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A solid cone of base radius $10 cm$ is cut into two parts through the midpoint of its height by a plane parallel to its base. What is the ratio of the volumes of the two parts of the cone?

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1:6

1:7

7:1

3:7

answer is B.

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Detailed Solution

Given that the radius of the base of the cone is 10cm.
Question Image

We have the following formulae,
Volume of the frustum cone

= 13 π H R 2 + r 2 + R r

Volume of the cone

= 13 π r 2

R = 10 cm Δ OAB ∼ Δ O CO O A O C = A B C D ⇒ r 10 = h 2 h ⇒ r = 5 cm

The cone is cut into two points through the mid points of its height,

H = h 2

Volume of the frustum of the cone will be,

= 13 × π × h 2 102 + 52 + 10 × 5 = 13 × π × h 2 [175] ⇒ 175 h π 6

Volume of the cone will be,

= 13 × π × 52 × h 2 ⇒ 25 h π 6

The ratio will be,

25 h π 6 175 π h 6 = 17

Thus, the ratio is 1:7.
Therefore the correct option is 2.

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