Question

A wire of resistance R is uniformly stretched until its radius is halved. Its new resistance is

• A. 2 R
• B. 4 R
• C. 8 R
• D. 16 R

Answer 1

Answer: (D)

As we know,
$R=\frac{\rho L}{A}.$
If you stretch the wire till it's radius is halved the previous value, then it's length will increase 4 times to keep the volume constant.
So, new values are,
$\begin{array}{l}{L}_2=4L\\ A_2=\frac{A}{4}\end{array}$
where, $L$ and $A$ are the values of original length and area
Resistivity remains the same as it depends upon the material of the wire. So, the new resistance will be,
$\begin{array}{l}{R}_1=\frac{\rho L}{A}\\ R_1=\frac{\rho 4L}{A/2}\\ R_1=\frac{16\rho L}{A}\\ R_1=16R\end{array}$