Momentum of a particle is increased by `50%`. By how much percentage kinetic energy of particle will increase?

Options

A. `25%`
B. `50%`
C. `100%`
D. `125%`

Answer

D

Explanation

We know that `K prop P^(2)` or `(K_(2))/(K_(1)) = (P_(2)^(2))/(P_(1)^(2))`
`(K_(2) - K_(1))/(K_(1)) xx 100 = ((P_(2)^(2) - P_(1)^(2))/(P_(1)^(2))) xx 100`
`= [((150)^(2) - (100)^(2))/((100)^(2))] xx 100 = 125%`