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A cubical vessel of height \mathrm{1 \mathrm{~m}} is full of water. What is the amount of work done in pumping water out of the vessel? \mathrm{Take \, \, g=10 \mathrm{~ms}^{-2}.}

Option: 1

1250 J


Option: 2

2500 J


Option: 3

5000 J


Option: 4

1000 J


Answers (1)

best_answer

Side of the cube l=1 \mathrm{~m}. Base area of the cubical vessel \mathrm{=l^2.} Mass of water contained in a height \mathrm{d x=l^2 d x \rho.} Weight of this water \mathrm{ =l^2 d x \rho g}. Therefore, work done in pumping out water up to a height x is

                                    \mathrm{ d W=\rho g l^2 x d x }
Therefore, total work done in pumping out water up to a height l is

                  \mathrm{ \begin{aligned} W & =\rho g l^2 \int_0^l x d x=\frac{1}{2} \rho g l^2\left|x^2\right|_0^l=\frac{1}{2} \rho g l^4 \\\\ & =\frac{1}{2} \times 1000 \times 10 \times(1)^4=5000 \mathrm{~J} \end{aligned} }

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shivangi.shekhar

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