In a moving coil galvanometer, two moving coils $M_{1}$ and $M_{2}$ have the following particulars : $R_{1}=5 \Omega, N_{1}=15, A_{1}=3.6 \times 10^{-3} \mathrm{~m}^{2}, B_{1}=0.25 \mathrm{~T}$ $R_{2}=7 \Omega, N_{2}=21, A_{2}=1.8 \times 10^{-3} \mathrm{~m}^{2}, B_{2}=0.50 \mathrm{~T}$ Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of $M_{1}$ and $M_{2}$ ?

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Current Electricity

In a moving coil galvanometer, two moving coils

M_{1}

and

M_{2}

have the following particulars :

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

R_{2} = 7Ω, 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}

Answer: (A)

Voltage sensitivity of a moving coil galvanometer is given by

\frac{θ}{V} = \frac{N A B}{c R}

, where

c

is the torsional constant. Since

c

is the same for both coils, the ratio of voltage sensitivities is:

\frac{( θ / V ) _{1}}{( θ / V ) _{2}} = \frac{N _{1} A _{1} B _{1}}{N _{2} A _{2} B _{2}} \times \frac{R _{2}}{R _{1}} = \frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.5} \times \frac{7}{5} = \frac{13.5}{18.9} \times 1.4 = 1

Therefore, the ratio is

1 : 1

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option