JEE Mains · Physics · STD 12 - 3. current electricity
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 :
- A \(63\)
- B \(70\)
- C \(66\)
- D \(35\)
Answer & Solution
Correct Answer
(C) \(66\)
Step-by-step Solution
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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