Question

If a wire of resistance 'R' is stretched twice its original length keeping volume constant then what will be its new resistance?
(let new resistance as R')

• A. 4R
• B. 2R
• C. R/3
• D. R/2

Answer 1

The correct option is A 4R

The volume of wire = $length\times crosssectionalarea$.

Since volume is constant,
Accordingly,
let length of wire = L
let area of wire = A
So resistance $R=\rho \frac{L}{A}$,
Now according to given condition length of stretched wire = 2L
therefore area of stretched wire = A/2
So resistance of stretched wire $R^{\prime }=\rho \frac{2L}{A/2}$ or,
$R^{\prime }=\rho \frac{2\times 2L}{A}$
$R^{\prime }=\rho \frac{4L}{A}$
$R^{\prime }=4\left(\frac{\rho L}{A}\right)$ or,
R' = 4 (R) or,
the new resistance (R') of the stretched wire is 4 times the value of resistance of the unstretched wire (R).