Question

A cylindrical conductor of length L and radius r has a resistance R. What will be the radius of another conductor of the same material of length 2L and resistance R?

• A. $2r$
• B. $r$
• C. $r/2$
• D. $\sqrt{2}$

Answer 1

The correct option is D $\sqrt{2}$

Given:
Resistance of the conductor $=R$
Length of the conductor $=L$
Area of the conductor $=\pi r^2$
Hence, resistance $R=\frac{\left(\rho L\right)}{\left(\pi r^2\right)}$
where ρ is the resistivity of the material.
$\Rightarrow \pi r^2=\frac{\left(\rho L\right)}{R}$
$r=\sqrt{\frac{\left(\rho L\right)}{\pi R}}$
Let the radius of the second conductor be r’
Now, the length of the conductor is made to 2L, rest of the parameters remaining the same.
Hence,
$R=\frac{\rho \left(2L\right)}{\left[\pi \right(r^{\prime 2}\left)\right]}$
$\pi \left(r^{\prime 2}\right)=2\frac{\left(\rho L\right)}{R}$
$\Rightarrow \left(r^{\prime 2}\right)=\frac{2\left(\rho L\right)}{\left(\pi R\right)}$
$\Rightarrow r^{\prime }=\sqrt{\frac{2\left(\rho L\right)}{\left(\pi R\right)}}$
$\Rightarrow r^{\prime }=\sqrt{2r}$
=>r’ = √2r
Thus, the new radius will be √2 times the original radius.