In a moving coil galvanometer, the voltage sensitivity is given by the formula:

$ \text{Voltage sensitivity} = \frac{\theta}{V} = \frac{N \cdot A \cdot B}{c \cdot R} $

Here, $ \theta $ is the deflection angle, $ V $ is the voltage, $ N $ is the number of turns, $ A $ is the coil area, $ B $ is the magnetic field, $ c $ is the torsional constant of the spring, and $ R $ is the resistance of the coil.

For the two moving coils $ \mathrm{M}_1 $ and $ \mathrm{M}_2 $, we want to determine the ratio of their voltage sensitivities. Given that the torsional constant $ c $ is the same for both coils, we can express the ratio as:

$ \text{Ratio of voltage sensitivities} = \left(\frac{N_1 \cdot A_1 \cdot B_1}{N_2 \cdot A_2 \cdot B_2}\right) \cdot \frac{R_2}{R_1} $

Plugging in the given values:

$ N_1 = 15 $

$ A_1 = 3.6 \times 10^{-3} \, \text{m}^2 $

$ B_1 = 0.25 \, \text{T} $

$ R_1 = 5 \, \Omega $

$ N_2 = 21 $

$ A_2 = 1.8 \times 10^{-3} \, \text{m}^2 $

$ B_2 = 0.50 \, \text{T} $

$ R_2 = 7 \, \Omega $

We calculate:

$ \text{Ratio} = \left(\frac{15 \times 3.6 \times 0.25}{21 \times 1.8 \times 0.5}\right) \times \frac{7}{5} = \frac{1}{1} $

Thus, the ratio of the voltage sensitivities of $ M_1 $ to $ M_2 $ is 1:1.