Current Electricity

JEE Main level, single correct, medium

286 of 380

Question 286 of 380

A wire of length 10cm10cm and radius 7×10−4m\sqrt{7}\times {10}^{-4}m connected across the right gap of a meter bridge. When a resistance of 4.5Ω4.5\Omega is connected on the left gap by using a resistance box, the balance length is found to be at 60cm60cm from the left end. If the resistivity of the wire is R×10−7ΩmR\times {10}^{-7}\Omega m, then value of RR is :

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Answer: (C) 6666

From the null point condition of a Meter Bridge, it follows that,
4.5Ω60=X100−60⇒X=4060×4.5Ω=3Ω\dfrac{4.5\Omega }{60}=\dfrac{X}{100-60} \Rightarrow X=\dfrac{40}{60}\times 4.5\Omega =3\Omega
Also,
X=ρlA=ρlπr2...(1)\begin{matrix}X & = & \dfrac{\rho l}{A} \\ & = & \dfrac{\rho l}{\pi {r}^{2}}...\left(1\right)\end{matrix}
From equation (1), it follows that
3=ρ×0.1π×(7×10−4)2⇒ρ=3×π×(7×10−4)20.1≈66×10−7Ωm3=\rho \times \dfrac{0.1}{\pi \times {\left(\sqrt{7}\times {10}^{-4}\right)}^{2}} \Rightarrow \rho =\dfrac{3\times \pi \times {\left(\sqrt{7}\times {10}^{-4}\right)}^{2}}{0.1} \approx 66\times {10}^{-7}\Omega m
Therefore, R=66R=66.

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