Question

A metallic sphere floats in an immiscible mixture of water \[\left( {{p_W} = {{10}^3}kg{m^{ - 3}}} \right)\] and a liquid \[\left( {{p_L} = 13.5 \times {{10}^3}} \right)\] with \[{\left( {\dfrac{1}{5}} \right)^{th}}\]portion by volume in the liquid and remaining in the water. The density of the metal is
A.\[4.5 \times {10^3}kg{m^{ - 3}}\]
B.\[4.0 \times {10^3}kg{m^{ - 3}}\]
C.\[3.5 \times {10^3}kg{m^{ - 3}}\]
D.\[1.9 \times {10^3}kg{m^{ - 3}}\]

Answer 1

Hint:As metallic sphere floats in mixture of water and liquid. The remaining was in liquid. Thus, the gravitational force is balanced by buoyant force. By equating the weight of a metallic sphere with the weight of two liquids. It gives the density of the metal.

Complete answer:
Buoyancy or upward force is the force that opposes the weight of a partially immersed object. It has the formula of \[F = \rho gV\] Where \[F\] is the buoyant force, \[\rho \] is the density of the metal, \[g\] is acceleration due to gravity, and \[V\] is the volume of sphere.
As the metallic sphere rests in the liquid mixture the gravitational force is balanced by buoyant force.
Weight of a metallic sphere is equal to the sum of the weight of liquid one and weight of liquid two.
\[mg = {m_1}g + {m_2}g\]
Density is represented by \[\rho \] and it is defined as the ratio of mass and volume. Thus, mass can be written in terms of density and volume.
\[V\rho g = \dfrac{4}{5}V \times {10^3}g + \dfrac{1}{5}V \times 13.5 \times {10^3}g\]
By simplifying the above value, the density of the metal is \[3.5 \times {10^3}kg{m^{ - 3}}\]

Note:
Gravitational force is given by the density of mass and acceleration due to gravity, when any object is placed in liquid and water. The buoyant force is balanced by the gravitational force. As the volume of the sphere is the same, it was cancelled out, and mass is taken as a product of density and volume.