Join now for JEE/NEET and also prepare for Boards Join now for JEE/NEET and also prepare for Boards. Register Now
Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
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 ............................................. 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 .............................................
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 .............................................
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
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
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -