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If kinetic energy of a body is increased...

If kinetic energy of a body is increased by 300%, then percentage change in momentum will be

A

1

B

2

C

`sqrt(300)`%

D

4

Text Solution

Verified by Experts

The correct Answer is:
A
Let initial K.E. `= 100 alpha`.
Finale `K.E. = (100+ 300) alpha = 400 alpha`
`therefore (p_(1)^(2))/(p_(1)^(2))= ( 2 mk_(1))/(2mk_(2))= (100 alpha)/(400 alpha) = (1)/(4)`
Or `(p_2)/(p_1)= (2)/(1)` or `p_2 = 2p_1`
`therefore ("Change in momentum")/("Original momentum")xx 100 = (p_2 - p_1)/( p_1) xx 100`
`=(2p_1 - p_1) /(p_1) xx 100=100%`
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