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

# An elevator accelerates upwards at a constant rate. A uniform string of length L and mass m supports a small block of mass M that hangs from the ceiling of the elevator. The tension at distance l from the ceiling is T. The acceleration of the elevator is

A
TM+mmlLg
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
T2M+mmlL+g
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
C
TM+mlLg
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
D
T2Mm+mlLg
No worries! Weâ€˜ve got your back. Try BYJUâ€˜S free classes today!
Open in App
Solution

## The correct option is A TM+m−mlL−gLet the acceleration of the lift be ′a′ Mass of the lower portion of the string is m(L−l)L Now using newton's 2nd law Fnet=msystema ⇒T−Mg−m(L−l)Lg=[M+m(L−l)L]a ⇒a=TM+m(L−l)L−g

Suggest Corrections
3
Join BYJU'S Learning Program
Related Videos
What's with All the Tension in the Room?
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program