In a moving coil galvanometer, two moving coils $$M_1$$ and $$M_2$$ have the following particulars :
$$R_1 = 5\,\Omega$$, $$N_1 = 15$$, $$A_1 = 3.6 \times 10^{-3}\,\text{m}^2$$, $$B_1 = 0.25\,\text{T}$$
$$R_2 = 7\,\Omega$$, $$N_2 = 21$$, $$A_2 = 1.8 \times 10^{-3}\,\text{m}^2$$, $$B_2 = 0.50\,\text{T}$$
Assuming that torsional constant of the springs are same for both coils, what will be the ratio of voltage sensitivity of $$M_1$$ and $$M_2$$ ?

Voltage sensitivity $$S_V$$ of a moving-coil galvanometer is defined as the deflection produced per unit applied voltage.
If the torsional constant of the suspension spring is $$k$$, then

Current sensitivity $$S_I$$ is $$\displaystyle S_I = \frac{NAB}{k} \quad -(1)$$
where
$$N$$ = number of turns, $$A$$ = area of each turn, $$B$$ = magnetic field.

Voltage sensitivity is obtained by dividing current sensitivity by the total resistance $$R$$ of the coil:

$$\displaystyle S_V = \frac{S_I}{R} = \frac{NAB}{kR} \quad -(2)$$

Because both coils have the same spring, the torsional constant $$k$$ is identical for them and cancels in the ratio.

For coil $$M_1$$:
$$N_1 = 15$$, $$A_1 = 3.6 \times 10^{-3}\,\text{m}^2$$, $$B_1 = 0.25\,\text{T}$$, $$R_1 = 5\,\Omega$$

For coil $$M_2$$:
$$N_2 = 21$$, $$A_2 = 1.8 \times 10^{-3}\,\text{m}^2$$, $$B_2 = 0.50\,\text{T}$$, $$R_2 = 7\,\Omega$$

Using $$-(2)$$ for each coil and taking the ratio:

$$\frac{S_{V1}}{S_{V2}} = \frac{\dfrac{N_1A_1B_1}{kR_1}}{\dfrac{N_2A_2B_2}{kR_2}} = \frac{N_1A_1B_1R_2}{N_2A_2B_2R_1}$$

Substituting the given values:

$$N_1A_1B_1R_2 = 15 \times 3.6\times10^{-3} \times 0.25 \times 7$$
$$= 15 \times 0.0036 \times 0.25 \times 7$$
$$= 0.0945$$

$$N_2A_2B_2R_1 = 21 \times 1.8\times10^{-3} \times 0.50 \times 5$$
$$= 21 \times 0.0018 \times 0.50 \times 5$$
$$= 0.0945$$

Therefore

$$\frac{S_{V1}}{S_{V2}} = \frac{0.0945}{0.0945} = 1$$

Thus, the voltage sensitivities of $$M_1$$ and $$M_2$$ are equal, giving the ratio

$$S_{V1} : S_{V2} = 1 : 1$$

Option A is correct.