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In a tangent galvanometer, a current of $1 \mathrm{~A}$ produces a deflection of $30^{\circ}$. The current required to produce a deflection of $60^{\circ}$ is
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The correct answer is:
$3 \mathrm{~A}$
Current in tangent galvanometer
$\mathrm{I}=\frac{\mathrm{H}}{\mathrm{G}} \tan \theta$
Where $\mathrm{G}=$ galvanometer constant
$\mathrm{H}=\text { earth's horizontal field }=\text { constant }$
$\therefore \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}} \Rightarrow \frac{1}{\mathrm{I}_{2}}=\frac{\tan 30^{\circ}}{\tan 60^{\circ}}$
$\begin{array}{l}
\Rightarrow \mathrm{I}_{2}=\frac{\tan 60^{\circ}}{\tan 30^{\circ}}=\frac{1.7321}{0.5774}=2.999 Å \\
\Rightarrow \mathrm{I}_{2} \simeq 3 \mathrm{~A}
\end{array}$
$\mathrm{I}=\frac{\mathrm{H}}{\mathrm{G}} \tan \theta$
Where $\mathrm{G}=$ galvanometer constant
$\mathrm{H}=\text { earth's horizontal field }=\text { constant }$
$\therefore \frac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\frac{\tan \theta_{1}}{\tan \theta_{2}} \Rightarrow \frac{1}{\mathrm{I}_{2}}=\frac{\tan 30^{\circ}}{\tan 60^{\circ}}$
$\begin{array}{l}
\Rightarrow \mathrm{I}_{2}=\frac{\tan 60^{\circ}}{\tan 30^{\circ}}=\frac{1.7321}{0.5774}=2.999 Å \\
\Rightarrow \mathrm{I}_{2} \simeq 3 \mathrm{~A}
\end{array}$
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