Question

A container is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the capacity and surface area of the container. Also, find the cost of milk which can completely filled the container at the rate of ₹ 25 per litre.
[Take π = 3.14].
(Class 10 Maths Sample Question Paper)

Answer 1

Given:

Radius (R) of upper end of the frustum of cone  = 20 cm
radius(r)of the lower end of the frustum of cone = 10 cm
Height of The frustum(h) = 30 cm

Volume of the frustum = 1/3πh[R² + r² + R×r]
= ⅓ × 3.14 × 30[20² + 10² + 20×10]
= 10 × 3.14[400 + 100 + 200]
= (31.4 ×700)
= 21980 cm³

= 21980/1000 = 21.980 L    [ 1L = 1000 cm³]

Now, Slant height 'l' of container= √(R - r)² + h²
l = √(20 - 10)² + 30²
l = √10² + 30²
l = √1000
l = 31.62 cm

Surface area of the container =curved surface area of the container + surface area of the bottom
= π(R + r)l + πr²  

= 3.14 × (20 + 10)× 31.62 + 3.14 × 10²
=  3.14 × (30)× 31.62 + 3.14 × 100
= 3.14 [ 30 × 31.62 + 100]
= 3.14 [ 948.60 + 100]
= 3.14 × 1048.60
= 3292.60 cm²

Cost of 1 L of  milk = ₹ 25
Total cost of 21.980  L  milk = 21.980 × 25
= ₹ 549.50

Answer 2

I hope this will help you