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A cubical container with side 2 m has a small hole with a cap at point C as shown. The water level is upto point D. (BC=0.5 m and BD=1.5 m). If container is given an acceleration of $8 m / s^{2}$ and the hole is opened simultaneously, find the amount of water that will spill out of the container (in litres).

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answer is 1200.

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Detailed Solution

$\tan θ = \frac{a}{g} = \frac{4}{5}$

$\Rightarrow RB = PQ - QBtan θ = 2 - 2 \times \frac{4}{5} = 0.4 m$

Volume of water

$contained = (\frac{PQ + BR}{2}) \times 4 m^{2}$

Initial volume of water $= BD \times 4 m^{2}$

$Amount of spill = V_{in} - V_{fin}$

$= 2 (2 BD - PQ - BR) \times 1000 litres$

$= 2 (3 - 2 - 0.4) \times 1000$

$= 1200 litres$

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