Q : 8    In a hot water heating system, there is a cylindrical pipe of length \small 28\hspace{1mm}m and diameter \small 5\hspace{1mm}cm. Find the total radiating surface in the system.
 

Answers (1)

Given,

Length of the cylindrical pipe = l = \small 28\hspace{1mm}m

Diameter =  \small d = 5\hspace{1mm}cm = 0.05\ m

The total radiating surface will be the curved surface of this pipe.

We know,

The curved surface area of a cylindrical pipe of radius r and length l = 2\pi r l

\therefore CSA of this pipe = 

= 2\times\frac{22}{7}\times\frac{0.05}{2}\times28

\\ = 22\times0.05\times4 \\ = 4.4\ m^2

Therefore, the total radiating surface of the system is 4.4\ m^2

Most Viewed Questions

Related Chapters

Preparation Products

Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 40000/-
Buy Now
Knockout NEET 2025

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 45000/-
Buy Now
NEET Foundation + Knockout NEET 2024

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 54999/- ₹ 42499/-
Buy Now
NEET Foundation + Knockout NEET 2024 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
NEET Foundation + Knockout NEET 2025 (Easy Installment)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 3999/-
Buy Now
Boost your Preparation for JEE Main with our Foundation Course
 
Exams
Articles
Questions