Two equal resistances are connected in the gaps of a meter bridge. If the resistance in the left gap is increased by $10\%$, the balancing point shift :

Hint: In a meter bridge, $\frac{R_{\text {left }}}{R_{\text {right }}}=\frac{l}{100-l}$, where $R_{\text {left }}$ and $R_{\text {right }}$ are the resistances on the left and the right gap of the meter bridge respectively and $l$ is the balancing length.

Correct option: D

Explanation for correct answer:

Step 1: Find the shift in balancing point.

Initially, $\frac{R}{R}=\frac{l}{100-l}$

$\Rightarrow l=50 \mathrm{~cm}$

After the increase of resistance in left gap by $10 \%$,

$\begin{array}{l} \frac{1.1 R}{R}=\frac{l}{100-l} \\ \Rightarrow l=52.38 \mathrm{~cm} \end{array}$

The shift in balancing point, $\frac{\Delta l}{l} \times 100=\frac{2.38}{50} \times 100=4.76 \approx 4.8 \%$

So, the shift in the balancing point is $4.8 \%$.