JEE Physics 2025 Question with Solution

Given: Two moving coil galvanometers $M_1$ and $M_2$ with $R_1 = 5 \, \Omega$, $N_1 = 15$, $A_1 = 3.6 \times 10^{-3} \, \text{m}^2$, $B_1 = 0.25 \, \text{T}$ and $R_2 = 7 \, \Omega$, $N_2 = 21$, $A_2 = 1.8 \times 10^{-3} \, \text{m}^2$, $B_2 = 0.50 \, \text{T}$. The torsional constant is the same for both.

Find: The ratio of voltage sensitivity of $M_1$ and $M_2$.

For a moving coil galvanometer, voltage sensitivity is

$$\frac{\theta}{V} = \frac{NAB}{kR}$$

where $k$ is the torsional constant.

Therefore,

$$\frac{S_{V1}}{S_{V2}} = \frac{N_1 A_1 B_1 / R_1}{N_2 A_2 B_2 / R_2} = \frac{N_1 A_1 B_1 R_2}{N_2 A_2 B_2 R_1}$$

Substituting the given values,

$$\frac{S_{V1}}{S_{V2}} = \frac{15 \times (3.6 \times 10^{-3}) \times 0.25 \times 7}{21 \times (1.8 \times 10^{-3}) \times 0.50 \times 5}$$

Now simplify:

$$\text{Numerator} = 15 \times 3.6 \times 0.25 \times 7 = 94.5$$ $$\text{Denominator} = 21 \times 1.8 \times 0.50 \times 5 = 94.5$$

Hence,

$$\frac{S_{V1}}{S_{V2}} = \frac{94.5}{94.5} = 1$$

Therefore, the ratio of voltage sensitivities is $1 : 1$. The correct option is A.

Given: Voltage sensitivity is to be compared for two galvanometers with the same torsional constant.

Find: $S_{V1}:S_{V2}$.

First use current sensitivity:

$$S_I = \frac{\theta}{I} = \frac{NAB}{k}$$

Since

$$S_V = \frac{\theta}{V} = \frac{\theta}{I} \cdot \frac{I}{V}$$

and

$$\frac{I}{V} = \frac{1}{R}$$

we get

$$S_V = \frac{NAB}{kR}$$

Now compare the two instruments:

$$\frac{S_{V1}}{S_{V2}} = \frac{15 \times 3.6 \times 10^{-3} \times 0.25 / 5}{21 \times 1.8 \times 10^{-3} \times 0.50 / 7}$$

Rearranging,

$$\frac{S_{V1}}{S_{V2}} = \frac{15 \times 3.6 \times 10^{-3} \times 0.25 \times 7}{21 \times 1.8 \times 10^{-3} \times 0.50 \times 5} = 1$$

So both galvanometers have the same voltage sensitivity. Therefore, the required ratio is $1 : 1$.