A conical flask contains sone water when the flask is oriented so that its base is horizontal and lies at the bottom (so that the vertex is at the top) the water is 1 cm deep when the flask is turned upsidedown so that the vertex is at the bottom the water is 2cm deep if h cm is the height of conical flasks then 3h^2-3h equals to ??
Answers
3h² - 3h = 7
Step-by-step explanation:
Let say Radius of Base of cone = r cm
Height of cone = h cm
Volume of cone = (1/3) πr²h
at height 1 cm , radius of base of cone would be
=> (h-1)/h = r'/r
=> r' = r (h-1)/h
Volume above 1 m height = (1/3) π (r (h-1)/h)²(h - 1)
Volume of Water when height is 1 cm = Volume of cone - Volume of cone above 1 cm height
= (1/3) πr²h - (1/3) π (r (h-1)/h)²(h - 1)
= (1/3) πr² ( h - (h-1)³/h²)
= (1/3) πr² (h³ - (h-1)³/)/h²
at height 2 cm from vertex , radius of base of cone would be
2/h = r'/r
=> r' = 2r/h
Volume of water = (1/3) π ( 2r/h)² * 2
(1/3) π ( 2r/h)² * 2 = (1/3) πr² (h³ - (h-1)³/)/h²
=> 8 = h³ - (h-1)³
=> 8 = h³ - ( h³ - 1 - 3h² + 3h)
=> 8 = 3h² - 3h + 1
=> 3h² - 3h = 7
3h² - 3h = 7
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