Q : 8    A hemispherical bowl is made of steel, \small 0.25\hspace{1mm}cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
 

Answers (1)

Given,

The inner radius of the bowl = r_1 = 5\ cm

The thickness of the bowl = \small 0.25\hspace{1mm}cm

\therefore Outer radius of the bowl = (Inner radius + thickness) = 

r_2 = 5+0.25 = 5.25\ cm

We know, Curved surface area of a hemisphere of radius r2\pi r^2

\therefore The outer curved surface area of the bowl = 2\pi r_2^2

= 2\times\frac{22}{7}\times (5.25)^2

= 2\times\frac{22}{7}\times5.25\times5.25 = 173.25\ cm^2

Therefore, the outer curved surface area of the bowl is 173.25\ cm^2

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