JEE Main 2025 Apr 2 Shift 2, Physics Q2: Moving Coil Galvanometer

JEE Main2025Apr 2, Shift 2Physics

In a moving coil galvanometer, two moving coils $M_{1}$ and $M_{2}$ have the following particulars :

$R_{1} = 5 Ω, N_{1} = 15, A_{1} = 3.6 \times 1 0^{- 3} m^{2}, B_{1} = 0.25 T$

$R_{2} = 7 Ω, N_{2} = 21, A_{2} = 1.8 \times 1 0^{- 3} m^{2}, B_{2} = 0.50 T$

Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of $M_{1}$ and $M_{2}$ ?

A
$1 : 1$
B
$1 : 4$
C
$1 : 3$
D
$1 : 2$

Solution

Voltage sensitivity of a moving coil galvanometer (with identical torsional constant $c$ ) is

$\frac{θ}{V} = \frac{N A B}{c R}$ .

The ratio for the two coils is

$\frac{( θ / V ) _{1}}{( θ / V ) _{2}} = \frac{N _{1} A _{1} B _{1}}{N _{2} A _{2} B _{2}} \cdot \frac{R _{2}}{R _{1}} = \frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.50} \times \frac{7}{5}$ .

Numerator: $15 \times 3.6 \times 0.25 = 13.5$ . Denominator: $21 \times 1.8 \times 0.5 = 18.9$ . Their ratio is $\frac{13.5}{18.9} = \frac{5}{7}$ .

Multiplying by $\frac{7}{5}$ gives exactly $1$ , so the voltage sensitivities are in the ratio $1 : 1$ .

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JEE Main 2025 Apr 2 Shift 2, Physics Q2: Moving Coil Galvanometer

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