Two solid cones A and B are placed in a cylin | vedclass

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(C) Let the volume of cone $A$ be $2V$ and the volume of cone $B$ be $V$. Let the height of cone $A$ be $h_1 \, cm$,then the height of cone $B$ is $(2

Vedclass · Accepted Answer

$396$

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(C) Let the volume of cone $A$ be $2V$ and the volume of cone $B$ be $V$. Let the height of cone $A$ be $h_1 \, cm$,then the height of cone $B$ is $(21 - h_1) \, cm$.Given,the diameter of the cones $= 6 \, cm$.Therefore,the radius of the cones $r = \frac{6}{2} = 3 \, cm$.Volume of cone $A = 2V = \frac{1}{3} \pi r^2 h_1 = \frac{1}{3} \pi (3)^2 h_1 = 3 \pi h_1$.Thus,$V = \frac{3}{2} \pi h_1$ ...... $(i)$Volume of cone $B = V = \frac{1}{3} \pi r^2 (21 - h_1) = \frac{1}{3} \pi (3)^2 (21 - h_1) = 3 \pi (21 - h_1)$ ...... $(ii)$Equating $(i)$ and $(ii)$:$\frac{3}{2} \pi h_1 = 3 \pi (21 - h_1)$$\frac{1}{2} h_1 = 21 - h_1$$\frac{3}{2} h_1 = 21$$h_1 = 14 \, cm$.Height of cone $A = 14 \, cm$,Height of cone $B = 21 - 14 = 7 \, cm$.Volume of cone $A = 3 \pi (14) = 3 	imes \frac{22}{7} 	imes 14 = 132 \, cm^3$.Volume of cone $B = 3 \pi (7) = 3 	imes \frac{22}{7} 	imes 7 = 66 \, cm^3$.Volume of the cylinder $= \pi r^2 H = \frac{22}{7} 	imes (3)^2 	imes 21 = 22 	imes 9 	imes 3 = 594 \, cm^3$.Volume of the remaining portion $=$ Volume of cylinder $-$ (Volume of cone $A$ $+$ Volume of cone $B$)$= 594 - (132 + 66) = 594 - 198 = 396 \, cm^3$.

Part $I$	Part $II$
$1.$ $CSA$ of a cylinder	$a.$ $3\pi r^2$
$2.$ $CSA$ of a sphere	$b.$ $2\pi rh$
$3.$ $CSA$ of a cone	$c.$ $4\pi r^2$
$4.$ $TSA$ of a hemisphere	$d.$ $\pi rl$

Two solid cones $A$ and $B$ are placed in a cylindrical tube as shown in the figure. The ratio of their capacities is $2:1$. Find the heights and capacities of the cones. Also,find the volume of the remaining portion of the cylinder (in $cm^{3}$).

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$1.$ $CSA$ of a cylinder $a.$ $3\pi r^2$

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