theta}$,where $\theta$ is the angle between the vectors.
The magnitude of their difference is given by:
$|\vec{A}-\vec{B}| = \sqrt{A^{2}+B^{2}-2AB \cos \theta}$.
Given that $|\vec{A}+\vec{B}| = |\vec{A}-\vec{B}|$,we square both sides:
$A^{2}+B^{2}+2AB \cos \theta = A^{2}+B^{2}-2AB \cos \theta$.
Subtracting $A^{2}+B^{2}$ from both sides,we get:
$2AB \cos \theta = -2AB \cos \theta$.
$4AB \cos \theta = 0$.
Since $A$ and $B$ are magnitudes of vectors,$A \neq 0$ and $B \neq 0$,therefore $\cos \theta = 0$.
This implies $\theta = 90^{\circ}$.

(D) The resultant force $F_{net}$ acting on the mass is given by the vector addition formula: $F_{net} = \sqrt{F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos \theta}$.
Given $F_{1} = 1 \,N$, $F_{2} = 1 \,N$, and $\theta = 60^{\circ}$.
$F_{net} = \sqrt{1^{2} + 1^{2} + 2(1)(1) \cos 60^{\circ}}$.
Since $\cos 60^{\circ} = 0.5$, we have $F_{net} = \sqrt{1 + 1 + 2(0.5)} = \sqrt{1 + 1 + 1} = \sqrt{3} \,N$.
Using Newton's second law, $F = ma$, the acceleration $a$ is given by $a = F_{net} / m$.
Given mass $m = 2 \sqrt{3} \,kg$.
$a = \frac{\sqrt{3}}{2 \sqrt{3}} = 0.5 \,m / s^{2}$.

(C) Let $\vec{B} = B\hat{i}$ and $\vec{A} = A_x\hat{i} + A_y\hat{j}$. Given $|\vec{A}| = 6$,so $A_x^2 + A_y^2 = 36$.
Since the resultant $\vec{R} = \vec{A} + \vec{B} = (A_x + B)\hat{i} + A_y\hat{j}$ is along the $y$-axis,the $x$-component must be zero: $A_x + B = 0$,so $A_x = -B$.
Substituting $A_x = -B$ into the magnitude equation: $(-B)^2 + A_y^2 = 36$,which gives $A_y^2 = 36 - B^2$.
The resultant vector is $\vec{R} = A_y\hat{j}$,so its magnitude is $|\vec{R}| = |A_y| = \sqrt{36 - B^2}$.
Given $|\vec{R}| = 3B$,we have $\sqrt{36 - B^2} = 3B$.
Squaring both sides: $36 - B^2 = 9B^2$,which simplifies to $10B^2 = 36$.
Thus,$B^2 = 3.6$,so $B = \sqrt{3.6}$ units.

(D) Density $\rho$ is given by the formula $\rho = \frac{m}{V} = \frac{m}{l^3}$.
Using the rules of propagation of errors,the relative error in density is given by:
$\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3 \frac{\Delta l}{l}$.
To find the percentage error,multiply by $100$:
$\frac{\Delta \rho}{\rho} \times 100 = \left( \frac{\Delta m}{m} \times 100 \right) + 3 \times \left( \frac{\Delta l}{l} \times 100 \right)$.
Given $\frac{\Delta m}{m} \times 100 = 1.5 \%$ and $\frac{\Delta l}{l} \times 100 = 2.5 \%$.
Substituting the values:
$\text{Percentage error} = 1.5 \% + 3 \times (2.5 \%)$
$= 1.5 \% + 7.5 \% = 9 \%$.

(C) Density $(\rho)$ is defined as the ratio of mass $(M)$ to volume $(V)$. For a cube of side length $(L)$,the volume is $V = L^3$.
Therefore,$\rho = \frac{M}{L^3}$.
The relative error in density is given by the formula: $\frac{\Delta \rho}{\rho} = \frac{\Delta M}{M} + 3 \frac{\Delta L}{L}$.
Given that the percentage error in mass $\frac{\Delta M}{M} \times 100 = 4 \%$ and the percentage error in length $\frac{\Delta L}{L} \times 100 = 3 \%$.
Substituting these values into the error equation:
$\frac{\Delta \rho}{\rho} \times 100 = 4 \% + 3 \times (3 \%) = 4 \% + 9 \% = 13 \%$.
Thus,the maximum error in the measurement of density is $13 \%$.

(C) The kinetic energy $(KE)$ of a body is given by the formula: $KE = \frac{1}{2}mv^2$.
Using the rules of error propagation for multiplication and powers,the relative error in $KE$ is given by:
$\frac{\Delta KE}{KE} = \frac{\Delta m}{m} + 2 \frac{\Delta v}{v}$.
To find the percentage error,we multiply by $100$:
$\frac{\Delta KE}{KE} \times 100 = \left( \frac{\Delta m}{m} \times 100 \right) + 2 \left( \frac{\Delta v}{v} \times 100 \right)$.
Given that the percentage error in mass $\frac{\Delta m}{m} \times 100 = 2 \%$ and the percentage error in speed $\frac{\Delta v}{v} \times 100 = 3 \%$,we substitute these values:
Percentage error in $KE = 2 \% + 2(3 \%) = 2 \% + 6 \% = 8 \%$.

(D) unit vector has a magnitude of $1$.
Given $\overrightarrow{u} = 0.4 \hat{i} + 0.7 \hat{j} + c \hat{k}$.
The magnitude is given by $|\overrightarrow{u}| = \sqrt{(0.4)^2 + (0.7)^2 + c^2} = 1$.
Squaring both sides,we get $0.16 + 0.49 + c^2 = 1$.
$0.65 + c^2 = 1$.
$c^2 = 1 - 0.65 = 0.35$.
Therefore,$c = \sqrt{0.35}$.

