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Question

The given plot shows the variation of U, the potential energy of interaction between two particles with the distance separating them be $$r$$, then which of the following is correct?
(1) B and D are equilibrium points
(2) C is a point of stable equilibrium
(3) The force of interaction between the two particles is attractive between points C and D and repulsive between points D and E on the curve.
(4) The force of interaction between the particles is repulsive between points E and F on the curve.

188714_7477f18cd4e144068005d43cdce9d116.png


A
1 and 3
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B
1 and 4
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C
2 and 4
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D
2 and 3
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Solution

The correct option is C 2 and 4
$$we\quad know\quad that\quad dU=-\vec { F } \cdot \vec { dr } \quad \Rightarrow F=-\frac { dU }{ dr } \\ For\quad equilibrium\quad position\quad F=0\quad \Rightarrow \frac { dU }{ dr } =0\\ So,\quad point\quad C\quad is\quad the\quad equilibrium\quad position\\ Also\quad \frac { { d }^{ 2 }U }{ d{ x }^{ 2 } } >0\quad at\quad point\quad C,\quad Therefore\quad point\quad C\quad is\quad the\quad point\quad of\quad stable\quad equilibrium\\ between\quad D\quad and\quad E,\quad dU>0\quad \Rightarrow \quad -\vec { F } \cdot \vec { dr } >0\quad \Rightarrow \vec { F } \cdot \vec { dr } <0\\ \Rightarrow \quad Both\quad \vec { F } \quad and\quad \vec { dr } \quad are\quad in\quad opposite\quad direction\\ Therefore\quad between\quad D\quad and\quad E\quad force\quad is\quad attractive\\ Similarly,\quad between\quad points\quad E\quad and\quad F\quad \vec { F } \cdot \vec { dr } >0\\ \Rightarrow \quad Both\quad \vec { F } \quad and\quad \vec { dr } \quad are\quad in\quad the\quad same\quad direction\\ Therefore\quad between\quad E\quad and\quad F\quad force\quad is\quad repulsive$$

Physics

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