Question

A liquid of density d is pumped by a pump p from situation (i) to situation (ii) as shown in the diagram. If the cross-section of each of the vessels is a, then the work done in pumping (neglecting friction effects) is

• A. $2dgh$
• B. $dgha$
• C. $2dg{h}^{2}a$
• D. $dg{h}^{2}a$

Answer 1

The correct option is D

$dg{h}^{2}a$

Potential energy of liquid column in first situation $=Vdg\frac{h}{2}+Vdg\frac{h}{2}=Vdgh=ahdgh=dg{h}^{2}a$

We know that, the centre of mass of liquid column lies at height $\frac{h}{2}$.

Potential energy of the liquid column in second situation $=Vdg\left(\frac{2h}{2}\right)=(A\times 2h)dgh=2dg{h}^{2}a$

Work done pumping = Change in potential energy $=2dg{h}^{2}a-dg{h}^{2}a=dg{h}^{2}a$