Question

The relative density of a material of a body is found by weighing it first in air and then in water. If the weight of the body in air is ${W_1} = 8.00 \pm 0.05\,N$ and the weight in water is ${W_2} = 6.00 \pm 0.05\,N$, then the relative density ${\rho _r} = \dfrac{{{W_1}}} {{{W_1} - {W_2}}}$ with the maximum permissible error is?
A) $4.00 \pm 0.62\%$

B) $4.00 \pm 0.82\%$

C) $4.00 + 3.2\%$

D) $4.00 \pm 5.62\%$

Answer 1

Here we have to use the given formula of relative density and multiply it by hundred to get the maximum permissible error.

Complete step by step solution:
Given,
Weight of the body in air, ${W_1} = 8.00 \pm 0.05\,N$

Weight of the body in air, ${W_2} = 6.00 \pm 0.05\,N$

Hence, ${W_1} - {W_2} = 2.00 \pm 0.1\,N$

Now, relative density
${\rho _r} = \dfrac{{{W_1}}} {{{W_1} - {W_2}}} \\\ = \dfrac{{8.00 \pm 0.05}} {{2.00 \pm 0.1}} \\\$

So, the maximum permissible error is:
$= \dfrac{8} {2} \pm \left( {\dfrac{{0.05}} {8} + \dfrac{{0.1}} {2}} \right) \times 100 \\\ = 4 \pm \left( {0.62 + 5} \right)\% \\\ = 4.00 \pm 5.62\% \\\$

Hence, option D is correct.

Additional information:
The ratio of the density of the product to the density of the reference material is known as relative density or specific gravity. When the water is at its densest then the relative density of liquids is almost always determined with respect to water.

If the relative density of a material is less than one, it is less dense than the reference, if it is greater than one than it is denser than the reference. If the relative density is exactly one then the densities are equal, i.e. equal concentrations of the two substances of the same mass. If the reference medium is water, a substance with a relative density of less than one can float in water. A material with a relative density greater than one is going to sink.

The density of the water varies by temperature and pressure, as does the density of the sample. It is also important to define the temperatures and pressures at which the densities of weights have been calculated. It is almost always the case that observations are made at $1$ nominal atmosphere.

Note: Here we have to pay attention while calculating the maximum permissible error as a change in value may result in a wrong answer. Also we have to multiply by hundred since it is given to find the error.