A small spherical ball of diameter $d$ falls from rest in any viscous liquid. During this, heat is produced due to viscous force. If the ball is moving with terminal velocity in the liquid then rate of production of heat is proportional to

Solution:

When a small spherical ball of radius $r$ (diameter $d=2r$) falls through a viscous liquid, it experiences a viscous drag force given by Stokes' Law:

$F_v = 6\pi\eta r v$

where $\eta$ is the coefficient of viscosity of the liquid and $v$ is the velocity of the ball.

The ball attains terminal velocity ($v_t$) when the viscous force balances the net downward force (gravitational force minus buoyant force).

Gravitational force: $F_g = mg = \rho_s V g = \rho_s (\frac{4}{3}\pi r^3) g$

Buoyant force: $F_b = \rho_l V g = \rho_l (\frac{4}{3}\pi r^3) g$

where $\rho_s$ is the density of the sphere and $\rho_l$ is the density of the liquid.

At terminal velocity, $F_v = F_g - F_b$

$6\pi\eta r v_t = (\rho_s - \rho_l) \frac{4}{3}\pi r^3 g$

From this, the terminal velocity is found to be:

$v_t = \frac{2 r^2 (\rho_s - \rho_l) g}{9\eta}$

Since $r = d/2$, we can substitute this into the expression for $v_t$:

$v_t = \frac{2 (d/2)^2 (\rho_s - \rho_l) g}{9\eta} = \frac{2 (d^2/4) (\rho_s - \rho_l) g}{9\eta} = \frac{d^2 (\rho_s - \rho_l) g}{18\eta}$

Thus, terminal velocity $v_t \propto d^2$.

The rate of production of heat is the power dissipated by the viscous force, which is given by:

$P = F_v \cdot v_t$

Substitute $F_v = 6\pi\eta r v_t$:

$P = (6\pi\eta r v_t) v_t = 6\pi\eta r v_t^2$

Now, substitute $r = d/2$:

$P = 6\pi\eta (d/2) v_t^2 = 3\pi\eta d v_t^2$

Finally, substitute the proportionality $v_t \propto d^2$ into the power equation:

$P \propto d (d^2)^2$

$P \propto d \cdot d^4$

$P \propto d^5$

Therefore, the rate of production of heat is proportional to $d^5$.