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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D

Potential energy of liquid column in first situation $= V d g \frac{h}{2} + V d g \frac{h}{2} = V d g h = a h d g h = d g h^{2} a$

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 $= V d g (\frac{2 h}{2}) = (A \times 2 h) d g h = 2 d g h^{2} a$

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

Suggest Corrections

Similar questions

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

Q. A liquid of density

d

is pumped by pump P from situation (i) to situation (ii) as shown in the figure. If the cross section of each of the vessel is

a

, then the work done in pumping (neglecting friction effects) is:

Q. A pump is used to pump a liquid of density

ρ^{'}

continuously through a pipe of cross section area

A

. If liquid is flowing with speed

V

, then power of pump is:

Q. A pump is used to pump a liquid of density

ρ

continuously through a pipe of cross-sectional area

A

. If the liquid is flowing with a speed

v

, find the power of the pump.

A pump is used to lift 500 kg of water from a depth of 80 m in 10 s.Calculate:

(a) the work done by the pump,

(b) the power at which the pump works, and

Join BYJU'S Learning Program