Oil is stored in a spherical vessel occupying $\dfrac{3}{4}$ of its full capacity. Radius of this spherical vessel is 28 cm. This oil is then poured into a cylindrical vessel with a radius of 21 cm. Find the height of the oil in the cylindrical vessel (correct to the nearest cm).

Take $\pi = \dfrac{22}{7}$

Given,

Radius of spherical vessel (r) = 28 cm

Volume of spherical vessel (v) = $\dfrac{4}{3}πr^3$

Volume of oil in vessel = $\dfrac{3}{4}v$

Substituting values we get :

$\Rightarrow v = \dfrac{4}{3} \times \dfrac{22}{7} \times 28^3 \\[1em] \Rightarrow \dfrac{3}{4}v = \dfrac{3}{4} \times \dfrac{4}{3} \times \dfrac{22}{7} \times 28^3 \\[1em] \Rightarrow \dfrac{3}{4}v = \dfrac{22}{7} \times 28^3.$

Radius of cylindrical vessel (R) = 21 cm

Let height of oil in cylindrical vessel be h cm.

Volume of oil = Volume of cylinder upto which oil is filled (πR²h)

$\Rightarrow \dfrac{22}{7} \times 28^3 = \dfrac{22}{7} \times 21^2 \times h \\[1em] \Rightarrow 28^3 = 21^2 \times h \\[1em] \Rightarrow h = \dfrac{28^3}{21^2}\\[1em] \Rightarrow h = \dfrac{21952}{441} \\[1em] \Rightarrow h = 49.77 ≈ 50 \text{ cm}.$

Hence, height of the oil in the cylindrical vessel = 50 cm.