Question

If maximum percentage errors in measurement of length, mass and time are 2%, 1% and 1% respectively then maximum percentage error in kinetic energy will be

• A. $1$
• B. $4$
• C. $7$
• D. $5$

Answer 1

The correct option is C $7$

$\begin{array}{c}\text{Given}\cdots \frac{dl}{l}\times 100\%=2\%\\ \frac{dt}{t}\times 100\%=1\%\end{array}$

$\frac{dm}{m}\times 100\%=1\%$

$v=\frac{d}{t}=\frac{L}{t}$

$\begin{array}{cc}\text{Kinetic energy}=k& =\frac{1}{2}m{v}^2\\ k& =\frac{1}{2}m{\left(\frac{L}{t}\right)}^2\\ \text{Taking log both side and differentiating}& \\ \frac{\delta k}{k}=\frac{\delta m}{m}+2\frac{\delta l}{l}-2\frac{\delta t}{t}& \end{array}$

$\begin{array}{c}\text{Error is always added}\\ \frac{\delta K}{k}\times 100\%=\frac{\delta m}{m}\times 100+2\frac{\delta l}{l}\times 100+2\frac{\delta t}{t}\times 100\\ =1+2\times 2+2\times 1\\ \frac{\delta K}{K}\times 100\%=7\%\end{array}$