A solid is hemispherical at the bottom and co | vedclass

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(D) Let the radius of the base be $r$ units and the height of the cone be $h$ units.The surface area of the hemispherical part is $2 \pi r^2$.The curv

Vedclass · Accepted Answer

$1: \sqrt{3}$

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(D) Let the radius of the base be $r$ units and the height of the cone be $h$ units.The surface area of the hemispherical part is $2 \pi r^2$.The curved surface area of the conical part is $\pi r l$,where $l = \sqrt{r^2 + h^2}$ is the slant height.According to the problem,the surface areas of the two parts are equal:$2 \pi r^2 = \pi r \sqrt{r^2 + h^2}$Dividing both sides by $\pi r$ (assuming $r 
eq 0$):$2r = \sqrt{r^2 + h^2}$Squaring both sides:$4r^2 = r^2 + h^2$$3r^2 = h^2$Taking the square root of both sides:$\sqrt{3}r = h$Therefore,the ratio of the radius to the height is:$\frac{r}{h} = \frac{1}{\sqrt{3}}$,which is $1: \sqrt{3}$.

$A$ solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal,then the ratio of the radius and height of its conical part is

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