A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain ?

Volume of hemisphere = $\dfrac{2}{3}πr^3$.

Putting values we get,

$\text{Volume of hemisphere} = \dfrac{2}{3}πr^3 \\[1em] = \dfrac{2}{3}π(3.5)^3 \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 42.875 \\[1em] = \dfrac{2 \times 22 \times 42.875}{3 \times 7} \\[1em] = \dfrac{1886.5}{21} \\[1em] = \dfrac{18865}{210} \\[1em] = \dfrac{539}{6} \\[1em] = 89\dfrac{5}{6} \text{ cm}^3$

Hence, the volume of water in the hemispherical bowl = $89\dfrac{5}{6} \text{ cm}^3$.