What will be the acceleration of the falling bar magnet which passes through the ring such that the ring is held horizontally and the bar magnet is dropped along the axis of the ring?

The correct option is "It is less than due to gravity".

Let us understand the physical principles behind this behavior step-by-step:
When a bar magnet is dropped vertically along the axis of a horizontally held conducting ring, the magnetic flux passing through the ring changes as the magnet falls towards and then passes through the ring.

According to Faraday's Law of Electromagnetic Induction, this changing magnetic flux induces an electromotive force (emf) and consequently an induced current in the conducting ring.
According to Lenz's Law, the direction of the induced current in the ring will be such that it opposes the change in magnetic flux that produced it.

This opposition manifests as an electromagnetic force acting on the falling magnet:
1. As the magnet approaches the ring, the ring exerts a repulsive magnetic force upwards on the falling magnet to oppose its approach.
2. As the magnet passes through the ring and begins to fall away, the ring exerts an attractive magnetic force upwards on the departing magnet to oppose its departure.
In both cases, the magnetic force exerted by the ring on the magnet is directed vertically upwards, opposing the downward gravitational pull.

Let the mass of the magnet be $m$ and the acceleration due to gravity be $g$. The downward gravitational force is $mg$. The upward electromagnetic force opposing the motion is $F_m$.
The net downward force $F_{\text{net}}$ acting on the magnet is given by:

$F_{\text{net}}=mg-F_m$

Therefore, the acceleration $a$ of the falling magnet is:

$a=\frac{F_{\text{net}}}{m}=g-\frac{F_m}{m}$

Since $F_m>0$, it follows that the acceleration $a$ of the falling bar magnet is less than the acceleration due to gravity ($a<g$).