Question

please answer asap......​

Answer 1

Thanks again

sorry for late

Answer 2

Let $r$ be the base radius of the cone and $l$ be slant height. So,

$\longrightarrow l^2=r^2+h^2$

Here $V$ is the volume of the cone. Then,

$\longrightarrow V=\dfrac{1}{3}\,\pi r^2h$

$\longrightarrow r^2=\dfrac{3V}{\pi h}\quad\quad\dots(1)$

Here $S$ is curved surface area of the cone. Then,

$\longrightarrow S=\pi rl$

$\longrightarrow S=\pi r\sqrt{r^2+h^2}$

$\longrightarrow S=\pi \sqrt{r^4+r^2h^2}$

From (1),

$\longrightarrow S=\pi \sqrt{\left(\dfrac{3V}{\pi h}\right)^2+\dfrac{3V}{\pi h}\cdot h^2}$

$\longrightarrow S=\pi \sqrt{\dfrac{9V^2}{\pi^2h^2}+\dfrac{3Vh}{\pi}}$

$\longrightarrow S=\dfrac{\pi\sqrt{3\pi Vh^3+9V^2}}{\pi h}$

$\longrightarrow Sh=\sqrt{3\pi Vh^3+9V^2}$

$\longrightarrow S^2h^2=3\pi Vh^3+9V^2$

$\longrightarrow\underline{\underline{3\pi Vh^3-S^2h^2+9V^2=0}}$

Hence (B) is the answer.