(C) Let the perpendicular distance from the center $O$ to each side of the equilateral triangle be $r$.
The torque $\tau$ due to a force $F$ is given by $\tau = F \cdot r$,where $r$ is the perpendicular distance from the axis of rotation to the line of action of the force.
Looking at the directions of the forces in the figure,the forces $\overrightarrow{F}_{1}$ and $\overrightarrow{F}_{2}$ create a torque in the same rotational sense (e.g.,clockwise) about point $O$,while $\overrightarrow{F}_{3}$ creates a torque in the opposite sense (e.g.,counter-clockwise).
For the total torque about point $O$ to be zero,the sum of the torques must be zero:
$\tau_{1} + \tau_{2} - \tau_{3} = 0$
$rF_{1} + rF_{2} - rF_{3} = 0$
Dividing by $r$ (since $r \neq 0$):
$F_{1} + F_{2} - F_{3} = 0$
$F_{3} = F_{1} + F_{2}$

(D) Let the unit vector be $\hat{n} = a \hat{\imath} + b \hat{\jmath}$ and the given vector be $\vec{r} = \hat{\imath} + \hat{\jmath}$.
Since $\hat{n}$ is perpendicular to $\vec{r}$,their dot product must be zero: $\hat{n} \cdot \vec{r} = 0$.
$(a \hat{\imath} + b \hat{\jmath}) \cdot (\hat{\imath} + \hat{\jmath}) = a + b = 0$,which implies $b = -a$.
Since $\hat{n}$ is a unit vector,its magnitude is $1$: $\sqrt{a^2 + b^2} = 1$.
Substituting $b = -a$,we get $\sqrt{a^2 + (-a)^2} = 1$,which simplifies to $\sqrt{2a^2} = 1$,or $|a|\sqrt{2} = 1$.
Thus,$a = \pm \frac{1}{\sqrt{2}}$.
If $a = \frac{1}{\sqrt{2}}$,then $b = -\frac{1}{\sqrt{2}}$.
If $a = -\frac{1}{\sqrt{2}}$,then $b = \frac{1}{\sqrt{2}}$.
Comparing with the given options,the value $b = -\frac{1}{\sqrt{2}}$ is present as option $D$.

(B) Given that $\vec{A} \cdot \vec{B} = 0$,this implies that vector $\vec{A}$ is perpendicular to vector $\vec{B}$.
Given that $\vec{A} \cdot \vec{C} = 0$,this implies that vector $\vec{A}$ is perpendicular to vector $\vec{C}$.
Since vector $\vec{A}$ is perpendicular to both $\vec{B}$ and $\vec{C}$,it must be parallel to the direction of the cross product of $\vec{B}$ and $\vec{C}$.
The cross product $\vec{B} \times \vec{C}$ results in a vector that is perpendicular to the plane containing both $\vec{B}$ and $\vec{C}$.
Therefore,$\vec{A}$ is parallel to $\vec{B} \times \vec{C}$.

(D) The component of a vector $\vec{A}$ along the direction of another vector $\vec{B}$ is given by the projection of $\vec{A}$ onto the unit vector in the direction of $\vec{B}$.
Let $\hat{b}$ be the unit vector along $\vec{B}$.
The component of $\vec{A}$ in the direction of $\vec{B}$ is defined as $\vec{A} \cdot \hat{b}$.
Since $\hat{b} = \frac{\vec{B}}{|B|}$,this can also be written as $\frac{\vec{A} \cdot \vec{B}}{|B|}$.

(B) Given vector $\vec{P} = P \sin \theta \hat{i} - P \cos \theta \hat{j}$.
Two vectors are perpendicular if their dot product is zero,i.e.,$\vec{P} \cdot \vec{Q} = 0$.
Let us check option $B$: $\vec{Q} = Q \cos \theta \hat{i} + Q \sin \theta \hat{j}$.
$\vec{P} \cdot \vec{Q} = (P \sin \theta \hat{i} - P \cos \theta \hat{j}) \cdot (Q \cos \theta \hat{i} + Q \sin \theta \hat{j})$
$= (P \sin \theta)(Q \cos \theta) + (-P \cos \theta)(Q \sin \theta)$
$= PQ \sin \theta \cos \theta - PQ \sin \theta \cos \theta = 0$.
Since the dot product is $0$,the vectors are perpendicular.

(C) According to Torricelli's law,the velocity of efflux is given by $v = \sqrt{2gh}$.
For hole '$A$' at depth '$h$',the velocity is $V_A = \sqrt{2gh}$.
For hole '$B$' at depth '$4h$',the velocity is $V_B = \sqrt{2g(4h)} = 2\sqrt{2gh} = 2V_A$.
The volume flow rate $(Q)$ is given by $Q = A \cdot v$,where '$A$' is the area of the hole.
Given that the flow rate is the same for both holes: $Q_A = Q_B$.
$A_A \cdot V_A = A_B \cdot V_B$.
Since hole '$A$' is a square of side '$L$',$A_A = L^2$.
Since hole '$B$' is a circle of radius '$R$',$A_B = \pi R^2$.
Substituting the values: $L^2 \cdot V_A = (\pi R^2) \cdot (2V_A)$.
$L^2 = 2\pi R^2$.
$L = \sqrt{2\pi} \cdot R$.

(B) According to Torricelli's law,the velocity of efflux is $v = \sqrt{2gh}$.
Let $A_1$ be the area of the square hole and $A_2$ be the area of the circular hole.
$A_1 = a^2$ and $A_2 = \pi r^2$.
The volume flow rate $Q$ is given by $Q = A \cdot v$.
Given that the volume flow rate is equal for both holes:
$A_1 \sqrt{2gy} = A_2 \sqrt{2g(16y)}$
$a^2 \sqrt{y} = (\pi r^2) \sqrt{16y}$
$a^2 \sqrt{y} = \pi r^2 (4 \sqrt{y})$
$a^2 = 4 \pi r^2$
$r^2 = \frac{a^2}{4 \pi}$
$r = \sqrt{\frac{a^2}{4 \pi}} = \frac{a}{2 \sqrt{\pi}}$

(C) According to the equation of continuity for an incompressible fluid,the product of the cross-sectional area and the velocity of flow remains constant at any point along the pipe: $A_1 V_1 = A_2 V_2$.
Here,$A_1$ is the area of the pipe,$V$ is the velocity in the pipe,$A_2$ is the area of the nozzle,and $V_1$ is the velocity at the nozzle.
Since the area $A = \pi (d/2)^2$,we have $A \propto d^2$.
Substituting this into the continuity equation: $d^2 V = d_2^2 V_1$.
Solving for the nozzle diameter $d_2$: $d_2^2 = d^2 \frac{V}{V_1}$.
Taking the square root on both sides: $d_2 = d \sqrt{\frac{V}{V_1}}$.

