Question

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

• A. 100%
• B. 200%
• C. 300%
• D. 400%

Answer 1

The correct option is C 300%

Since error in measurement of momentum is + 100%
$\therefore p_1=p,p_2=2p$
$K_1=\frac{P^2}{2m},K_2=\frac{(2p{)}^2}{2m}$
$\%\text{error in K}=\left(\frac{K_2-K_1}{K_1}\right)\times 100$
$=\left(\frac{4-1}{1}\right)\times 100$
$=300\%$