Question

A cubical vessel of height 1 m is full of water. Find the work done in pumping out whole water

• A. 49 J
• B. 4900 J
• C. 490 J
• D. 49000 J

Answer 1

The correct option is B 4900 J

Work done in pumping out whole water is equal to the change in potential energy.

Work done = change in potential energy

Or,

$workdone=\Delta U$

$workdone=U_f-U_i$

$=mgL-\frac{mgL}{2}$

$=\frac{mgL}{2}$

Since,

$density=\frac{mass}{volume}$

$m=\rho V$

$m=\rho \times A\times L$

So, work becomes,

$W=\frac{\rho A{L}^{2}g}{2}$

$W=\frac{1\times {10}^3\times 9.8}{2}$

$W=4900J$