JEE MAIN - Physics (2024 - 29th January Morning Shift - No. 1)
The resistance $$R=\frac{V}{I}$$ where $$\mathrm{V}=(200 \pm 5) \mathrm{V}$$ and $$I=(20 \pm 0.2) \mathrm{A}$$, the percentage error in the measurement of $$\mathrm{R}$$ is :
5.5%
3%
7%
3.5%
Explanation
$$\mathrm{R}=\frac{\mathrm{V}}{1}$$
According to error analysis
$$\begin{aligned} & \frac{\mathrm{dR}}{\mathrm{R}}=\frac{\mathrm{dV}}{\mathrm{V}}+\frac{\mathrm{dI}}{\mathrm{I}} \\ & \frac{\mathrm{dR}}{\mathrm{R}}=\frac{5}{200}+\frac{0.2}{20} \\ & \frac{\mathrm{dR}}{\mathrm{R}}=\frac{7}{200} \\ & \% \text { error } \frac{\mathrm{dR}}{\mathrm{R}} \times 100=\frac{7}{200} \times 100=3.5 \% \end{aligned}$$
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