The given plot shows the variation of U, the potential energy of interaction between two particles w.r.t. the distance of separation, r.
1.
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
B and D are equilibrium points.
2.
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
C is a point of stable equilibrium.
3. The force of interaction between the two particles is attractive between points
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
C and D and repulsive between
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
D and E .
4. The force of interaction between particles is repulsive between points
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
E and F .
Which of the above statements are correct?
The given plot shows the variation of U, the potential energy of interaction between two particles w.r.t. the distance of separation, r.
1.
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
B and D are equilibrium points.
2.
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
C is a point of stable equilibrium.
3. The force of interaction between the two particles is attractive between points
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
C and D and repulsive between
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
D and E .
4. The force of interaction between particles is repulsive between points
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
E and F .
Which of the above statements are correct?
The correct option is C 2 and 4
In stable equilibrium, PE is minimum (local minima),
At C, potential energy is minimum. So it is a stable equilibrium position.
Further, the relation between force and the potential energy is given by,
F=−dUdr=−[slope of (U-r) graph].
Negative force means attraction and positive force means repulsion.
Clearly, from the graph, the slope of (U−r) graph is negative between E to F. So, the force will be positive.
Hence, the force of interaction between particles is repulsive between points
<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}-->
E and F .
Hence, option (c) is the correct answer.
Key concepts-
In a conservative force field, the force and potential energy at a position are related as:
→F=−dUdr^r [ along position vector r]
In this expression at the state of equilibrium of body we use
→F=0⇒dUdr=0
Stable euilibrium (local minima): When a particleis displaced slightly from a equilibrium position and a force acting on it brings it back to its initial position, it is said to be stable equilibrium position.
Unstable euilibrium (local maxima): When a particleis displaced slightly from the equilibrium position and a force acting on it pushes it further it is said to be unstable equilibrium position.
Necessary condition :
For local minima :
Just after the equilibrium point slope of dUdr is =+ve and ddr(slope(U−r))=d2Udr2=+ve
F=−ddrdUdr=d2Udr2=−ve
For local maxima :
Just after the equilibrium point slope of dUdr is =−ve and ddr(slope(U−r))=d2Udr2=−ve
F=−ddrdUdr=d2Udr2=+ve
Hence, C, E and F are equilibrium points.
C, is stable equilibrium point , E is unstable equilibrium point and F is neutral equilibrium point.
In stable equilibrium, PE is minimum (local minima),
At C, potential energy is minimum. So it is a stable equilibrium position.
Further, the relation between force and the potential energy is given by,
F=−dUdr=−[slope of (U-r) graph].
Negative force means attraction and positive force means repulsion.
Clearly, from the graph, the slope of (U−r) graph is negative between E to F. So, the force will be positive.
Hence, the force of interaction between particles is repulsive between points <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> E and F .
Hence, option (c) is the correct answer.
| Key concepts- In a conservative force field, the force and potential energy at a position are related as: →F=−dUdr^r [ along position vector r] In this expression at the state of equilibrium of body we use →F=0⇒dUdr=0 Stable euilibrium (local minima): When a particleis displaced slightly from a equilibrium position and a force acting on it brings it back to its initial position, it is said to be stable equilibrium position. Unstable euilibrium (local maxima): When a particleis displaced slightly from the equilibrium position and a force acting on it pushes it further it is said to be unstable equilibrium position. Necessary condition : For local minima : Just after the equilibrium point slope of dUdr is =+ve and ddr(slope(U−r))=d2Udr2=+ve F=−ddrdUdr=d2Udr2=−ve For local maxima : Just after the equilibrium point slope of dUdr is =−ve and ddr(slope(U−r))=d2Udr2=−ve F=−ddrdUdr=d2Udr2=+ve Hence, C, E and F are equilibrium points. C, is stable equilibrium point , E is unstable equilibrium point and F is neutral equilibrium point. |