(A) According to the equation of continuity,$A_{1}V_{1} = A_{2}V_{2}$.
Since the flow rate is constant,if the cross-sectional area $A$ decreases,the velocity $V$ must increase.
According to Bernoulli's principle for a horizontal pipe,$P + \frac{1}{2}\rho V^{2} = \text{constant}$.
This implies that as the velocity $V$ increases,the pressure $P$ must decrease.
Therefore,in the narrowest part of the pipe,the velocity is maximum and the pressure is minimum.

(B) According to Bernoulli's principle for a fluid in motion,the sum of pressure energy,kinetic energy,and potential energy per unit volume remains constant.
For the fluid inside the pipe,the initial pressure is $P_{1}$ when the velocity is zero.
When the valve is opened,the pressure drops to $P_{2}$ and the fluid gains a velocity $v$.
Applying Bernoulli's equation between the inside of the pipe and the outlet:
$P_{1} + 0 = P_{2} + \frac{1}{2} \rho v^{2}$
Rearranging the terms to solve for $v$:
$P_{1} - P_{2} = \frac{1}{2} \rho v^{2}$
$v^{2} = \frac{2(P_{1} - P_{2})}{\rho}$
$v = \sqrt{\frac{2(P_{1} - P_{2})}{\rho}}$
Thus,the correct option is $B$.

(C) Let $n$ capacitors be connected in parallel,each of capacitance $C = 2 \mu F$. The equivalent capacitance of this parallel group is $C_p = nC = 2n \mu F$.
Let $m$ such parallel groups be connected in series. The total number of capacitors is $N = n \times m = 7$.
The equivalent capacitance of $m$ such groups in series is given by $\frac{1}{C_{eq}} = \frac{1}{C_p} + \frac{1}{C_p} + ... (m \text{ times}) = \frac{m}{C_p}$.
Thus,$C_{eq} = \frac{C_p}{m} = \frac{2n}{m}$.
Given $C_{eq} = \frac{10}{11} \mu F$,we have $\frac{2n}{m} = \frac{10}{11}$,which simplifies to $\frac{n}{m} = \frac{5}{11}$.
This implies $11n = 5m$. Since $n \times m = 7$,this does not yield integer solutions for $n$ and $m$ directly. Let us re-evaluate the combination: If we have $n$ parallel branches each containing $m$ capacitors in series,the equivalent capacitance is $C_{eq} = \frac{n}{m} C$.
Substituting $C = 2 \mu F$ and $C_{eq} = \frac{10}{11} \mu F$,we get $\frac{10}{11} = \frac{n}{m} \times 2$,so $\frac{n}{m} = \frac{5}{11}$.
Given $n+m$ is not necessarily $7$,but the total capacitors $n \times m = 7$ is not correct. The total capacitors $N = n \times m = 7$ is not possible for these ratios. Checking option $C$: $5$ in parallel and $2$ in series would mean $n=5, m=2$,$C_{eq} = \frac{5}{2} \times 2 = 5 \mu F$. Checking option $D$: $4$ in parallel and $3$ in series would mean $n=4, m=3$,$C_{eq} = \frac{4}{3} \times 2 = 2.66 \mu F$.
Wait,if we have $n$ capacitors in series and $m$ such rows in parallel,$C_{eq} = \frac{m}{n} C = \frac{m}{n} \times 2 = \frac{10}{11} \implies \frac{m}{n} = \frac{5}{11}$. Total capacitors $m \times n = 7$ is not possible.
Re-reading: If $5$ capacitors are in series and $2$ are in parallel,$C_{eq} = \frac{2}{5} \times 2 = 0.8 \mu F$. If $2$ capacitors are in series and $5$ are in parallel,$C_{eq} = \frac{5}{2} \times 2 = 5 \mu F$.
Actually,for $C_{eq} = \frac{10}{11} \mu F$ with $C=2 \mu F$,we need $\frac{m}{n} = \frac{5}{11}$. This is not possible with $7$ capacitors. However,if the question implies a mixed grouping,the closest match is $5$ in series and $2$ in parallel.

(D) From the circuit diagram,capacitors $C_{1}$,$C_{2}$,and $C_{3}$ are in series with each other,and this combination is in parallel with capacitor $C_{4}$.
Given: $C_{1} = C$,$C_{2} = 2C$,$C_{3} = 3C$,and $C_{4} = 4C$.
The equivalent capacitance of the series combination of $C_{1}$,$C_{2}$,and $C_{3}$ is given by:
$\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} = \frac{1}{C} + \frac{1}{2C} + \frac{1}{3C} = \frac{6+3+2}{6C} = \frac{11}{6C}$
$\Rightarrow C_{eq} = \frac{6}{11}C$
The charge on the series combination (which is the same for each capacitor $C_{1}$,$C_{2}$,and $C_{3}$) is:
$Q_{series} = C_{eq} V = \frac{6}{11}CV$
Since $C_{2}$ is in this series branch,the charge on $C_{2}$ is $Q_{2} = \frac{6}{11}CV$.
The charge on capacitor $C_{4}$ (which is in parallel with the battery) is:
$Q_{4} = C_{4} V = (4C)V = 4CV$
The ratio of the charges on $C_{2}$ and $C_{4}$ is:
$\frac{Q_{2}}{Q_{4}} = \frac{\frac{6}{11}CV}{4CV} = \frac{6}{11 \times 4} = \frac{6}{44} = \frac{3}{22}$

(B) $1$. First, find the equivalent capacitance of the parallel combination of $C$ and $2C$. Since they are in parallel, $C_p = C + 2C = 3C$.
$2$. Now, this combination $(C_p = 3C)$ is in series with the third capacitor of capacity $3C$.
$3$. The total equivalent capacitance $C_{eq}$ of the series combination is given by $\frac{1}{C_{eq}} = \frac{1}{3C} + \frac{1}{3C} = \frac{2}{3C}$, so $C_{eq} = \frac{3C}{2}$.
$4$. The total charge $Q$ supplied by the source is $Q = C_{eq} \times V = \frac{3CV}{2}$.
$5$. In a series circuit, the charge on each branch is the same. Thus, the charge on the $3C$ capacitor is $\frac{3CV}{2}$, and the charge on the parallel combination $(C_p = 3C)$ is also $\frac{3CV}{2}$.
$6$. For the parallel combination, the voltage across both capacitors ($C$ and $2C$) is the same. Let this voltage be $V'$. $V' = \frac{Q_{parallel}}{C_p} = \frac{3CV/2}{3C} = \frac{V}{2}$.
$7$. The charge on the capacitor of capacity $C$ is $q = C \times V' = C \times \frac{V}{2} = \frac{CV}{2}$.

