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  4. The galvanometer deflection, when key K1 is closed but K2 is open, equals θ 0 (see figure). On closing K2 also and adjusting R2 to 5 Ω, the deflection in galvanometer becomes (θ0/5). The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :

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Q. The galvanometer deflection, when key $K_1$ is closed but $K_2$ is open, equals $\theta _0$ (see figure). On closing $K_2$ also and adjusting $R_2$ to $5 \Omega$, the deflection in galvanometer becomes $\frac{\theta_0}{5}$. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery] :Physics Question Image

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Solution:

case I $i_{g} =\frac{E}{220 +R_{g}} =C\theta_{0} $ ....(i)
case II $ i_{g} = \left(\frac{E}{220 + \frac{5R_{g}}{5+R_{g}}}\right)\times\frac{5}{\left(R_{g} +5\right)} = \frac{C\theta_{0}}{5} $ ....(ii)
$ \Rightarrow \frac{5E}{ 225R_{g} +1100} = \frac{C\theta_{0}}{5} $
$ \frac{E}{220 +R_{g}} =C\theta $
$ \Rightarrow \frac{225R_{g} +1100}{1100+5R_{g}} = 5$
$ \Rightarrow 5500 +25R_{g} = 225R_{g} +1100 $
$200R_{g} = 4400 $
$R_{g} =22\Omega $