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(D) The ice cream cone is a combination of a hemisphere and a cone.Given,radius of the hemisphere $(r)$ $= 5 \, cm$.Volume of the hemisphere $= \frac{

Vedclass · Accepted Answer

$327.4$

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(D) The ice cream cone is a combination of a hemisphere and a cone.Given,radius of the hemisphere $(r)$ $= 5 \, cm$.Volume of the hemisphere $= \frac{2}{3} \pi r^{3} = \frac{2}{3} 	imes \frac{22}{7} 	imes (5)^{3} = \frac{5500}{21} \approx 261.90 \, cm^{3}$.Now,the radius of the cone $(r)$ $= 5 \, cm$.The total height of the ice cream cone is $10 \, cm$. Since the hemisphere has a radius of $5 \, cm$,the height of the conical part $(h)$ $= 10 - 5 = 5 \, cm$.Volume of the cone $= \frac{1}{3} \pi r^{2} h = \frac{1}{3} 	imes \frac{22}{7} 	imes (5)^{2} 	imes 5 = \frac{2750}{21} \approx 130.95 \, cm^{3}$.Total volume of the ice cream cone $= 261.90 + 130.95 = 392.85 \, cm^{3}$.Since $\frac{1}{6}$ part is left unfilled,the volume of ice cream filled is $\frac{5}{6}$ of the total volume.Required volume of ice cream $= \frac{5}{6} 	imes 392.85 = 5 	imes 65.475 = 327.375 \approx 327.4 \, cm^{3}$.

An ice cream cone full of ice cream has a radius of $5 \, cm$ and a total height of $10 \, cm$ as shown in the figure. Calculate the volume of the ice cream,provided that its $\frac{1}{6}$ part is left unfilled with ice cream. (in $cm^{3}$)

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