Resistance are connected in a meter bridge circuit as shown in the figure. The balancing length $l_{1}$ is $40 c m$ . Now an unknown resistance $x$ is connected in series with $P$ and new balancing length is found to be $80 c m$ measured from the same end. Then the value of $x$ will be _____ $Ω$ .

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Applying balanced condition for meter bridge
Initially, $\frac{P}{Q} = \frac{40 c m}{60 c m} = \frac{2}{3} ... (1)$
Finally, $\frac{P + x}{Q} = \frac{80 c m}{20 c m} = \frac{4}{1} ... (2)$
Dividing equation $(2)$ by ( $(1)$ ,
$\frac{P + x}{P} = 4 \times \frac{3}{2} = 6$
$\Rightarrow 1 + \frac{x}{P} = 6 \Rightarrow \frac{x}{P} = 5$
$∴ x = 5 P = 5 \times 4 = 20Ω$

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