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Q. In a meter bridge experiment, the ratio of the left gap resistance to right gap resistance is $ 2:3, $ the balance point from left is

J & K CETJ & K CET 2007Current Electricity

Solution:

In a meter bridge
$\frac{R_{1}}{R_{2}}=\frac{l}{100-l}$
$\frac{2}{3}=\frac{l}{(100-l)}$
$\Rightarrow 200-2 l=3 l$
$\Rightarrow 5 l=200$
$\therefore l=40\, cm$