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 :

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Answer & Solution

Correct Answer

(C) \(66\)

Detailed explanation

For null point,

\( \frac{4.5}{60}=\frac{R}{40} \)

\( \text { Also, } R=\frac{\rho \ell}{A}=\frac{\rho \ell}{\pi r^2} \)

\( 4.5 \times 40=\rho \times \frac{0.1}{\pi \times 7 \times 10^{-8}} \times 60 \)

\( \rho=66 \times 10^{-7} \Omega \times \mathrm{m}\)

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