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

Answer

` (8 pi g ^ 2 (rho _S- rho _L)^2 r^5)/(27 eta ) `

Explanation

` (8 pi g ^ 2 ( rho _S- rho_L) ^ 2 r ^ 5 )/(27 eta ) ` Terminal velocity ` v = (2r ^ 2 (rho _S- rho _L)g )/( 9 eta ) , ` Viscous force , ` F = 6 pi eta r v `
Since viscous force is dissipative, ` therefore `energy is dissipated in the form of heat.
Rate of production of heat = Fv, ` = (6 pi eta r v ) v = 6pi eta r v ^ 2 = 6pi eta r ((2)/(9) (r ^ 2 (rho _ S rho _L)g)/(eta )) ^2 = (8 pi g ^2 (rho _S- rho _L)^2 r ^5 )/(27 eta ) `