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(C) The volume of a right circular cone is given by the formula $V = \frac{1}{3} \pi r^2 h$.For the first cone:Radius $r_1 = 1.6 \ cm$,Height $h_1 = 3

Vedclass · Accepted Answer

$6.4$

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(C) The volume of a right circular cone is given by the formula $V = \frac{1}{3} \pi r^2 h$.For the first cone:Radius $r_1 = 1.6 \ cm$,Height $h_1 = 3.6 \ cm$.Volume $V_1 = \frac{1}{3} \pi (1.6)^2 (3.6) = \pi 	imes 1.6 	imes 1.6 	imes 1.2 \ cm^3$.When the cone is melted and recast,the volume remains constant. Let the height of the new cone be $H$.For the second cone:Radius $r_2 = 1.2 \ cm$,Height $H$.Volume $V_2 = \frac{1}{3} \pi (1.2)^2 H$.Equating the volumes $(V_1 = V_2)$:$\frac{1}{3} \pi (1.2)^2 H = \pi (1.6)^2 (1.2)$$H = \frac{1.6 	imes 1.6 	imes 1.2 	imes 3}{1.2 	imes 1.2}$$H = \frac{2.56 	imes 3}{1.2} = \frac{7.68}{1.2} = 6.4 \ cm$.

$A$ right circular cone is $3.6 \ cm$ high and the radius of its base is $1.6 \ cm$. It is melted and recast into a right circular cone with a base radius of $1.2 \ cm$. The height of the new cone (in $cm$) is:

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