Physics - Fluid Question with Solution | TestHub

Take downward as positive. Let r be the sphere radius and v its downward speed. In the Stokes model:

$$m=\frac{4}{3}\pi r^3\rho$$

$$B=\frac{4}{3}\pi r^3\sigma g$$

$$F_{\mathrm{visc}}=6\pi \eta rv$$

$$ma=mg-B-F_{\mathrm{visc}}$$

$$a=g\left(1-\frac{\sigma }{\rho }\right)-\frac{9\eta v}{2\rho r^2}$$

Just after entry, the sphere has nonzero speed from its fall. Drag therefore makes its acceleration smaller than the value stated in A. A is incorrect.

The relaxation time is:

$$\tau =\frac{m}{6\pi \eta r}$$

$$\tau =\frac{2\rho r^2}{9\eta }$$

This time scale depends on the sphere density, so B is incorrect. Exact terminal speed is approached asymptotically; time to attain it means a practical approach time.

Terminal force balance gives:

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

If entry speed exceeds terminal speed, the sphere slows down and the viscous force decreases. If entry speed is below terminal speed, the viscous force increases. Thus terminal viscous force need not be the maximum; C is incorrect.

At terminal speed acceleration and net force are zero, so D is correct. The incorrect statements are A, B and C.