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: 10
        Steel ball is dropped from a height of 1 M on two horizontal nonconducting surface. Every time it bounces it reaches 80% of its previous height. Nearly by how much will the temperature of the ball rise after 4 bounces? Specific heat capacity of the ball is 0.1calorie per gram degree Celsius. Neglect loss in  heat to the surroundings and the floor
11 months ago

Answers : (3)

Arun
15903 Points
							
\DeltaH = mg × 1 × 0.2 + mg × 1 × 0.8 × 0.2 + mg × 0.8 × 0.8 × 0.2 .....(1)
\DeltaH = m × 500 \DeltaT .....(2)
2 + 1.6 + 1.28 = 500 \DeltaT
\DeltaT = 0.01°C
11 months ago
Sanju
106 Points
							Let `m` be the mass of the steel b
						
11 months ago
Arun
15903 Points
							
Delta H = mg* 0.1 *0.2 + mg*1 *0.8*0.2 +mg*0.8*0.8*0.2
Delta H = m *500 (delta T)
 
2 + 1.6 + 1.28 = 500 (delta T)
Delta T = 0.01 °c
 
 
Regards
Arun (askIITians forum expert)
11 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Get Extra Rs. 1,590 off

COUPON CODE: SELF10


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


Get Extra Rs. 64 off

COUPON CODE: SELF10

Ask Experts

Have any Question? Ask Experts

Post Question

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