Q76 / 90 · JEE Main 2024 (06 Apr Shift 1) · Physics

To find the spring constant $(k)$ of a spring experimentally, a student commits $2 \%$ positive error in the measurement of time and $1 \%$ negative error in measurement of mass. The percentage error in determining value of $k$ is :

  • Correct answer
    A
    $5 \%$
  • B
    $1 \%$
  • C
    $3 \%$
  • D
    $4 \%$

Solution

$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \\ & \mathrm{T}^2 \propto \frac{\mathrm{m}}{\mathrm{k}} \\ & \frac{2 \Delta \mathrm{T}}{\mathrm{T}} \%=\frac{\Delta \mathrm{m}}{\mathrm{m}} \%-\frac{\Delta \mathrm{k}}{\mathrm{k}} \% \\ & \frac{\Delta \mathrm{k}}{\mathrm{k}} \%=\frac{\Delta \mathrm{m}}{\mathrm{m}} \%-\frac{2 \Delta \mathrm{T}}{\mathrm{T}} \% \\ & \frac{\Delta \mathrm{k}}{\mathrm{k}} \%=(-1) \%-2(2) \%=|-5 \%|=5 \%\end{aligned}$

Hence, the correct answer is option: 'A'.

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