Tag: motion of system of particles and rigid bodies

Questions Related to motion of system of particles and rigid bodies

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

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

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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 $M _3$

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

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

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