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A wire of resistance $2 R$ is stretched such that its length is doubled. Then the increases in its resistance is
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Verified Answer
The correct answer is:
$6 \mathrm{R}$
Resistance of wire, $R_1=2 R$
Length of wire, $L_2=2 L_1$
$$
\begin{aligned}
& A_1 L_1=A_2 L_2 \\
& A_1 L_1=A_2 2 L_1 \\
& A_2=A_1 / 2 \\
& R_1=\frac{\rho L_1}{A_1} \\
& R_2=\frac{\rho L_2}{A_2} \\
& \frac{R_2}{R_1}=\frac{\rho L_2}{A_2} \times \frac{A_1}{\rho L_1}=\frac{2 L_1 \times 2}{A_1} \times \frac{A_1}{L_1} \\
& R_2=4 R_1=8 R
\end{aligned}
$$
The increase in resistance $=R_2-R_1$ $=8 \mathrm{R}-2 \mathrm{R}=6 \mathrm{R}$.
Length of wire, $L_2=2 L_1$
$$
\begin{aligned}
& A_1 L_1=A_2 L_2 \\
& A_1 L_1=A_2 2 L_1 \\
& A_2=A_1 / 2 \\
& R_1=\frac{\rho L_1}{A_1} \\
& R_2=\frac{\rho L_2}{A_2} \\
& \frac{R_2}{R_1}=\frac{\rho L_2}{A_2} \times \frac{A_1}{\rho L_1}=\frac{2 L_1 \times 2}{A_1} \times \frac{A_1}{L_1} \\
& R_2=4 R_1=8 R
\end{aligned}
$$
The increase in resistance $=R_2-R_1$ $=8 \mathrm{R}-2 \mathrm{R}=6 \mathrm{R}$.
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