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(A) Let the resistance of each of the $n$ identical resistors be $R_0$.When $\frac{n}{2}$ resistors are connected in series in the left gap,the equiva

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$100 \cdot \frac{n^2}{n^2+4}$

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(A) Let the resistance of each of the $n$ identical resistors be $R_0$.When $\frac{n}{2}$ resistors are connected in series in the left gap,the equivalent resistance $R_L$ is:$R_L = \frac{n}{2} \cdot R_0$When $\frac{n}{2}$ resistors are connected in parallel in the right gap,the equivalent resistance $R_R$ is:$\frac{1}{R_R} = \frac{1}{R_0} + \frac{1}{R_0} + \dots (\frac{n}{2} 	ext{ times}) = \frac{n}{2R_0} \Rightarrow R_R = \frac{2R_0}{n}$For a metre bridge,the balancing condition is given by:$\frac{R_L}{R_R} = \frac{l}{100-l}$Substituting the values:$\frac{\frac{n R_0}{2}}{\frac{2 R_0}{n}} = \frac{l}{100-l}$$\frac{n^2}{4} = \frac{l}{100-l}$$n^2(100-l) = 4l$$100n^2 - n^2l = 4l$$100n^2 = l(n^2+4)$$l = \frac{100n^2}{n^2+4} 	ext{ cm}$

$n$ identical resistors are taken,in which $\frac{n}{2}$ resistors are joined in series in the left gap and the remaining $\frac{n}{2}$ resistors are joined in parallel in the right gap of a metre bridge. The balancing length in $cm$ is

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