A wire of length $10 \mathrm{~cm}$ and radius $\sqrt{7} \times 10^{-4} \mathrm{~m}$ connected across the right gap of a meter bridge. When a resistance of $4.5 \Omega$ is connected on the left gap by using a resistance box, the balance length is found to be at $60 \mathrm{~cm}$ from the left end. If the resistivity of the wire is $\mathrm{R} \times 10^{-7} \Omega \mathrm{m}$, then value of $\mathrm{R}$ is :
Solution
<p>For null point,</p>
<p>$$\begin{aligned}
& \frac{4.5}{60}=\frac{R}{40} \\
& \text { Also, } R=\frac{\rho \ell}{A}=\frac{\rho \ell}{\pi r^2} \\
& 4.5 \times 40=\rho \times \frac{0.1}{\pi \times 7 \times 10^{-8}} \times 60 \\
& \rho=66 \times 10^{-7} \Omega \times \mathrm{m}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Wheatstone Bridge
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