Let the area of cross-section of the solid block be A and its height be h. Its volume V will be Ah. Let the density of the solid block be .

It is immersed in a liquid of density . It is given that

> . The block is tied up to the container with a string. Free-body diagram of the block will be as shown below.

We will draw free-body diagram of the block in two situations: one when the container is at rest and two when the container is under upward acceleration a.

Let be the tension in the string. As the block is in equilibrium the net force in the vertical direction on it has to be zero. That is

is the weight of the block and is equal to . is the buoyant force on the block exerted by the liquid and is equal to the weight of the liquid displaced by the block. As the gauge pressure in a stationary liquid is , We, therefore, have the relation

In this situation the container, the liquid, and the block are under upward acceleration a.

As the liquid is under acceleration, the gauge pressure in it will vary with depth h as . Therefore, the buoyant force on the block in this situation will be

As the string remains stretched when the block is under upward acceleration, tension T in it can be found by applying the Newton’s second law of motion on the accelerated motion of the block. The equation of motion of the block in this situation is

In the above expression by substituting

we get