If a cone of maximum volume is inscribed in a given sphere, then the ratio of the he the cone to the diameter of the sphere is : A. 3/4 B. 1/3 ​

Question

If a cone of maximum volume is inscribed in a given sphere, then the ratio of the he the cone to the diameter of the sphere is :

A. \(\frac{3}{4}\)

B. \(\frac{1}{3}\)

C. \(\frac{1}{4}\)

D. \(\frac{2}{3}\)

Answer 1

Option : (D)

In the figure, 

∆PQR represents the 2D view of the cone and the circle represents the sphere. 

PA is perpendicular to QR and PS is diameter of circle. 

C will lie on PA due to symmetry.

Let the radius and height of cone be r and h and the radius of sphere be R. 

Also, 

The semi vertical angle of cone is α.

In ∆PAR,

tanP = tanα

= \(\frac{AR}{PA}\) = \(\frac{r}{h}\) - (1)

∠APR = α 

∠PAR= 90° 

Hence, 

∠PRA = 180°- 90°- α = 90°- α 

Also, 

∠PRS = 90° (Angle in a semicircle) 

Hence, 

∠ARS =∠PRS - ∠PRA = α 

In ∆RAS 

AS = PS - PA = 2R - h

tanR = tanα = \(\frac{2r-h}{r}\) - (2)

From (1) and (2), we get 

r²=2Rh - h² 

The volume of cone will be

V = \(\frac{1}{3}\)πr²h 

= \(\frac{1}{3}\)πh (2Rh - h²)

Differentiating V with respect to h, we get

V' = \(\frac{1}{3}\)π(4Rh - 3h²)

Differentiating V’ with respect to h, we get

V'' = \(\frac{1}{3}\)π(4R - 6h)

For maxima at h = c, 

V’(c) = 0 and v’’(c)<0

V' = 0

⇒ \(\frac{h}{2R}\) = \(\frac{2}{3}\)

V''(h = \(\frac{4R}{3}\))

= \(-\frac{4}{3}\)πR < 0

Hence, the ratio of height of cone to diameter of sphere is \(\frac{2}{3}\).