In a meter bridge, a standard resistor of $R Ω$ is connected in the left gap and the two wires A and B are connected one after the other in the right gap. The balancing length measured from the left is $50 c m$ for either of them. If the two wires are connected in series and put in the right gap, balancing length measured from left would be (in $c m$ )

25

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33.3

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66.7

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100

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Solution

The correct option is B $33.3$
Assuming, $x$ is the resistance of each wire.
$\frac{l_{1}}{100 - l_{1}} = \frac{R}{x}; l_{1} = 50 c m \Rightarrow R = x$

$∴$ Each wire A, B should have resistance $R$ .

When wire, A, B connected in series:
$\frac{l_{2}}{100 - l_{2}} = \frac{R}{2 R}$

$2 l_{2} = 100 - l_{2} \Rightarrow l_{2} = \frac{100}{3} = 33.3 c m$

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The correct option is B 33.3Assuming, x is the resistance of each wire. l1100−l1=Rx; l1=50 cm⇒ R=x ∴ Each wire A, B should have resistance R. When wire, A, B connected in series: l2100−l2=R2R 2l2=100−l2⇒l2=1003=33.3 cm