English, asked by kavanapoojari012, 2 months ago

the charges
34. A resistance wire of length 0.85m and radius0.2mm is
connected in the left gap of the metre bridge. A standard
resistance of 50 is connected in the right gap. The balancing
length is 0.485m. Calculate the resistivity of the material of​

Answers

Answered by allysia
0

Answer:

6.96\times 10^{-6} \Omega m

Explanation:

Using relation,

\\\tt \dfrac{R_{1}}{R_{2}} = \dfrac{l}{100-l}

Where R1 is the resistance just above the wire of l units, and R2 is above the remaining length of 100-l units (since the wire length is 100cm) in a meter bridge.

Let the resistance of unknown wire be R

Now using the values from here :

\\\tt \dfrac{R}{50 \Omega} = \dfrac{48.5}{51.5} \\\\\tt R =\dfrac{48.5}{51.5} x 50 \Omega =  47.08 \Omega

Now since,

\\\tt R= \rho\dfrac{l}{A} \\\\\tt 47.08 \Omega =  \rho\dfrac{0.85 m}{ (0.2x10^{-3})^{2}\pi m^{2} }\\\\\tt \roh = 47.08 \Omega x \ \dfrac{( (0.2x10^{-3})^{2}\pi m^{2}) }{0.85m} \\\\\tt = 6.96x 10^{-6} \Omega m

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