Question

A cubical vessel of height $1m$ is full of water. What is the amount of work done in pumping water out of the vessel?
(Take $g=10m/s^2$)

• A. $1250J$
• B. $5000J$
• C. $1000J$
• D. $2500J$

Answer 1

The correct option is B $5000J$

$\text{Step 1: Calculate potential energy of water}$

COM of water will lie at the center of vessel.

Assuming potential energy as $0$ at bottom of the vessel.

Potential energy $=mgh$

$\Rightarrow (PE{)}_{initial}=\frac{mgh}{2}$ $....\left(1\right)$

and $(PE{)}_{finaal}=0$ $...\left(2\right)$

$\text{Step 2: Apply energy conservation}$

From energy conservation, we can say

$Work=-\left(\Delta PE\right)$

$\Rightarrow Work=-\left[\right(PE{)}_{final}-\left(PE{)}_{initial}\right]$

From equation $\left(1\right)$ and $\left(2\right)$

$\Rightarrow Work=-(0-\frac{mgh}{2})=\frac{mgh}{2}$

Volume of vessel $=lbh=1m\times 1m\times 1m=1m^3$ and $m=\rho V$

$\Rightarrow Work=\frac{\rho Vgh}{2}=\frac{\left(1000Kg/m^3\right)\left(1m^3\right)\left(10m/s^2\right)\left(1m\right)}{2}$ $\rho =1000Kg/m^3$

$\Rightarrow Work=5000Joule$

Hence option B is correct.