Question

If the error in measuring the radius of the sphere is$\;2\%$and that in measuring its mass is$\;3\% ,$ then the error in measuring the density of the sphere is

$$A.\;\;\;\;\;5\% \\\ B.\;\;\;\;\;7\% \\\ C.\;\;\;\;\;9\% \\\ D.\,\,\,\,\,11\% \\\$$

Answer 1

Hint The data of the physical quantities are already given in terms of percentage errors. For division of physical quantities that is the ratio of mass to volume, the total percentage error is the sum of the individual percentage errors.

Complete step-by-step answer
Density of the sphere is given by
$\rho = \dfrac{M}{V}$
Where, M is the mass and V is the volume.
To find the error in density, it is to find the error in division of quantities. This is given by,
$x = \dfrac{a}{b} \\\ \% error(x) = \% error(a) + \% error(b) \\\ x = \dfrac{{{a^n}}}{{{b^n}}} \\\ \% x = n \times \% a + n \times \% b \\\$
It is given that Percentage error in mass is 3% and radius is 2%.
Volume of the sphere is $\dfrac{4}{3}\pi {r^3}$
$\Rightarrow \% V = 3 \times \% r \\\ \Rightarrow \% V = 3 \times 2 \\\ \Rightarrow \% V = 6\% \\\$
The constants don’t take part in error calculations.
$\% \rho = \% M + \% V \\\ = 3 + 6 \\\ = 9\% \\\$

Hence, the error in measuring the density of the sphere is 9% and the correct option is C.

Note: When the values of the physical quantities are given in terms of absolute errors, first find the relative error,
$x = \dfrac{a}{{{b^n}}} \\\ \dfrac{{\Delta x}}{x} = \dfrac{{\Delta a}}{a} + n \times \dfrac{{\Delta b}}{b} \\\$
Now multiply by 100 to get the percentage error.