JEE MAIN - Physics (2024 - 27th January Morning Shift - No. 18)
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 :
63
70
66
35
Explanation
For null point,
$$\begin{aligned} & \frac{4.5}{60}=\frac{R}{40} \\ & \text { Also, } R=\frac{\rho \ell}{A}=\frac{\rho \ell}{\pi r^2} \\ & 4.5 \times 40=\rho \times \frac{0.1}{\pi \times 7 \times 10^{-8}} \times 60 \\ & \rho=66 \times 10^{-7} \Omega \times \mathrm{m} \end{aligned}$$
Comments (0)
