Question

A solid weighs $50\,gf$ in air (where $gf$ is gram force) and $44\,gf$ when completely immersed in water. Calculate: (Given Density of water=$1000\,Kg/{m^3}$)
(i) The Upthrust
(ii) The volume of the solid
(iii) The relative density of the solid.

Answer 1

Upthrust is the force exerted by the liquid when any object is immersed in the liquid. This upthrust force is also called buoyancy. Which is given by the difference between the apparent weight and the actual weight of the solid immersed. The volume could be obtained by the formula $W = {\text{ }}V\rho g$. And hence the relative density could be calculated

Complete step by step answer:
We have to find the Upthrust exerted by the liquid on the solid, the volume of the solid and its relative density. We have given that,
-Weight of the object in air which is its actual weight, ${W_1} = {\text{ }}50gf$
-Weight of the object when it is immersed in liquid, which is its apparent weight ${W_2} = {\text{ }}44gf$
-Density of water = $1000\,Kg/{m^3}$

(i) Upthrust is given by the difference between the actual weight and apparent weight
$Upthrust{\text{ (U)}} = {\text{ }}{W_1}{\text{ }}-{\text{ }}{W_2}$
Putting the respective values we get,
$U = {\text{ }}50gf{\text{ }}-{\text{ }}44gf$
$\Rightarrow U\, = {\text{ }}6gf$
For conversion in S.I Unit. Considering the value of acceleration due to gravitation$g{\text{ }} = {\text{ }}10{\text{ }}m{s^{ - 2}}$ and converting from gram to kilogram we have
$6gf = \dfrac{{6 \times 10}}{{1000}}N = 0.06N$

Hence $Upthrust{\text{ (U)}} = {\text{ }}0.06{\text{ }}N$.

(ii) Now when the solid is immersed in water buoyancy/ Upthrust $U$ is given by (Density of water is$1000\,Kg/{m^3}$)
$U{\text{ }} = {\text{ }}V\rho g$
Putting the values of$U,\,\rho \,\,\& \,g$, we get:

\Rightarrow V\,{\text{ = }}\,\dfrac{{{\text{0}}{\text{.06}}}}{{1000 \times 10}}{m^3} \\\ \Rightarrow V\, = \,6\, \times {10^{ - 6}}{m^3} \\\ \therefore V\, = \,6\,c{m^3} \\\ $$ **Hence the volume of the solid is $$V\, = \,6\,c{m^3}$$.** (iii) Relative density is the ratio of the density of the solid to the ratio of the liquid in which it is immersed and is given by $$\text{Relative Density (R.D.)} = \,\dfrac{{{W_1}}}{{{W_1} - {W_2}}}$$ Where the symbols have the meaning stated above Now on substituting the values we get $$R.D.\, = \,\dfrac{{50gf}}{{50gf - 44gf}}$$ On simplifying we get $$R.D.\, = \,\dfrac{{50gf}}{{6gf}}$$ $$\therefore R.D.\, = \,8.333$$ **Hence the relative density of the solid is 8.333.** **Note:** Density of the solid and relative density of the solid are two different things. Density determines the mass to volume ratio whereas relative density gives the comparative density with respect to the liquid in which the substance is immersed. The result must be obtained in the S.I. Unit for precise.