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(C) For a meter bridge,the balance condition is given by $\frac{P}{Q} = \frac{l_1}{l_2}$,where $P$ is the resistance in the left gap,$Q$ is the resist

Vedclass · Accepted Answer

$66$

Vedclass · Answer

(C) For a meter bridge,the balance condition is given by $\frac{P}{Q} = \frac{l_1}{l_2}$,where $P$ is the resistance in the left gap,$Q$ is the resistance in the right gap,$l_1 = 60 \ cm$,and $l_2 = 100 - 60 = 40 \ cm$.Given $P = 4.5 \ \Omega$,we have $\frac{4.5}{Q} = \frac{60}{40} = 1.5$.Thus,$Q = \frac{4.5}{1.5} = 3 \ \Omega$.The resistance of the wire is given by $Q = \frac{ho L}{A} = \frac{ho L}{\pi r^2}$,where $L = 10 \ cm = 0.1 \ m$ and $r = \sqrt{7} 	imes 10^{-4} \ m$.Substituting the values: $3 = \frac{ho 	imes 0.1}{\pi 	imes (\sqrt{7} 	imes 10^{-4})^2} = \frac{ho 	imes 0.1}{\pi 	imes 7 	imes 10^{-8}}$.Solving for $ho$: $ho = \frac{3 	imes \pi 	imes 7 	imes 10^{-8}}{0.1} = 210 \pi 	imes 10^{-8} \ \Omega \ m$.Using $\pi \approx 3.14$,$ho = 210 	imes 3.14 	imes 10^{-8} = 659.4 	imes 10^{-8} \approx 66 	imes 10^{-7} \ \Omega \ m$.Therefore,$R = 66$.

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