23 Jan - Work, Energy and PowerContact Number: 9667591930 / 8527521718

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A particle is moving along the x-axis under a conservative force and its potential energy U varies with x co-ordinate as shown in the figure. Then force is positive at:

(1) A

(2) C, D

(3) B

(4) D, E

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 $\left(\frac{1}{\mathrm{x}}\right)$

If U is the potential energy of a particle and x is its displacement, then in the position of stable equilibrium,

1. $\frac{\mathrm{dU}}{\mathrm{dx}}=0\mathrm{and}\frac{{\mathrm{d}}^{2}\mathrm{U}}{{\mathrm{dx}}^{2}}0$

2. $\frac{\mathrm{dU}}{\mathrm{dx}}=0\mathrm{and}\frac{{\mathrm{d}}^{2}\mathrm{U}}{{\mathrm{dx}}^{2}}0$

3. $\frac{\mathrm{dU}}{\mathrm{dx}}=0\mathrm{and}\frac{{\mathrm{d}}^{2}\mathrm{U}}{{\mathrm{dx}}^{2}}=0$

4. All of these

The given plot shows the variation of U, the potential energy of interaction between two particles with the distance separating them, r

1. B and D are equilibrium points

2. C is a point of stable equlibrium

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 attractive between points E and F on the curve.

A particle located in a one-dimensional potential field has its potential energy function as $U\left(x\right)=\frac{a}{{x}^{4}}-\frac{b}{{x}^{2}}$, where a and b are positive constants. The position of equilibrium x corresponds to

1. $\frac{b}{2a}$

2. $\sqrt{\frac{2a}{b}}$

3. $\sqrt{\frac{2b}{a}}$

4. $\frac{a}{2b}$

In the figure potential energy of a particle varies with \(\mathrm{x}\)-co-ordinates. The particle moves under the effect of conservative force along the \(\mathrm{x}\)-axis.

Statement-I: |
If the particle is released at origin it will move in the negative x-direction. |

Statement-II: |
\(\mathrm{x}=-5\) and \(\mathrm{x}=+5\) both are stable equilibrium positions of the particle. |

1. | Statement-I is correct only. |

2. | Statement-II is correct only. |

3. | Both statements are correct. |

4. | Both statements are wrong. |

The points of maximum and minimum attraction in the curve between potential energy (*U*) and distance (*r*) of a diatomic molecules are respectively -

(1) *S *and *R*

(2) *T* and *S*

(3) *R* and *S*

(4) *S* and *T*

A particle free to move along the *x*-axis has potential energy given by $U\left(x\right)=k[1-{e}^{-{x}^{2}}]$ for $-\infty \le x\le +\infty $, where *k* is a positive constant of appropriate dimensions. Then

(1) At point away from the origin, the particle is in unstable equilibrium

(2) For any finite non-zero value of *x*, there is a force directed away from the origin

(3) If its total mechanical energy is *k*/2, it has its minimum kinetic energy at the origin

(4) For small displacements from *x *= 0, the motion is simple harmonic

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