(1)

Stoke's law relates the backward dragging force F acting on a small sphere of radius $r$ moving through a viscous medium of viscosity $\eta$ with velocity $v$. 

$F=6\pi r\eta v$

$=\left(6\right)(3.14)\left(\frac{{\displaystyle 1\times {10}^{-3}}}{{\displaystyle 2}}\right)(0.8)\left(2\right)=15\times {10}^{-3}$ $\mathrm{N}$
 

(2)

Initially, when the ball starts falling down, due to the force of gravity, the speed of ball increases and the viscous drag force increases also (it depends on speed $F=6\pi \eta rv$). As the viscous force balances the force of gravity, the net acceleration of the ball is zero, and it starts moving with constant velocity called terminal velocity.
So, option (b) is correct.

(2)

Given, Mass of ball = M

Density of ball = $d$

Density of given wire = $\frac{d}{2}$

Viscous force = weight - buoyant force 

$=vdg-v\left(\frac{d}{2}\right)g$

$=vdg(1-\frac{1}{2})=\frac{vdg}{2}$                            ...(i)

Also mass, $M=vd$

So, from equation (i), 

$\text{Viscous force}=\frac{Mg}{2}$

(4)

Terminal velocity, $v=\frac{2r^2(\rho -\sigma )g}{9\eta }$

Ratio of terminal velocity of spherical metal balls, 

$\frac{v_1}{v_2}=\frac{\frac{2}{9}{{\displaystyle (}{\displaystyle 1}{\displaystyle )}}^2(8{\rho }_2-0.1{\rho }_2)}{\frac{2}{9}{{\displaystyle (}{\displaystyle 2}{\displaystyle )}}^2({\rho }_2-0.1{\rho }_2)}\Rightarrow \frac{v_1}{v_2}=\frac{7.9{\rho }_2}{4(0.9{\rho }_2)}=\frac{79}{36}$

(3)

The viscous drag force, $F=6\pi \eta rv;$

where $v=$ terminal velocity

$\therefore$   The rate of production of heat = power

              $=\text{force}\times \text{terminal velocity}$

$\Rightarrow$ $\text{Power}=6\pi \eta rv·v=6\pi \eta r{v}^2$

$\text{∵}$  Terminal velocity$v=\frac{2r^2(\rho -\sigma )g}{9\eta };$

$\text{Now, power}=6\pi \eta r\left[\frac{{\displaystyle 4r^4{(\rho -\sigma )}^2}}{{\displaystyle 81{\eta }^2}}{g}^2\right]$

or        Power $\propto r^5$