Resistance of a wire is 8 Omega. It is drawn | vedclass

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(B) Given,initial resistance $R_1 = 8 \Omega$.Longitudinal strain $\varepsilon = \frac{\Delta l}{l_1} = 400 \% = 4$.The final length $l_2 = l_1 + \Del

Vedclass · Accepted Answer

$200$

Vedclass · Answer

(B) Given,initial resistance $R_1 = 8 \Omega$.Longitudinal strain $\varepsilon = \frac{\Delta l}{l_1} = 400 \% = 4$.The final length $l_2 = l_1 + \Delta l = l_1 + 4l_1 = 5l_1$.Since the volume of the wire remains constant during stretching,the resistance $R$ is proportional to the square of the length: $R \propto l^2$.Therefore,$\frac{R_2}{R_1} = \left(\frac{l_2}{l_1}ight)^2$.Substituting the values: $\frac{R_2}{8} = \left(\frac{5l_1}{l_1}ight)^2 = 5^2 = 25$.Thus,$R_2 = 25 	imes 8 = 200 \Omega$.

Resistance of a wire is $8 \Omega$. It is drawn in such a way that it experiences a longitudinal strain of $400 \%$. The final resistance of the wire is (in $Omega$)

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