A test tube consists of a hemisphere and a cylinder of the same radius. The volume of the water required to fill the whole tube is $\dfrac{5159}{6}$ cm³, and $\dfrac{4235}{6}$ cm³ of water is required to fill the tube to a level which is 4 cm below the top of the tube. Find the radius of the tube and the length of its cylindrical part.

Given,

Let radius of hemisphere be r cm and height of cylindrical portion be h cm.

Total volume of test tube = Volume of hemisphere + Volume of cylinder

$\Rightarrow \dfrac{5159}{6} = \dfrac{2}{3}πr^3 + πr^2h \\[1em] \Rightarrow \dfrac{5159}{6} = πr^2\Big(\dfrac{2}{3}r + h\Big) \\[1em] \Rightarrow \dfrac{5159}{6} = πr^2\Big(\dfrac{2r + 3h}{3}\Big) \\[1em] \Rightarrow πr^2(2r + 3h) = \dfrac{5159}{6} \times 3 \\[1em] \Rightarrow πr^2(2r + 3h) = \dfrac{5159}{2} …………..(1)$

Given,

$\dfrac{4235}{6}$ cm³ of water is required to fill the tube to a level which is 4 cm below the top of the tube.

$\therefore \dfrac{4235}{6} = \text{Vol. of hemisphere + Vol. of cylinder[upto (h - 4) cm]} \\[1em] \Rightarrow \dfrac{4235}{6} = \dfrac{2}{3}πr^3 + πr^2(h - 4) \\[1em] \Rightarrow \dfrac{4235}{6} = πr^2\Big(\dfrac{2}{3}r + h - 4\Big) \\[1em] \Rightarrow \dfrac{4235}{6} = πr^2\Big(\dfrac{2r + 3h - 12}{3}\Big) \\[1em] \Rightarrow \dfrac{4235 \times 3}{6} = πr^2(2r + 3h) - 12πr^2 \\[1em] \Rightarrow \dfrac{4235}{2} = πr^2(2r + 3h) - 12πr^2$

Substituting value of πr²(2r + 3h) from equation 1 in above equation.

$\Rightarrow \dfrac{4235}{2} = \dfrac{5159}{2} - 12πr^2 \\[1em] \Rightarrow 12πr^2 = \dfrac{5159}{2} - \dfrac{4235}{2} \\[1em] \Rightarrow 12πr^2 = \dfrac{924}{2} \\[1em] \Rightarrow r^2 = \dfrac{924}{2 \times 12π} \\[1em] \Rightarrow r^2 = \dfrac{924}{24 \times \dfrac{22}{7}} \\[1em] \Rightarrow r^2 = \dfrac{924 \times 7}{24 \times 22} \\[1em] \Rightarrow r^2 = \dfrac{6468}{528} \\[1em] \Rightarrow r^2 = 12.25 \\[1em] \Rightarrow r = \sqrt{12.25} \\[1em] \Rightarrow r = 3.5 \text{ cm}.$

Substituting value of r in equation 1, we get :

$\Rightarrow πr^2(2r + 3h) = \dfrac{5159}{2} \\[1em] \Rightarrow \dfrac{22}{7} \times (3.5)^2 \times (2 \times 3.5 + 3h) = \dfrac{5159}{2} \\[1em] \Rightarrow 22 \times 0.5 \times 3.5 \times (7 + 3h) = \dfrac{5159}{2} \\[1em] \Rightarrow 38.5 \times (7 + 3h) = \dfrac{5159}{2} \\[1em] \Rightarrow 7 + 3h = \dfrac{5159}{2 \times 38.5} \\[1em] \Rightarrow 7 + 3h = \dfrac{5159}{77} \\[1em] \Rightarrow 7 + 3h = 67 \\[1em] \Rightarrow 3h = 67 - 7 \\[1em] \Rightarrow 3h = 60 \\[1em] \Rightarrow h = \dfrac{60}{3} \\[1em] \Rightarrow h = 20 \text{ cm}.$

Hence, radius of cylindrical part = 3.5 cm and height = 20 cm.