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

Two identical thin rings each of radius R are coaxially placed at a separation of R. If the rings have a uniform mass distribution and masses m1 and m2 respectively, then the work done in moving a mass m from centre of the ring having mass m1 to center of the other ring is

A
0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Gm(22)(m1m2)2R
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Gm(22)(m2m1)2R
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Gm(22)(m1+m2)2R
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B Gm(22)(m1m2)2R
Work done will be equal to the change in the potential energy of the two rings and one mass system due to rearrangement.

When mass is at the centre of ring m1,
Initial potential energy,

Ui=U(between two rings)+U(between ring of mass m1 and mass m)
+U(between ring of mass m2 and mass m)

Ui=Gm1m2R2+R2Gm1mRGm2mR2+R2

Ui=Gm1m22RGm1mRGm2m2R

When mass is at the centre of ring m2,
Final potential energy,

Uf=U(between two rings)+U(between ring of mass m1 and mass m)
+U(between ring of mass m2 and mass m)

Uf=Gm1m2R2+R2Gm1mR2+R2Gm2mR

Uf=Gm1m22RGm1m2RGm2mR

So, work done,
W=ΔU=UfUi=Gm1m(22)2RGm2m(22)2R

W=Gm(22)(m1m2)2R

Hence, option (b) is the correct answer.

​​​​
Why this question?

The gravitational potential of a ring of radius R and mass M at a distance r from its centre on the axis is given by V=GMR2+r2;0r

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon