Question

Find the density of a sphere the percentage error in mass is $0.26\% $ and the percentage error in measuring radius is $0.38\% $, then find the percentage error?

Answer 1

Hint: In this question, the data of the physical quantities are given in terms of percentage errors. For division of physical quantities, which in this case is the ratio of mass to volume, the percentage error is defined as the sum of the individual percentage errors of mass and volume.

Complete step by step solution:
The density of the sphere is given by the following formula,
$\rho = \dfrac{m}{v}$
So, we need to have the values of both mass and volume, in order to calculate the density. The volume of the sphere is given by the expression,
$V = \dfrac{4}{3}\pi {r^3}$
To find the error in density, we need to find the error in the division of mass and volume. This is given by,
$x = \dfrac{a}{b}$
$\% error(x) = \% error(a) + \% error(b)$
$x = \dfrac{{{a^n}}}{{{b^n}}}$
$\% error(x) = n \times \% error(a) + n \times \% error(b)$
It is given that the percentage error in the mass of the sphere is $0.26\% $ and the percentage error in measuring of radius of the sphere is $0.38\% $.
We know that the volume of a sphere is given by the expression,
$V = \dfrac{4}{3}\pi {r^3}$
We also know that the constants do not take part in error calculation. So,
$\% error(V) = 3 \times \% error(R)$
$\% error(V) = 3 \times 0.38$
$\% error(V) = 1.14\% $
Now, we will calculate the percentage error in calculating the density,
$\% error(\rho ) = \% error(m) + \% error(v)$
$\% error(\rho ) = 1.14 + 0.26$
$\% error(\rho ) = 1.4\% $
So, the percentage error in calculating the density is $\% error(\rho ) = 1.4\% $.

Additional information: Error is defined as the difference between the actual value and the calculated value in case of any physical quantity. Broadly, there are three types of errors, random errors, blunders, and systematic errors. This error occurs due to the non-calibrated equipment, the observer's careless behaviour and also due to some physical conditions. If we come to know the origin of the errors, we can easily reduce the errors by rectifying those problems.

Note:
When the values of the various physical quantities are given as absolute errors and not in terms of percentage, first find the relative error,
$x = \dfrac{a}{{{b^n}}}$
$\dfrac{{\Delta x}}{x} = \dfrac{{\Delta a}}{a} + n\dfrac{{\Delta b}}{b}$
Now we will multiply by 100 to get the percentage error.