Question

a container opened at the top and made up of a metal sheet,is in the form of a frustum of a cone of height 16 cm with radii of each lower and upper ends as 8cm and 20cm respectively. Find the cost of milk which can completely fill the container, at the rate of rupees 50 per litre. Also find the cost of metsl sheet used to make the container, if it costs rupees 10per 100 cm square.​ (take π = 3.14)

Answer 1

u can do this by

step by step explanation

Answer 2

Answer:

Step-by-step explanation:

Given the frustum we need to visualize that we have two cones.

The larger cone and the smaller cone.

We have larger radii and the smaller radii hence the linear scale factor is given by:

LSF = 20/8 = 2.5

Let the height of the smaller cone be x.

The height of the larger cone is given by :

Height = x + 16

Using the linear scale factor we can get the value of x as follows:

(x + 16)/x = 2.5

x + 16 = 2.5x

2.5x - x = 16

1.5x = 16

x = 10.67cm

Height of the larger cone = 16 + 10.67 = 26.67 cm

Volume of a cone = 1/3πr²h

Volume of the larger cone = 1/3 × 3.142 × 20² × 26.67 = 11172.952 cm³

Volume of smaller cone = 3.142 × 1/3 × 8² × 10.67 = 715.20 cm³

Volume of the frustum = 11172.952 - 715.20 = 10457.752 cm³

1 liter = 1000 cm³

10457.752 / 1000 = 10.457752 l

1 liter = rs 50

Total cost = 50 × 10.457752 = 522.8876

Rs 522.8876

For the surface area we need the lateral height of the cones.

By pythagoras theorem we get the heights as follows:

The larger cone = √(20² + 26.67²) = 33.34 cm

The smaller cone = √(8² + 10.67²) = 13.34 cm

Lateral are of the cone = 2πrl

Smaller cone = 2 × 3.142 × 8 × 13.34 = 670.63 cm²

Larger cone = 2 × 3.142 × 20 × 33.34 = 4190.171 cm²

Area of the frustum = 4190.171 - 670.63 = 3519.54 cm²

Area of the bottom of the container = 3.142 × 8² = 201.088 cm²

Total area = 201.088 + 3519.54 = 3720.628 cm²

Total cost of painting is :

3720.628/100 × 10 = Rs 372.062