Specifying the total charge inside the surface is a sufficient condition for finding the electric flux \( \Phi \) through a closed surface.

According to Gauss's law:
\[
\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}
\]

The flux depends only on the enclosed charge \( q_{\text{enclosed}} \), regardless of the surface's shape or the charge distribution.

Yes, direct formulas can simplify the solution for problems like this involving conducting plates. Here's how:

Formula for charge redistribution in conductors:
For three parallel conducting plates with charges \( Q_1, Q_2, Q_3 \):
1. The charge on the inner faces is:
\[
Q_{\text{inner}} = \frac{Q_1 - Q_3}{2}
\]
2. The charge on the  outer faces remains equal to the net charge on the respective plates:
- Outer left face: \( Q_1 \)
- Outer right face: \( Q_3 \)

---

Applying the formula:
- Given charges:
\[
Q_1 = -2Q, \, Q_2 = Q, \, Q_3 = -Q
\]
- Charge on the inner faces of plate \( b \):
\[
Q_{\text{inner}} = \frac{Q_1 - Q_3}{2} = \frac{-2Q - (-Q)}{2} = \frac{-2Q + Q}{2} = \frac{-Q}{2}
\]

- Since the **right face of \( b \)** contributes to the inner faces:
\[
Q_{\text{right face of \( b \)}} = 0
\]

Final Answer:
\[
{Zero (0)}
\]

After closing the switch:

1. Potential equality: \(\frac{Q_1}{R} = \frac{Q_2}{3R} \Rightarrow Q_2 = 3Q_1\).

2. Charge conservation: \(Q_1 + Q_2 = 36Q \Rightarrow Q_1 + 3Q_1 = 36Q \Rightarrow Q_1 = 9Q\).

Final charge on the smaller sphere = 9Q.

To determine the direction of the resultant force on the positive charge \(q_0\) placed at the center due to the four charges on the clock face, we analyze the forces exerted by each charge.

Solution Steps:

1. Forces by Opposing Pairs:
- The positive charges at 12 and 3 exert a repulsive force on \(q_0\), while the negative charges at 6 and 9 exert an attractive force on \(q_0\).
- Each pair of opposing charges creates forces with equal magnitudes but different directions.

2. Force Components:
- The force due to the charges at 12 and 6 (along the vertical line) will cancel each other out in the vertical direction, as they are equal and opposite.
- Similarly, the forces due to the charges at 3 and 9 (along the horizontal line) will cancel each other out in the horizontal direction.

3. Resultant Force:
- Since the charges at 12 and 9 are positive and repulsive, they push \(q_0\) away from them.
- The vector sum of these forces results in a net force pointing diagonally between the 6 and 9 positions.

4. Direction:
- The resultant force points toward the 7:30 position, as this is the direction of the net vector resulting from the combination of all four forces.

Thus, the direction of the resultant force on \(q_0\) points to 7:30.

To find the time period of oscillation of the third charge \(-q\) when displaced perpendicular to the line joining the two charges of magnitude \(+Q\), we can follow these steps:

1. Restoring Force: When the charge \(-q\) is displaced by a small distance \(y\) perpendicular to the line joining the two charges, the net force on it due to the two fixed charges \(+Q\) has a restoring nature and acts toward the equilibrium position.

2. Approximation: For small \(y\), the restoring force \(F\) is proportional to \(y\), making it a simple harmonic motion (SHM) problem.

3. Electric Field: The electric field at the midpoint due to each charge \(+Q\) is \(E = \frac{Q}{4\pi\epsilon_0 (a^2 + y^2)}\). Using \(y \ll a\), we approximate the force on \(-q\) by expanding for small \(y\), resulting in a force \(F = -k y\), where \(k = \frac{2Qq}{4\pi \epsilon_0 a^3}\).

4. Angular Frequency: For SHM, the angular frequency \(\omega\) is \(\sqrt{\frac{k}{m}} = \sqrt{\frac{2Qq}{4\pi \epsilon_0 a^3 m}}\).

5. Time Period: The time period \(T\) is given by:
\[
T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{2ma^3 \pi \epsilon_0}{Qq}}
\]

When the outer shell is grounded, its potential becomes 0. The potential difference between the shells remains the same as before grounding (since grounding doesn't change the relative configuration of charges).

Initial potential difference:
Inner shell - Outer shell = \( 15 \, \text{V} - 10 \, \text{V} = 5 \, \text{V} \).

After grounding, the inner shell’s potential relative to the grounded outer shell becomes \( 0 + 5 \, \text{V} = 5 \, \text{V} \).

Thus, the inner shell’s potential is 5 V.

Pointed ends have more surface charge density so,

\[\sigma_{A}>\sigma_{B} > \sigma_{C}\]

This statement applies when a particle moves against a force field (e.g., an electron moving against an electric field). As it moves, work is done on it, leading to:

- Decrease in kinetic energy: The particle slows down as it loses speed.
- Increase in potential energy: The work done on the particle increases its potential energy.

This satisfies the conservation of energy principle.

The kinetic energy gained by the electron is equal to the work done by the electric field:

\[
eV = \frac{1}{2} m v^2
\]

Rearranging for \(v\):

\[
v = \sqrt{\frac{2eV}{m}}
\]

This shows that \(v \propto \sqrt{V}\).

For an electric dipole in a uniform electric field, the torque \( \tau \) on the dipole is given by:

\[
\tau = pE \sin \theta
\]

where:
- \( p \) is the dipole moment,
- \( E \) is the electric field strength,
- \( \theta \) is the angle between the dipole and the electric field.

Analysis:

1. Torque Variation:
- The torque \( \tau \) is maximum when \( \theta = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \) (odd multiples of \( \frac{\pi}{2} \)), as \( \sin \theta = \pm 1 \).
- The torque \( \tau \) is zero when \( \theta = 0, \pi, 2\pi, \dots \), as \( \sin \theta = 0 \).

2. Graph Interpretation:
- The correct graph of \( \tau \) versus \( \theta \) will have positive and negative peaks at \( \theta = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \), and cross zero at \( \theta = 0, \pi, 2\pi, \dots \).

Conclusion:

The correct answer is (b) as it represents the sinusoidal variation of torque with respect to \( \theta \) with alternating positive and negative values, matching the behavior of \( \tau = pE \sin \theta \).