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

Options

A. `g(1-K)`
B. `g(1+K)`
C. `g(1-(1)/(K))`
D. `g(1+(1)/(K))`

Answer

C

Explanation

The net force acting on the stone in a downward direction would be equal to `F = v sigmag-vrhog,` where m is the mass of the stone, the force acting downwards would be equal to `Vsigmag` due to its weight and the force acting upwards would be equal to `Vrhog` due to the buoyant force.
`"Hence, we get "=F=Vsigma g- Vrhog=Vsigmag(1-rho/sigma)`
`=mg(1-rho/sigma)=mg(1-1/K)`
`"Thus," "alpha=g(1-1/K)`