Question

The more is the density of liquid than a solid, the more the volume of a solid that will submerge in it. State whether true or false.
A) True
B) False

Answer 1

Hint: In this solution, we will use the concepts of buoyant force acting on the body. To hold the object in a medium, the buoyant force pushes the body outside the liquid while the weight of the object will push the object downwards.

Formula used: In this solution, we will use the following formula
-Buoyant force $ {F_B} = \rho Vg $ where $ \rho $ is the density of the plastic body, $ V $ is the volume of the object, and $ g $ is the gravitational acceleration .

Complete step by step answer
When the solid is to be held, the weight of the body must be less than the buoyant force on the object otherwise the object will sink. This can be implied to say that if the weight of the body is more than the buoyant force, it will sink
So, $ W > {F_B} $
The weight of the object $ W = mg = {\rho _{object}}{V_1}g $ where $ {V_1} $ is the object’s total volume has to be more than the buoyant force $ {F_B} = {\rho _{liquid}}{V_2}g $ where $ {V_2} $ is the volume of the object immersed in the object.
Now, for the object to be held stable, it must have sufficiently high density so that it doesn’t sink in the water which implies $ {\rho _{solid}} > {\rho _{liquid}} $ (As the object will have submerged completely, $ {V_1} = {V_2} $ ).
Hence if the density of the liquid increases, the buoyant force acting on the object increases according to the relation $ {F_B} = {\rho _{liquid}}{V_2}g $ . Hence the object will submerge less since the buoyant force will be higher and push the object outside it.
Hence the statement is false and option (B) is correct.

Note
The amount of object that would be immersed in the liquid is decided based on its density. In a case where the density of the liquid increases after the system has achieved equilibrium, the increase in density would push the solid upwards and less amount of it would be submerged in the liquid.