Surface Tension and Viscosity: Practice Problem & Solution
An object of mass m and density \(\rho\) is falling in a viscous liquid of density \(\frac{\rho}{4}\). When it attains terminal velocity, then viscous force acting on it will be:
Solution Explained:
To solve this problem, we apply the core principles of Surface Tension and Viscosity. Understanding the underlying formula is key to arriving at the correct answer below:
At terminal velocity, the upward forces balance the downward force: \(F_v + F_b = mg\). Since buoyant force is \(F_b = V \rho_l g = \left(\frac{m}{\rho}\right)\left(\frac{\rho}{4}\right)g = \frac{mg}{4}\), we get \(F_v = mg - \frac{mg}{4} = \frac{3mg}{4}\).
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