Question 56

A bucket in the shape of the frustum of a cone has its top and bottom radii as 20 cm and 10 cm, respectively. The depth of the bucket is 24 cm. The capacity of the bucket is: (Take $$\pi=\frac{22}{7}$$)

Solution

Given,

Capacity / Volume of bucket =

$$\frac{1}{3}\times\pi\times\left(R^2+r^2+Rr\right)\times\ H$$

$$=\frac{1}{3}\times\pi\times\left(20^2+10^2+20\times\ 10\right)\times\ 24$$

$$=\frac{1}{3}\times\frac{22}{7}\times\left(700\right)\times\ 24$$

$$=17600\ cm^3$$

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