A conical vessel of radius 6cm and height 8cm is completely filled with water.A sphere is lowered into the water and its size is such that when it touches the sides it is just immersed. What fraction of water overflows?
Answers
Hello mate,
Here is your answer,
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Radius (R) of conical vessel = 6 cm
Height (H) of conical vessel = 8 cm
volume of conical vessel (Vc)
= 1/3πr^2h
= 1/3π × 36 × 8
= 96π
Let the radius of the sphere be r cm.
In right ΔPO'R, by Pythagoras theorem:
L = √(64 + 36)
L = 10 cm
Hence sin@ = O'P / PR = 6/10 = 3/5
In right triangle MRO
Sin@ = OM /OR = r / OR
3/5 = r / (8 - r)
⇒ 24 – 3r = 5r
⇒ 8r = 24
⇒ r = 3 cm
∴ Volume of sphere (Vs)
Now,
Volume of the water = Volume of cone (Vc) = 96 π cm3
Clearly, Volume of the water that flows out of cone is same as the volume of the sphere i.e. Vs.
∴ Fraction of the water that flows out
Vs/Vc = 36π / 96π = 3/8
Vs : Vc = 3 : 8
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HOPE THIS HELPS YOUU :)
AND STAY BLESSED.
Radius (R) of conical vessel = 6 cm
Height (H) of conical vessel = 8 cm
volume of conical vessel (Vc)
= 1/3πr^2h
= 1/3π × 36 × 8
= 96π
Let the radius of the sphere be r cm.
In right ΔPO'R, by Pythagoras theorem:
L = √(64 + 36)
L = 10 cm
Hence sin@ = O'P / PR = 6/10 = 3/5
In right triangle MRO
Sin@ = OM /OR = r / OR
3/5 = r / (8 - r)
⇒ 24 – 3r = 5r
⇒ 8r = 24
⇒ r = 3 cm
∴ Volume of sphere (Vs)
Now,
Volume of the water = Volume of cone (Vc) = 96 π cm3
Clearly, Volume of the water that flows out of cone is same as the volume of the sphere i.e. Vs.
∴ Fraction of the water that flows out
Vs/Vc = 36π / 96π = 3/8
Vs : Vc = 3 : 8
