a container opened at the top and made up of metal sheet, is in the shape of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively.find the cost of milk which can completely fill the container , at the rate of ₹ 50 per liter .also find the cost of metal sheet used to make the container ,if it cost ₹ 10 per 100 cm².
Answers
QUESTION IS NOT COMPLETE. PIE IS 3.14 . IT SHOULD BE MENTION IN THE QUESTION
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 :