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

The Correct Option is C

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, \(R = 66\).