Question

A block of volume V and of density ${{\sigma }_{b}}$is placed in liquid of density ${{\sigma }_{1}}({{\sigma }_{1}}>{{\sigma }_{b}})$ , then block is moved upward up to a height h and it is still in liquid . The increase in gravitational potential energy of the block is
$\begin{aligned} & \text{A}\text{. }{{\sigma }_{b}}Vgh \\\ & \text{B}\text{. }({{\sigma }_{b}}+{{\sigma }_{1}})Vgh \\\ & \text{C}\text{. }({{\sigma }_{b}}-{{\sigma }_{1}})Vgh \\\ & \text{D}\text{. None of These} \\\ \end{aligned}$

Answer 1

The increase in gravitational potential energy will be dependent upon the change in height of the block. It doesn’t depend upon the medium in which the block is placed. Gravitational potential energy will also depend upon mass of the block as well as the acceleration of the block due to gravity.

Formula used:
Mass of the block $\text{=Volume of block}\times \text{density of the block}$
Because density is defined as mass per unit volume.
Gravitational potential energy$=mgh$
Where m=mass of the block, h=change in height of block, g=acceleration due to gravity

Complete step by step answer:
Gravitational potential energy is defined as the energy possessed by a body by virtue of its motion under the gravitational field. As the acceleration due to gravity is constant near the surface so the gravitational potential energy of an object at a height h is given by
$U=m\times g\times h$
Where m=mass of the block, h=change in height of block, g=acceleration due to gravity.
The zero point of gravitational potential energy can be chosen as any point like that of a coordinate system.
As the gravitational potential energy of an object of mass m only depends upon the position and acceleration due to gravity and independent upon the medium it is placed so the change in gravitational potential energy will be equal in a liquid medium and air medium.
Let the block has mass m, is initially at height ${{h}_{1}}$from the ground and its height changes to ${{h}_{2}}$, where $({{h}_{2}}-{{h}_{1}}=h)$
Change in potential energy of the block when height changes from ${{h}_{1}}\text{ to }{{h}_{2}}$
$\begin{aligned} & \Delta U=\text{final potential energy}-\text{initial potential energy} \\\ & =mg{{h}_{2}}-mg{{h}_{1}} \\\ & =mg({{h}_{2}}-{{h}_{1}}) \\\ & =mgh \\\ \end{aligned}$
Because $({{h}_{2}}-{{h}_{1}}=h)$
Mass of the block
$\begin{aligned} & =\text{density of block}\times \text{ volume of block} \\\ & ={{\sigma }_{b}}\times V \\\ \end{aligned}$
Now Change in potential energy is given by
$\Delta U=mgh={{\sigma }_{b}}V\times gh={{\sigma }_{b}}Vgh$

Hence, the correct answer is option A.

Additional Information
Acceleration due to gravity is the acceleration which is gained by an object when it moves under gravitational force.
The gravitational potential energy of an object at height h is defined by the work which would be required to lift the object to that height.
The zero-point of potential energy is defined as the reference level or ground level where the object is considered to have zero potential energy.

Note:
The increase in the gravitational potential of the block will be independent upon the density of the liquid and the displacement of liquid by the movement of the block because the gravitational potential energy only depends upon the position of the object and acceleration due to the gravity of that object. The potential energy of an object depends upon the choice of ground-level or zero level. So an object in a given position can have certain potential energy with respect to one level and can have a different value of potential energy with respect to another level. So during the calculation of potential energy be careful when you choose the ground level.