Principle involved

The galvanometer shows current only when the magnetic flux through the coil changes.

By Faraday’s law, $\mathcal{E} = -\frac{d\Phi}{dt}$ so induced current exists only when $\Phi$ is changing.

Case (a): Magnet entering the coil

As the bar magnet enters the coil, the magnetic flux linked with the coil increases.

Therefore, $\frac{d\Phi}{dt} \neq 0$ and the galvanometer shows a deflection in one direction.

Case (b): Magnet completely inside the coil

When the magnet is fully inside and moving in such a way that the flux linked with the coil is effectively not changing appreciably (the standard textbook observation for this setup), the induced emf becomes zero.

Hence, $\frac{d\Phi}{dt} = 0$ and the galvanometer returns to zero.

Case (c): Magnet leaving the coil

As the magnet exits the coil, the magnetic flux through the coil decreases.

Therefore the sign of $\frac{d\Phi}{dt}$ reverses, so the induced current also reverses.

Thus the galvanometer shows a deflection in the opposite direction.

Overall observation

The sequence is:

• entry: deflection in one direction,
• fully inside: zero,
• exit: deflection in opposite direction.

This corresponds to Option C.

Comparison with stored answer

Stored correct answer = C.

My derived answer also = C.

So they agree.