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
Menu
Grade: 11
         A steel wire of original length 1 m and cross-sectional area 4.00 mm2 is clamped at the two ends so that it lies horizontally and without tension. If a load of 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression?

Y of the steel = 2.0 × 1011 N m-2. Take g = 10 m s-2.
5 years ago

Answers : (1)

Navjyot Kalra
askIITians Faculty
654 Points
							Sol. From figure cos θ = x/√x2 + l2 = x/l [1+ x2/l2]-1/2
= x / l  ….(1)
Increase in length ∆L = (AC + CB) – AB
Here, AC = (l2 + x2)1/2
So, ∆L = 2(l2 + x2)1/2 – 100 …(2)
Y = F/A l/∆l
From equation (1), (2) and (3) and the freebody diagram,
2l cosθ = mg.

						
5 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

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


Course Features

  • 110 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

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details