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 \ldots (i) $
Now, $l_{2}=3 l_{1}$
$\because$ Volume remains same.
$\therefore V_{1}=V_{2} $
$\Rightarrow A_{1} l_{1}=A_{2} l_{2}$
$A_{1} l_{1}=A_{2} \times 3 l_{1}$
$ \Rightarrow A_{2}=\frac{A_{1}}{3}$
So, final resistance,
$R_{2}=\frac{\rho l_{2}}{A_{2}}=\frac{\rho \times 3 l_{1}}{\left(\frac{A_{1}}{3}\right)}$
$=\frac{9 \rho l_{1}}{A_{1}}=9 \times 20\, \Omega$ by [Eq. (i) ]
$=180\, \Omega$