Question

surface area of sphere is increased by 1, 1 forth times the original . find yhe percentage increase in diameter ana volume​

Answer 1

$$\begin{gathered} Hi, \\ \mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{sphere}=4{\mathrm{\pi r}}^2 \\ \mathrm{new}\mathrm{surface}\mathrm{area}=1\frac{1}{4}\mathrm{old}\mathrm{surface}\mathrm{area}=\frac{5}{4}\mathrm{old}\mathrm{surface}\mathrm{area} \\ 4{\mathrm{\pi R}}^2=\frac{5}{4}\times 4{\mathrm{\pi r}}^2 \\ \mathrm{R}=\frac{\sqrt{5}}{2}\mathrm{r} \\ \mathrm{so}\mathrm{new}\mathrm{diameter}=2\mathrm{R}=\sqrt{5}\mathrm{r} \\ \mathrm{so}\mathrm{percentage}\mathrm{increase}=\frac{\sqrt{5}\mathrm{r}-2\mathrm{r}}{2\mathrm{r}}\times 100=11.8\mathrm{percent} \\ \mathrm{now}\mathrm{new}\mathrm{volume}=\frac{4}{3}\mathrm{\pi }{\left(\frac{\sqrt{5}}{2}\mathrm{r}\right)}^3 \\ \mathrm{so}\mathrm{percenateg}\mathrm{increase}=\frac{\frac{4}{3}\mathrm{\pi }{\left(\frac{\sqrt{5}}{2}\mathrm{r}\right)}^3-\frac{4}{3}\mathrm{\pi }{\left(\mathrm{r}\right)}^3}{\frac{4}{3}\mathrm{\pi }{\left(\mathrm{r}\right)}^3}\times 100 \\ =\frac{5\sqrt{5}-8}{8}\times 100 \\ =39.75\mathrm{percentage} \\ \mathrm{Regards} \end{gathered}$$