Question

A wire of resistance R and length L is pulled and stretched to a length 2L. What is its new resistance?

• A. 2R
• B. 4R
• C. $\frac{R}{2}$
• D. R

Answer 1

The correct option is B 4R

On first thought one may conclude that, hey!! The length has increased by two times and $\rho =\frac{l}{A}$ so R must also increase two times. Right? Seems easy! BUT NO!!
What have we forgotten?
As the wire is stretched, the cross-sectional area also decreases!! :D :D
The volume obviously remains constant.
i.e. A x l remains constant!!
Just observe how I am going to manipulate the relation $R=\rho \frac{l}{A}$
Multiplying and dividing by length l we get
$R=\rho \frac{l^2}{Al}=\rho =\frac{l^2}{v}$ where V is the volume of the wire.
Now this equality has everything constant on the RHS except the length l As the length l increases to two times, the resistance increases by four times!