In a meter bridge,as shown in the figure,it i | vedclass

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(B) For a balanced meter bridge,the condition is $\frac{X}{l_1} = \frac{Y}{100 - l_1}$.Given $Y = 12.5 \, \Omega$ and $l_1 = 39.5 \, cm$,we have:$\fra

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$8.16 \, \Omega$ and $60.5 \, cm$

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(B) For a balanced meter bridge,the condition is $\frac{X}{l_1} = \frac{Y}{100 - l_1}$.Given $Y = 12.5 \, \Omega$ and $l_1 = 39.5 \, cm$,we have:$\frac{X}{39.5} = \frac{12.5}{100 - 39.5} = \frac{12.5}{60.5}$.$X = \frac{12.5 	imes 39.5}{60.5} \approx 8.16 \, \Omega$.When resistances $X$ and $Y$ are interchanged,the new balance condition is $\frac{Y}{l_2} = \frac{X}{100 - l_2}$.Substituting $X = \frac{Y 	imes l_1}{100 - l_1}$,we get $\frac{Y}{l_2} = \frac{Y 	imes l_1}{(100 - l_1)(100 - l_2)}$.This simplifies to $100 - l_2 = l_1 	imes \frac{l_2}{100 - l_1}$,which implies $l_2 = 100 - l_1$.Therefore,$l_2 = 100 - 39.5 = 60.5 \, cm$.

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