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Q.

The internal and external radii of a hemispherical bowls are $r_{1}$ and $r_{2}$ respectively, then the surface area of the bowl is

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a

$2 π (r_{1}^{2} + r_{2}^{2})$

b

$π (r_{1}^{2} + r_{2}^{2})$

c

$π (r_{1}^{2} - r_{2}^{2})$

d

$2 π (r_{1}^{2} - r_{2}^{2})$

answer is D.

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Detailed Solution

Surface area of bowls = outer surface − inner surface

$= 2 π r_{1}^{2} - 2 π r_{2}^{2}$

$= 2 π (r_{1}^{2} - r_{2}^{2})$

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