A small sphere falls from rest in a viscous liquid. Due to friction, heat is prodced. Find the relation between the rate of production of heat and the radius of the sphere at terminal velocity.
Correct Answer – C
Rate of heat produced `=` power loss against vicous force
`implies(dQ)/(dt)=F_vxxv_T`
where the terminal velocity `v_T=(2r^2)/(9eta)(rho-sigma)g`
and viscous force `F_v=6pietarv_T`
Hence `(dQ)/(dt)=7pietarv_Txxv_T` or `v_Talphar^2`
`implies(dQ)/(dt)proprv_T^2` but `v_Tpropr^2`
Hence `(dQ)/(dt)propr^5`
Viscous force on a falling sphere in a liquid `F=pietav_(T)`
where `v_(T)=(2.9)r^(2)(rho-sigma)g.eta` is terminal velocity `rho=` density of sphere `sigma=`density of liquid
Rate of production of heat `=(=”power”)=Fv_(T)=6pietarv_(1)^(2)`
`=6pietar[2/9(r^(2)(rho-sigma)g)/eta]^(2)=8/27(pig^(2)(rho-sigma)^(2))/etar^(5)`
clearly
`(dQ)/(dt)propr^(5)`