A test tube consists of a hemisphere and a cylinder of the same radius. The volume of water required to fill the whole tube is $\dfrac{2849}{3}$ cm³ and $\dfrac{2618}{3}$ cm³ of water is required to fill the tube to a level which is 2 cm below the top of the tube. Find the radius of the tube and the length of its cylindrical part.

Let length of cylindrical part be h cm and radius of hemispherical and conical part be r cm.

Given,

Volume of water required to fill whole test tube = $\dfrac{2849}{3}$ cm³.

Volume of water required to fill test tube upto a level which is 2 cm below the top of tube = $\dfrac{2618}{3}$ cm³.

∴ Volume of water above 2 cm part of the test tube = $\dfrac{2849}{3} - \dfrac{2618}{3} = \dfrac{231}{3}$ = 77 cm³.

$\Rightarrow π \times r^2 \times 2 = 77 \\[1em] \Rightarrow \dfrac{44}{7}r^2 = 77 \\[1em] \Rightarrow r^2 = \dfrac{77 \times 7}{44} \\[1em] \Rightarrow r^2 = \dfrac{539}{44} \\[1em] \Rightarrow r^2 = \dfrac{49}{4} \\[1em] \Rightarrow r = \sqrt{\dfrac{49}{4}} \\[1em] \Rightarrow r = \dfrac{7}{2} = 3.5 \text{ cm}.$

Volume of water in test tube = Volume of cylindrical part + Volume of hemispherical part = $\dfrac{2849}{3}$ cm³.

$\therefore πr^2h + \dfrac{2}{3}πr^3 = \dfrac{2849}{3} \\[1em] \Rightarrow πr^2\Big(h + \dfrac{2r}{3}\Big) = \dfrac{2849}{3} \\[1em] \Rightarrow \dfrac{22}{7}r^2\Big(h + \dfrac{2r}{3}\Big) = \dfrac{2849}{3} \\[1em] \Rightarrow \dfrac{22}{7} \times (3.5)^2 \times \Big(h + \dfrac{2 \times 3.5}{3}\Big) = \dfrac{2849}{3} \\[1em] \Rightarrow 38.5 \times \Big(h + \dfrac{7}{3}\Big) = \dfrac{2849}{3} \\[1em] \Rightarrow h + \dfrac{7}{3} = \dfrac{2849}{3 \times 38.5} \\[1em] \Rightarrow h + \dfrac{7}{3} = \dfrac{2849}{115.5} \\[1em] \Rightarrow h + \dfrac{7}{3} = \dfrac{28490}{1155} \\[1em] \Rightarrow h + \dfrac{7}{3} = \dfrac{74}{3} \\[1em] \Rightarrow h = \dfrac{74}{3} - \dfrac{7}{3} \\[1em] \Rightarrow h = \dfrac{67}{3} = 22\dfrac{1}{3} \text{ cm}.$

Hence, radius of tube = 3.5 cm and length of cylindrical part = $22\dfrac{1}{3}$ cm.