1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Two masses M and m (with M > m) are connected by means of a pulley as shown in the figure. The system is released from rest. At the instant when mass M has fallen through a distance h, the velocity of mass m will be

A

2gh

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
B

2ghMm

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C

2gh(Mm)(M+m)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

2gh(M+m)(Mm)

No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is C √2gh(M−m)(M+m) If mass m falls through a distance h, mass m rises up through the same distance h. Let v be the common velocity of the masses when this happens. Now, Loss in PE = Gain in KE ⇒Mgh−mgh=12(M+m)v2 ⇒v=√2gh(M−m)(M+m) Hence, the correct choice is (c).

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Conservative Forces and Potential Energy
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program