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3.Current Electricity
hard
The resistance of the meter bridge $AB$ in given figure is $4\,\Omega $. With a cell of emf $\varepsilon \, = 0.5\,\,V$ and rheostat resistance $R_h = 2\,\Omega $ the null point is obtained at some point $J.$ When the cell is replaced by another one of emf $\varepsilon \, = {\varepsilon _2}$ the same null point $J$ is found for $R_h = 6\,\Omega .$ The $emf$ ${\varepsilon _2}$ is ................. $V$
A
$0.4$
B
$0.3$
C
$0.6$
D
$0.5$
(JEE MAIN-2019)
Solution
$0.5=\frac{6}{(2+\lambda L)} \lambda x$ ……$(1)$
$E_{2}=\frac{6}{(6+\lambda L)} \lambda x$ …….$(2)$
So dividing equation $( 1)$ and $( 2)$
$\frac{E_{2}}{0.5}=\frac{2+4}{6+4}=\frac{3}{5}$
$\Rightarrow \quad E_{2}=0.3$ $volt$
Standard 12
Physics
