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A small sphere falls from rest in a visc...

A small sphere falls from rest in a viscous liquid. Due to frication, heat is produced. Find the relation between the rate of production of heat and the radius of the sphere at terminal velocity.

Text Solution

Verified by Experts

The correct Answer is:
`(dQ)/(dt)propr^(5)`
Terminal velocity `v_(T) = (2r^(2)g)/(9eta)(rho_(s) - rho_(L))`
and viscous force `F = 6pirv_(T)`
Viscous force is the dissipative force. Hence.
`(dQ)/(dt) = Fv_(T) = (6pietarv_(T))(v_(T)) = 6pietarv_(T)^(2)`
`= 6pietar{(2)/(9)(r^(2)g)/(eta)(rho_(s) - rho_(L))}^(2) = (8pig^(2))/(27eta)(rho_(s) - rho_(L))^(2)r^(5) = (dQ)/(dt)propr^(5)`
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Knowledge Check

  • 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

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  • When a sphere falling in a viscous fluid attains a terminal velocity, then

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    B
    the drag force balances the buoyant force
    C
    the drag force balances the weight of the sphere
    D
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