# In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

**Solution:**

Since the well is cylindrical its curved surface area with base radius 'r' and height 'h' is 2πrh.

Diameter of the pipe, d = 5 cm

Radius of the pipe, r = d/2 = 5/2 cm = 2.5 cm = 2.5/100 m = 0.025 m

Length of the pipe, h = 28 m

The total radiating surface area of the pipe = 2πrh

= 2 × 22/7 × 0.025 m × 28 m

= 4.4 m²

Thus, the total radiating surface is 4.4 m².

**Video Solution:**

## In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

### Class 9 Maths NCERT Solutions - Chapter 13 Exercise 13.2 Question 8:

**Summary:**

It is given that in a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. We have found that the total radiating surface is 4.4 m².