A dome of a building is in the form of a hemi | vedclass

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(B) Total cost of white-washing $= Rs. 4989.60$; Rate of white-washing $= Rs. 20$ per $m^2$.$	herefore 	ext{Inner Surface Area} = \frac{	ext{Total

Vedclass · Accepted Answer

$523.9$

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(B) Total cost of white-washing $= Rs. 4989.60$; Rate of white-washing $= Rs. 20$ per $m^2$.$	herefore 	ext{Inner Surface Area} = \frac{	ext{Total cost}}{	ext{Rate}} = \frac{4989.60}{20} = 249.48 \, m^2$.Let $r$ be the radius of the hemispherical dome.$	herefore$ Inner Surface Area $= 2 \pi r^2 = 249.48$.$2 	imes \frac{22}{7} 	imes r^2 = 249.48 \Rightarrow r^2 = \frac{249.48 	imes 7}{2 	imes 22} = \frac{1746.36}{44} = 39.69$.$r = \sqrt{39.69} = 6.3 \, m$.Volume of air inside the dome $= \frac{2}{3} \pi r^3 = \frac{2}{3} 	imes \frac{22}{7} 	imes (6.3)^3$.$= \frac{2}{3} 	imes \frac{22}{7} 	imes 6.3 	imes 6.3 	imes 6.3 = 2 	imes 22 	imes 0.3 	imes 6.3 	imes 6.3 = 523.908 \, m^3$.Thus,the volume of air inside the dome is approximately $523.9 \, m^3$.

$A$ dome of a building is in the form of a hemisphere. From inside,it was white-washed at the cost of $Rs. 4989.60$. If the cost of white-washing is $Rs. 20$ per square metre,find the volume of the air inside the dome. (in $, m^3$)

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