A child stands at the centre of a turn table with his two arms out stretched. The turntable is set rotating with an angular speed of 40 rev/min. Now, the child folds his hands back and thereby reduces his moment of inertia to `(2)/(5)` times the initial value. The new kinetic energy of rotation is x times the initial kinetic energy of rotation. The value of x is :

Options

A. 2.5
B. 5
C. 1
D. 6.756

Answer

A

Explanation

According to law of conservation of angular momentum, we get
`therefore I_(1)omega_(1)=I_(2)omega_(2)`
Now it is given that
`I_(2)=(2)/(5)I_(1)`
`therefore omega_(1)=5//2omega_(1)`
Also `K.E.=(1)/(2)((2)/(5)I_(1))(5//2omega_(1))^(2)`
`E_(k)=(1)/(2)xx5//2xxI_(1)omega_(1)^(2)`
`=2.5xx(1)/(2)I_(1)omega_(1)^(2)`
`therefore` value of `x=2.5`