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 $39.5\, cm$ 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. $8.16 \, \Omega$ and $60.5\, cm$
• B. $19.15 \, \Omega$ and $39.5\, cm$
• C. $8.16 \, \Omega$ and $39.5\, cm$
• D. $19.15 \, \Omega$ and $60.5\, cm$

Answer 1

Answer: (A)

The balanced condition of Wheatstone bridge is
$X\left(100-l_{1}\right)=Y \times l_{1}$
Given $l_{1}=39.5 cm ; Y=12.5 \square$. Therefore,
$X(100-39.5)=12.5(39.5)$
$\Rightarrow X(60.5)=12.5(39.5)$
$\Rightarrow X=\frac{12.5 \times 39.5}{60.5}=8.16 \Omega$
Now, if $X$ and $Y$ are interchanged then balanced condition of Wheatstone bridge becomes
$Y\left(100-l_{2}\right)=X l_{2}$
In this condition, $Y=12.5 \square ; X=8.16$. Therefore,
$12.5\left(100-l_{1}\right)=8.16 l_{2}$
$\Rightarrow 1250-12.5 l_{2}=8.16 l_{2}$
$\Rightarrow 1250=20.66 l_{2}$
$\Rightarrow l_{2}=\frac{1250}{20.66}=60.6\, cm$