Question

Two bodies are in equilibrium when suspended in water from arms of a balance. The mass of one body is $36\,g$ and its density is $9\,g/cc$ . If the mass of the other is $48\,g$ , its density in $g/cc$ is
A. $4$
B. $\dfrac{3}{2}$
C. $3$
D. $5$

Answer 1

Hint:In this question, we will first calculate the volume of the body for which both the mass and density is given using the relation $\rho = \dfrac{m}{V}$ where m is the mass of the object, $\rho $ is the density of the object and V is the volume occupied by the object. Then we shall use the concept of Archimedes principle and calculate the apparent weight for both the objects. It is mentioned that the bodies are in equilibrium. So, we need to balance the apparent weights of both the bodies and calculate the volume of the other body. Finally, we shall calculate the density of the other object again using the density relation $\rho = \dfrac{m}{V}$.

Complete step by step answer:
The density and mass of an object are related as $\rho = \dfrac{m}{V}$ where m is the mass of the object, $\rho $ is the density of the object and V is the volume occupied by the object.
This can be rewritten as $V = \dfrac{m}{\rho }\,\,\,\,\,\,\,\,\,\,\,\,.......(1)$

For the body A:
The mass is given by ${m_A} = 36\,g$
The density is given as ${\rho _A} = 9\,g/cc$
The volume occupied by the object will be given as ${V_A} = \dfrac{{{m_A}}}{{{\rho _A}}}$
Substituting the appropriate values, we get
${V_A} = \dfrac{{36}}{9}$
$ \Rightarrow {V_A} = 4\,cc$

For the body B:
The mass is given by ${m_B} = 48\,g$. The density is given as ${\rho _B}$.The volume occupied by the object will be given as,
${V_B} = \dfrac{{{m_B}}}{{{\rho _B}}}$
Archimedes' principle states that when a body is submerged in a fluid it experiences a buoyant force which is equal to the volume it displaces in the fluid.
The apparent weight of body A is equal to ${W_A} = {m_A}g - {V_A}{\rho _{water}}g$
The apparent weight of body B is equal to ${W_B} = {m_B}g - {V_B}{\rho _{water}}g$
It is given in the question that the two bodies are in equilibrium. This means that the apparent weight must nullify each other.
Hence, we can say that,
${W_A} = {W_B}$
This can be rewritten as
${m_A}g - {V_A}{\rho _{water}}g = {m_B}g - {V_B}{\rho _{water}}g$

Cancelling out the common terms we have,
${m_A} - {V_A}{\rho _{water}} = {m_B} - {V_B}{\rho _{water}}$
Substituting the known values we have,
$36 - 4 \times 1 = 48 - {V_B}$
$ \Rightarrow {V_B} = 16\,cc$
Now using relation (1), we can calculate the density of the body B.
The mass is given by ${m_B} = 48\,g$
The density is given as ${\rho _B}$
The volume occupied by the object will be given as ${V_B} = \dfrac{{{m_B}}}{{{\rho _B}}}$
This can be written as $16 = \dfrac{{48}}{{{\rho _B}}}$
$ \therefore {\rho _B} = 3g/cc$

Therefore, option C is the correct answer.

Note:Mass and weight are synonymous terms and often are misinterpreted by the students. Mass is the measure of the matter contained in an object. It is fixed and is a fundamental property. However, weight is a force that is experienced by an object by the virtue of its mass. It can be changed under different experimental conditions and so is not fixed.