A rectangular water tank has dimensions 80, m | vedclass

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(C) Speed of water $= 10\, km/hr = 10 	imes \frac{5}{18}\, m/s = \frac{25}{9}\, m/s$.Area of pipe opening $= 40\, cm^{2} = \frac{40}{10000}\, m^{2} =

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$\frac{5}{8}$

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(C) Speed of water $= 10\, km/hr = 10 	imes \frac{5}{18}\, m/s = \frac{25}{9}\, m/s$.Area of pipe opening $= 40\, cm^{2} = \frac{40}{10000}\, m^{2} = \frac{1}{250}\, m^{2}$.Volume of water flowing in per second $= 	ext{Area} 	imes 	ext{Speed} = \frac{1}{250} 	imes \frac{25}{9} = \frac{1}{90}\, m^{3}/s$.Time $= 30\, 	ext{minutes} = 30 	imes 60 = 1800\, s$.Total volume of water in $30\, 	ext{minutes} = \frac{1}{90} 	imes 1800 = 20\, m^{3}$.Let the rise in water level be $h\, m$. The volume of water in the tank is $80 	imes 40 	imes h = 3200h\, m^{3}$.Equating the volumes: $3200h = 20 \implies h = \frac{20}{3200} = \frac{1}{160}\, m$.Converting to $cm$: $h = \frac{1}{160} 	imes 100 = \frac{10}{16} = \frac{5}{8}\, cm$.

$A$ rectangular water tank has dimensions $80\, m \times 40\, m$. Water flows into it through a pipe with an opening area of $40\, cm^{2}$ at a speed of $10\, km/hr$. By how much will the water level rise in the tank in half an hour (in $cm$)?

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