A stone of relative density k is released from rest on the surface of a lake. If viscous effects are ignored, the stone sinks in water with an acceleration of

g (1 - k)

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g (1 + k)

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g (1 - \frac{1}{k})

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g (1 + \frac{1}{k})

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Solution

The correct option is D $g (1 - \frac{1}{k})$
let the mass of the is stone is $M$ and volume $V$
$k = \frac{ρ_{s}}{ρ_{w}}$ $∵$ given
net force acting on the stone $F_{n e t} = M g - F_{B}$
where $F_{B} =$ weight of the water displaced $= V ρ_{w} g = \frac{V ρ_{s} g}{k} = \frac{M g}{k}$
$\Rightarrow F_{n e t} = M g - \frac{M g}{k} = M g (1 - \frac{1}{k})$
$\Rightarrow a = \frac{F_{n e t}}{M} = \frac{M g}{M} (1 - \frac{1}{k}) = g (1 - \frac{1}{k})$

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A stone of relative density k is released from rest on the surface of a lake. If viscous effects are ignored, the stone sinks in water with an acceleration of

The correct option is D g(1−1k)let the mass of the is stone is M and volume Vk=ρsρw ∵ givennet force acting on the stone Fnet=Mg−FBwhere FB= weight of the water displaced =Vρwg=Vρsgk=Mgk⇒Fnet=Mg−Mgk=Mg(1−1k)⇒a=FnetM=MgM(1−1k)=g(1−1k)