An iron sphere of radius 'a' units is immersed completely in water contained in aright circular cone of semi vertical angle 30 degree ,water is drained off from the cone till it's surface toches the sphere.Find the volume of water remaining in cone.

pls help me frnds

Here is the answer to your question.
Let R and H be the radius and height of the cone.
In ∆BOC,


  • 43

find the surface area of a cube whose edge=15 cm?

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give me answer

  • -4

 height of cone = h

sin 30 = a/h

1/2 = a / h

h = 2a

vol = 1/3 pi a2 x 2a

a frustum will be formed when immersed

r of frustum be a and radius of sphere

height of frustum of diameter of sphere

rest u do by yourself

  • -5

SA of cube = 6 x 15 x 15 = 1350 cm2

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raghu here how sin 30 = a/h

sin 30 = a / l where l is the slant height on the point of contact of sphere and cone 

  • -3

no buddy.....l is tangent so it the angle it subtends = 90

height is the opp side that is hypotenuse...hence i used sin 30 ....can u understand now.....

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okay thanks now i got my mistake

how  height of frustum is diameter of sphere  where sphere is not touching top of cone

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