If the length and diameter of a conductor both are halved then the resistance of conductor will be...............

The formula of the resistance of a conductor is-

$\boxed{R=\rho(\frac{L}{A})}$

Here,

$R\rightarrow$ resistance of the conductor

$\rho\rightarrow$ resistivity of the conductor

$L\rightarrow$length of the conductor

$A\rightarrow$area of the cross-section of the conductor

We know the formula of the area of a circle, $A=\pi r^2$

Or $A=\pi (\frac{d}{2})^2$            [$d$ is the diameter of the circle and $d=2r$]

Therefore, resistance of a conductor $R=\rho(\frac{L}{\pi(\frac{d}{2})^2})$

Or $R=\rho(\frac{4L}{\pi d^2})$    ........ $(i)$

If both the length and diameter are halved, then 

$R'=\rho(\frac{4(\frac{L}{2})}{\pi (\frac{d}{2})^2})$

Or $R'=\rho(\frac{8L}{\pi d^2})$      .......$(ii)$

On comparing $(i)$ and $(ii)$, we get to know that $R'=2R$

Therefore, the resistance of a conductor gets doubled if the length and diameter of a conductor are halved.