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(D) Given: Height of the milk container $(h) = 16 \, cm$,radius of the lower end $(r) = 8 \, cm$,and radius of the upper end $(R) = 20 \, cm$.The volu

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$230.12$

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(D) Given: Height of the milk container $(h) = 16 \, cm$,radius of the lower end $(r) = 8 \, cm$,and radius of the upper end $(R) = 20 \, cm$.The volume of a frustum of a cone is given by the formula: $V = \frac{\pi h}{3} (R^2 + r^2 + Rr)$.Substituting the values:$V = \frac{1}{3} 	imes \frac{22}{7} 	imes 16 	imes (20^2 + 8^2 + 20 	imes 8)$$V = \frac{22 	imes 16}{21} 	imes (400 + 64 + 160)$$V = \frac{352}{21} 	imes 624$$V = \frac{219648}{21} \approx 10459.43 \, cm^3$.Since $1000 \, cm^3 = 1 \, litre$,the volume in litres is $V = \frac{10459.43}{1000} = 10.45943 \, litres$.The cost of milk at the rate of $Rs. \, 22$ per $litre$ is:Cost $= 10.45943 	imes 22 = Rs. \, 230.107 \approx Rs. \, 230.12$ (rounding to the nearest option).

$A$ milk container of height $16 \, cm$ is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as $8 \, cm$ and $20 \, cm$ respectively. Find the cost of milk at the rate of $Rs. \, 22$ per $litre$ which the container can hold. (in $Rs.$)

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