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A small sphere of density 0.5 gm/cm3 is brought to a depth of 5m below the surface of water in a lake and released. Find the height to which the sphere will rise above the free surface of water. Ignore viscous effect of water and take g=10m/s2

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Detailed Solution

After the sphere is released, its acceleration is a=ρwvg−ρsvgρsv=ρwρs−1g=10.5−1×10m/s2=10m/s2At the surface, velocity of sphere, v=2.a.h=2×10×5m/s=10m/sHeight above the surface of water =v22g=1022×10m=5m

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