Current Electricity Question 7
Question 7 - 2024 (27 Jan Shift 1)
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 :
(1) 63
(2) 70
(3) 66
(4) 35
Show Answer
Answer: (3)
Solution:
For null point,
$\frac{4.5}{60}=\frac{R}{40}$
Also, $\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}=\frac{\rho \ell}{\pi \mathrm{r}^{2}}$
$4.5 \times 40=\rho \times \frac{0.1}{\pi \times 7 \times 10^{-8}} \times 60$
$\rho=66 \times 10^{-7} \Omega \times \mathrm{m}$
