Surface Areas & Volumes | Exercise 13.2 | Q7

Question 7: The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) Its inner curved surface area
(ii) The cost of plastering this curved surface at the rate of Rs.40 per m²

Explanatory Answer

Curved surface area of cylinder = 2πrh
Inner diameter of the circular well = 3.5 m
So, inner radius of the base of the cylinder, r = \\frac{3.5}{2}) = \\frac{7}{4}) m
'h' is the height of the cylinder. In this question, the height of the cylinder is the depth of the well = 10 m
Take π = \\frac{22}{7})

(i) Inner curved surface area of well = 2 × \\frac{22}{7}) × \\frac{7}{4}) × 10
\\frac{44 × 10}{4}) = 110 m²

(ii) Cost of plastering @ Rs.40 per m² = 110 × 40 = Rs.4400

Exercise 13.1 | Surface Area of Cubes & Cuboids

1. Class 9 Exercise 13.1 Question 3 | Height of hall ▶
2. Class 9 Exercise 13.1 Question 6 | TSA of Herbarium ▶
3. Class 9 Exercise 13.1 Question 7 | Shanti Sweets ▶

Exercise 13.2 | Surface Area of Cylinders

1. Class 9 Exercise 13.2 Question 3 | CSA TSA of Metal Pipe ▶
2. Class 9 Exercise 13.2 Question 4 | Roller & Playground ▶
3. Class 9 Exercise 13.2 Question 9 | Cylindrical Petrol Tank ▶
4. Class 9 Exercise 13.2 Question 10 | CSA of Decorative Lampshade ▶

Exercise 13.3 | Surface Area of Cones

1. Class 9 Exercise 13.3 Question 5 | Length of Tarpaulin 3 m wide ▶
2. Class 9 Exercise 13.3 Question 8 | Cost of painting conical barricades ▶

Exercise 13.4 | Surface Area of Spheres & Hemispheres

1. Class 9 Exercise 13.4 Question 8 | Outer CSA of hemispherical bowl ▶
2. Class 9 Exercise 13.4 Question 9 | Surface area of sphere & Cylinder ▶