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Question
The diameter of the cross-section of a water pipe is 5 cm. Water flows through it at 10km/hr into a cistern in the form of a cylinder. If the radius of the base of the cistern is 2.5 m, find the height to which the water will rise in the cistern in 24 minutes.
Solution
Area of the cross-section of the water pipe = πr2
= 3.142 x `5/2 xx 5/2`
Speed of the water = 10 km/hr
= `(10 xx 1000 xx 100)/60` cm/minutes
∴ Quantity of water supplied in 24 minutes
= `3.142 xx 5/2 xx 5/2 xx (10,00,000)/6 xx 24`
= 78,55,000 cm3
Let the height of water in the cistern be h cm.
The quantity of water collected in the cistern
= 3.142 x 250 x 250 x h cm3
Both the above quantities must be equal
∴ 3.142 x 250 x 250 x h = 78,55,000
h = 78,55,000 x `1/(3.142 xx 250 xx 250)`
h = 40 cm.
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