Question

The surface areas of the sphere and a cube are equal and if their volumes are $V_1$ and$V_2$ respectively then $\frac{V_1}{V_2}$

• A. $3:\sqrt{2}$
• B. $\sqrt{6}:\sqrt{\pi }$
• C. $\sqrt{3}:2\pi$
• D. None

Answer 1

The correct option is B $\sqrt{6}:\sqrt{\pi }$

Let the radius of the sphere be $r$ and
length of a side of a cube be $l$
$\therefore$ Surface area of sphere=$4\pi r^2$
and surface area of cube=$6l^2$
Given,
$4\pi r^2=6l^2$
$=>\frac{r^2}{l^2}=\frac{6}{4k}$
$=>\frac{r}{l}=\sqrt{\frac{3}{2\pi }}$
Now,
$\frac{volumeofsphere}{volumeofcube}=\frac{\sqrt{1}}{\sqrt{2}}$
$=>\frac{\sqrt{1}}{\sqrt{2}}=\frac{\frac{4}{3}\pi r^3}{l^3}$
$=>\frac{4}{3}\pi (\frac{r}{l}{)}^3$
$=>\frac{4}{3}\pi (\sqrt{\frac{3}{2\pi }}{)}^3$
$=>\frac{4}{3}\pi \times \left(\frac{3}{2\pi }\right)\times \sqrt{\frac{3}{2\pi }}$
$=>2\sqrt{\frac{3}{2\pi }}$
$=>\sqrt{\frac{2\times 2\times 3}{2\pi }}$
$=>\sqrt{\frac{6}{\pi }}$