# What is rate of heat produced in terms of radius $r$ in viscous liquid under gravity?

what is the dependency of the rate of heat produced in terms of radius $$r$$ after the drop attains terminal velocity when A small ball of radius $$r$$ is falling in a viscous liquid under gravity?

My efforts include:

The forces acting on the sphere are its weight $$\dfrac 4 3πr^3ρg$$ downwards, buoyancy force $$\dfrac 4 3 \pi r^3\sigma g$$ upwards, and viscous force $$6π\eta rv$$ upwards. The sphere attains the terminal velocity it when the resultant force on it is zero i.e.,$$\dfrac 4 3πr^3ρg=\dfrac43πr^3σg+6πηrvt$$

I don't know how to Solve the above equation to get the terminal

• Hi Aakash Thoriya. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. – Qmechanic Jun 1 at 10:18
• @Qmechanic okay – Aakash Thoriya Jun 1 at 10:28
• @Qmechanic guide me, how can I solve this question? – Aakash Thoriya Jun 1 at 10:29
• The heat generation is equal to the integral of the drag force integrated with respect to the differential displacement. – Chet Miller Jun 1 at 14:48