Solution

$$\text{For moving coil meter } M_1$$: Resistance, $$R_1 = 10\ Ω$$ Number of turns, $$N_1 = 30$$ Area of cross-section, $$A_1 = 3.6 × 10^{-3} m^2$$ Magnetic field strength, $$B_1 = 0.25\ T$$ Spring constant $$K_1 = K$$ $$\text{For moving coil meter } M_2$$: Resistance, $$R_2 = 14\ Ω$$ Number of turns, $$N_2 = 42$$ Area of cross-section, $$A_2 = 1.8 × 10^{-3} m^2$$ Magnetic field strength, $$B_2 = 0.50\ T$$ Spring constant, $$K_2 = K$$ (a) Current sensitivity of $$M_1$$ is given as: $$\implies I_{s1}=\dfrac{N_1 B_1 A_1}{K_1}$$ And, Current sensitivity of $$M_2$$ is given as: $$\implies I_{s2}=\dfrac{N_2 B_2 A_2}{K_2}$$ $$\therefore$$ Ratio, $$\dfrac{I_{s2}}{I_{s1}}= \dfrac{N_2 B_2 A_2K_1}{N_1 B_1 A_1K_2}=\dfrac{42 \times 0.5 \times 1.8 \times 10^{-3}\times K}{K\times 30 \times 0.25 \times 3.6 \times 10^{-3}}=1.4$$ Hence, the ratio of current sensitivity of $$M_2$$ to $$M_1$$ is $$1.4$$. (b) Voltage sensitivity for $$M_2$$ is given as: $$\implies v_{s2}=\dfrac{N_2B_2A_2}{K_2R_2}$$ And, voltage sensitivity for $$M_1$$ is given as: $$\implies v_{s1}=\dfrac{N_1B_1A_1}{K_1R_1}$$ $$\therefore$$ Ratio of voltage sensitivity of $$M_2$$ to $$M_1$$:- $$\implies \dfrac{v_{s2}}{v_{s1}}= \dfrac{N_2 B_2 A_2K_1R_1}{N_1 B_1 A_1K_2R_2}=\dfrac{42 \times 0.5 \times 1.8 \times 10^{-3}\times10\times K}{K\times14\times 30 \times 0.25 \times 3.6 \times 10^{-3}}=1$$