Question

The ratio of height of a cone having maximum volume which can be inscribed in a sphere with the diameter of sphere is

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\frac{3}{4}$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{2}{3}$

Answer 2

Let $OM = x$

Then height of cone i.e., $h = x + a$

(where a is radius of sphere)

Radius of base of cone = $\sqrt{a^{2} - x^{2}}$

Therefore, volume $V = \frac{1}{3}\pi(a^{2} - x^{2})(x + a)$

⇒ $\frac{dV}{dx} = \frac{\pi}{3}(a + x)(a - 3x)$

Now, $\frac{dV}{dx} = 0$ ⇒ $x = - a,\frac{a}{3}$

But $x \neq - a,$ So, $x = \frac{a}{3}$

The volume is maximum at $x = \frac{a}{3}$

Height of a cone $h = a + \frac{a}{3} = \frac{4}{3}a$

Therefore ratio of height and diameter = $\frac{\frac{4}{3}a}{2a} = \frac{2}{3}$.