A streamlined body falls through air from a height h on the surface of a liquid. If d and D(D>d) represent the densities of the material of the body and liquid respectively, then the time after which the body will be instantaneously at rest, is:

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2hg

2hg⋅Dd

2hg⋅dD

2hg⋅(dD−d)

answer is D.

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Detailed Solution

Up thrust – Weight of body = apparent weight VDg−Vdg=Vda Where a= retardation of body=(D−dd)g The velocity gained after fall from height h in air, v=2gh Hence, time taken to come to restt=va=2ghd(D−d)g=2hg×d(D−d)

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