Question

a hollow copper ball has an external diameter of 12 cm and a thickness of 0.1 cm
find
outer surface area of the ball
the weight of the ball if 1cm*3 of copper weighs 8.8g(take pie to be 3.14)

Answer 1

$$\begin{gathered} \mathrm{Expternal}\mathrm{radius}=\frac{12}{2}=6\mathrm{cm} \\ \mathrm{So}\mathrm{internal}\mathrm{radius}=6-0.1\mathrm{cm}=5.9\mathrm{cm} \\ \mathrm{Outer}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{spherical}\mathrm{ball}=4{{\mathrm{\pi r}}_{\mathrm{ext}}}^2 \\ =4\times \frac{22}{7}\times 36=452.57{\mathrm{cm}}^2\left[\mathrm{Answer}\right] \\ \mathrm{Now}\mathrm{volume}\mathrm{of}\mathrm{hollow}\mathrm{ball} \\ =\frac{4}{3}\mathrm{\pi }{\left({\mathrm{r}}_{\mathrm{ext}}-{\mathrm{r}}_{\mathrm{int}}\right)}^3=\frac{4}{3}\times 3.14\times 0.1\times 0.1\times 0.1 \\ =0.004187{\mathrm{cm}}^3 \\ \mathrm{So}\mathrm{here}\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{weight}\mathrm{of}1{\mathrm{cm}}^3=8.8\mathrm{g} \\ \mathrm{So}\mathrm{weight}\mathrm{of}0.004187{\mathrm{cm}}^3=8.8\times 0.004187\mathrm{g} \\ =0.0368456\mathrm{g} \end{gathered}$$