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Question : 35 of 40

Marks: +1, -0

The diameters of the ends of a dustbin 24 cm high, which is in the shape of a frustum of a cone are 30 cm and 10 cm. Determine its capacity and surface area.

(Use π = 22/7)

8000 $\text{cm}^3$ ; 2000 $\text{cm}^2$
81100 $\text{cm}^3$ ; 2200 $\text{cm}^2$
81100 $\text{cm}^3$ ; 2400 $\text{cm}^2$
8171.43 $\text{cm}^3$ ; 2420 $\text{cm}^2$

Solution: 👈: Video Solution

Slant height, l =

\sqrt{l^2+(R-r)^2}

=

\sqrt{(24)^2+(15-5)^2}

=

\sqrt{676}

= 26 cm

Now, Capacity of the dustbin =

\frac{\pi h}{3} (R^2+r^2+Rr)

=

\frac{22 \times 24}{7 \times 3} (15^2+5^2+15 \times 5)

=

\frac{22 \times 8}{7}

(225 + 25 + 75)

=

\frac{2 \times 8}{7} \times 325

=

\frac{57200}{7}

= 8171.43

\text{cm}^3

Surface area of the dustbin =

\pi [R^2+r^2)+l(R+r)]

=

\frac{22}{7}[(15)^2+(5)^2+26(15+5)]

=

\frac{22}{7} [225 + 25 + 520]

=

\frac{22}{7} \times 770

= 2420

\text{cm}^2

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