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: 12
        

A THIN ROD OF LENGTH 1M IS FIXED IN A VERTICAL POSITION INSIDE A TRAIN, WHICH IS MOVING HORIZONTALLY WITH CONSTANT ACCELERATION 4M/S2. A BEAD CAN SLIDE ON THE ROAD AND FRICTION COEFFICIENT BETWEEN THEM IS 0.5. IF THE BEAD IS RELEASED FROM REST AT THE TOP OF THE ROD, IT WILL REACH THE BOTTOM IN ............................................. 

8 years ago

Answers : (1)

vikas askiitian expert
509 Points
							

accleration of train = 4m/s2

bead will tend to move in horizontal direction due to seudo force = ma

normal reaction bw rod & bead = ma                  (a is accleration of train)

friction = u(N) = uma

weight = mg

net downward force = mg-uma

 m(net accleration) = m(g-ua)

     net accleration(A) = g - ua = 10-4*0.5 = 8m/s2

time taken by particle to reach the bottom is t

 s = At2/2                                (s=L =1m (given)

 1 = 8t2/2

t = 1/2 sec

8 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