Question

A cubical vessel of height $1m$ is full of water. The minimum work done in taking water out of vessel will be

• A. $5000J$
• B. $10000J$
• C. $5J$
• D. $10J$

Answer 1

The correct option is

A

$5000J$

total water volume$=1m\times 1m\times 1m=1m^3$

So mass $=1m^3\times \frac{1000kg}{m^3}=1000kg$

height $=1m$

So centre of mass will be at a

$\frac{1}{2}=0.5m$ from ground

So potential energy $=mg\times 0.5=1000\times 10\times 0.5=5000J$

water spread on ground $\Rightarrow PE=0$

$△PE=0-\left(5000J\right)=-5000J$

work done $=-△PE=5000J$