If the error in the measurement of momentum of a particle is (+ 100%), then the error in the measurement of kinetic energy is

Options

A. 1
B. 2
C. 3
D. 4

Answer

C

Explanation

Since error in measurement of momentum is +100%.`p_(1)=p,p_(2)=2p`
`K_(1)=(p^(2))/(2m),K_(2)=((2p)^(2))/(2m)`
`%"in" K=((K_(2)-K_(1))/(K_(1)))xx100`
`=((4-1)/(1))xx100=300%`