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(B) Let the resistance of the wire be $R$. In a metre bridge,the balancing condition is given by $\frac{R}{S} = \frac{l}{100-l}$,where $S$ is the resi

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$\frac{800}{11} \ cm$

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(B) Let the resistance of the wire be $R$. In a metre bridge,the balancing condition is given by $\frac{R}{S} = \frac{l}{100-l}$,where $S$ is the resistance in the right gap and $l$ is the balancing length from the left end.Initially,$l_1 = 40 \ cm$,so $\frac{R}{S} = \frac{40}{60} = \frac{2}{3}$,which implies $S = 1.5R$.When the wire is stretched to double its length,its new length $l' = 2l$ and its new area of cross-section $A' = \frac{A}{2}$ (since volume $V = Al$ remains constant).The new resistance $R'$ is given by $R' = \rho \frac{l'}{A'} = \rho \frac{2l}{A/2} = 4 \rho \frac{l}{A} = 4R$.Now,let the new balancing length be $l_2$. The new balancing condition is $\frac{R'}{S} = \frac{l_2}{100-l_2}$.Substituting $R' = 4R$ and $S = 1.5R$,we get $\frac{4R}{1.5R} = \frac{l_2}{100-l_2}$.$\frac{4}{1.5} = \frac{8}{3} = \frac{l_2}{100-l_2}$.$8(100 - l_2) = 3l_2 \implies 800 - 8l_2 = 3l_2 \implies 11l_2 = 800$.$l_2 = \frac{800}{11} \ cm$.

$Sl. No.$	$R \, (\Omega)$	$l \, (cm)$
$1.$	$1000$	$60$
$2.$	$100$	$13$
$3.$	$10$	$1.5$
$4.$	$1$	$1.0$

Similar Questions

In the meter bridge experiment,for given values of $R_1$ and $R_2$,the balance point is $20 \ cm$ from point $A$. When $R_2$ is shunted by a $20 \ \Omega$ resistance,the balancing point shifts to $50 \ cm$ from point $A$. Find the value of $R_2$ in $\Omega$.

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$Sl. No.$ $R \, (\Omega)$ $l \, (cm)$

$1.$ $1000$ $60$

$2.$ $100$ $13$

$3.$ $10$ $1.5$

$4.$ $1$ $1.0$

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