Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
28.4k views
in Mathematics by (30.0k points)

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.

1 Answer

0 votes
by (130k points)
selected by
 
Best answer

Here, r = 5/ 2 cm = 2.5 cm = 0.025 m, h = 28 m. 

Total radiating surface in the system = total surface area of the cylinder 

= 2π r(h + r) 

= 2 × 22/ 7 × 0.025 (28 + 0.025) m

= 44x 0 025x 28. 025/ 7 m2 = 4.4 m2 (approx) An

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...