A stream-lined body falls through air from a height h on the surface of a liquid. Let d and D denote the densities of the material of the body and the liquid respectively. If D > d, then the time after which the body will be instantaneously at rest is:
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a
2hg
b
2h Dg d
c
2h dg D
d
2hg(dD-d)
answer is D.
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Detailed Solution
Velocity u of the body when it enters the liquid is given by mgh = 12mu2 or u = 2ghLet Volume of the body = V Mass of the body = Vd Weight of the body = Vdg Mass of liquid displaced = VD Net upward force = VDg-VDg = Vg (D-d) Retardation = net weightmass = V(D-d)gVd = (D-dd)gAcceleration a = -(D-dd)gFinal velocity, v in the liquid when the body is instantaneosuly at rest is zero. Let the time taken be t. v = u+at 0 = 2gh-(D-dd)gt.(D-dd)gt = 2gh t = [dD-d]2hg