Question

Suppose a bodyweight$200gm$ in the air $180gm$ in liquid and $175gm$in water. Calculate the density of the material of the body-
A. $8000\dfrac{{kg}}{{{m^3}}}$
B. $800\dfrac{{kg}}{{{m^3}}}$
C. $8\dfrac{{kg}}{{{m^3}}}$
D. none

Answer 1

In order to answer this question we have to know about the Archimedes principle. According to this principle, apparent loss of weight is equal to the buoyant force and weight of the liquid. By using this principle first we will find the total volume of the liquid and then we will find its density.

Complete step by step solution:
According to the question given the body weight in the air$200gm$, i.e. $W \Rightarrow \dfrac{{200}}{{1000}} \times g \Rightarrow 0.2g$
Again the body weights in water$175gm$ i.e. $W' \Rightarrow \dfrac{{175}}{{1000}} \times g \Rightarrow 0.175g$
Now putting the above value on Archimedes principle, we get –

$$W - W' = {\rho _{liquid}} \times {V_{solid}} \times g \\\ \Rightarrow 0.2g - 0.175g = 1000 \times {V_{solid}} \times g \\\ \Rightarrow 0.025g = 1000 \times Vsolid \times g \\\ \Rightarrow 0.025 = 1000{V_{solid}} \\\ \Rightarrow {V_{solid}} = \dfrac{{0.025}}{{1000}} \\\$$

Hence we get the volume of the solid as${V_{solid}} = \dfrac{{0.025}}{{1000}}$, now as we know that the density of a material is given by the formula as $D = \dfrac{M}{V}$
Therefore the density of the material of the body will be

$$D = \dfrac{M}{V} \\\ \Rightarrow D = \dfrac{{200 \times 1000}}{{0.025 \times 1000}} \\\ \Rightarrow D = 8000 \\\$$

So, option A) $8000\dfrac{{kg}}{{{m^3}}}$ is correct.
Note:
The buoyant force on a body drifting in a fluid or gas is equivalent in magnitude to the weight of the floating object and is inverse of course; the body neither rises nor sinks. Some applications of Archimedes' principle are submarines, hot air balloons and hydrometers.