(C) Given,resultant capacity,$C_{eq} = 2 \mu F$.
From the figure,all five capacitors are connected in series.
The equivalent capacity of capacitors in series is given by:
$\frac{1}{C_{eq}} = \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C} + \frac{1}{C}$
$\frac{1}{C_{eq}} = \frac{5}{C}$
Substituting the value of $C_{eq}$:
$\frac{1}{2 \mu F} = \frac{5}{C}$
$C = 5 \times 2 \mu F = 10 \mu F$
Therefore,the capacity of each capacitor is $10 \mu F$.

(A) Let $C_1 = 2 \mu F$,$C_2 = 4 \mu F$,and $C_3 = 6 \mu F$.
$C_1$ and $C_2$ are in parallel,so their equivalent capacitance is $C_p = C_1 + C_2 = 2 + 4 = 6 \mu F$.
Now,$C_p$ and $C_3$ are in series. The equivalent capacitance of the whole circuit is $C_{eq} = \frac{C_p \times C_3}{C_p + C_3} = \frac{6 \times 6}{6 + 6} = 3 \mu F$.
The total charge supplied by the battery is $Q = C_{eq} \times V = 3 \mu F \times 12 \text{ V} = 36 \mu C$.
Since $C_p$ and $C_3$ are in series,the charge on the parallel combination $(C_p)$ is also $36 \mu C$.
This charge $Q$ is divided between $C_1$ and $C_2$ in proportion to their capacitances:
$Q_1 = Q \times \left( \frac{C_1}{C_1 + C_2} \right) = 36 \mu C \times \left( \frac{2}{2 + 4} \right) = 36 \times \frac{2}{6} = 12 \mu C$.
Thus,the charge on the $2 \mu F$ capacitor is $12 \mu C$.

(A) The circuit consists of a $3 \mu F$ capacitor in series with a parallel combination of two $1.5 \mu F$ capacitors,which is then in series with another $3 \mu F$ capacitor.
$1$. First,calculate the equivalent capacitance of the two $1.5 \mu F$ capacitors in parallel: $C_p = 1.5 \mu F + 1.5 \mu F = 3 \mu F$.
$2$. Now,the circuit simplifies to three capacitors of $3 \mu F$ each,connected in series.
$3$. The equivalent capacitance $C_{eq}$ for capacitors in series is given by: $\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$.
$4$. Substituting the values: $\frac{1}{C_{eq}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 \mu F^{-1}$.
$5$. Therefore,$C_{eq} = 1 \mu F$.

(C) repeater is a combination of a receiver and a transmitter.
It receives the signal from the transmitter,amplifies it,and retransmits it to the destination.
This process compensates for the signal loss (attenuation) that occurs over long distances,thereby extending the range of the communication system.

(A) Digital signals are signals that represent data as a sequence of discrete values at any given time. Unlike analog signals,which are continuous,digital signals consist of discrete stepwise values,typically represented by binary digits ($0$ and $1$). Therefore,they are characterized by discrete stepwise values.

(B) In a standard superheterodyne receiver block diagram,the signal from the receiving antenna first passes through an amplifier (often an $RF$ amplifier) to boost the weak signal.
Then,it goes to the Intermediate Frequency $(IF)$ stage,where the frequency is converted to a lower,fixed intermediate frequency.
After the $IF$ stage,the signal passes through a detector (demodulator) to extract the original message signal.
Finally,the signal is passed through an audio amplifier to increase its power for the output device.
Looking at the provided diagram:
- The first box '$Y$' represents an amplifier ($RF$ amplifier).
- The second box '$X$' represents the $IF$ stage.
- The third box is the detector.
- The fourth box '$Y$' represents an amplifier (audio amplifier).
Therefore,'$X$' represents the $IF$ stage and '$Y$' represents an amplifier.

(B) repeater is a device that receives a signal,amplifies it,and then retransmits it to extend the range of communication. Therefore,it essentially consists of a receiver to pick up the signal and a transmitter to send it further. Thus,it is a combination of a receiver and a transmitter.

(B) Given:
Frequency of the modulating signal,$f_m = 3 \text{ kHz} = 0.003 \text{ MHz}$.
Frequency of the carrier wave,$f_c = 2.5 \text{ MHz} = 2500 \text{ kHz}$.
The upper sideband frequency $(f_{USB})$ is given by $f_c + f_m = 2500 \text{ kHz} + 3 \text{ kHz} = 2503 \text{ kHz}$.
The lower sideband frequency $(f_{LSB})$ is given by $f_c - f_m = 2500 \text{ kHz} - 3 \text{ kHz} = 2497 \text{ kHz}$.
Thus,the sideband frequencies are $2503 \text{ kHz}$ and $2497 \text{ kHz}$.

(D) The range $d$ for line-of-sight propagation is given by the formula $d = \sqrt{2Rh}$,where $R$ is the radius of the Earth and $h$ is the height of the antenna.
From this formula,we can see that $d \propto \sqrt{h}$.
Let the initial range be $d_1 = \sqrt{2Rh}$ and the initial height be $h_1 = h$.
Let the new height be $h_2 = 2h$.
The new range $d_2$ is given by $d_2 = \sqrt{2Rh_2} = \sqrt{2R(2h)}$.
$d_2 = \sqrt{2} \times \sqrt{2Rh} = \sqrt{2} d_1$.
Therefore,if the height is doubled,the new range becomes $\sqrt{2} d$.

(D) In amplitude modulation,the amplitude of the carrier wave is varied in accordance with the instantaneous value of the modulating signal (information signal),while the frequency and phase of the carrier wave remain constant.

(D) The modulation index $(m_a)$ is defined as the ratio of the peak voltage of the modulating signal $(V_m)$ to the peak voltage of the carrier wave $(V_c)$.
Formula: $m_a = \frac{V_m}{V_c}$
Given:
Peak voltage of carrier wave $(V_c)$ = $16 \text{ V}$
Modulation index $(m_a)$ = $75 \% = 0.75$
Calculation:
$V_m = m_a \times V_c$
$V_m = 0.75 \times 16 \text{ V}$
$V_m = 12 \text{ V}$
Therefore, the peak voltage of the modulating signal is $12 \text{ V}$.

