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3.Current Electricity
medium
An unknown resistance $R_1$ is connected in series with a resistance of $10 \,\Omega$. This combinations is connected to one gap of a meter bridge while a resistance $R_2$ is connected in the other gap. The balance point is at $50\, cm$. Now, when the $10 \,\Omega$ resistance is removed the balance point shifts to $40\, cm$. The value of $R_1$ is (in $ohm$)
A
$60$
B
$40$
C
$20$
D
$10$
Solution
For first balancing condition $\frac{{10 + {R_1}}}{{{R_2}}} = \frac{{50}}{{50}}$
$ \Rightarrow $ ${R_2} = 10 + {R_1}$.
For second balancing condition
$\frac{{{R_1}}}{{{R_2}}} = \frac{{40}}{{60}}$ $ \Rightarrow $ $\frac{{{R_1}}}{{10 + {R_1}}} = \frac{2}{3}$
$ \Rightarrow $ ${R_1} = 20\,\Omega $
Standard 12
Physics
