- Home
- Standard 12
- Physics
3.Current Electricity
hard
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
$35$
(JEE MAIN-2024)
Solution
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}$
Standard 12
Physics
