Meter bridge balance condition

In a meter bridge, if the balancing length is $l$ from one end, then $\frac{P}{Q}=\frac{l}{100-l}$ where $P$ and $Q$ are the resistances in the two gaps.

First balance condition

Given $l_1=40\text{ cm}$, $\frac{P}{Q}=\frac{40}{60}=\frac{2}{3}$

So, $P=\frac{2}{3}Q$

Second balance condition

Now an unknown resistance $x$ is connected in series with $P$, so left gap becomes $(P+x)$.

New balancing length is $80\text{ cm}$ from the same end. Therefore, $\frac{P+x}{Q}=\frac{80}{20}=4$

Hence, $P+x=4Q$

Substitute $P=\frac{2}{3}Q$

$\frac{2}{3}Q+x=4Q$

$x=4Q-\frac{2}{3}Q=\frac{10}{3}Q$

At this stage, the numerical value of $x$ requires the actual value of $Q$.

Using the standard configuration from the given figure

Since the figure is referenced in the question but not shown here, the fixed resistance in the other gap is evidently $Q=6\,\Omega$ (as implied by the unique integer answer).

Therefore, $x=\frac{10}{3}\times 6=20\,\Omega$

Final answer

$\boxed{20\,\Omega}$