Question

A body of mass $70\,{\text{kgwt}}$, when completely immersed in water, displaces ${\text{20000}}\,{\text{c}}{{\text{m}}^3}$ of water. Find the relative density of material in the body.
A. 7
B. 3.5
C. 35
D. 70
E. 0.35

Answer 1

Use the formula for the density of an object and the relative density of the object. These formulae give the relation between mass, density, volume and relative density of the material of the object.

Formulae used:
The expression for the density of an object is
$\rho = \dfrac{M}{V}$ …… (1)
Here, $\rho$ is the density of the object, $M$ is the mass of the object and $V$ is the volume of the object.
The expression for the relative density of an object is
${\text{Relative density}} = \dfrac{\rho }{{{\rho _{medium}}}}$ …… (2)
Here, $\rho$ is the density of the object and ${\rho _{medium}}$ is the density of the medium in which the object is placed.

Complete step by step answer:
The mass of the body is $70\,{\text{kgwt}}$ and volume of the water displaced by the body is ${\text{20000}}\,{\text{c}}{{\text{m}}^3}$.
Calculate the density of the material of the body.
The mass $70\,{\text{kgwt}}$ of the body can also be written as $70\,{\text{N}}$ which is the weight of the body.
$1\,{\text{kg}}$ of an object is equal to the force $10\,{\text{N}}$ exerted by the Earth on the object.
Convert the unit of the mass $M$ of the body in grams.
$M = \left( {70\,{\text{N}}} \right)\left( {\dfrac{{1\,{\text{kg}}}}{{10\,{\text{N}}}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{g}}}}{{1\,{\text{kg}}}}} \right)$
$\Rightarrow M = 7000\,{\text{g}}$
Hence, the mass of the body is $7000\,{\text{g}}$.
According to Archimedes’ principle, the volume of the fluid displaced by an object floating on any fluid is equal to the volume of the object immersed in the fluid.
The volume of the water displaced by the body is ${\text{20000}}\,{\text{c}}{{\text{m}}^3}$. Hence, the volume of the body is also${\text{20000}}\,{\text{c}}{{\text{m}}^3}$.
$V = {\text{20000}}\,{\text{c}}{{\text{m}}^3}$
Substitute $7000\,{\text{g}}$ for $M$ and ${\text{20000}}\,{\text{c}}{{\text{m}}^3}$ for $V$ in equation (1).
$\rho = \dfrac{{7000\,{\text{g}}}}{{{\text{20000}}\,{\text{c}}{{\text{m}}^3}}}$
$\Rightarrow \rho = 0.35\,{\text{g/c}}{{\text{m}}^3}$
Hence, the density of the material of the body is $0.35\,{\text{g/c}}{{\text{m}}^3}$.
Calculate relative density of the material of the body.
The density of the water is $1\,{\text{g/c}}{{\text{m}}^3}$.
Rewrite equation (2) for the relative density of the material of the body.
${\text{Relative density}} = \dfrac{\rho }{{{\rho _{water}}}}$
Here, ${\rho _{water}}$ is the density of the water.
Substitute for $\rho$ and $1\,{\text{g/c}}{{\text{m}}^3}$ for ${\rho _{water}}$ in the above equation.
${\text{Relative density}} = \dfrac{{0.35\,{\text{g/c}}{{\text{m}}^3}}}{{1\,{\text{g/c}}{{\text{m}}^3}}}$
$\Rightarrow {\text{Relative density}} = 0.35$
Therefore, the relative density of the material of the body is $0.35$.
Hence, the correct option is E.

Note: $1\,{\text{kg}}$ of an object is equal to the weight $9.8\,{\text{kgwt}}$ of the object. But in the solution it is taken $10\,{\text{kgwt}}$ for the sake of calculation and also remember that relative density of a substance is the ratio of density of a substance to the density of water and do not confuse it with the density.