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Ex 13.3, 4 - A well of diameter 3 m is dug 14 m deep. The earth - Ex 13.3

  1. Chapter 13 Class 10 Surface Areas and Volumes
  2. Serial order wise
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Ex 13.3, 4 A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. Both well and embankment are in the from of cylinder. Let well be cylinder A and embankment be cylinder B. Since mud of well is distributed in embankment Volume of well = Volume of embankment Volume of well For cylinder A, Diameter = 3 m So, radius = r = 3/2 m = 1.5 m Height = 14 m Volume of cylinder A = ๐œ‹๐‘Ÿ2โ„Ž = ฯ€ร—(1.5)2ร—(14) = ฯ€ร—2.25ร—14 = 31.5 ๐œ‹ m3 So, Volume of well = Volume of cylinder A = 31.5 ๐œ‹ m3 Volume of the embankment For cylinder B Cylinder B is a hollow cylinder with Inner Diameter = Diameter of well = 3 m Internal radius = r1 = 3/2 = 1.5 m External radius = r2 = internal radius + width = 1.5 + 4 = 5.5 m Volume of cylinder with internal radius = ๐œ‹๐‘Ÿ12โ„Ž = ๐œ‹โ„Ž(1.5)2 Volume of cylinder with external radius = ๐œ‹๐‘Ÿ22โ„Ž = ๐œ‹โ„Ž(5.5)2 Volume of cylinder B = Volume of cylinder with external radius โ€“ Volume of cylinder with internal radius = ๐œ‹โ„Ž(5.5)2โˆ’ฯ€h(1.5)2 = ๐œ‹โ„Ž((5.5)2โˆ’(1.5)2) = ๐œ‹โ„Ž(30.25โˆ’2.25) = ๐œ‹โ„Ž(28) = 28๐œ‹โ„Ž m3 So, Volume of embankment = 28๐œ‹โ„Ž m3 Now, Volume of well = Volume of embankment 31.5 ฯ€ = 28 ฯ€h 28 ๐œ‹h = 31.5 ๐œ‹ h = 31.5/28 h = 1.125 Hence, the height of the embankment = 1.125 m.

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