Question

A rectangular water reservoir id 10.8m by 3.75m at the base . Water flows into it at the rate of 18m/s through a pipe having cross-section 7.5cm × 4.5 cm. Find the height in which water will rise in 30 minutes

Answer 1

area of the base of the reservoir=10.8m*3.75m
=40.5sq.m

the pipe is cuboidal in shape:
cross-section area of the pipe=0.075m*0.045m
=.0.003375sq.m

volume of water through the pipe in 1sec=cross-sectional area * rate of water
=0.003375sq.m*18
=0.06075cu.m

This is the volume filled in the reservoir by the pipe in 1sec
so volume filled by the pipe in 30 mins=0.06075*30*60(1 min =60sec)
=109.35 cu.m

This is the total volume in the reservoir after 30 mins.

so height of water level=volume/area of base of the reservoir
=109.35/33.75
=3.24 m

Answer 2

Answer:

Water that flows into the reservoir in one second forms a cuboid of dimensions 7.5cm × 4.5cm × 1800 cm.

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Volume of the water that flows in one second

$\sf = (18 \times 0.075 \times 0.045) {m}^{3} \\ \\ = \sf18 \times \frac{75}{1000} \times \frac{45}{1000} {m}^{3} = \frac{243}{4000}$

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Volume of the water that flows in 30 minutes

$\sf = \frac{243}{4000} \times 60 \times 30 \: {m}^{3}$

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Area of the base of the reservoir = (10.8 × 3.75) m²

$\therefore$ Height of water

$\sf = \frac{volume}{area \: of \: the \: base} = \sf\frac{2187}{20} \times \frac{1}{10.8 \times 3.75} m \\ \\ =\sf 2.7m$