Consider a man standing on the spring balance such that the reading on the spring balance is 60kg. If the man jumps out of the spring balance what will be the final reading of the spring balance?

The correct option is "First increases then decrease to zero".

To understand why the reading behaves this way, let us analyze the forces acting on the man and the spring balance during the process of jumping.

Initially, when the man is standing still on the spring balance, he is in static equilibrium. The reading on the spring balance represents the normal force (or contact force) $R$ exerted by the balance on the man, which is equal to his actual weight:
$R=mg=60\text{kg}$

When the man decides to jump upwards off the spring balance, he must accelerate his body vertically upwards. According to Newton's second law of motion, to produce an upward acceleration $a$, the net vertical force acting on the man must be upwards.
Thus, the equation of motion for the man during the push-off phase is:
$R-mg=ma$
Rearranging this gives:
$R=mg+ma$

Since $ma>0$ during the upward push-off, the normal force $R$ exerted by the balance must exceed the man's static weight ($mg$). By Newton's third law, the man exerts an equal and opposite downward force on the spring balance. Therefore, during the initial phase of the jump, the reading on the spring balance increases (becomes greater than 60 kg).

Once the man loses contact with the spring balance and is airborne, there is no contact force acting between him and the scale:
$R=0$
Consequently, the reading on the spring balance immediately decreases to zero.

Combining these stages, we find that the reading on the scale first increases then decreases to zero.