If the radius of the base of a right circular cylinder is halved, keeping the height same, the ratio of the volume of th

Question

If the radius of the base of a right circular cylinder is halved, keeping the height same, the ratio of the volume of the reduced cylinder to that of the original cylinder is:

Accepted Answer

To find the ratio of the volume of the reduced cylinder to the original cylinder, let us use the formula for the volume of a cylinder, which is $V = \pi r^2 h$. Let the original radius of the cylinder be $r$ and the original height be $h$. Therefore, the original volume ($V1$) is: $$V1 = \pi r^2 h$$ According to the question, the radius is halved while the height remains the same. So, the new radius ($r'$) = $\frac{r}{2}$ and the new height ($h'$) = $h$. Therefore, the volume of the reduced cylinder ($V2$) is: $$V2 = \pi (r')^2 h' = \pi \left(\frac{r}{2}ight)^2 h = \pi \frac{r^2}{4} h = \frac{1}{4} \pi r^2 h$$ Now, find the ratio of the volume of the reduced cylinder to the original cylinder ($V2 : V1$): $$\frac{V2}{V1} = \frac{\frac{1}{4} \pi r^2 h}{\pi r^2 h} = \frac{1}{4}$| 
Thus, the ratio is $1:4$, which makes option B the correct answer.

Surface Areas and Volumes

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