Tag: motion of system of particles and rigid bodies

Quizzes (12) Questions (32)

Questions Related to motion of system of particles and rigid bodies

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 $M _3$

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

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

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