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Two cells are connected with 2 $Ω$ resistance and a TG(tangent galvanometer ). If the deflection in TG is 30°.Then the resistance of the coil.
Given that coil has 5 turns and radius = 0.08 and $B_{H} = 5 \times 10^{- 5} T$

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$5 .5 Ω$

$7 .5 Ω$

$3 .5 Ω$

$1 .5 Ω$

answer is B.

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Detailed Solution

$\begin{array}{l} i = \frac{5}{5 + R} \\ B = B_{H} \tan θ \\ \frac{μ_{0} Ni}{2 r} = B_{H} \tan θ \\ i = \frac{2 {rB}_{H}}{μ_{0} N} \tan θ = \frac{5}{5 + R} \end{array}$
Solving $R = 3 .5 Ω$

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