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Question: Answered & Verified by Expert
A physical quantity $X$ is given by $X=\frac{2 k^3 l^2}{m \sqrt{n}}$ The percentage error in the measurements of $k$, $l, m$ and $n$ are $1 \%, 2 \%, 3 \%$ and $4 \%$ respectively. The value of $X$ is uncertain by
PhysicsMathematics in PhysicsAIIMSAIIMS 2017
Options:
  • A $8 \%$
  • B $10 \%$
  • C $12 \%$
  • D none of these
Solution:
1773 Upvotes Verified Answer
The correct answer is: $12 \%$
$\begin{aligned} & \text { : Given : } X=\frac{2 k^3 l^2}{m \sqrt{n}} \\ & \frac{\Delta X}{X}=3 \frac{\Delta k}{k}+2 \frac{\Delta l}{l}+\frac{\Delta m}{m}+\frac{1}{2} \frac{\Delta n}{n}\end{aligned}$
Percentage error in $X$
$\begin{aligned} \frac{\Delta X}{X} & \times 100=\left(3 \frac{\Delta k}{k}+2 \frac{\Delta l}{l}+\frac{\Delta m}{m}+\frac{1}{2} \frac{\Delta n}{n}\right) \times 100 \\ & =3 \times 1 \%+2 \times 2 \%+3 \%+\frac{1}{2} \times 4 \% \\ & =3 \%+4 \%+3 \%+2 \%=12 \%\end{aligned}$
Hence, the value of $X$ is uncertain by $12 \%$.

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