A hemispherical bowl is made of steel,0.25 , | vedclass

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Question

(D) Inner radius $(r) = 5 \, cm$Thickness $= 0.25 \, cm$Outer radius $(R) = r + 	ext{thickness} = 5 + 0.25 = 5.25 \, cm$The outer curved surface area

Vedclass · Accepted Answer

$173.25$

Vedclass · Answer

(D) Inner radius $(r) = 5 \, cm$Thickness $= 0.25 \, cm$Outer radius $(R) = r + 	ext{thickness} = 5 + 0.25 = 5.25 \, cm$The outer curved surface area of a hemispherical bowl is given by the formula $2 \pi R^2$.Outer curved surface area $= 2 	imes \frac{22}{7} 	imes (5.25)^2 \, cm^2$$= 2 	imes \frac{22}{7} 	imes 5.25 	imes 5.25 \, cm^2$$= 2 	imes \frac{22}{7} 	imes \frac{525}{100} 	imes \frac{525}{100} \, cm^2$$= 2 	imes 22 	imes \frac{75}{100} 	imes 5.25 \, cm^2$$= 44 	imes 0.75 	imes 5.25 \, cm^2$$= 173.25 \, cm^2$

$A$ 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. (in $, cm^2$)

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