Question

The heights of two cones are in the ratio 3:2 and their radii are in the ratio 2:7.

Find the ratio of their volume.​

Answer 1

GIVEN :-

• Ratio of Height of cone = 3:2.
• Ratio of Radii of cone = 2:7.

TO FIND :-

• The ratio of volume.

SOLUTION :-

Let the ratio constant be "x".

Let the volume of first cone be "v" and the volume of second cone be "v'".

$\\ \underline{ \bigstar \: \textsf{According to the Question}}$

$: \implies \displaystyle \sf \: \frac{v}{v'} = \frac{\pi r ^{2} \dfrac{h}{3} }{\pi r'^{2} \dfrac{h'}{3} }$

$: \implies \displaystyle \sf \: \frac{v}{v'} = \frac{\pi \times (2x) ^{2} \times \dfrac{3x}{3} }{\pi \times (7x) ^{2} \times \dfrac{2x}{3} }$

$: \implies \displaystyle \sf \: \frac{v}{v'} = \frac{ 4x^{2} \times \dfrac{3x}{3} }{ 49x^{2} \times \dfrac{2x}{3} }$

$: \implies \displaystyle \sf \: \frac{v}{v'} = \frac{ 4 \times \dfrac{3x}{3} }{ 49 \times \dfrac{2x}{3} }$

$: \implies \displaystyle \sf \: \frac{v}{v'} = 4 \times \frac{3x}{3} \times \frac{1}{49} \times \frac{3}{2x}$

$: \implies \displaystyle \sf \: \frac{v}{v'} = 4 \times 3x \times \frac{1}{49} \times \frac{1}{2x}$

$: \implies \displaystyle \sf \: \frac{v}{v'} = \frac{12x}{98x}$

$: \implies \displaystyle \sf \: \frac{v}{v'} = \frac{12}{98}$

$: \implies \displaystyle \sf \: \frac{v}{v'} = \frac{6}{49}$

$: \implies \underline{ \boxed{\displaystyle \sf \: {v}:{v'} = 6:49}}$

Answer 2

Let r1 and r2 be cone radius and similarly h1 and h2 be heights.

r1/r2= 2/3, i.e r1= 2/3*r2

Similarly,

h1= 3/2h1.

Apply them into formula of volume of cone and find the answer.(solve it using pen and paper you will see)

Answer should be- V1/V2= 2/3.