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Question

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

A
g(1K)
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B
g(1+K)
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C
g(11K)
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D
g(1+1K)
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Solution

The correct option is D g(11K)
The net force acting on the stone in a downward direction would be equal to F=VσgVρg, where m is the mass of the stone, the force acting downwards would be equal to Vσg due to its weight and the force acting upwards would be equal to Vρg due to the buoyant force.
Hence, we get:
F=VσgVρg
=Vσg(1ρσ)
=mg(1ρσ)
=mg(11K)

Thus, a=g(11K)

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