In a meter bridge $30\Omega$  resistance is connected in the left gap and a pair of resistance$20\Omega$ and Q  in(Ohm) are in right gap and they are connected in series, then the balancing length is $50cm$ . When they were connected in parallel ,the balancing length will be

Balancing length $l=50cm$

Resistance $R=30\Omega$

Balancing condition for meter bridge

$\frac{R}{S}=\frac{l}{100-l}$

$20\Omega$  and $Q$ are in right gap $\left(S\right)$  and they are connected in series

$\Rightarrow \frac{30}{20+Q}=\frac{50}{50}$

     $Q=10\Omega$

$20\Omega$  and $Q$ are connected in parallel

$\frac{30\left(20+Q\right)}{20Q}=\frac{l^1}{100-l^1}$

$\frac{30\times 30}{200}=\frac{l^1}{100-l^1}$

$\Rightarrow \frac{9}{2}=\frac{l^1}{100-l^1}$

$900=11l^1$

$l^1=\frac{900}{11}$

$\therefore l^1=81.81cm$