To solve this problem use π=22/7

In this problem a bucket is in the form of a frustum of a cone.

This question was asked in 2005 cbse board exam. Question from RS Aggarwal book chapter volume and surface area of solid, page number 823, exercise 17C, problem number 7

solution

given data

Depth of the bucket = height of frustum = h = 15 cm

Diameter of top of bucket = 56 cm

Radius of top = R = 56/2 = 28 cm

Diameter of bottom of bucket = 42 cm

Radius of bottom = r = 42/2 = 21 cm

Volume of frustum of cone = [1/3] π h(R

^{2}+ r^{2}+ Rr)= [1/3] * [22/7] * 15(28

^{2}+ 21^{2}+ 28×21)= 22 * 5* 259

= 28490

Volume of water bucket can hold = volume of bucket which is in form of frustum

∴volume of water bucket can hold is 28490 cm

^{3}∴ volume of water bucket can hold is 28.49 liter