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Question: Answered & Verified by Expert
A current of $\left(\frac{2}{\sqrt{3}}\right)$ A, produces a deflection of $60^{\circ}$ in a tangent galvanometer. The reduction factor is
PhysicsMagnetic Properties of MatterKCETKCET 2011
Options:
  • A $\left(\frac{2}{\sqrt{3}}\right) \mathrm{A}$
  • B $\left(\frac{2}{3}\right) \mathrm{A}$
  • C $2 \mathrm{~A}$
  • D $\left(\frac{3}{2}\right) \mathrm{A}$
Solution:
2711 Upvotes Verified Answer
The correct answer is: $\left(\frac{2}{3}\right) \mathrm{A}$
For tangent galvanometer
$$
\begin{aligned}
I &=K \tan \theta \\
K &=\frac{I}{\tan \theta} \\
K &=\frac{2 / \sqrt{3}}{\tan 60^{\circ}} \\
K &=\frac{2}{3} \mathrm{~A}
\end{aligned}
$$

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