Question

A wire of length 10 cm and radius $\sqrt{7} \times 10^{-4} \, m$ is 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 the value of $R$ is:

• A. 63
• B. 70
• C. 66
• D. 35

Answer 1

Answer: (C)

66

Answer 2

For a balanced Wheatstone bridge in a meter bridge setup:

$\frac{4.5}{R} = \frac{60}{40}$

Solving for $R$:

$R = \frac{4.5 \times 40}{60} = 3 \, \Omega$

Now, using the formula for resistance:

$R = \frac{\rho \ell}{A} = \frac{\rho \ell}{\pi r^2}$

where $\ell = 10 \, \text{cm} = 0.1 \, \text{m}$ and $r = \sqrt{7} \times 10^{-4} \, \text{m}$.

Substitute values:

$3 = \frac{\rho \times 0.1}{\pi (\sqrt{7} \times 10^{-4})^2}$

$\rho = 66 \times 10^{-7} \, \Omega \, \text{m}$

Thus, $R = 66$.