In a meter bridge, as shown in the figure, it is given that resistance $Y=12.5 \Omega$ and that the balance…
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 $39.5 \mathrm{~cm}$ 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 $l_2$ ?
$19.15 \Omega$ and $39.5 \mathrm{~cm}$
$8.16 \Omega$ and $60.5 \mathrm{~cm}$
$19.15 \Omega$ and $60.5 \mathrm{~cm}$
$8.16 \Omega$ and $39.5 \mathrm{~cm}$
Solution
For a balanced meter bridge,
$
\begin{aligned}
&\frac{\mathrm{X}}{39.5}=\frac{\mathrm{Y}}{(100-39.5)} \\
&\Rightarrow \mathrm{Y}=39.5=\mathrm{X} \times(100-39.5) \\
&or \ldots \mathrm{X}=\frac{12.5 \times 39.5}{60.5}=8.16 \Omega
\end{aligned}
$
When $\mathrm{X}$ and $\mathrm{Y}$ are interchanged $l_1$ and (100 $-l_1$ ) will also interchange so, $l_2=60.5 \mathrm{~cm}$
