Question

A particle moving along x-axis is being acted upon by one dimensional conservative force F. In the f - x curve shown, four points J, K, L, M are marked on the curve. Column II gives different type of equilibriums for the particle at different positions. Column I gives certain positions on the force position graph. Match the position in Column - I with the corresponding nature of equilibriums in Column - II.

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{(A) Point J is position of}& \text{(p) Neutral equilibrium}\\ \text{(B) Point K is position of}& \text{(q) Unstable equilibrium}\\ \text{(C) Point L is position of}& \text{(r) Stable equilibrium}\\ \text{(D) Point M is position of}& \text{(s) No equilibrium}\end{array}$

• A. $A\to r;B\to q;C\to s;D\to p$
• B. $A\to q;B\to s;C\to r;D\to p$
• C. $A\to s;B\to q;C\to r;D\to p$
• D. $A\to s;B\to p;C\to r;D\to q$

Answer 1

The correct option is C $A\to s;B\to q;C\to r;D\to p$

At $J:$
Force and displacement are changing continuously, so its not a state of equlibrium.

At $K:$
We see that, the slope of $dF/dx$ graph is positive, which means that $dU/dx=+ve$ at that point. So at $K$, the particle will have unstable equilibrium.

At $L:$
We see that, the slope of $dF/dx$ graph is negative, which means that $dU/dx=-ve$ at that point. So at $L$, the particle will have stable equilibrium.

At $M:$
We see that, the force acting is zero, hence there will be a neutral equilibrium at this point.

$\therefore A\to s;B\to q;C\to r;D\to p$