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Question

A wooden block is floating on water kept in beaker. 40% of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to g/2. The block will then

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Solution

The correct option is **A** float with 40% above the water surface

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.

The buoyant force acting on an immersed body is equal to the weight of the fluid displaced by it, if the fluid is in rest. In this case, the fluid is accelerating upwards so the buoyant force must also provide the displaced fluid force to accelerate. Therefore, buoyant force will be fVρgeff where V = volume of body and f = fraction of volume of body immersed in fluid and geff=g+a=1.5g.

When fluid is at rest: f=1−0.4=0.6, so 0.6Vρg=Vρbg where ρb is the density of the block.

⇒ρb=0.6ρ.

In the second case: 1.5fVρg=Vρbg+Vρba=1.5Vρbg⇒fρ=ρb⇒f=0.6.

Thus, the fraction of immersed volume remains the same.

Body will float with 40% of the volume above water surface.

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