Q 12-04-244JEE MainJEE Main 2025 (2 Apr, Shift 2)Easy
In a moving coil galvanometer, two moving coils $M_1$ and $M_2$ have the following particulars:
$R_1 = 5\ \Omega$, $N_1 = 15$, $A_1 = 3.6\times10^{-3}\ \text{m}^2$, $B_1 = 0.25\ \text{T}$
$R_2 = 7\ \Omega$, $N_2 = 21$, $A_2 = 1.8\times10^{-3}\ \text{m}^2$, $B_2 = 0.50\ \text{T}$
Assuming that the torsional constants of the springs are the same for both coils, what will be the ratio of voltage sensitivity of $M_1$ and $M_2$?
Answer: (A) $1 : 1$
Voltage sensitivity $= \dfrac{\theta}{V} = \dfrac{NAB}{kR}$.
$$\frac{S_1}{S_2} = \frac{N_1A_1B_1}{N_2A_2B_2}\cdot\frac{R_2}{R_1} = \frac{15\times3.6\times0.25}{21\times1.8\times0.5}\times\frac75 = \frac{13.5}{18.9}\times\frac75 = 1$$
The ratio is $1 : 1$.
Solution by Sreeraj P, M.Sc Physics