(C) Low frequency audio signals cannot be transmitted directly over long distances because they require antennas of impractical size and suffer from high attenuation. To overcome this,the low frequency audio signal (baseband signal) is superposed on a high frequency wave,known as the carrier wave. This process is called modulation.

(A) In amplitude modulation,the modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier wave. If $\mu > 1$,it leads to over-modulation,which causes severe distortion in the signal. Therefore,to avoid distortion,the modulation index $\mu$ is always kept less than or equal to $1$ (i.e.,$\mu \le 1$). Thus,the statement that $\mu$ is kept greater than one is incorrect.

(A) For a linear antenna of length $l$ (where $l \ll \lambda$),the power radiated $P$ is given by the relation $P \propto \frac{1}{\lambda^2}$.
This implies that the power radiated is inversely proportional to the square of the wavelength.
Therefore,as the wavelength $\lambda$ increases,the power radiated decreases.
Thus,the correct proportionality is $\lambda^{-2}$.

(A) modem stands for Modulator-Demodulator.
In the process of space communication,at the transmitting end,the modem acts as a modulator to convert digital signals into analog signals for transmission.
At the receiving end,the modem acts as a demodulator to convert the received analog signals back into digital signals.
Therefore,the modem acts as a modulator during transmission and as a demodulator during reception.

(B) In the given circuit,the two cells are connected in series but with opposing polarities.
Applying Kirchhoff's voltage law $(KVL)$ to the loop,we start from the $200 \ V$ cell and move in the direction of the current:
$200 - I(38) - 10 = 0$
$190 - 38I = 0$
$38I = 190$
$I = \frac{190}{38} = 5 \ A$
Therefore,the current flowing through the circuit is $5 \ A$.

(B) The $3 \mu F$ and $6 \mu F$ capacitors are connected in parallel. Their equivalent capacitance is $C_p = 3 \mu F + 6 \mu F = 9 \mu F$.
This $C_p$ is in series with the $4.5 \mu F$ capacitor. The equivalent capacitance of the circuit is $C_{eq} = \frac{4.5 \times 9}{4.5 + 9} = \frac{40.5}{13.5} = 3 \mu F$.
The total charge supplied by the $12 \text{ V}$ battery is $Q = C_{eq} V = 3 \mu F \times 12 \text{ V} = 36 \mu C$.
Since the $4.5 \mu F$ capacitor is in series with the combination,the charge on it is equal to the total charge $Q = 36 \mu C$.
The potential difference across the $4.5 \mu F$ capacitor is $V_{4.5} = \frac{Q}{C} = \frac{36 \mu C}{4.5 \mu F} = 8 \text{ V}$.

(D) The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} = 20 \ \Omega + 10 \ \Omega = 30 \ \Omega$.
The current flowing through the circuit is $I = \frac{E}{R_{total}} = \frac{3 \ V}{30 \ \Omega} = 0.1 \ A$.
The potential difference across the wire is $V_{wire} = I \times R_{wire} = 0.1 \ A \times 20 \ \Omega = 2 \ V$.
The potential gradient along the wire is defined as the potential drop per unit length: $\text{Potential gradient} = \frac{V_{wire}}{L} = \frac{2 \ V}{10 \ m} = 0.2 \ V/m$.

(D) Applying Kirchhoff's Current Law $(KCL)$ at the junctions:
At the right junction: $I_{1} + I_{2} = I_{4}$
Given $I_{1} = -0.4 \text{ A}$ and $I_{4} = 1 \text{ A}$,we have:
$-0.4 + I_{2} = 1 \implies I_{2} = 1.4 \text{ A}$
At the bottom-left junction: $I_{5} = I_{3} + I_{4}$
Given $I_{5} = 0.4 \text{ A}$ and $I_{4} = 1 \text{ A}$,we have:
$0.4 = I_{3} + 1 \implies I_{3} = -0.6 \text{ A}$
At the top-left junction: $I_{6} = I_{5} + I_{3}$ (Wait,looking at the diagram,$I_{6}$ is the current entering the top-right junction from the left branch,so $I_{6} = I_{5}$ is not correct. Let's re-evaluate the circuit).
Actually,applying $KCL$ at the top-right junction: $I_{6} = I_{1} + I_{2} + I_{3}$.
Substituting the values: $I_{6} = -0.4 + 1.4 + (-0.6) = 0.4 \text{ A}$.
Thus,$I_{2} = 1.4 \text{ A}$,$I_{3} = -0.6 \text{ A}$,and $I_{6} = 0.4 \text{ A}$.

(C) The current $I$ is defined as the rate of flow of charge,given by $I = \frac{q}{T}$,where $q$ is the charge and $T$ is the time period of revolution.
For an electron moving in a circular orbit of radius $r$ with orbital velocity $v$,the distance covered in one revolution is the circumference $2 \pi r$.
Thus,the time period $T$ is given by $T = \frac{2 \pi r}{v}$.
Substituting this into the expression for current:
$I = \frac{e}{T} = \frac{e}{(2 \pi r / v)} = \frac{ev}{2 \pi r}$.
Since $2 \pi$ is a constant,we have $I \propto e^{1} v^{1} r^{-1}$.
Therefore,the correct option is $C$.

(A) The galvanometer has a resistance $G$ and a full-scale deflection voltage $V_g$. The full-scale current $I_g$ is given by $I_g = \frac{V_g}{G}$.
To convert the galvanometer into a voltmeter of range $V$,a series resistance $R$ must be connected in series with the galvanometer.
The total resistance of the circuit becomes $R + G$.
According to Ohm's law,for the new range $V$,the current $I_g$ remains the same:
$V = I_g(R + G)$
Substituting $I_g = \frac{V_g}{G}$ into the equation:
$V = \left(\frac{V_g}{G}\right)(R + G)$
$\frac{V}{V_g} = \frac{R+G}{G}$
$\frac{V}{V_g} = \frac{R}{G} + 1$
$\frac{R}{G} = \frac{V}{V_g} - 1$
$R = G\left(\frac{V}{V_g} - 1\right)$

(D) Let the initial current through the galvanometer be $I$. The deflection is proportional to the current,so the new current through the galvanometer is $I' = I/2$.
By the principle of current division in a parallel circuit,the total current $I_{total}$ is divided between the galvanometer $G$ and the shunt $X$.
Since $I_G = I/2$,the remaining current must flow through the shunt: $I_X = I_{total} - I_G = I - I/2 = I/2$.
Since the galvanometer and shunt are in parallel,the potential difference across them is equal: $I_G \cdot G = I_X \cdot X$.
Substituting the values: $(I/2) \cdot G = (I/2) \cdot X$.
Therefore,$G = X$ or $X = G$.

