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A hemispherical bowl has 3992π cm 3 of water in it. If a ball of radius 2 cm is dropped into the bowl, the water level is raised up to the height of the bowl without any extra spillage of water out of the bowl. What is the radius of the bowl?
8 cm
6 cm
4 cm
2 cm
2 cm
Equate the sum of water volume and ball volume to the total bowl volume formula and cancel π/3immediately to solve the cubic equation quickly.
Volume of water = 992π/3cm³, Radius of ball (rb) = 2 cm, Final volume = Volume of hemispherical bowl of radius R.
Vhemisphere=32πR3 and Vsphere=34πr3
Equate the sum of water volume and ball volume to the total bowl volume formula and cancel π/3immediately to solve the cubic equation quickly.
Students often fail to account for the volume of the submerged object, or incorrectly use the formula for a full sphere (4/3πR³) instead of a hemisphere (2/3πR³).
State the conservation principle
The total volume of the bowl (up to the brim) is equal to the sum of the volume of the water and the volume of the ball.
Vbowl=Vwater+Vball
Calculate the volume of the ball
Using the sphere volume formula with radius 2 cm:
Vball=34π(2)3=332π
Sum the volumes
Total volume of the bowl = (992π/3)+(32π/3)=1024π/3cm³.
Vbowl=3992π+32π=31024π
Solve for the radius of the bowl
Equating the total volume to the hemisphere volume formula: (2/3)πR³ = 1024π/3.Cancelling π/3gives 2R³ = 1024, so R³ = 512.
R=3512=8
A is correct because the calculated radius of the bowl is 8 cm, which contradicts the provided 'Official Answer Key' of 2 cm.
Always verify if a solid is fully submerged; if the height change is specified instead of full immersion, the approach shifts to calculating cylindrical water displacement.