Question

# A cubical vessel of height $1\mathrm{m}$ is full of water. What is the work done in pumping water out of the vessel?

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Solution

## Step 1: Given:Height of the vessel, h$=1\mathrm{m}$The centre of mass of the water in the cubical vessel lies at CM, i.e $\mathrm{h}\text{'}=\frac{\mathrm{h}}{2}=\frac{1}{2}\mathrm{m}$The volume water, $\mathrm{V}={\mathrm{h}}^{3}={1}^{3}=1{\mathrm{m}}^{3}$∴ Mass of water,$\mathrm{m}=\mathrm{\rho V}=1000×1=1000\mathrm{kg}$Step 2: Formula & SolutionWork done for pumping out water is equal to negative of change in its potential energy.$\mathrm{W}=-\mathrm{\Delta P}.\mathrm{E}\phantom{\rule{0ex}{0ex}}=-\left(\mathrm{P}.{\mathrm{E}}_{\mathrm{f}}-\mathrm{P}.{\mathrm{E}}_{\mathrm{i}}\right)\phantom{\rule{0ex}{0ex}}=-\left(0-\mathrm{mgh}\text{'}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{mgh}\text{'}\phantom{\rule{0ex}{0ex}}=1000×10×\frac{1}{2}\phantom{\rule{0ex}{0ex}}=5000\mathrm{J}$Hence,the work done in pumping water out of the vessel is $5000\mathrm{J}$.

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