Question

A small sphere of radius \(r\) is dropped in a viscous liquid. When it is moving with terminal velocity in the liquid, the relation between the rate of heat produced \[ \left(\frac{dQ}{dt}\right) \] and \(r\) is

• A. \(\dfrac{dQ}{dt}\propto r^2\)
• B. \(\dfrac{dQ}{dt}\propto r^5\)
• C. \(\dfrac{dQ}{dt}\propto \dfrac{1}{r^2}\)
• D. \(\dfrac{dQ}{dt}\propto \dfrac{1}{r^5}\) \bigskip

Hint:

For a sphere falling through a viscous liquid, \[ v_t\propto r^2. \] Since \[ F_{\text{viscous}}\propto r\,v_t, \] we get \[ F_{\text{viscous}}\propto r^3. \] Therefore, \[ \frac{dQ}{dt}=F_{\text{viscous}}v_t \propto r^5. \]

Answer 1

The Correct Option is B

Concept: At terminal velocity, the loss of gravitational potential energy per second is completely converted into heat due to viscous drag. Hence, \[ \frac{dQ}{dt} = F_{\text{viscous}}\,v_t. \] According to Stokes' law, \[ F_{\text{viscous}} = 6\pi\eta r v_t. \]

Step 1: Write the expression for terminal velocity. For a sphere moving in a viscous liquid, \[ v_t = \frac{2r^2(\rho-\sigma)g}{9\eta}. \] Therefore, \[ v_t\propto r^2. \]

Step 2: Express the viscous force in terms of \(r\). Since \[ F_{\text{viscous}} = 6\pi\eta r v_t, \] and \[ v_t\propto r^2, \] we get \[ F_{\text{viscous}} \propto r^3. \]

Step 3: Calculate the rate of heat production. \[ \frac{dQ}{dt} = F_{\text{viscous}}\,v_t. \] Substituting the proportionalities, \[ \frac{dQ}{dt} \propto r^3\times r^2. \] \[ \frac{dQ}{dt} \propto r^5. \]

Step 4: Write the final answer. \[ \boxed{\frac{dQ}{dt}\propto r^5} \] \[ \boxed{\text{Answer = (B)}} \]