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In the experimental set up of metre bridge shown in the figure, the null point is obtaine data distance of $40 \mathrm{~cm}$ from
A. If a $10 \Omega$ resistor is connected in series with $\mathrm{R}_{1},$ the null point shifts by $10 \mathrm{~cm}$. The resistance that should be connected in parallel with $\left(\mathrm{R}_{1}+10\right) \Omega$ such that the null point shifts back to its initial position is
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A. If a $10 \Omega$ resistor is connected in series with $\mathrm{R}_{1},$ the null point shifts by $10 \mathrm{~cm}$. The resistance that should be connected in parallel with $\left(\mathrm{R}_{1}+10\right) \Omega$ such that the null point shifts back to its initial position is
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The correct answer is:
$60 \Omega$
Initially at null deflection $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{2}{3}$$\ldots(\mathrm{i})$
Finally at null deflection, when null point is shifted $\frac{\mathrm{R}_{1}+10}{\mathrm{R}_{2}}=1 \Rightarrow \mathrm{R}_{1}+10=\mathrm{R}_{2}$...(ii)
Solving equations (i) and (ii) we get $\frac{2 R_{2}}{3}+10=R_{2}$
$10=\frac{\mathrm{R}_{2}}{3} \Rightarrow \mathrm{R}_{2}=30 \Omega$
$\& R_{1}=20 \Omega$
Now if required resistance is $\mathrm{R}$ then $\frac{\frac{30 \times \mathrm{R}}{30+\mathrm{R}}}{30}=\frac{2}{3}$
$\mathrm{R}=60 \Omega$
Finally at null deflection, when null point is shifted $\frac{\mathrm{R}_{1}+10}{\mathrm{R}_{2}}=1 \Rightarrow \mathrm{R}_{1}+10=\mathrm{R}_{2}$...(ii)
Solving equations (i) and (ii) we get $\frac{2 R_{2}}{3}+10=R_{2}$
$10=\frac{\mathrm{R}_{2}}{3} \Rightarrow \mathrm{R}_{2}=30 \Omega$
$\& R_{1}=20 \Omega$
Now if required resistance is $\mathrm{R}$ then $\frac{\frac{30 \times \mathrm{R}}{30+\mathrm{R}}}{30}=\frac{2}{3}$
$\mathrm{R}=60 \Omega$
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