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Two tangent galvanometers, which are identical except in their number of turns, are connected in parallel. The ratio of their resistances of the coils is 1 : 3 . If the deflections in the two tangent galvanometers are $30^{\circ}$ and $60^{\circ}$ respectively, then the ratio of their number of turns is
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$1: 6$
We know that the current passing through galvanometer coil is
$I=K \tan \theta$ where $K=\frac{2 r B_{H}}{n \mu_{0}}$
$\therefore$ Here, $\quad I_{1}=K_{1} \tan \theta_{1}$ and $I_{2}=K_{2} \tan \theta_{2}$
or $\quad \frac{l_{1}}{I_{2}}=\frac{K_{1} \tan \theta_{1}}{K_{2} \tan \theta_{2}}=\frac{\frac{2 r B_{H}}{n_{1} \mu_{0}} \tan \theta_{1}}{\frac{2 r B_{H}}{n_{2} \mu_{0}} \tan \theta_{2}}$
$\Rightarrow \quad \frac{R_{2}}{R_{1}}=\frac{n_{2} \tan 30^{\circ}}{n_{1} \tan 60^{\circ}}$
$\Rightarrow \quad \frac{3}{1}=\frac{n_{2}}{n_{1}} \cdot \frac{\sqrt{3}}{\sqrt{3}} \quad\left(\because I=\frac{V}{R}\right)$
$\Rightarrow \quad \frac{n_{1}}{n_{2}}=\frac{1}{3}$
$I=K \tan \theta$ where $K=\frac{2 r B_{H}}{n \mu_{0}}$
$\therefore$ Here, $\quad I_{1}=K_{1} \tan \theta_{1}$ and $I_{2}=K_{2} \tan \theta_{2}$
or $\quad \frac{l_{1}}{I_{2}}=\frac{K_{1} \tan \theta_{1}}{K_{2} \tan \theta_{2}}=\frac{\frac{2 r B_{H}}{n_{1} \mu_{0}} \tan \theta_{1}}{\frac{2 r B_{H}}{n_{2} \mu_{0}} \tan \theta_{2}}$
$\Rightarrow \quad \frac{R_{2}}{R_{1}}=\frac{n_{2} \tan 30^{\circ}}{n_{1} \tan 60^{\circ}}$
$\Rightarrow \quad \frac{3}{1}=\frac{n_{2}}{n_{1}} \cdot \frac{\sqrt{3}}{\sqrt{3}} \quad\left(\because I=\frac{V}{R}\right)$
$\Rightarrow \quad \frac{n_{1}}{n_{2}}=\frac{1}{3}$
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