Question

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

• A. $200\Omega$
• B. $160\Omega$
• C. $120\Omega$
• D. $180\Omega$

Answer 1

Answer: (D)

Initial resistance,
$R_1=20\Omega$
$\Rightarrow \frac{\rho l_1}{A_1}=20\dots (i)$
Now, $l_2=3l_1$
$\because$ Volume remains same.
$\therefore V_1=V_2$
$\Rightarrow A_1{l}_1=A_2{l}_2$
$A_1{l}_1=A_2\times 3l_1$
$\Rightarrow A_2=\frac{A_1}{3}$
So, final resistance,
$R_2=\frac{\rho l_2}{A_2}=\frac{\rho \times 3l_1}{\left(\frac{A_1}{3}\right)}$
$=\frac{9\rho l_1}{A_1}=9\times 20\Omega$ by [Eq. (i) ]
$=180\Omega$