Question

In a meter bridge, as shown in the figure, it is given that resistance $Y=12.5\Omega$ and that the balance is obtained at a distance $39.5\text{cm}$ from end A (by Jockey $J$). After interchanging the resistances $X$ and $Y$, a new balance point is found at a distance $l_2$ from end $A$. What are the values of $X$ and $l_2$?

• A. $19.15\Omega \text{and}39.5\text{cm}$
• B. $8.16\Omega \text{and}60.5\text{cm}$
• C. $8.16\Omega \text{and}39.5\text{cm}$
• D. $19.15\Omega \text{and}60.5\text{cm}$

Answer 1

The correct option is B $8.16\Omega \text{and}60.5\text{cm}$

Given:
$Y=12.5\Omega$
$l_1=39.5\text{cm}$

Using balance condition of wheat-stone bridge

$\frac{P}{Q}=\frac{R}{S}$

In $1^{st}$ case

$\frac{l_1}{100-l_1}=\frac{X}{Y}$ ,

$\Rightarrow \frac{39.5}{60.5}=\frac{X}{12.5}$

$\Rightarrow X=\frac{39.5\times 12.5}{60.5}$

$X=8.16\Omega$

In $2^{nd}$ case
When X and Y are interchanged and again apply wheat stone bridge condition

$\frac{l_2}{100-l_2}=\frac{12.5}{8.16}$

$l_2=60.5\text{cm}$

Hence, Option $\left(B\right)$ is the correct asnwer.