What, roughly, is the percent uncertainty in the volume of a spherical beach ball of radius \({\bf{r = 0}}{\bf{.84 \pm 0}}{\bf{.04 m}}\)?

Volume is defined as the amount of space that a particular object or body occupies in the space.

The radius of the spherical beach ball, \(r = 0.84 \pm 0.0{\rm{4 m}}\).

The formula of the volume of a sphere is

\(V = \frac{4}{3}\pi {r^3}\) … (i)

From equation (i), the volume is

\(V = \frac{4}{3}\pi {\left( {0.84} \right)^3}\).

Thus, the final value of the multiplication is 0.6%.

Differentiating equation (i) w.r.t to radius,

\(\frac{{{\rm{d}}V}}{{{\rm{d}}r}} = \frac{4}{3}\pi {\rm{3}}{r^2}{\rm{d}}r\).

The formula of the small change in the volume of a sphere is

\({\rm{d}}V = 4\pi {r^2}{\rm{d}}r\). … (ii)

From equation (ii), the small change in the volume is

\({\rm{d}}V = 4\pi {\left( {0.84} \right)^2}\left( {0.04} \right)\).

The formula of percent uncertainty is

\(V' = \frac{{{\rm{d}}V}}{V} \times 100\% \) . … (iii)

From equation (iii), the value of percent uncertainty is

\(\begin{aligned}{c}V' = \frac{{4\pi {{\left( {0.84} \right)}^2}\left( {0.04} \right)}}{{\frac{4}{3}{\rm{\pi }}{{\left( {0.84} \right)}^3}}} \times 100\% \\ \approx 10\% \end{aligned}\).

Thus, the final value is 10%.