In a moving coil galvanometer, two moving coils M1 and M2 have the following particulars : R1 = 5 Ω, N1 = 15, A1 = 3.6 x 10-3 m2, B1 = 0.25 T R2 = 7 Ω, N2 = 21, A2 = 1.8 x 10-3 m2, B2 = 0.50 T Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of M1 and M2 ?

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q45 – JEE-MAIN 2025

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Home Physics Moving ChargesQuestion #357

In a moving coil galvanometer, two moving coils

M_{1}

and

M_{2}

have the following particulars :

R_{1} = 5 Ω

N_{1} = 15

A_{1} = 3.6 \times 10^{- 3} m^{2}

B_{1} = 0.25 T

R_{2} = 7 Ω

N_{2} = 21

A_{2} = 1.8 \times 10^{- 3} m^{2}

B_{2} = 0.50 T

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}

Clue 1 of 3: Understanding Galvanometer Sensitivity

Recall the definitions of current sensitivity and voltage sensitivity for a moving coil galvanometer.

🧠 Full Solution Path

Step 1: Recall the formula for voltage sensitivity✦ Active

The voltage sensitivity ( $S_{V}$ ) of a moving coil galvanometer is defined as the deflection per unit voltage. It is given by the formula $S_{V} = \frac{N A B}{k R}$ , where $N$ is the number of turns, $A$ is the area, $B$ is the magnetic field, $k$ is the torsional constant, and $R$ is the resistance.

S_{V} = \frac{N A B}{k R}

Step 2: Set up the ratio of voltage sensitivities○ Expand

Given that the torsional constant $k$ is the same for both coils, the ratio of voltage sensitivities for $M_{1}$ and $M_{2}$ can be written as:

\frac{S_{V 1}}{S_{V 2}} = \frac{\frac{N_{1} A_{1} B_{1}}{k R_{1}}}{\frac{N_{2} A_{2} B_{2}}{k R_{2}}} = \frac{N_{1} A_{1} B_{1}}{R_{1}} \times \frac{R_{2}}{N_{2} A_{2} B_{2}}

Step 3: Substitute the given values and calculate the ratio○ Expand

Substitute the provided values: $N_{1} = 15, A_{1} = 3.6 \times 10^{- 3}, B_{1} = 0.25, R_{1} = 5$ and $N_{2} = 21, A_{2} = 1.8 \times 10^{- 3}, B_{2} = 0.50, R_{2} = 7$ .

\frac{S_{V 1}}{S_{V 2}} = \frac{15}{21} \times \frac{3.6 \times 10^{- 3}}{1.8 \times 10^{- 3}} \times \frac{0.25}{0.50} \times \frac{7}{5}

Simplify the terms and multiply:

\frac{S_{V 1}}{S_{V 2}} = (\frac{5}{7}) \times (2) \times (\frac{1}{2}) \times (\frac{7}{5}) = 1

The ratio of voltage sensitivity of $M_{1}$ and $M_{2}$ is $1 : 1$ .

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Physics Question 45 – JEE-MAIN 2025

Clue 1 of 3: Understanding Galvanometer Sensitivity

🧠 Full Solution Path