Question

Two equal resistance are connected in the gaps of a metre bridge. If the resistance in the left gap is increased by $20\text{\%}$. The shift in the balancing point is-

• A. $15\text{\%}$ to right
• B. $13\text{\%}$ to right
• C. $9\text{\%}$ to right
• D. $12\text{\%}$ to right

Answer 1

The correct option is C $9\text{\%}$ to right

Let the two resistance be $p$ and $Q$

Given, $P=Q;l_1=50\text{cm}\left(\text{initial balancing length}\right)P^{{}^{\prime }}=\frac{120}{100}P$

At the balancing condition,

$\frac{P^{{}^{\prime }}}{Q}=\frac{l_2}{100-l_2}\Rightarrow \frac{120}{100}=\frac{l_2}{100-l_2}$

$1200-12l_2=10l_2$

$1200=22l_2$

$\Rightarrow l_2=\frac{1200}{22}=54.54\text{cm}$

Change in percentage of the balancing length is,

$(\frac{l_2}{l_1}-1)\times 100\text{\%}=(\frac{54.54}{50}-1)100\text{\%}$

$=(1.09-1)100\text{\%}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.

$=9\text{\%}$ to right.