The resistance of a wire is 20 Omega. It is s | vedclass

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(D) Initial resistance,$R_1 = 20 \Omega \Rightarrow \frac{\rho l_1}{A_1} = 20 \quad \dots (i)$Now,the length becomes $l_2 = 3 l_1$.Since the volume of

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$180$

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(D) Initial resistance,$R_1 = 20 \Omega \Rightarrow \frac{ho l_1}{A_1} = 20 \quad \dots (i)$Now,the length becomes $l_2 = 3 l_1$.Since the volume of the wire remains constant during stretching:$V_1 = V_2 \Rightarrow A_1 l_1 = A_2 l_2$$A_1 l_1 = A_2 	imes 3 l_1 \Rightarrow A_2 = \frac{A_1}{3}$So,the final resistance $R_2$ is given by:$R_2 = \frac{ho l_2}{A_2} = \frac{ho 	imes 3 l_1}{(A_1 / 3)} = 9 	imes \frac{ho l_1}{A_1}$Substituting the value from Eq. $(i)$:$R_2 = 9 	imes 20 \Omega = 180 \Omega$

The resistance of a wire is $20 \Omega$. It is stretched,so that the length becomes three times,then the new resistance of the wire will be (in $\Omega$)

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