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:
- 63
- 70
- 66
- 35
The Correct Option is C
Approach Solution - 1
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.
Approach Solution -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, \(R = 66\).
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