Equilibrium and Its Types

Q.

A particle moves in a potential region given by $U=8x^2-4x+400J$. Its state of equilibrium will be

1. x = 25 m

2. x = 0.25 m

3. x = 0.025 m

4. x = 2.5 m

Q.

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

1. 

2. 

3. 

4. 

Q.

The potential energy between two atoms in a molecule is gives by, U (x) = $\frac{a}{x^{12}}$ - $\frac{b}{x^6}$ , where a and b are postive constant and x is the distant between the atome. The system is in stable equilibrium when

1. x = 0

2. x =

3. x = ()

4. x =

Q. The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $U\left(x\right)=\frac{a}{x^{12}}-\frac{b}{x^6}$, where $a$ and $b$ are constants and $x$ is the distance between the atoms. If the dissociation energy of the molecule is $D=[U_{x=\infty }-U_{\text{at equilibrium}}]$, $D$ is

1. $\frac{b^2}{2a}$
2. $\frac{b^2}{12a}$
3. $\frac{b^2}{4a}$
4. $\frac{b^2}{6a}$

Q. The given plot shows the variation of the potential energy $\left(U\right)$ of interaction between two particles with respect to the distance $\left(r\right)$ separating them. Then,

1. $\text{B}$ and $\text{D}$ are equilibrium points
2. $\text{C}$ is a point of unstable equilibrium
3. The force of interaction between the two particles is attractive between points $\text{C}$ and $\text{D}$ and repulsive between $\text{D}$ and $\text{E}$
4. The force of interaction between the particles is repulsive between points $\text{E}$ and $\text{F}$.

Q. The potential energy between two atoms in a molecule is given by $U\left(x\right)=\frac{a}{x^{12}}-\frac{b}{x^6}$ where $a$ and $b$ are positive constants and $x$ is the distance between the atoms. The atom is in stable equilibrium when

1. $x={\left(\frac{2a}{b}\right)}^{\frac{1}{6}}$
2. $x={\left(\frac{11a}{5b}\right)}^{\frac{1}{6}}$
3. $x=0$
4. $x={\left(\frac{a}{2b}\right)}^{\frac{1}{6}}$

Q. The graph of potential energy ($U$) Vs position ($X$) is shown in the figure given below. Select the correct order regarding the equilibrium state of $\left(\text{a}\right),\left(\text{b}\right)$ & $\left(\text{c}\right)$.

1. Stable, Unstable, Neutral
2. Unstable, Stable, Neutral
3. Neutral, Stable, Unstable
4. Neutral, Unstable, Stable

Q. The given plot shows the variation of the potential energy $\left(U\right)$ of interaction between two particles with respect to the distance $\left(r\right)$ separating them. Then,

1. $\text{B}$ and $\text{D}$ are equilibrium points
2. $\text{C}$ is a point of unstable equilibrium
3. The force of interaction between the two particles is attractive between points $\text{C}$ and $\text{D}$ and repulsive between $\text{D}$ and $\text{E}$
4. The force of interaction between the particles is repulsive between points $\text{E}$ and $\text{F}$.

Q. The potential energy of a particle is given by relation $U=x^3-6x^2$ (In SI units). The nature of equilibrium at $x=0$ is

1. Stable
2. Unstable
3. Neutral
4. Undefined

Q.

A particle located in a one - dimentional potential field has its potential energy functions as U(x) = $\frac{a}{x^4}-\frac{b}{x^2}$, 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. 
   $\frac{a}{2b}$

Q. what is the meaning of equilibrium point ? the eqm point of sin ^2omega t is 1/2 means ? how to find out the equilibrium point ?

Q. The potential energy variation between two atoms as a function of the distance between them is given by $U\left(x\right)=U_0[{\left(\frac{a}{x}\right)}^{12}-2{\left(\frac{a}{x}\right)}^6],U_0,a>0$.
Find the equilibrium position and state the nature of equilibrium (stable or unstable).

1. $x_0=2a$ and stable
2. $x_0=a$ and stable
3. $x_0=2a$ and unstable
4. $x_0=a$ and unstable

Q. The potential energy between two atoms in a molecule is given by $U\left(x\right)=\frac{a}{x^{12}}-\frac{b}{x^6}$ where a and b are positive constants and x is the distance between the atoms. The atom is in stable equilibrium when

1. $x=\sqrt[6]{6\frac{11a}{5b}}$
2. $x=\sqrt[6]{6\frac{a}{2b}}$
3. $x=0$
4. $x=\sqrt[6]{6\frac{2a}{b}}$

Q.

A particle located in a one - dimentional potential field has its potential energy functions as U(x) = $\frac{a}{x^4}-\frac{b}{x^2}$, 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. 
   $\frac{a}{2b}$

Q. A thin circular ring of mass $M$ and radius $r$ is rotating about its axis with constant angular velocity $\omega$. Two objects each of mass $m$ are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with angular velocity given by
$\text{[Main-2010]}$

1. $\frac{(M+2m)\omega }{2m}$
2. $\frac{M\omega }{M+2m}$
3. $\frac{(M+2m)\omega }{M}$
4. $\frac{2M\omega }{M+2m}$

Q. Potential energy of a particle moving along $x-$ axis is given by $U=(\frac{x^3}{3}-4x+6)$.
Here, $U$ is in joules and $x$ in metres. Find the position of stable and unstable equilibriums, respectively.

1. $x=2\text{m}$ and $x=-2\text{m}$
2. $x=-2\text{m}$ and $x=2\text{m}$
3. $x=0$ and $x=0$
4. $x=1\text{m}$ and $x=-1\text{m}$