Tag: rotational equilibrium

Questions Related to rotational equilibrium

An object is said to be in equilibrium when :
1. There is no resultant force acting on the object.
2. The total clockwise moments about any point is equal to the total anti-clockwise moments about the same point.
3. The object has no energy.

rotational equilibrium option b: engineering physics motion of system of particles and rigid bodies equilibrium physics

  1. 1 only

  2. 2 only

  3. 1 and 2 only

  4. 1, 2 and 3

Show answer
Correct Option: C
Explanation:

An object is said to be in equilibrium when there is no resultant force acting on the object, the total clockwise moments about any point is equal to the total anticlockwise moments about the same point.

Which of the following object is/are in equilibrium?

rotational equilibrium option b: engineering physics motion of system of particles and rigid bodies equilibrium physics

  1. A sleeping cat

  2. A flask resting on a table

  3. A rocking see-saw

  4. A man walking

Show answer
Correct Option: A,B
Explanation:

A sleeping cat, a flask resting on a table and winding a clock are the objects is / are in equilibrium.

The bob of simple pendulum having length l, is displaced from mean position to an angular position with respect to vertical. If it is released, then velocity of bob at equilibrium position : 

rotational equilibrium option b: engineering physics motion of system of particles and rigid bodies equilibrium physics

  1. $\sqrt { 2g\ell (1-cos\theta ) }$

  2. $\sqrt { 2g\ell (1+cos\theta ) } $

  3. $\sqrt { 2g\ell cos\theta } $

  4. $\sqrt { 2g\ell } $

Show answer
Correct Option: A
Explanation:

Potential energy at extreme position = kinetic energy at mean position

$mg\ell (1-cos\theta )=\frac { 1 }{ 2 } m{ v }^{ 2 }$

If the potential energy of the molecule is given by $U = \dfrac {A}{r^{6}} - \dfrac {B}{r^{12}}$. Then at equilibrium position its potential energy is equal to

rotational equilibrium option b: engineering physics motion of system of particles and rigid bodies equilibrium physics

  1. $-A^{2}/ 4B$

  2. $A^{2}/ 4B$

  3. $2A/B$

  4. $A/2B$

Show answer
Correct Option: B

For the equilibrium condition shown. the cords are strong enough to withstand a maximum tension $100 N.$ The largest value of $W $ (in newton) that can be suspended is $W _1$ Find $W _0$ :

rotational equilibrium option b: engineering physics motion of system of particles and rigid bodies equilibrium physics

  1. $25$

  2. $35$

  3. $20$

  4. $60$

Show answer
Correct Option: B

A rope ladder with a length $\ell$ carrying a man ofmass $m$ at its end is attached to the basket ofballoon with a mass $\mathrm { M }$ . The entire system is in equilibrium in the air. As the man climbs up theladder into the balloon, the balloon descends bya height h. Then the potential energy of the man: 

rotational equilibrium option b: engineering physics motion of system of particles and rigid bodies equilibrium physics

  1. Increases by mg $( \ell - h )$

  2. Increases by mge

  3. Increases by mgh

  4. Increases by mg $( 2 \ell - h )$

Show answer
Correct Option: C

A wheel rolling on a horizontal surface is an example of

rotational equilibrium option b: engineering physics motion of system of particles and rigid bodies equilibrium physics

  1. Stable equilibrium

  2. Unstable equilibrium 

  3. Neutral equilibrium 

  4. All of the above

Show answer
Correct Option: B
Explanation:

Neutral equilibrium means that, with a small deviation, the body remains in equilibrium. An example is a wheel rolling on a horizontal surface. If you stop it at any point, the wheel will be in a state of equilibrium. A ball lying on a flat horizontal surface is in a state of neutral equilibrium.

In a stable equilibrium, the line of action of weight of the object lies _____ the base area of the object

rotational equilibrium option b: engineering physics motion of system of particles and rigid bodies equilibrium physics

  1. Inside

  2. outside

  3. cant say

  4. can be both

Show answer
Correct Option: A

Three copper blocks of masses ${ M } _{ 1 }$, ${ M } _{ 2 }$, and ${ M } _{ 3 }$, kg respectively are brought into thermal contact till they reach equilibrium. Before contact, they were at ${ T } _{ 1 }$,${ T } _{ 2 }$,${ T } _{ 3 }$ $\left( { T } _{ 1 }{ >T } _{ 2 }>{ T } _{ 3 } \right)$. Assuming there is no heat loss to the surroundings, the equilibrium temperature T is (s is specific heat of copper)     

rotational equilibrium option b: engineering physics motion of system of particles and rigid bodies equilibrium physics

  1. $T=\dfrac { { T } _{ 1 }{ +T } _{ 2 }+{ T } _{ 3 } }{ 3 } $

  2. $T=\dfrac { { { { M } _{ 1 }T } _{ 1 }{ +{ M } _{ 2 }T } _{ 2 }+{ M } _{ 3 }{ T } _{ 3 } } }{ { M } _{ 1 }+{ M } _{ 2 }+{ M } _{ 3 } } $

  3. $T=\dfrac { { { M } _{ 1 }T } _{ 1 }{ +{ M } _{ 2 }T } _{ 2 }+{ M } _{ 3 }{ T } _{ 3 } }{ 3\left( { M } _{ 1 }+{ M } _{ 2 }+{ M } _{ 3 } \right) } $

  4. $T=\dfrac { { { M } _{ 1 }T } _{ 1 }s{ +{ M } _{ 2 }T } _{ 2 }s+{ M } _{ 3 }{ T } _{ 3 }s }{ { M } _{ 1 }+{ M } _{ 2 }+{ M } _{ 3 } }$

Show answer
Correct Option: B
Explanation:
Let us assume that $T _1>T _2,T _3$ and $T _1>T>T _2,T _3$

Now heat loss by $M _1=$ Heat gained by $M _2$ and $M _3$

$M _1S(T _1-T)=M _2S(T-T _1)+M _3S(T-T _3)$

$\implies M _1T _1+M _2T _2+M _3T _3=(M _1+M _2+M _3)T$

$\implies T=\dfrac{M _1T _1+M _2T _2+M _3T _3}{M _1+M _2+M _3}$