Consider the situation shown in figure. A spring of spring constant 400 N/m is attached at one end to a wedge fixed rigidly with the horizontal part. A 40g mass is released from rest while situated at a height 5m on the curved track. The maximum deformation in the spring is nearly equal to?

Correct option is B. 9.8cm
The loss in gravitational potential energy would be equal to the gain in elastic potential energy.
At the point of maximum deformation, all the gravitational potential energy would convert to elastic potential energy.
mgh = (1/2)kx²
x = √(2mgh/k)
Put the value into the formula
x = √(2 × 0.04 × 9.8 × 5 / 400)
x = 0.098 m
x = 9.8 cm
So, The minimum deformation in the spring is 9.8 cm.
Hence, B is the correct option