Volumes Practice Test 1

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 $cm^3$ ; 2000 $cm^2$
81100 $cm^3$ ; 2200 $cm^2$
81100 $cm^3$ ; 2400 $cm^2$
8171.43 $cm^3$ ; 2420 $cm^2$

Validate

Solution:

Slant height, l =

√{l^2+(R-r)^2}

√{(24)^2+(15-5)^2}

√{676}

= 26 cm
Now, Capacity of the dustbin =

{πh}/3 (R^2+r^2+Rr)

{22×24}/{7×3} (15^2+5^2+15×5)

{22×8}/7

(225 + 25 + 75)
=

{2×8}/7 × 325

{57200}/{7}

= 8171.43

cm^3

Surface area of the dustbin =

π[R^2+r^2)+l(R+r)]

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

{22}/7 [225 + 25 + 520]

{22}/7 × 770

= 2420

cm^2

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