Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 12th pass

                        

A piece of copper tubing (αcoper = 17 × 10−6/ ◦C) used in plumbing has a length of 60.0 cm and an inner diameter of 1.50 cm at 20◦ C. When hot water at 85◦ C flows through the tube, what are (a) the tube’s new length and (b) the change in its cross sectional area?

A piece of copper tubing (αcoper = 17 × 10−6/ ◦C) used in plumbing has a length of 60.0 cm and an inner diameter of 1.50 cm at 20◦ C. When hot water at 85◦ C flows through the tube, what are (a) the tube’s new length and (b) the change in its cross sectional area?

2 months ago

Answers : (1)

Pawan Prajapati
askIITians Faculty
5328 Points
							Answer ::

a) Use L = L0*(1 + α*ΔT) = 60.0cm*(1 + 1.7x10-6*(85 - 20)) = 60.0063cm

b) Use ΔA = A0*(2*α*ΔT) = π*(0.75cm)2(2*1.7x10-6*(85-20)) = 0.0003903 cm2


Thanks
2 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 18 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts
 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details