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:

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 $g (1 - \frac{1}{K})$
The net force acting on the stone in a downward direction would be equal to $F = V σ g - V ρ g$ , where $m$ is the mass of the stone, the force acting downwards would be equal to $V σ g$ due to its weight and the force acting upwards would be equal to $V ρ g$ due to the buoyant force.
Hence, we get:
$F = V σ g - V ρ g$
$= V σ g (1 - \frac{ρ}{σ})$
$= m g (1 - \frac{ρ}{σ})$
$= m g (1 - \frac{1}{K})$

Thus, $a = g (1 - \frac{1}{K})$

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

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: