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

To find the value of \( R \) given the scenario of a meter bridge, we will use the principle of the Wheatstone bridge. The balance point condition in the meter bridge can be expressed as: 

\[\frac{R_1}{R_2} = \frac{l_1}{l_2}\]

where:

• \( R_1 \) is the resistance in the left gap (4.5 Ω in this case).
• \( R_2 \) is the resistance in the right gap (resistance of the wire).
• \( l_1 = 60 \, \text{cm} \) is the length from the left end to the balance point.
• \( l_2 = 100 - 60 = 40 \, \text{cm} \) is the length from the right end to the balance point.

Substituting the known values into the balance condition:

\[\frac{4.5}{R} = \frac{60}{40}\]

Solving for \( R \):

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

The resistance of the wire can also be expressed in terms of its physical dimensions:

\(R = \frac{\rho L}{A}\)

where:

• \( \rho \) is the resistivity.
• \( L = 0.1 \, \text{m} \) is the length of the wire.
• \( A = \pi (\text{radius})^2 = \pi \left(\sqrt{7} \times 10^{-4}\right)^2 \) is the cross-sectional area.

Substituting the known resistance of the wire and solving for resistivity:

\[3 = \frac{\rho \times 0.1}{\pi (7 \times 10^{-8})}\]

Simplifying gives:

\[\rho = 3 \times \pi \times 7 \times 10^{-8} \times 10\]\[\rho = 3 \times 7 \times \pi \times 10^{-7}\]

Comparing with \( R \times 10^{-7} \, \Omega \, m \), we find:

\(R = 3 \times 7 \approx 66\)

Therefore, the value of \( R \) is 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, \(R = 66\).