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In the experimental setup of meter bridge shown in the figure, the null point is obtained at a distance of $40\,cm$ from $A$. If a $10\,\Omega $ resistor is connected in series with $R_1$, the null point shifts by $10\,cm$. The resistance that should be connected in parallel with $\left( {{R_1} + 10} \right)\,\Omega $ such that the null point shifts back to its initial position is .............. $\Omega$

$20$
$40$
$60$
$30$
Solution
$\frac{R_{1}}{R_{2}}=\frac{2}{3}$ …..$(i)$
$\frac{R_{1}+10}{R_{2}}=1$
$\Rightarrow \quad R_{1}+10=R_{2}$ …..$(ii)$
$\frac{2 \mathrm{R}_{2}}{3}+10=\mathrm{R}_{2} \quad ; \quad 10=\frac{\mathrm{R}_{2}}{3}$
$\Rightarrow \quad \mathrm{R}_{2}=30\, \Omega \quad \& \quad \mathrm{R}_{1}=20\, \Omega$
$\frac{\frac{30 \times \mathrm{R}}{30+\mathrm{R}}}{30}=\frac{2}{3}$
$\mathrm{R}=60 \,\Omega$