wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

The given plot shows the variation of the potential energy (U) of interaction between two particles with respect to the distance (r) separating them. Then,


A
B and D are equilibrium points
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
C is a point of unstable equilibrium
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
The force of interaction between the two particles is attractive between points C and D and repulsive between D and E
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
The force of interaction between the particles is repulsive between points E and F.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D The force of interaction between the particles is repulsive between points E and F.
At B and D, dUdr0, so they are not equilibrium points.
At C, dUdr=0 and potential energy is minimum. So, it is a stable equilibrium position.
Now, F=dUdr= − (Slope of Ur curve)
In the region CD, the slope of the graph is positive, therefore by the relation, F is negative i.e, force is attractive in nature.
In the region DE, the slope of the graph is again positive, therefore by the relation, F is negative i.e, force is attractive in nature in the region DE too.
But in the region EF, the slope of the graph is negative which means F is positive i.e, force is repulsive in nature.

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Conservative Forces and Potential Energy
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon