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A right circular cone is divided by a plane parallel to its base into small cone of volume $V 1$ at the top and a frustum of volume $V 2$ as second part at the bottom. If $V 1 : V 2 = 1 : 3$ , then what is the ratio of the height of the altitude of small cone and that of frustum?

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213 − 1 1

412 − 1 1

413 − 1 1

413 − 1 2

answer is C.

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Detailed Solution

Given that

V 1 : V 2 = 1 : 3

.
We have the following formula,
Volume of the cone

= 13 π r 2 h

Volume of the frustum

= π 3 H (R 2 + R r + r 2)

Then,

Volume of small cone Volwone of frustum = 13 ....... (i) V 1 V 2 = 13 π r 2 h π 3 H (R 2 + r 2 + R r) ........ (i i)

Since both the cones are similar, their sides will be proportional,

⇒ r R = h P ..... (i i i) P = h + H ...... (i v)

We divide both the sides of equation (iv) by

h

P h = 1 + H h

Assuming that,

R r = x

Then, by comparing equation (iii) and (v) we get,

⇒ R r = 1 + H h ⇒ H h = x − 1 [∴ R r = x] ⇒ 1 (x − 1) (x 2 + x + 1) = 13 (x − 1) (x 2 + x + 1) = 3 x (x 2 + x + 1) − (x 2 + x + 1) = 3 ⇒ x 3 − 1 = 3 ⇒ x 3 = 4 x = 413

We substitute the above value in equation (vi) and get,

R r = 1 + H h ⇒ H h = 413 − 1 ⇒ 413 − 1 1

Thus, the ratio of the heights is

413 − 1 1

.
Therefore the correct option is 3.

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