The radius of a cylinder is doubled and the h | vedclass

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(C) Let the initial radius and height of the cylinder be $r$ and $h$,respectively.The initial volume $V_{1}$ is given by the formula $V_{1} = \pi r^{2

Vedclass · Accepted Answer

$2:1$

Vedclass · Answer

(C) Let the initial radius and height of the cylinder be $r$ and $h$,respectively.The initial volume $V_{1}$ is given by the formula $V_{1} = \pi r^{2} h$.According to the problem,the new radius $r' = 2r$ and the new height $h' = \frac{h}{2}$.The new volume $V_{2}$ is given by $V_{2} = \pi (r')^{2} h' = \pi (2r)^{2} \left(\frac{h}{2}\right)$.Simplifying the expression for $V_{2}$,we get $V_{2} = \pi (4r^{2}) \left(\frac{h}{2}\right) = 2 \pi r^{2} h$.The ratio of the new volume to the previous volume is $\frac{V_{2}}{V_{1}} = \frac{2 \pi r^{2} h}{\pi r^{2} h} = \frac{2}{1}$.Therefore,the ratio is $2:1$.

The radius of a cylinder is doubled and the height is halved. What is the ratio between the new volume and the previous volume?

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