A metallic sphere of radius 21 cm is dropped into a cylindrical vessel, which is partly filled with water. The diameter of the vessel is 1.68m. If the sphere is completely submerged, find by how much the surface of water will rise?
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Not very sure of the answer.
volume of sphere = 4/3 π r^3
here r = 21 cm
radius of cylinder = 1.68 / 2 = 0.84 m = 840cm
volume of cylinder = π r^2 h
initial volume = π r^2 h1
final volume = π r^2 h2
increase volume = π r^2 (h2 - h1)
this increase volume will be equal to volume of sphere.
4/3 π (21)^3 = π (840)^2 (h2 - h1)
calculate value of h2 - h1
this should be = 0.0175 cm
volume of sphere = 4/3 π r^3
here r = 21 cm
radius of cylinder = 1.68 / 2 = 0.84 m = 840cm
volume of cylinder = π r^2 h
initial volume = π r^2 h1
final volume = π r^2 h2
increase volume = π r^2 (h2 - h1)
this increase volume will be equal to volume of sphere.
4/3 π (21)^3 = π (840)^2 (h2 - h1)
calculate value of h2 - h1
this should be = 0.0175 cm
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Hope this will help you simple and easy as it could be
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