The adjacent diagram represents a frustum of a cone whose bottom and top radii are 6 cm and 3 cm, its height is 8 cm. There is a conical cavity to a height of 3 cm at the bottom. The amount of material in the solid is

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132 π c m^{3}

168 π c m^{3}

100 π c m^{3}

135 π c m^{3}

answer is A.

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Detailed Solution

Given a frustum of a cone whose bottom and top radii are 6 cm and 3 cm and height is 8 cm.
Let R, r and H be top radius, bottom radius and height of frustum of a cone respectively.
Also,

{r_{1} and}^{} h

be radius and height of cone.
Question Image

Volume of material (V)

=

Volume of frustum

-

Vol. of cavity

\Rightarrow V = \frac{1}{3} ​ π (R^{2} + r^{2} + R . r) H - \frac{1}{3} ​ π {r_{1}}^{2} ​h

\Rightarrow V = \frac{1}{3} ​ π [6^{2} + 3^{2} + 6 \times 3] \times 8 - \frac{1}{3} ​ π \times 6 \times 6 \times 3

\Rightarrow V = \frac{1}{3} ​ π [63 \times 8 - 108]

\Rightarrow V = \frac{π ​}{3} ​ \times [504 - 108]

\Rightarrow V = \frac{π ​}{3} ​ \times 396

\Rightarrow V = 132 πc m^{3}

Therefore, amount of material in the solid is

132 πc m^{3}

.
Hence, the correct option is 1.

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