Two moving coil galvanometer, $X$ and $Y$ have coils with resistance $10 \Omega$ and $14 \Omega$…
Two moving coil galvanometer, $X$ and $Y$ have coils with resistance $10 \Omega$ and $14 \Omega$ cross-sectional areas $4.8 \times 10^{-3} \mathrm{~m}^2$ and $2.4 \times 10^{-3} \mathrm{~m}^2$, number of turns 30 and 45 respectively. They are placed in magnetic field of $0.25 \mathrm{~T}$ and $0.50 \mathrm{~T}$ respectively. Then, the ratio of their current sensitivities and the ratio of their voltage sensitivities are respectively
2 : 3, 14 : 15
5 : 7, 2 : 1
2 :13, 1 : 2
14 : 15, 2 : 9
Solution
There is no information regarding spring constant.
So, let's assume their spring constants are same.
Current sensitivity, $I_S=\frac{N B A}{k}$
So,
$
\begin{aligned}
\frac{I_{S_1}}{I_{S_2}} & =\frac{N_1 B_1 A_1 k_2}{N_2 B_2 A_2 k_1} \\
& =\frac{30 \times 0.25 \times 4.8 \times 10^{-3}}{45 \times 0.5 \times 24 \times 10^{-3}} \times\left(\frac{1}{1}\right) \\
\frac{I_{S_1}}{I_{S_2}} & =2 / 3
\end{aligned}
$
Voltage sensitivity, $V_S=N B A / R k$
So,
$
\begin{aligned}
\frac{V_{S_1}}{V_{S_2}} & =\frac{N_1 B_1 A_1 k_2 R_2}{N_2 B_2 A_2 k_1 R_1} \\
& =\frac{2}{3} \times\left(\frac{R_2}{R_1}\right)=\frac{2}{3} \times \frac{14}{10}=\frac{14}{15}
\end{aligned}
$