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

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.

  1. 1 only

  2. 2 only

  3. 1 and 2 only

  4. 1, 2 and 3
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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 : 

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

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

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

  4. $\sqrt { 2g\ell } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Potential energy at extreme position = kinetic energy at mean position

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

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

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

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

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

  3. $2A/B$

  4. $A/2B$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At equilibrium, the force F = -dU/dr = 0. Differentiating U = A/r^6 - B/r^12 gives -6A/r^7 + 12B/r^13 = 0, so r^6 = 2B/A. Substituting this back into U gives A/(2B/A) - B/(2B/A)^2 = A^2/2B - B * A^2/4B^2 = A^2/2B - A^2/4B = A^2/4B.

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

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: 

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

  2. Increases by mge

  3. Increases by mgh

  4. Increases by mg $( 2 \ell - h )$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice rotational equilibrium option b: engineering physics motion of system of particles and rigid bodies equilibrium physics

A wheel rolling on a horizontal surface is an example of

  1. Stable equilibrium

  2. Unstable equilibrium 

  3. Neutral equilibrium 

  4. All of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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

  1. Inside

  2. outside

  3. cant say

  4. can be both
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an object to be in stable equilibrium, its center of gravity must be positioned such that the vertical line passing through it falls within the base of support.

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

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)     

  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. $T=\dfrac { { { M } _{ 1 }T } _{ 1 }s{ +{ M } _{ 2 }T } _{ 2 }s+{ M } _{ 3 }{ T } _{ 3 }s }{ { M } _{ 1 }+{ M } _{ 2 }+{ M } _{ 3 } }$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $M _3$

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

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

$\implies T=\dfrac{M _1T _1+M _2T _2+M _3T _3}{M _1+M _2+M _3}$
Multiple choice physics units and measurement: error analysis significant figures significant figures and rounding of digits units and measurements

Write the number of significant digits in 0.6464 :

  1. 4

  2. 3

  3. 5

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

All non-zero numbers are always significant.All zeroes before a non zero number are insignificant. All zeroes which are simultaneously to the right of the decimal point and at the end of the number are significant. The only non significant digit here is the zero before the decimal point.
Hence, number of significant digits is 4.