A vessel in the shape of an inverted cone surmounted by a cylinder has common radius
7 cm. This was filled with liquid till it covered 1/3rd height of the cylinder. If height of each pair (conical and cylindrical) be 9 cm and the vessel is turned upside down, find the volume
liquid and to what height will it reach in the cylindrical part.
Please Ans it fast along with the procedure.
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The volume of liquid is 924 cm^3 and the height it will reach is 6 cm.
Step-by-step explanation:
We have radius of base of cone = radius of base of cylinder = r = 7 cm
- Now, height of cone = height of cylinder = h = 9 cm
- Now, at first the liquid is completely filled in cone and up to 13 of the height of cylinder.
- Now, height of the water level inside the cylinder, h1 = 13 of 9 cm = 3 cm
- Now, volume of liquid = volume of cone + volume of cylinder of height h1
=1/3πr^2h + πr^2h1
=1/3πr2 × 9 + πr^2 × 3
=6πr^2 = 6 × 227×7×7 = 924 cm^3
- Now, volume of cylindrical part of height h2 = volume of liquid⇒πr2h2 = 924
22/7×49 × h2 = 924
154 h2 = 924
h2 = 6 cm
The volume of liquid is 924 cm^3 and the height it will reach is 6 cm.
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