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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 $l_2$ ?
$19.15\,\Omega $ and $39.5\,cm$
$8.16\,\Omega $ and $60.5\,cm$
$19.15\,\Omega $ and $60.5\,cm$
$8.16\,\Omega $ and $39.5\,cm$
Solution
For a balanced meter bridge,
$\frac{X}{39.5}=\frac{Y}{(100-39.5)}$
$\Rightarrow Y=39.5=X \times(100-39.5)$
or, $\quad {\rm{X}} = \frac{{12.5 \times 39.5}}{{60.5}} = 8.16\,\Omega $
${\text { When } X \text { and } Y \text { are interchanged } l_{1} \text { and }(100} $
${\left.-l_{1}\right) \text { will also interchange so, } l_{2}=60.5\, \mathrm{cm}}$
