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 flows out?
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The radius of the sphere is 3cm which could be solved.
Therefore volume of the sphere = (4/3)Pi*3^3 cm =113.0973 cm^3........(1) which is the spilled volume of water. (Remember the ARCHTMDE's principle by the sphere.)
The volume of the cone= (1/3) pi.(base radius)^2 * cones height = (1/3)(Pi)(6)^2*(8cm) = 301.5929 ...........(2)
Therefore (2) - ( 1) = water remaing in the cone after spilled by sphere = 301.5929-113.0973=188.4956cm^3 is the water spilled.
Radius of the sphere.
Let r be the radius of the sphere. Then the base of the circular plane of the cone and the slant side or surface of the cone are tangential to the sphere. Therefore, base radius of the cone and the slant length from the base to the touching point of the sphere are tangents to the sphere. So distance from the base of the cone to the touching point the slant length is 6cm, since the two tangents drawn from an exteral point to the circle(or sphere) are of equal length.The Entire slant length of the cone =(base radius^2+height^2)^(1/2) = (6^2+8^2)^(0.5) = 10cm. Therefore, the distance between the tip A, of the cone to the touching point T of the sphere =AT = 10-6 = 4cm.
Also, if the shphere is of radius r cm, and O is the center of the sphere in the cone as per the given discription, the distance between the tip, A of the cone and the centre of the sphere is (8-r)cm. Now,the Tip A of the cone , the centre O of the sphere and the touching point T of the sphere with slant surface, make a right angled triangle with right angle at T, AT being the tangent to the sphere. So,
AT^2 +OT^2= AO^2 or
4^2+r^2= (8-r)^2. Solving this , we get r=3cm
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