The resistance of a wire is 5 Omega. Its new | vedclass

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(C) The initial resistance is given by $R_{	ext{initial}} = \frac{ho \ell}{A} = 5 \Omega$.Since the volume of the wire remains constant during stre

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$125$

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(C) The initial resistance is given by $R_{	ext{initial}} = \frac{ho \ell}{A} = 5 \Omega$.Since the volume of the wire remains constant during stretching:$V_i = V_f$$A_i \ell_i = A_f \ell_f$Given $\ell_f = 5 \ell_i$,we have $A_i \ell_i = A_f (5 \ell_i)$,which implies $A_f = \frac{A_i}{5}$.The new resistance $R_f$ is:$R_f = \frac{ho \ell_f}{A_f} = \frac{ho (5 \ell_i)}{\left(\frac{A_i}{5}ight)}$$R_f = 25 \left(\frac{ho \ell_i}{A_i}ight)$$R_f = 25 	imes 5 = 125 \Omega$.

The resistance of a wire is $5 \Omega$. Its new resistance in ohm,if stretched to $5$ times of its original length,will be:

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