Mitchell pushes a coin of diameter 3 cm into a coin with diameter 9 cm and height 12 cm. How far into the cone can Mitchell push the coin before it gets stuck
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Step-by-step explanation:
Diameter of base of cone = r cm
Radius of base of cone = r/2 cm
Radius of base of cylinder = r cm
Height of cone =12cm
Height of water in cylinder before cone was taken out = 12cm
∴Volume of water left in cylinder when cone is removed out = Volume of water - Volume of cone
= πr2h = (1/3)πr2h
= 12πr2 – (1/3)π(r/2)2(12)
= 12πr2 - πr2
= 11πr2
Thus, height to which the water level will fall = 11 cm, which is the present height of water left in cylinder.
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