K.E. of body is increased by 300 percent, then percentage increase in linear momentum will be:

Options

A. `300%`
B. `200%`
C. `100%`
D. `150%`

Answer

C

Explanation

Here `E_(k)=K.E=(L^(2))/(2I)`
Where L = angular momentum
`therefore L=(2IxxE_(k))^(1//2)`
% age increase = `((L_(2)-L_(1))/(L_(1)))xx100`
`=(sqrt(E_(k_(2)))-sqrt(E_(k_(1))))/(sqrt(E_(k_(1))))xx100`
`=(sqrt(400)-sqrt(100))/(sqrt(100))xx100`
`=((20-10)/(10))xx100`
`=100%`