(C) In a moving coil galvanometer,a strong radial magnetic field is produced by using concave-shaped pole pieces of a permanent magnet. This internal magnetic field is significantly stronger than any external stray magnetic field. Due to this high intensity,the effect of external stray magnetic fields becomes negligible,ensuring that the deflection of the coil depends only on the current flowing through it.

(D) The sensitivity of a potentiometer is defined as the smallest potential difference that can be measured by it.
Sensitivity is inversely proportional to the potential gradient $(k = V/L)$.
$A$ smaller potential gradient means a higher sensitivity.
Therefore, to decrease the sensitivity, we must increase the potential gradient along the wire.
Since the potential gradient $k = I \cdot R/L$, increasing the current $(I)$ through the wire increases the potential gradient, thereby decreasing the sensitivity.

(B) Given: Resistance of galvanometer $G = 100 \Omega$, Shunt resistance $S = 1 \Omega$, Full-scale deflection current $I_g = 10 \mu A = 10 \times 10^{-6} A$.
To convert a galvanometer into an ammeter, the shunt resistance $S$ is connected in parallel.
The total current $I$ required for full-scale deflection is given by the formula: $I = I_g \left( \frac{G+S}{S} \right)$.
Substituting the values: $I = 10 \mu A \left( \frac{100 + 1}{1} \right)$.
$I = 10 \mu A \times 101 = 1010 \mu A$.
Converting to milliamperes: $I = 1.01 \text{ mA}$.

(D) Let the potential difference across the wire be $V$. The potential gradient $k$ is given by $k = \frac{V}{L_{total}}$.
For the first case,the balancing length $l_1 = \frac{L}{5}$. The e.m.f. $E$ is given by $E = k_1 l_1 = \frac{V}{L} \cdot \frac{L}{5} = \frac{V}{5}$.
When the length of the wire is increased by $\frac{L}{2}$,the new total length becomes $L' = L + \frac{L}{2} = \frac{3L}{2}$.
The new potential gradient is $k_2 = \frac{V}{L'} = \frac{V}{3L/2} = \frac{2V}{3L}$.
Let the new balancing length be $l_2$. Then $E = k_2 l_2$.
Equating the two expressions for $E$: $\frac{V}{5} = \frac{2V}{3L} \cdot l_2$.
Solving for $l_2$: $l_2 = \frac{V}{5} \cdot \frac{3L}{2V} = \frac{3L}{10}$.

(D) In a potentiometer,the balancing length $\ell$ is directly proportional to the terminal potential difference $V$ across the cell.
For an open circuit,the balancing length $\ell_{1}$ corresponds to the e.m.f. $E$ of the cell: $E \propto \ell_{1}$.
When the cell is shunted with an external resistance $R$,the terminal potential difference $V$ is given by $V = E - Ir$,where $I = \frac{E}{R+r}$.
Thus,$V = E - \left(\frac{E}{R+r}\right)r = E\left(1 - \frac{r}{R+r}\right) = E\left(\frac{R}{R+r}\right)$.
The new balancing length $\ell_{2}$ corresponds to the terminal potential difference $V$: $V \propto \ell_{2}$.
Given $\ell_{2} = \frac{\ell_{1}}{2}$,we have $\frac{V}{E} = \frac{\ell_{2}}{\ell_{1}} = \frac{1}{2}$.
Substituting the expression for $V/E$: $\frac{R}{R+r} = \frac{1}{2}$.
Solving for $r$: $2R = R + r$,which gives $r = R$.

(C) Let $I$ be the total current and $I_g$ be the current passing through the galvanometer (ammeter coil).
Given that $I_g = 2 \% \text{ of } I = \frac{2}{100} I = \frac{1}{50} I$.
The formula for current division in a shunt circuit is $\frac{I_g}{I} = \frac{S}{S+R}$,where $S$ is the shunt resistance and $R$ is the coil resistance.
Substituting the values: $\frac{1}{50} = \frac{S}{S+R}$.
Cross-multiplying gives: $S + R = 50S$.
Rearranging the terms: $R = 50S - S = 49S$.
Therefore,the shunt resistance is $S = \frac{R}{49}$.

(D) In a potentiometer experiment to find the internal resistance $(r)$ of a cell,the cell $E_1$ is connected in series with a resistance box $R$.
When the circuit is open,the balancing length $l_1$ is proportional to the $EMF$ $(E_1)$: $E_1 = k l_1$.
When the circuit is closed (resistance $R$ is connected),the balancing length $l_2$ is proportional to the terminal potential difference $(V)$: $V = k l_2$.
The internal resistance is given by the formula: $r = R \left( \frac{E_1}{V} - 1 \right)$.
Here,$E_1$ is the $EMF$ of the cell,$V$ is the terminal potential difference,and $R$ is the external resistance.

(D) The resistance $R$ of a wire is given by the formula $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length of the wire,and $A$ is the cross-sectional area.
Since the coil of a galvanometer is made of a wire of a certain length,the total length $L$ is directly proportional to the number of turns $N$ $(L \propto N)$.
If the number of turns $N$ is decreased,the total length $L$ of the wire used in the coil decreases.
Since $R \propto L$,a decrease in the length $L$ leads to a decrease in the resistance $R$ of the galvanometer.

(D) Let $k$ be the potential gradient of the potentiometer wire.
When the cells are connected in series with the same polarity,the total e.m.f. is $(E_{1} + E_{2})$. The balancing condition is $(E_{1} + E_{2}) = k \ell_{1}$,where $\ell_{1} = 64 \ cm$.
When the polarity of $E_{2}$ is reversed,the total e.m.f. is $(E_{1} - E_{2})$. The balancing condition is $(E_{1} - E_{2}) = k \ell_{2}$,where $\ell_{2} = 32 \ cm$.
Dividing the two equations: $\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{64}{32} = 2$.
$E_{1} + E_{2} = 2(E_{1} - E_{2})$.
$E_{1} + E_{2} = 2E_{1} - 2E_{2}$.
$3E_{2} = E_{1}$.
Therefore,$\frac{E_{1}}{E_{2}} = 3: 1$.

