Q 11-01-032NEETJEE MainEasy
The momentum of a body is measured with an error of $2\%$ and its mass with an error of $1\%$. The maximum percentage error in its kinetic energy, calculated using $K = \dfrac{p^2}{2m}$, is
Answer: (C) $5\%$
$$\frac{\Delta K}{K}\times 100 = 2\,\frac{\Delta p}{p}\times 100 + \frac{\Delta m}{m}\times 100 = 2(2) + 1 = 5\%$$
The constant $\tfrac12$ has no error, so it contributes nothing.
Solution by Sreeraj P, M.Sc Physics