A metal rod of mass 10 kg and length 5 m is present horizontally on the ground. The work done in lifting it up so that it remain perpendicular to the ground is $\left(\text{g}=10{\text{m/s}}^{\text{2}}\right)$

Given mass of rod  m = 10 kg,  L = 5 m

When it is lifted up so that it remain perpendicular to the ground

The force required F = mg

And displacement of centre of mass = 2.5 m from the ground

Workdone  $=\text{F}\times \text{S}$

$=\text{mg}\times \text{S}$

$=\text{10}\times \text{10}\times \text{2}\text{.5}=\text{250}\text{J}$