The potential energy \(\mathrm{U}\) of a system is given by $\mathrm{U}=$ $\mathrm{A}$ $-$ ${\mathrm{Bx}}^2$ (where \(\mathrm{x}\) is the position of its particle and \(\mathrm{A},\) \(\mathrm{B}\) are constants). The magnitude of the force acting on the particle is:
1. constant

2. proportional to \(\mathrm{x}\)

3. proportional to ${\mathrm{x}}^2$

4. proportional to $\left(\frac{1}{\mathrm{x}}\right)$

The potential energy of a particle varies with distance \(r\) as shown in the graph. The force acting on the particle is equal to zero at:

1. \(P\)
2. \(S\)
3. both \(Q\) and \(R\)
4. both \(P\) and \(S\)

A particle is moving such that the potential energy U varies with position in metre as U (x) = ($4{\mathrm{x}}^2$ - 2x + 50) J. The particle will be in equilibrium at: 
1. x = 25 cm
2. x = 2.5 cm
3. x = 25 m
4. x = 2.5 m

The figure shows the potential energy function U(x) for a system in which a particle is in a one-dimensional motion. What is the direction of the force when the particle is in region AB? (symbols have their usual meanings)

1.  The positive direction of x

2.  The negative direction of X

3.  Force is zero, so direction not defined

4.  The negative direction of y

A particle of mass 'm' is moving in a horizontal circle of radius 'r' under a centripetal force equal to –K/r², where K is a constant. The total energy of the particle will be:

1. $\frac{K}{2r}$

2. $-\frac{K}{2r}$

3. $-\frac{K}{r}$

4. $\frac{K}{r}$