Question

The relative density of a material is found by weighing the body first in air then water. If the weight in air is $\left( 10.0\pm 0.1 \right)\text{ gf}$ and the weight in water is $\left( 5.0\pm 0.1 \right)\text{ gf}$ then the maximum permissible percentage error in the relative density is
a) 1
b) 2
c) 3
d) 5

Answer 1

Hint: The error in a given physical quantity is the sum of all the errors in the individual quantities relating to the derived physical quantity. Hence we can find the percentage error in any quantity by using the expression for relative error in the given physical quantity.

Complete step by step answer:
When a physical quantity is written in this format i.e. (X$\pm $ Y) unit
X represents the actual value of the quantity,
Y represents the possible or mean or absolute error in the physical quantity.
Before we begin with the solution to the problem, let us understand relative density of a substance.
$\text{Relative density=}\dfrac{\text{density of substance}}{\text{density of water}}\text{, at }{{4}^{\centerdot }}C$
If the mass of the of the substance and the water are contained in equal volumes then the relative density can be written as R= $\dfrac{m}{M}$ let us denote this as equation 1
Where m is the mass of the substance in the given volume and M is the mass of water contained in the same volume.
Multiplying equation 1 by g i.e. acceleration due to gravity
Relative density= $\dfrac{mg}{Mg}$
\[\text{Relative density=}\dfrac{\text{Weight of the substance}}{\text{Weight of water}}\]
If we dip the substance in water the volume of the displaced water will be equal to volume of the substance. From Archimedes principle we know the weight of the displaced liquid is equal to the thrust generated on the substance. Since the above equation is for a given volume of both water and the substance, hence we can write the expression for relative density as,
\[\text{Relative density=}\dfrac{\text{Weight of the substance}}{\text{Up thrust }\!\!~\!\!\text{ generated when the substance is immersed in water}}\text{ }\!\!~\!\!\text{ }\]
By definition,
\[~\text{Upthrust }\!\!~\!\!\text{ =weight of substance in air - weight of substance in water}\]
Let us denote,
Weight of the substance in air as ${{W}_{A}}$
Weight of the substance in water as ${{W}_{X}}$
Hence relative density i.e. ${{R}_{D}}$ of a substance becomes,
${{R}_{D}}=\dfrac{{{W}_{A}}}{{{W}_{A}}-{{W}_{X}}}$
Let us denote ${{W}_{A}}-{{W}_{X}}$ as w.
Percentage error in relative density is given by
$\dfrac{\Delta {{R}_{D}}}{{{R}_{D}}}\times 100=\left( \dfrac{\Delta {{W}_{A}}}{{{W}_{A}}}\times +\dfrac{\Delta w}{w} \right)100$
In the above expression,
$\dfrac{\Delta {{R}_{D}}}{{{R}_{D}}}$ is the relative error in ${{R}_{D}}$
$\dfrac{\Delta {{W}_{A}}}{{{W}_{A}}}$ is the relative error in${{W}_{A}}$
$\dfrac{\Delta w}{w}$ is the relative error in $w$
It is to be noted that $\dfrac{\Delta {{R}_{D}}}{{{R}_{D}}}\times 100$ is the percentage error in relative density.
Using the expression of percentage error,
$\dfrac{\Delta {{R}_{D}}}{{{R}_{D}}}\times 100=\left( \dfrac{\Delta {{W}_{A}}}{{{W}_{A}}}\times +\dfrac{\Delta w}{w} \right)100$
Since $\Delta {{W}_{A}}=0.1$ and $\Delta w=\Delta {{W}_{A}}+{{W}_{X}}=0.1+0.1=0.2$
$\dfrac{\Delta {{R}_{D}}}{{{R}_{D}}}\times 100=\left( \dfrac{0.1}{10}+\dfrac{0.2}{5} \right)100$
Multiplying and dividing two in the second term inside the bracket,
$\dfrac{\Delta {{R}_{D}}}{{{R}_{D}}}\times 100=\left( \dfrac{0.1}{10}+\dfrac{0.4}{10} \right)100$
Taking the denominator common,
$\dfrac{\Delta {{R}_{D}}}{{{R}_{D}}}\times 100=\left( \dfrac{0.5}{10} \right)100$
$\dfrac{\Delta {{R}_{D}}}{{{R}_{D}}}\times 100=5$

Note:
The above quantity does not have units as it is the ratio of the same physical quantity. If there exists a physical quantity in the derived quantity whose percentage error has to be determined, has a power to it then the power gets multiplied to the relative error of the quantity and not added to it.