Question

When body falls in liquid with terminal velocity, the ratio of resistive force of liquid to its weight is-

• A. $\frac{2r^{2}\rho_{s}g}{9\eta^{2}}$
• B. $\frac{2r^{2}(\rho_{s} - \rho_{m})g}{9\eta}$
• C. $\frac{2r^{2}\rho_{m}g}{9\eta}$
• D. 1

Answer 1

Answer: (D)

1

Answer 2

F_v + F_u = Mg; so $\frac{F_{v} + F_{u}}{Mg}$ = 1