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A current of $\left(\frac{2}{\sqrt{3}}\right)$ A, produces a deflection of $60^{\circ}$ in a tangent galvanometer. The reduction factor is
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Verified Answer
The correct answer is:
$\left(\frac{2}{3}\right) \mathrm{A}$
For tangent galvanometer
$$
\begin{aligned}
I &=K \tan \theta \\
K &=\frac{I}{\tan \theta} \\
K &=\frac{2 / \sqrt{3}}{\tan 60^{\circ}} \\
K &=\frac{2}{3} \mathrm{~A}
\end{aligned}
$$
$$
\begin{aligned}
I &=K \tan \theta \\
K &=\frac{I}{\tan \theta} \\
K &=\frac{2 / \sqrt{3}}{\tan 60^{\circ}} \\
K &=\frac{2}{3} \mathrm{~A}
\end{aligned}
$$
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