Oil is stored in a spherical vessel occupying $\frac{3}{4}$ 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 $\pi = \frac{22}{7}$

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Given,
Radius of spherical vessel (r) = 28 cm
Volume of spherical vessel (v) = $\frac{4}{3}πr^3$
Volume of oil in vessel = $\frac{3}{4}v$
Substituting values we get :
$$\begin{aligned}\Rightarrow v = \frac{4}{3} \times \frac{22}{7} \times 28^3 \\[1em] \Rightarrow \frac{3}{4}v = \frac{3}{4} \times \frac{4}{3} \times \frac{22}{7} \times 28^3 \\[1em] \Rightarrow \frac{3}{4}v = \frac{22}{7} \times 28^3.\end{aligned}$$
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)
$$\begin{aligned}\Rightarrow \frac{22}{7} \times 28^3 = \frac{22}{7} \times 21^2 \times h \\[1em] \Rightarrow 28^3 = 21^2 \times h \\[1em] \Rightarrow h = \frac{28^3}{21^2}\\[1em] \Rightarrow h = \frac{21952}{441} \\[1em] \Rightarrow h = 49.77 ≈ 50 \text{ cm}.\end{aligned}$$
Hence, height of the oil in the cylindrical vessel = 50 cm.