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

A monkey of mass 12 kg climbs up a light rope as shown in figure. The rope passes over a light and frictionless pulley and it is attached to a bunch of bananas having mass 16 kg. What will be maximum acceleration of monkey without lifting the bunch of bananas? (Take g=10 m/s2)


A
2.33 m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3.33 m/s2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
4.33 m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
5.33 m/s2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 3.33 m/s2
Mass of monkey (m)= 12 kg
Mass of banana bunch (M)= 16 kg
From FBD of banana bunch:


The banana bunch will just leave the ground when N=0
Applying equilibrium condition in vertical direction and putting N=0 for just leaving the contact surface

T+N=Mg
T=Mg
T=16×10=160 N .........(i)

From F.B.D. of monkey, the maximum acceleration will correspond to T=160 N in string.


From Newton's 2nd law:
Tmg=ma
T12g=12×a
T120=12×a ...........(ii)

By equation (i) and (ii):
160120=12×a
a=4012=3.33 m/s2
This is the maximum acceleration of monkey for which banana bunch will not be lifted.

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon