In a meter bridge experiment,when a nichrome | vedclass

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(C) Let $R_L$ be the resistance in the left gap and $R_R$ be the resistance of the nichrome wire in the right gap. The balancing condition is $\frac{R

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$51$

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(C) Let $R_L$ be the resistance in the left gap and $R_R$ be the resistance of the nichrome wire in the right gap. The balancing condition is $\frac{R_L}{R_R} = \frac{l}{100-l}$.Given $l = 60 \ cm$,so $\frac{R_L}{R_R} = \frac{60}{40} = 1.5$.When the wire is stretched by $20 \%$,its new length $l' = 1.2l_0$. Since volume $V = A \cdot l$ is constant,the new area $A' = \frac{A}{1.2}$.The new resistance $R_R' = \rho \frac{l'}{A'} = \rho \frac{1.2l_0}{A/1.2} = (1.2)^2 R_R = 1.44 R_R$.Let the new balancing length be $l_{new}$. Then $\frac{R_L}{R_R'} = \frac{l_{new}}{100-l_{new}}$.Substituting $R_L = 1.5 R_R$ and $R_R' = 1.44 R_R$:$\frac{1.5 R_R}{1.44 R_R} = \frac{l_{new}}{100-l_{new}} \Rightarrow \frac{1.5}{1.44} = \frac{l_{new}}{100-l_{new}}$.$1.04167 = \frac{l_{new}}{100-l_{new}} \Rightarrow 104.167 - 1.04167 l_{new} = l_{new}$.$2.04167 l_{new} = 104.167 \Rightarrow l_{new} \approx 51 \ cm$.

In a meter bridge experiment,when a nichrome wire is in the right gap,the balancing length is $60 \ cm$. When the nichrome wire is uniformly stretched to increase its length by $20 \%$,and again connected in the right gap,the new balancing length is nearly: (in $cm$)

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