Question

Two solid cones A and B are placed in a cylinderical tube as shown in the figure. The ratio of their capacities are 2:1. Find the heights and capacities of cones. Also, find the volume of the remaining portion of the cylinder

Answer 1

Let volume of cone A be 2V and volume of cone B be V.

Again, let height of the cone A = h₁ cm,

Then height of cone B = (21 – h₁) cm

Given, diameter of the cone = 6 cm

∴ Radius of the cone = `6/2` = 3 cm

Now, volume of the cone,

A = 2V

= `1/3 pi"r"^2"h"`

= `1/3 pi(3)^2"h"_1`

⇒ V = `1/6 pi9"h"_1`

= `3/2 "h"_1pi`  ...(i)

And volume of the cone,

B = V

= `1/3 pi(3)^2(21 - "h"_1)`

= 3π(21 – h₁)  ...(ii)

From equations (i) and (ii),

`3/2 "h"_1 pi` = 3π(21 – h₁)

⇒ h₁ = 2(21 – h₁)

⇒ 3h₁ = 42

⇒ h₁ = `42/3` = 14 cm

∴ Height of cone,

B = 21 – h₁

= 21 – 14

= 7 cm

Now, volume of the cone,

A = `3 xx 14 xx 22/7` = 132 cm²   ...[Using equation (i)]

And volume of the cone,

B = `1/3 xx 22/7 xx 9 xx 7` = 66 cm³   ...[Using equation (ii)]

Now, volume of the cylinder

= πr²h

= `22/7 (3)^2 xx 21`

= 594 cm³

∴ Required volume of the remaining portion

= Volume of the cylinder – (Volume of cone A + Volume of cone B)

= 594 – (132 + 66)

= 396 cm³