The ratio of diameters of two right circular cones is 3 : 7 and that of their heights is 14 : 9, then their volumes are in ratio:

1. 3 : 7

2. 2 : 7

3. 3 : 2

4. 9 : 49

Let the diameters of the two right circular cones be d₁ and d₂, their heights be h₁ and h₂ and their radius be r₁ and r₂.

Given,

⇒ d₁ : d₂ = 3 : 7

Let d₁ and d₂ be 3x and 7x respectively.

⇒ r₁ = $\dfrac{3x}{2}$

⇒ r₂ = $\dfrac{7x}{2}$

Given,

⇒ h₁ : h₂ = 14 : 9

Let h₁ and h₂ be 14a and 9a respectively.

⇒ h₁ : h₂ = 14a : 9a

By formula,

Volume of cone = $\dfrac{1}{3}\pi r^2h$

$\Rightarrow V$1 : V2 = \dfrac{1}{3}\pi r1^2h1:\dfrac{1}{3}\pi r2^2h2 \\[1em] = \dfrac{\dfrac{1}{3}\pi r1^2h1}{\dfrac{1}{3} \pi r2^2h2} \\[1em] = \dfrac{r1^2h1}{r2^2h2} \\[1em] = \dfrac{\Big(\dfrac{3x}{2}\Big)^2 \times 14a}{\Big(\dfrac{7x}{2}\Big)^2 \times 9a} \\[1em] = \dfrac{\Big(\dfrac{9x^2}{4}\Big) \times 14}{\Big(\dfrac{49x^2}{4}\Big) \times 9} \\[1em] = \dfrac{9x^2 \times 14 \times 4}{49x^2 \times 9 \times 4} \\[1em] = \dfrac{9 \times 14}{49 \times 9} \\[1em] = \dfrac{14}{49} \\[1em] = \dfrac{2}{7} = 2 : 7.

Hence, option 2 is the correct option.