Question

A ball of mass $m$ and density $\rho$ is immersed in a liquid of density $3\rho$ at a depth of $h$ and is released. To what height will the ball jump up above the surface of the liquid?
(Neglect the resistance of water and air)

Answer 1

Density is defined as the relation between the mass and volume of a substance. It varies directly with the mass but inversely with the volume of the substance. If the mass of an object or liquid is constant, the volume increases, and the density will be small.

Complete step by step answer:
Step I:
Recall that buoyancy is the upward force exerted by a fluid and it opposes the weight of the object immersed in it. Since the ball is immersed in a liquid so it will have an upward buoyant force. Also, the Archimedes principle is used to calculate the buoyant force.
Step II:
As per Archimedes Principle, the formula is
${F_{buoyant}} = \rho Vg$------------(i)
Where $\rho$ is the density
$V$ is the volume and
$g$ is the acceleration due to gravity
For ball the buoyant force will be
${F_{buoyant}} = \dfrac{m}{\rho }.3\rho .g$-----------(ii)
Where $3\rho$ is the density given, $\dfrac{m}{\rho }$ is the volume.
Step III:
Acceleration of the ball inside the liquid is given by
$a = \dfrac{{{F_{net}}}}{m}$-----------(iii)
Net force on the ball will be
$\Rightarrow {F_{net}} = {F_{buoyant}} - Weight$
$\Rightarrow a = \dfrac{{{F_{buoyant}} - Weight}}{m}$
Also weight of the ball is $W = mg$
Step IV:
Substitute all the values in equation (iii) and solve,
$\Rightarrow a = \dfrac{{\dfrac{m}{\rho }(3\rho )(g) - mg}}{m}$
$\Rightarrow a = \dfrac{{3mg - mg}}{m}$
$\Rightarrow a = 2g$
Step V:
Any object held above the ground level will have a potential energy and since the ball is made to jump so its velocity can be calculated by using the formula
$\Rightarrow v = \sqrt {2gh}$----------(iv)
Here $g$ is acceleration due to gravity.
$\Rightarrow v = \sqrt {2ah} = \sqrt {2 \times 2g \times h} = \sqrt {4gh}$
Step VI:
Let $H$ be the height to which a ball jumps. So the formula for height can be known using the equation.
$\Rightarrow H = \dfrac{{{v^2}}}{{2g}}$
Substituting the values and solving for h,
$\Rightarrow H = \dfrac{{4gh}}{{2g}}$
$\Rightarrow H = 2h$

Therefore, the ball will jump to the height $2h$.

Note:
It is to be remembered that the mass and density of the object immersed in the liquid will not affect the buoyancy. The density of the object is of great importance as it helps in determining whether the object will float or sink. If the density of the object is less than that of water then it will float. If the density of the object is more than that of water then it will sink.