A uniform chain of length l and mass m lies on a smooth table. A very small part of this chain hangs from the table. It begins to fall under the weight of the hanging end. Find the velocity of the chain when the length of the hanging part becomes y.
A
y√gl
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2y√gl
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
y√2gl
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
y√g2l
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Ay√gl Let, mass of chain =m When length of the hanging part =y, mass of hanging part of chain =myl According to principle of energy conservation : Loss in gravitational potential energy of hanging part = Gain in KE of the whole chain ⇒(myl)g(y2)=12mv2 (because COM of hanging part has fallen by a distance of y2) ⇒v2=gy2l⇒v=y√gl