JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2025
- Medium
- MCQ
In a moving coil galvanometer, two moving coils \(M_1\) and \(M_2\) have the following particulars :
\(\begin{aligned}
& \mathrm{R}_1=5 \Omega, \mathrm{~N}_1=15, \mathrm{~A}_1=3.6 \times 10^{-3} \mathrm{~m}^2, \mathrm{~B}_1=0.25 \mathrm{~T} \\
& \mathrm{R}_2=7 \Omega, \mathrm{~N}_2=21, \mathrm{~A}_2=1.8 \times 10^{-3} \mathrm{~m}^2, \mathrm{~B}_2=0.50 \mathrm{~T}
\end{aligned}\)
Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of \(M_1\) and \(M_2\) ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(1: 1\)
Step-by-step Solution
Detailed explanation
\(\text { Voltage sensitivity }=\frac{\theta}{V}=\frac{N A B}{c R} \)
\( \text { Ratio }==\left(\frac{\mathrm{N}_1 \mathrm{~A}_1 \mathrm{~B}_1}{\mathrm{~N}_2 \mathrm{~A}_2 \mathrm{~B}_2}\right) \frac{\mathrm{R}_2}{\mathrm{R}_1}=\frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.5} \times \frac{7}{5}=\frac{1}{1}\)
\( \text { Ratio }==\left(\frac{\mathrm{N}_1 \mathrm{~A}_1 \mathrm{~B}_1}{\mathrm{~N}_2 \mathrm{~A}_2 \mathrm{~B}_2}\right) \frac{\mathrm{R}_2}{\mathrm{R}_1}=\frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.5} \times \frac{7}{5}=\frac{1}{1}\)
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