In a meter bridge experiment,resistances are | vedclass

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(C) For the given meter bridge,the balancing condition is given by $\frac{P}{Q} = \frac{l_1}{100 - l_1}$.Given $l_1 = 55 \, cm$,so $100 - l_1 = 45 \,

Vedclass · Accepted Answer

$\frac{48}{11} \, \Omega$

Vedclass · Answer

(C) For the given meter bridge,the balancing condition is given by $\frac{P}{Q} = \frac{l_1}{100 - l_1}$.Given $l_1 = 55 \, cm$,so $100 - l_1 = 45 \, cm$.Given $P = 3 \, \Omega$,we have $\frac{3}{Q} = \frac{55}{45} = \frac{11}{9}$.Thus,$Q = 3 	imes \frac{9}{11} = \frac{27}{11} \, \Omega$.When an unknown resistance $x$ is connected in series with $P$,the new balancing length $l_1' = 75 \, cm$,so $100 - l_1' = 25 \, cm$.The new balance condition is $\frac{P + x}{Q} = \frac{75}{25} = 3$.Substituting the values,$3 + x = 3 	imes Q = 3 	imes \frac{27}{11} = \frac{81}{11}$.Therefore,$x = \frac{81}{11} - 3 = \frac{81 - 33}{11} = \frac{48}{11} \, \Omega$.

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

In a meter bridge experiment,resistances are connected as shown in the figure. The balancing length $l_1$ is $55 \, cm$. Now,an unknown resistance $x$ is connected in series with $P$ and the new balancing length is found to be $75 \, cm$. The value of $x$ is

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In a meter bridge experiment, the circuit diagram and the corresponding observation table are shown in the figure.

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

$1.$ $1000$ $60$

$2.$ $100$ $13$

$3.$ $10$ $1.5$

$4.$ $1$ $1.0$

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