Two tangent galvanometers,which are identical | vedclass

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(D) For a tangent galvanometer,the current $I$ is given by $I = K 	an 	heta$,where $K = \frac{2r B_{H}}{n \mu_{0}}$.Since the galvanometers are conn

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$1: 6$

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(D) For a tangent galvanometer,the current $I$ is given by $I = K 	an 	heta$,where $K = \frac{2r B_{H}}{n \mu_{0}}$.Since the galvanometers are connected in parallel,the voltage $V$ across them is the same. Thus,$I_1 R_1 = I_2 R_2$,which implies $\frac{I_1}{I_2} = \frac{R_2}{R_1} = \frac{3}{1}$.Using the formula $I = K 	an 	heta$,we have $\frac{I_1}{I_2} = \frac{n_2}{n_1} \cdot \frac{	an 	heta_1}{	an 	heta_2}$.Substituting the given values: $\frac{3}{1} = \frac{n_2}{n_1} \cdot \frac{	an 30^{\circ}}{	an 60^{\circ}}$.$\frac{3}{1} = \frac{n_2}{n_1} \cdot \frac{1/\sqrt{3}}{\sqrt{3}} = \frac{n_2}{n_1} \cdot \frac{1}{3}$.Therefore,$\frac{n_1}{n_2} = \frac{1}{3 	imes 3} = \frac{1}{9}$.Wait,re-evaluating: $\frac{3}{1} = \frac{n_2}{n_1} \cdot \frac{1}{3} \implies \frac{n_2}{n_1} = 9 \implies \frac{n_1}{n_2} = \frac{1}{9}$.Given the options,let's re-check the ratio: $\frac{R_2}{R_1} = 3$. $\frac{I_1}{I_2} = 3$. $\frac{n_2}{n_1} = \frac{I_1 	an 	heta_2}{I_2 	an 	heta_1} = 3 \cdot \frac{\sqrt{3}}{1/\sqrt{3}} = 3 \cdot 3 = 9$. The ratio $n_1:n_2$ is $1:9$. Since $1:9$ is not an option,let's re-verify the resistance ratio logic. If $R_1:R_2 = 1:3$,then $I_1:I_2 = 3:1$. The correct ratio is $1:9$.

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