Question

If the percentage error in the measurement of velocity of a particle 50% then the percentage error in the measurement of its kinetic energy will be

• A. 225%
• B. 125%
• C. 100%
• D. 200%

Answer 1

The correct option is C 100%

$\begin{array}{c}kE=\frac{1}{2}m{v}^2\\ \text{Differentiating both sides}\\ d\left(kE\right)=\frac{1}{2}m\times 2vdv\end{array}$

$\begin{array}{c}\text{Dividing both side by}k\cdot E\\ \frac{d\left(kE\right)}{kE}=\frac{mvdv}{kE}\\ \frac{d\left(kE\right)}{kE}=\frac{mvdv}{\frac{1}{2}m{v}^2}\end{array}$

$\begin{array}{c}\text{\%error:-}\\ \begin{array}{cc}\frac{d\left(KE\right)}{KE}\times 100\%& =2\frac{dv}{v}\times 100\\ & =2\times 50\%\\ & =100\%\text{Ans.}\end{array}\end{array}$