Current Electricity Question 7
Question 7 - 2024 (27 Jan Shift 1)
A wire of length $10 cm$ and radius $\sqrt{7} \times 10^{-4} m$ 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 $60 cm$ from the left end. If the resistivity of the wire is $R \times 10^{-7} \Omega m$, then value of $R$ 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, $R=\frac{\rho \ell}{A}=\frac{\rho \ell}{\pi r^{2}}$
$4.5 \times 40=\rho \times \frac{0.1}{\pi \times 7 \times 10^{-8}} \times 60$
$\rho=66 \times 10^{-7} \Omega \times m$