(A) Let $I_g$ be the full-scale deflection current of the galvanometer.
For the first case,the total resistance is $(200 + G)$,so the voltage range is $V = I_g(200 + G)$.
For the second case,the range is tripled,so $V' = 3V = I_g(2000 + G)$.
Dividing the two equations: $\frac{3V}{V} = \frac{I_g(2000 + G)}{I_g(200 + G)}$.
$3 = \frac{2000 + G}{200 + G}$.
$3(200 + G) = 2000 + G$.
$600 + 3G = 2000 + G$.
$2G = 1400$.
$G = 700 \Omega$.

(B) Given:
Current sensitivity,$S_I = \frac{d\theta}{dI} = 10 \text{ div./mA} = 10 \times 10^3 \text{ div./A}$.
Voltage sensitivity,$S_V = \frac{d\theta}{dV} = 4 \text{ div./mV} = 4 \times 10^3 \text{ div./V}$.
We know that $V = IR$,so $dV = dI \times R$,which implies $R = \frac{dV}{dI}$.
Also,$S_V = \frac{d\theta}{dV} = \frac{d\theta}{dI \cdot R} = \frac{S_I}{R}$.
Therefore,$R = \frac{S_I}{S_V} = \frac{10 \text{ div./mA}}{4 \text{ div./mV}} = \frac{10 \times 10^{-3} \text{ A}^{-1}}{4 \times 10^{-3} \text{ V}^{-1}} = \frac{10}{4} \Omega = 2.5 \Omega$.

(B) Total resistance $R_{total} = 495 \, \Omega + 5 \, \Omega = 500 \, \Omega$.
Current $I$ flowing through the circuit is given by $I = \frac{V}{R_{total}} = \frac{4 \, V}{500 \, \Omega} = 8 \times 10^{-3} \, A$.
The potential difference $V_{wire}$ across the potentiometer wire is $V_{wire} = I \times R_{wire} = 8 \times 10^{-3} \, A \times 5 \, \Omega = 40 \times 10^{-3} \, V$.
The potential gradient $k$ is defined as the potential drop per unit length: $k = \frac{V_{wire}}{L} = \frac{40 \times 10^{-3} \, V}{4 \, m} = 10 \times 10^{-3} \, V/m = 0.01 \, V/m$.

(D) The torque acting on the coil of a moving coil galvanometer is given by $\tau = nABi \sin \theta$. For small angles,$\sin \theta \approx \theta$. The restoring torque is $\tau_r = C \theta$,where $C$ is the torsional constant of the suspension wire.
Equating the torques: $C \theta = nABi$.
The current sensitivity $S_i$ is defined as the deflection per unit current: $S_i = \frac{\theta}{i} = \frac{nAB}{C}$.
From this expression,it is clear that the sensitivity $S_i$ is inversely proportional to the torsional constant $C$ of the suspension wire.

(D) The potential gradient $k$ of a potentiometer wire is given by $k = V/L$,where $V$ is the potential difference across the wire and $L$ is the total length of the wire.
When the length of the wire is doubled $(L' = 2L)$,the potential gradient becomes $k' = V/(2L) = k/2$.
The balancing condition for an unknown e.m.f. $E$ is $E = k \cdot l$,where $l$ is the balancing length.
Since $E$ remains constant,$k \cdot l = k' \cdot l'$.
Substituting $k' = k/2$,we get $k \cdot l = (k/2) \cdot l'$,which simplifies to $l' = 2l$.
Given the initial balancing length $l = 1.5 \,m$,the new balancing length is $l' = 2 \times 1.5 \,m = 3 \,m$.

(C) The potential difference between $A$ and $B$ is $V_{AB} = E_{1}$. The null point length is $L_{AB} = 0.9 \ m$.
Since $V \propto L$,we have $E_{1} = k \times 0.9$,where $k$ is the potential gradient.
The potential difference between $A$ and $C$ is $V_{AC} = E_{1} - E_{2}$. The null point length is $L_{AC} = 0.3 \ m$.
Thus,$E_{1} - E_{2} = k \times 0.3$.
Dividing the two equations: $\frac{E_{1}}{E_{1} - E_{2}} = \frac{0.9}{0.3} = 3$.
$E_{1} = 3(E_{1} - E_{2})$
$E_{1} = 3E_{1} - 3E_{2}$
$3E_{2} = 2E_{1}$
$\frac{E_{2}}{E_{1}} = \frac{2}{3}$.

(B) The sensitivity of a galvanometer is defined as the deflection produced per unit current,i.e.,$S = \frac{\theta}{I}$.
Given that both galvanometers produce the same deflection $\theta$,the sensitivity is inversely proportional to the current required: $S \propto \frac{1}{I}$.
For $G_{1}$,$I_{1} = 2 \ mA$,so $S_{1} = \frac{\theta}{2 \ mA}$.
For $G_{2}$,$I_{2} = 3 \ mA$,so $S_{2} = \frac{\theta}{3 \ mA}$.
Comparing the two,since $I_{1} < I_{2}$,it follows that $S_{1} > S_{2}$.
Therefore,$G_{1}$ is more sensitive than $G_{2}$.

(A) The resistance of the ammeter is $G = 20 \Omega$ and the full-scale deflection current is $i_g = 1 \text{ mA} = 10^{-3} \text{ A}$.
Four resistors of $16 \Omega$ each are connected in parallel. The equivalent shunt resistance $S$ is given by $\frac{1}{S} = \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4} \Omega^{-1}$.
Thus,$S = 4 \Omega$.
For an ammeter,the shunt resistance $S$ is connected in parallel to the galvanometer $G$. The current $i$ to be measured is divided into $i_g$ through the galvanometer and $i_s$ through the shunt.
The voltage across the parallel combination is $i_s S = i_g G$.
$(i - i_g) S = i_g G$.
$(i - 10^{-3}) \times 4 = 10^{-3} \times 20$.
$i - 10^{-3} = \frac{20 \times 10^{-3}}{4} = 5 \times 10^{-3}$.
$i = 5 \times 10^{-3} + 1 \times 10^{-3} = 6 \times 10^{-3} \text{ A} = 6 \text{ mA}$.

