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Physics · Current Electricity

JEE Main 2024 — 5 April, Shift 1 — Question 57

In the experiment to determine the galvanometer resistance by half-deflection method, the plot of 1θ\frac{1}{\theta} vs the resistance (R)(\mathrm{R}) of the resistance box is shown in the figure. The figure of merit of the galvanometer is \qquad ×10−1 A/\times 10^{-1} \mathrm{~A} / division. [The source has emf 2V]

Question figure

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

i=Kθ\mathrm{i}=\mathrm{K} \theta

2G+R=Kθ\frac{2}{G+R}=K \theta

⇒1θ=(G+R)K2=R(K2)+KG2\Rightarrow \frac{1}{\theta}=\frac{(\mathrm{G}+\mathrm{R}) \mathrm{K}}{2}=\mathrm{R}\left(\frac{\mathrm{K}}{2}\right)+\frac{\mathrm{KG}}{2}

Slope =K2=14⇒ K=0.5=5×10−1 A=\frac{\mathrm{K}}{2}=\frac{1}{4} \Rightarrow \mathrm{~K}=0.5=5 \times 10^{-1} \mathrm{~A}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
5 April, Shift 1
Subject
Physics
Chapter
Current Electricity
Topic
Electrical Measuring Instruments

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