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A small sphere of radius `r' falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to

  • Option 1)

    r5

  • Option 2)

    r2

  • Option 3)

    r3

  • Option 4)

    r4

 

Answers (2)

best_answer

As we have learned

Net force on the body -

F_{B}+F_{v}= W


ightarrow 6pi eta rv+frac{4}{3}pi r^{3}sigma g= frac{4}{3}pi r^{3}
ho g


ightarrow 6pi eta rv=frac{4}{3}pi r^{3} gleft ( 
ho -sigma 
ight )


ightarrow v_{t}=frac{2}{9}frac{ r^{2} left ( 
ho -sigma 
ight )}{eta }g

 

- wherein

F_{B}-Buoyant : force

F_{v}-viscous : force

w-weight


ho 
ightarrow density : of : ball

sigma 
ightarrow density : of : water

V_{T} =terminal : velocity

 

 Rate of heat production = \vec{F}\cdot \vec{V}

= (6\pi \eta rv)\cdot v

= 6\pi \eta( rv^{2})

v\propto r^{2}

\therefore rate of heat \therefore \propto r^{5}

 

 

 

 

 


Option 1)

r5

This is correct

Option 2)

r2

This is incorrect

Option 3)

r3

This is incorrect

Option 4)

r4

This is incorrect

Posted by

prateek

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3

Posted by

Shital

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