Question:
Note Take $\pi=\frac{22}{7}$, unless stated otherwise.
A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
Solution:
Inner radius of the bowl, r = 5 cm
Let the outer radius of the bowl be R cm.
Thickness of the bowl = 0.25 cm (Given)
∴ R − r = 0.25 cm
⇒ R = 0.25 + r = 0.25 + 5 = 5.25 cm
$\therefore$ Outer curved surface area of the bowl $=2 \pi r^{2}=2 \times \frac{22}{7} \times(5.25)^{2}=173.25 \mathrm{~cm}^{2}$
Thus, the outer curved surface area of the bowl is 173.25 cm2.
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