Question

A solid sphere with centre 'O' of certain radius fits exactly into a right circular cone as shown in the diagram given below. If the volume of the sphere is 4186.67 ${cm}^3$, find the total surface area of the cone. (Take $\pi$ = 3.14). It is known that slant height of the cone is two times the radius of the base of the cone and the ratio of AO to DO is 2:1.

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• A. 400$\pi$ ${cm}^2$
• B. 500$\pi$ ${cm}^2$
• C. 900$\pi$ ${cm}^2$
• D. 600$\pi$ ${cm}^2$

Answer 1

The correct option is C

900$\pi$ ${cm}^2$

From the given volume of the sphere, radius of sphere is 10 cm i.e., OD = 10 cm.

Now,
$\frac{AO}{OD}$ = 2
$\Rightarrow$ AO = 2 OD = 20 cm.

Therefore, AD = 30 cm.

Now, since slant height is two times the radius at the base, it means that diameter is equal to the slant height. So, the cone is basically an equilateral triangle if we visualise it in two dimensions.

Let the side of the triangle be $a$ cm.

If we write the area of the triangle, we have
$\frac{1}{2}$ BC $\times$ AD = $\frac{\sqrt{3}}{4}$ $a^2$

$\Rightarrow \frac{1}{2}$ $\times a$ $\times$ 30 = $\frac{\sqrt{3}}{4}$ $a^2$

$\Rightarrow$ $a$ = $\frac{60}{\sqrt{3}}$ = 20 $\sqrt{3}$ cm

Slant height of the cone = 20 $\sqrt{3}$ cm.

Radius of the base = 10 $\sqrt{3}$ cm.

$\therefore$ Total surface area of the cone
= $\pi$ $r^2$ + $\pi$ r l
= $\pi$ ${\left(10\sqrt{3}\right)}^2$ + $\pi$ (10 $\sqrt{3}$) (20 $\sqrt{3}$)
= 900 $\pi$ ${cm}^2$