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(A) The volume of water collected in the tank in $8\,hours$ is given by $V = 	ext{length} 	imes 	ext{breadth} 	imes 	ext{height}$.$V = 30\,m 	im

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$30$

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(A) The volume of water collected in the tank in $8\,hours$ is given by $V = 	ext{length} 	imes 	ext{breadth} 	imes 	ext{height}$.$V = 30\,m 	imes 20\,m 	imes 1\,m = 600\,m^3$.Therefore,the volume of water collected in $1\,hour$ is $\frac{600}{8} = 75\,m^3$.The cross-sectional area of the square pipe is $A = 5\,cm 	imes 5\,cm = 25\,cm^2$.Converting the area to square meters: $A = \frac{25}{10000}\,m^2 = 0.0025\,m^2$.The volume of water flowing through the pipe in $1\,hour$ is equal to the product of the cross-sectional area and the speed of water $(v)$: $V = A 	imes v$.$75\,m^3 = 0.0025\,m^2 	imes v$.$v = \frac{75}{0.0025}\,m/h = 30000\,m/h$.Since $1\,km = 1000\,m$,the speed is $v = \frac{30000}{1000}\,km/h = 30\,km/h$.

$A$ rectangular tank is $30\,m$ long and $20\,m$ broad. Water is being flown into it through a square pipe of side $5\,cm$. What is the speed of water if the level of water in the tank rises by $1\,m$ in $8\,hours$? (in $km/h$)

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