A spherical ball of radius 1 mm and density 10.5 g cc$$^{-1}$$ is dropped in glycerine of coefficient of viscosity 9.8 poise and density 1.5 g cc$$^{-1}$$. Viscous force on the ball when it attains constant velocity is $$3696 \times 10^{-x}$$ N. The value of $$x$$ is
(Given, $$g = 9.8$$ m s$$^{-2}$$ and $$\pi = \frac{22}{7}$$)

A spherical ball has radius $$r = 1$$ mm $$= 10^{-3}$$ m and density $$\rho_b = 10.5$$ g/cc $$= 10500$$ kg/m³. The glycerine has coefficient of viscosity $$\eta = 9.8$$ poise $$= 0.98$$ Pa·s and density $$\rho_g = 1.5$$ g/cc $$= 1500$$ kg/m³. Taking $$g = 9.8$$ m/s² and $$\pi = \frac{22}{7}$$.

Since at terminal velocity the net force is zero, the weight of the ball equals the sum of the buoyant force and the viscous force, so the viscous force is given by

$$\text{Viscous force} = \text{Weight} - \text{Buoyant force}.$$

Substituting the expressions for weight and buoyancy into this relation yields

$$F_{\text{viscous}} = \frac{4}{3}\pi r^3 (\rho_b - \rho_g) g$$

and hence

$$F_{\text{viscous}} = \frac{4}{3} \times \frac{22}{7} \times (10^{-3})^3 \times (10500 - 1500) \times 9.8 = \frac{4}{3} \times \frac{22}{7} \times 10^{-9} \times 9000 \times 9.8.$$

Now evaluating each factor:

$$\frac{4}{3} \times \frac{22}{7} = \frac{88}{21},\qquad 9000 \times 9.8 = 88200,$$

so that

$$F = \frac{88}{21} \times 88200 \times 10^{-9} = \frac{88 \times 88200}{21} \times 10^{-9} = \frac{7761600}{21} \times 10^{-9} = 369600 \times 10^{-9} = 3696 \times 10^{-7}\,\text{N}.$$

Therefore, comparing with the given form $$3696 \times 10^{-x}\,\text{N}$$ shows that $$x = \mathbf{7}.$$