X and Y are two cylinders of the same height. | vedclass

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(C) Let the heights of cylinders $X$ and $Y$ be $h_1 = h_2 = h$.Let the diameter of $X$ be $d_1$ and the diameter of $Y$ be $d_2$. Given $d_1 = \frac{

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half the volume of $Y$

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(C) Let the heights of cylinders $X$ and $Y$ be $h_1 = h_2 = h$.Let the diameter of $X$ be $d_1$ and the diameter of $Y$ be $d_2$. Given $d_1 = \frac{1}{2} d_2$,which implies $d_2 = 2d_1$.The volume of a cylinder is given by $V = \pi r^2 h = \pi (d/2)^2 h = \frac{\pi d^2 h}{4}$.Volume of $X$ $(V_X)$ = $\frac{\pi d_1^2 h}{4}$.Volume of $Y$ $(V_Y)$ = $\frac{\pi d_2^2 h}{4} = \frac{\pi (2d_1)^2 h}{4} = \frac{4 \pi d_1^2 h}{4} = 4 V_X$.If the height of $X$ is doubled,the new volume $V_X' = \frac{\pi d_1^2 (2h)}{4} = 2 	imes \frac{\pi d_1^2 h}{4} = 2 V_X$.Since $V_Y = 4 V_X$,then $V_X = \frac{1}{4} V_Y$.Therefore,$V_X' = 2 	imes (\frac{1}{4} V_Y) = \frac{1}{2} V_Y$.Thus,the volume of $X$ becomes half the volume of $Y$.

$X$ and $Y$ are two cylinders of the same height. The base of $X$ has a diameter which is equal to half the diameter of the base of $Y$. If the height of $X$ is doubled,the volume of $X$ becomes

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