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: 12th pass
        
A ball of relative density of .8 had fell in water from 2 metre. Find the depth of water
 
one year ago

Answers : (1)

Arun
23013 Points
							

Speed of the ball

 V = sqrt (2 gh)

= 6.32 m/s

Buoyancy force by water try to stop the ball. 

Buoyancy force = weight of displaced water

 = (d V)g

where d = density of water

V = volume of the ball  

g = 10 m/s2

deceleration of the body by buoyancy force, a = (dVg)/ m

where m= d'V

d' = density of block

a = dVg/(d' V)

= dg/d'

=(d/d')*g

=g/(0.8)

= 10/0.8  (Given, d'/d = 0.8)

= 12.5 m/s

Net deceleration of ball,a' = a-g = 2.5 m/s2

Final speed of ball v' = 0

Use v' = v2 + 2a's 

s= depth of ball in the water

=> 40 = 0 + 2*2.5*s

=> s = 8m 

one year 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

Post Question

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