The charges34. A resistance wire of length 0.85m and radius0.2mm iscon

1.	The charges34. A resistance wire of length 0.85m and radius0.2mm isconnected in the left gap of the metre bridge. A standardresistance of 50 is connected in the right gap. The balancinglength is 0.485m. Calculate the resistivity of the material of
Answer» 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, $\\\tt R= \rho\dfrac{l}{A} \\\\\tt 47.08 \Omega = \rho\dfrac{0.85 m}{ (0.2x10^{-3})^{2}\pi m^{2} }\\\\\tt \roh = 47.08 \Omega x \ \dfrac{( (0.2x10^{-3})^{2}\pi m^{2}) }{0.85m} \\\\\tt = 6.96x 10^{-6} \Omega m$

The charges34. A resistance wire of length 0.85m and radius0.2mm isconnected in the left gap of the metre bridge. A standardresistance of 50 is connected in the right gap. The balancinglength is 0.485m. Calculate the resistivity of the material of

Answer»

Answer:

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

Explanation:

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

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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