Two moving coil meters M_1 and M_2 have the f | vedclass

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(D) The voltage sensitivity $V_s$ of a moving coil galvanometer is given by the formula:$V_s = \frac{NAB}{kR}$where $N$ is the number of turns,$A$ is

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$1$

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(D) The voltage sensitivity $V_s$ of a moving coil galvanometer is given by the formula:$V_s = \frac{NAB}{kR}$where $N$ is the number of turns,$A$ is the area,$B$ is the magnetic field,$k$ is the spring constant,and $R$ is the resistance.Given that the spring constants are identical $(k_1 = k_2 = k)$,the ratio of voltage sensitivity of $M_2$ to $M_1$ is:$\frac{V_{s2}}{V_{s1}} = \frac{N_2 B_2 A_2}{k R_2} 	imes \frac{k R_1}{N_1 B_1 A_1} = \frac{N_2 B_2 A_2 R_1}{N_1 B_1 A_1 R_2}$Substituting the given values:$\frac{V_{s2}}{V_{s1}} = \frac{42 	imes 0.50 	imes 1.8 	imes 10^{-3} 	imes 10}{30 	imes 0.25 	imes 3.6 	imes 10^{-3} 	imes 14}$$\frac{V_{s2}}{V_{s1}} = \frac{42 	imes 0.50 	imes 1.8 	imes 10}{30 	imes 0.25 	imes 3.6 	imes 14}$$\frac{V_{s2}}{V_{s1}} = \frac{378}{378} = 1$Thus,the ratio of the voltage sensitivity of $M_2$ and $M_1$ is $1$.

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