Quantity 1: Height of the tank if the volume | vedclass

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Question

(A) For $Quantity$ $1$: Let radius $r = 7x$ and height $h = 10x$.
Volume of cylinder $= \pi r^2 h = 12320 \, cm^3$.
$\frac{22}{7} 	imes (7x)^2 	im

Vedclass · Accepted Answer

Quantity $I >$ Quantity $II$

Vedclass · Answer

(A) For $Quantity$ $1$: Let radius $r = 7x$ and height $h = 10x$.
Volume of cylinder $= \pi r^2 h = 12320 \, cm^3$.
$\frac{22}{7} 	imes (7x)^2 	imes (10x) = 12320$.
$\frac{22}{7} 	imes 49x^2 	imes 10x = 12320$.
$22 	imes 7x^3 	imes 10 = 12320$.
$1540x^3 = 12320$.
$x^3 = \frac{12320}{1540} = 8$.
$x = 2$.
Height $h = 10x = 10 	imes 2 = 20 \, cm$.
For $Quantity$ $2$: Volume of cone $= \frac{1}{3} \pi r^2 h = \frac{1}{3} 	imes \pi 	imes (2)^2 	imes 3 = 4\pi \, cm^3$.
This is recast into a cylinder of radius $2 \, cm$. Let the height be $H$.
Volume of cylinder $= \pi R^2 H = \pi 	imes (2)^2 	imes H = 4\pi H$.
Equating volumes: $4\pi H = 4\pi$.
$H = 1 \, cm$.
Comparing $Quantity$ $1$ $(20 \, cm)$ and $Quantity$ $2$ $(1 \, cm)$: $Quantity$ $I > Quantity$ $II$.

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