Question

How many times do the volume and surface area of a cylinder increase if its radius doubled and height remains same

Answer 1

Answer

$\bf \large \underline{Formula \: to \: be \: used : }$

$\tt \bullet \: \: volume \: of \: a \: cylinder = \pi {r}^{2} h$

$\bullet \tt \: \: surface \:area \: of \: cylinder = 2\pi {r}( h + {r})$

$\bf \large \underline{Solution : }$

Let us consider the radius of the cylinder in different cases be r₁ and r₂

where r₂ = 2r₁

and height of the cylinder remains constant

Therefore , calculating the ratios of volume of cylinder

$\sf \implies \dfrac{V_{1}}{V_{2}}=\dfrac{\pi r_{1}^{2} h }{\pi r_{2}^{2} h} \\\\ \sf\implies \dfrac{V_{1}}{V_{2}} = \dfrac{r_{1}^{2}}{r_{2}^{2}} \\\\ \sf\implies \dfrac{V_{1}}{V_{2}} = \dfrac{r_{1}^{2}}{4r_{1}^{2}}\\\\ \sf\implies \dfrac{V_{1}}{V_{2}} = \dfrac{1}{4} \\\\ \sf\implies V_{1} : V_{2} = 1 : 4$

Thus , when radius is doubled volume increases by 4 times

Now , calculating the ratios of surface area of cylinder to find out the change ,

$\sf \dfrac{S.A_{1}}{S.A_{2}} = \dfrac{2\pi {r}_{1}( h + r_{1} )}{2\pi {r}_{2}(h + r_{2})} \\\\ \sf\implies \dfrac{S.A_{1}}{S.A_{2}} = \dfrac{r_{1}(h + r_{1})}{r_{2}(h + r_{2})} \\\\ \sf\implies \dfrac{S.A_{1}}{S.A_{2}} = \dfrac{r_{1}(h + r_{1})}{2r_{1}(h + 2r_{1})} \\\\ \sf\implies \dfrac{S.A_{1}}{S.A_{2}} = \dfrac{h + r_{1}}{h + 2r_{1}}$