In the adjoining diagram, a tilted right circular cylindrical vessel with base diameter 7 cm contains a liquid. When placed vertically, the height of the liquid in the vessel is the mean of two heights shown in the diagram. Find the area of wet surface, when the cylinder is placed vertically on a horizontal surface. (Use $\pi = \frac{22}{7}$ )
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When vertically placed,
Height of liquid (h) = $\frac{1 + 6}{2} = \frac{7}{2}$ cm
Diameter of base = 7 cm
Radius (r) = $\frac{7}{2}$ cm
$$\begin{aligned}\text{Area of wet surface} = πr^2 + 2πrh \\[1em] = πr(r + 2h) \\[1em] = \frac{22}{7} \times \frac{7}{2}\Big(\frac{7}{2} + 2\times \frac{7}{2}\Big) \\[1em] = 11 \times (3.5 + 7) \\[1em] = 11 \times 10.5 \\[1em] = 115.5 \text{ cm}^2.\end{aligned}$$
Hence, area of wet surface = 115.5 cm2.