Two identical spherical masses of mass M are kept at some distance. Potential energy of the system when a mass M is taken from the surface of one of the sphere to the other
A
increases continuously.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
decreases continuously.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
first increases, then decreases.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
first decreases, then increases.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C first increases, then decreases. Suppose the identical spheres are at distance d.
Let at any time t, mass M is at distance r from one of the sphere.
Potential energy of the system at this time, U=U(between two spheres)+U(between left sphere and mass M) +U(between right sphere and mass M)
⇒U=−GM2d−GM2r−GM2(d−r)
⇒U=−GM2d−GM2(1r+1d−r)
⇒U=−GM2d−GM2(dr(d−r))
Now let us take different cases.
(i) When mass is at point r=d4, U1=−6.3×GM2d ...........(1)
(ii) When mass is at point r=d2, U2=−5.0×GM2d ...........(2)
(iii) When mass is at point r=3d4, U3=−6.3×GM2d ...........(3)
Hence, option (c) is the correct alternative.
From (1), (2) and (3), we can see that potential energy of the system first increases then decreases.