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A bucket is in the form of a frustum of a cone with a capacity of 12308.8 ${cm}^{3}$ of water. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of the metal sheet used in its making. (Use π = 3.14)

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12 cm

2420.32 {cm}^{3}

10 cm

2030.32 {cm}^{3}

15 cm

2160.32 {cm}^{3}

None of these

answer is C.

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Detailed Solution

Let the height of bucket = h cm
Question Image

From the question, we have Radius of top circular end(

r_{1}) = 20 cm

Radius of bottom circular end(

r_{2}) = 12 cm

Capacity of the bucket

= 12308.8 {cm}^{3}

As we know that,
Volume of frustum cone

= \frac{1}{3} π (r_{1}^{2} + r_{2}^{2} + r_{1} r_{2}) h

= \frac{1}{3} π (20^{2} + 12^{2} + 20 \times 12) h = \frac{784 πh}{3} {cm}^{3}

According to question,

\frac{784 π h}{3} = 12308.8

\Rightarrow h = \frac{12308.8 \times 3}{784 \times 3.14} = 15 cm

∴

height of the bucket

(h) = 15 cm

Let the slant height of bucket =

l cm

l^{2} = {(r_{1} - r_{2})}^{2} + h^{2}

\Rightarrow l = \sqrt{{(r_{1} - r_{2})}^{2} + h^{2}}

\Rightarrow l = \sqrt{{(20 - 12)}^{2} + 15^{2}} = \sqrt{289} = 17 cm

∴

Curved surface area

=

π (r_{1} + r_{2}) l + π r_{2}^{2}

= 3.14 \times (20 + 12) 17 + 3.14 \times 12^{2}

= 3.14 \times (9248 + 144) = 2160.32 {cm}^{3}

Hence, the height of the bucket and the area of the metal sheet are

15 cm

and

2160.32 {cm}^{3}

respectively.

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