wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A body of mass 2 kg is attached to a light rod of length 5 m and is allowed to move in a vertical circle. The body is given an initial velocity of u m/s at the bottom-most point, such that the final velocity at the top-most point is zero. The value of u is
(Take g=10 m/s2)

A
52m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
10m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
102m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
202m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 102m/s
Here at the top most point velocity v=0,

From the law of conservation of energy, let's say the bottom-most point is A and the top-most point is B

(K.E)A+(U)A=(K.E)B+(U)B

Let potential at the bottom-most point be UA=0
12mu2+0=0+mg(2l)

where u is the initial velocity and 2l is the height of the mass from the bottom.
u=4gl
u=4×10×5
u=200
u=102m/s

Hence, option (C) is the correct answer.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon