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JEE MAIN 2024
05-04-2024 S1
Question
In the experiment to determine the galvanometer resistance by half-deflection method, the plot of $\frac{1}{\theta}$ vs the resistance $(R)$ of the resistance box is shown in the figure. The figure of merit of the galvanometer is $\_\_\_\_$ $\times 10^{-1} \mathrm{~A}$ /division. [The source has emf 2 V ]
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Solution
$\begin{aligned} & \mathrm{i}=\mathrm{K} \theta \\ & \frac{2}{\mathrm{G}+R}=K \theta \\ & \Rightarrow \frac{1}{\theta}=\frac{(\mathrm{G}+\mathrm{R}) \mathrm{K}}{2}=\mathrm{R}\left(\frac{\mathrm{K}}{2}\right)+\frac{\mathrm{KG}}{2} \\ & \text { Slope }=\frac{\mathrm{K}}{2}=\frac{1}{4} \Rightarrow \mathrm{~K}=0.5=5 \times 10^{-1} \mathrm{~A}\end{aligned}$
Question Tags
JEE Main
Physics
Medium
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