(C) The potential gradient $k$ of the potentiometer wire is given by the ratio of the total potential difference $V$ to the total length $L$ of the wire.
$k = \frac{V}{L} = \frac{3 \,V}{4 \,m} = 0.75 \,V/m$.
We are given that the cell balances at a length $l = 100 \,cm = 1 \,m$.
The e.m.f. $E$ of the cell is equal to the potential drop across the balancing length $l$, which is given by $E = k \times l$.
Substituting the values, we get $E = 0.75 \,V/m \times 1 \,m = 0.75 \,V$.

(C) To convert a moving coil galvanometer into an ammeter,a small resistance called a shunt $(S)$ is connected in parallel with the galvanometer $(G)$.
The equivalent resistance $(R)$ of the ammeter is the parallel combination of $G$ and $S$.
The formula for parallel resistance is:
$\frac{1}{R} = \frac{1}{G} + \frac{1}{S}$
Taking the common denominator:
$\frac{1}{R} = \frac{S + G}{GS}$
Therefore,the resistance of the ammeter is:
$R = \frac{GS}{G + S}$
Comparing this with the given options,the correct expression is $\frac{SG}{S+G}$,which corresponds to option $(C)$.

(D) direct reading instrument provides the value of the measured quantity directly on a scale or display without requiring any manual calculation or balancing process.
$A$,$B$,and $C$ are direct reading instruments because they show the result immediately.
$A$ potentiometer is a null-type instrument. It requires a balancing process (finding the null point) to measure potential difference or $EMF$,and the final value is calculated based on the balancing length.
Therefore,the potentiometer is not a direct reading instrument.
Correct option is $D$.

(C) The total resistance of the circuit is $R_{total} = R_{wire} + R_{series} + r_{internal} = 3 \ \Omega + 8 \ \Omega + 1 \ \Omega = 12 \ \Omega$.
The current flowing through the potentiometer wire is $I = \frac{V}{R_{total}} = \frac{4 \ V}{12 \ \Omega} = \frac{1}{3} \ A$.
The potential difference across the potentiometer wire is $V_{wire} = I \times R_{wire} = \frac{1}{3} \ A \times 3 \ \Omega = 1 \ V$.
The potential gradient $K$ of the wire is $K = \frac{V_{wire}}{L} = \frac{1 \ V}{100 \ cm} = 0.01 \ V/cm$.
The e.m.f. $E$ of the cell balanced at length $\ell = 50 \ cm$ is $E = K \times \ell = 0.01 \ V/cm \times 50 \ cm = 0.50 \ V$.

(D) Current sensitivity $I_{s} = \frac{NBA}{k}$.
Voltage sensitivity $V_{s} = \frac{I_{s}}{R} = \frac{NBA}{kR}$.
Given $k_{1} = k_{2} = k$.
Ratio of current sensitivity $\frac{I_{s2}}{I_{s1}} = \frac{N_{2}B_{2}A_{2}}{N_{1}B_{1}A_{1}} = \frac{42 \times 0.50 \times 1.8 \times 10^{-2}}{30 \times 0.25 \times 3.6 \times 10^{-3}} = \frac{21 \times 1.8 \times 10^{-2}}{7.5 \times 3.6 \times 10^{-3}} = \frac{37.8 \times 10^{-2}}{27 \times 10^{-3}} = 1.4$.
Ratio of voltage sensitivity $\frac{V_{s2}}{V_{s1}} = \frac{I_{s2}}{I_{s1}} \times \frac{R_{1}}{R_{2}} = 1.4 \times \frac{10}{14} = 1.4 \times \frac{1}{1.4} = 1$.
Thus,the ratio is $1.4: 1$ and $1: 1$.

(D) In a potentiometer circuit,when two cells are connected to assist each other,the balancing length $\ell_{1}$ is proportional to $(E_{1} + E_{2})$. Thus,$E_{1} + E_{2} = k \ell_{1}$,where $k$ is the potential gradient.
When they are connected to oppose each other,the balancing length $\ell_{2}$ is proportional to $(E_{1} - E_{2})$. Thus,$E_{1} - E_{2} = k \ell_{2}$.
Dividing the two equations,we get: $\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{\ell_{1}}{\ell_{2}}$.
Using the given values $\ell_{1} = 490 \ cm$ and $\ell_{2} = 90 \ cm$:
$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{490}{90} = \frac{49}{9}$.
Applying componendo and dividendo:
$\frac{E_{1}}{E_{2}} = \frac{49 + 9}{49 - 9} = \frac{58}{40} = 1.45$.

(A) The sensitivity $S$ of a galvanometer is defined as the deflection per unit current,given by $S = \frac{\theta}{I}$.
Given that both galvanometers produce the same deflection $\theta = 20$ divisions.
For galvanometer $A$,$S_{A} = \frac{\theta}{I_{A}} = \frac{20}{4 \ mA} = 5 \ \text{div/mA}$.
For galvanometer $B$,$S_{B} = \frac{\theta}{I_{B}} = \frac{20}{7 \ mA} \approx 2.86 \ \text{div/mA}$.
Comparing the two,$S_{A} > S_{B}$.
Alternatively,since $S \propto \frac{1}{I}$ for a constant deflection,$\frac{S_{A}}{S_{B}} = \frac{I_{B}}{I_{A}} = \frac{7}{4} = 1.75$.
Thus,$S_{A} = 1.75 S_{B}$,which implies $S_{A} > S_{B}$.

(A) To convert a milliammeter into a voltmeter,a high resistance $R$ must be connected in series with the milliammeter.
Given:
Resistance of milliammeter,$G = 40 \Omega$
Full-scale deflection current,$I_g = 30 \text{ mA} = 30 \times 10^{-3} \text{ A} = 0.03 \text{ A}$
Required voltage range,$V = 15 \text{ V}$
The formula for the series resistance $R$ is:
$R = \frac{V}{I_g} - G$
Substituting the values:
$R = \frac{15}{30 \times 10^{-3}} - 40$
$R = \frac{15}{0.03} - 40$
$R = 500 - 40$
$R = 460 \Omega$
Therefore,a resistance of $460 \Omega$ should be connected in series.