Question.
A hemispherical bowl is made of steel, $0.25 \mathrm{~cm}$ thick. The inner radius of the bowl is $5 \mathrm{~cm}$. Find the outer curved surface area of the bowl. $\left[\right.$ Assume $\left.\pi=\frac{22}{7}\right]$
Solution:
Inner radius of hemispherical bowl = 5 cm
Thickness of the bowl = 0.25 cm
$\therefore$ Outer radius $(r)$ of hemispherical bowl $=(5+0.25) \mathrm{cm}$
$=5.25 \mathrm{~cm}$
Outer CSA of hemispherical bowl $=2 \pi r^{2}$
$=2 \times \frac{22}{7} \times(5.25 \mathrm{~cm})^{2}=173.25 \mathrm{~cm}^{2}$
Therefore, the outer curved surface area of the bowl is $173.25 \mathrm{~cm}^{2}$.
Inner radius of hemispherical bowl = 5 cm
Thickness of the bowl = 0.25 cm
$\therefore$ Outer radius $(r)$ of hemispherical bowl $=(5+0.25) \mathrm{cm}$
$=5.25 \mathrm{~cm}$
Outer CSA of hemispherical bowl $=2 \pi r^{2}$
$=2 \times \frac{22}{7} \times(5.25 \mathrm{~cm})^{2}=173.25 \mathrm{~cm}^{2}$
Therefore, the outer curved surface area of the bowl is $173.25 \mathrm{~cm}^{2}$.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.