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If the momentum of a body is increased b...

If the momentum of a body is increased by 50%, then the percentage increase in its kinetic energy is

A

`125%`

B

`50%`

C

`250%`

D

`sqrt50%`

Text Solution

Verified by Experts

The correct Answer is:
A
Let original momentum `p_(1) = 100 alpha`,
Find momentum `= 150 alpha`
`(p_(1)^(2))/(p_(2)^(2))= (2mk_(1))/(2mk_(2))= (k_(1))/(k_(2))`
`((100 alpha)/(150 alpha))^(2)= (k_(1))/(k_(2))`
`therefore ((2)/(3))^(2) = (k_(1))/(k_(2)) rArr (4)/(9) = (k_(1))/(k_(2))`
`therefore k_(2)= (9k_(1))/(4)`
`therefore` Percentage increament in `k= ("Change in KE")/("Original KE") xx 100`
`=(k_(2)- k_(1))/(k_(1)) xx 100`
`=((9)/(4) k_(1)-k_(1))/(k_(1)) xx 100= (5)/(4) xx 100= 125%`
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