Question

A spherical metal ball (of density \( \rho_s \) and diameter \( D \)), attached to a string, is exposed to a crossflow (of velocity \( U_\infty \)) of a viscous fluid (of viscosity \( \mu \) and density \( \rho_f \)). Due to the crossflow, the string makes an angle of inclination \( \theta \) with the top surface as shown in the figure. The acceleration due to gravity is denoted by \( g \). For this flow, Reynolds number, \( \text{Re} = \frac{\rho_f U_\infty D}{\mu} \ll 1 \) and buoyancy force in the fluid is negligible compared to viscous force. Assuming the string to be weightless and offering negligible drag, the expression for \( \theta \) is
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• A. \( \tan^{-1} \left[ \frac{1}{18} \frac{D^2 \rho_s^2 g}{\mu U_\infty \rho_f} \right] \)
• B. \( \tan^{-1} \left[ \frac{1}{18} \frac{D^2 \rho_f g}{\mu U_\infty} \right] \)
• C. \( \sin^{-1} \left[ \frac{2 D^2 \rho_s g}{9 \mu U_\infty} \right] \)
• D. \( \tan^{-1} \left[ \frac{D^2 \rho_s g}{18 \mu U_\infty} \right] \)

Hint:

For a spherical ball in a viscous fluid flow, the angle of inclination of the string is a result of the balance between the drag force and the gravitational force acting on the sphere.

Answer 1

The Correct Option is D

In this problem, we are dealing with the drag force on a spherical object (a spherical ball) in a viscous fluid flow. The drag force is influenced by both the gravitational force and the viscous force acting on the object. The drag force on a small spherical particle in a low Reynolds number flow can be described by Stokes' law, which is: \[ F_d = 6 \pi \mu r U \] where \( r \) is the radius of the sphere, \( \mu \) is the dynamic viscosity of the fluid, and \( U \) is the relative velocity of the fluid with respect to the particle. However, in this case, the spherical ball is attached to a string, and the system is subjected to both the drag force and the gravitational force. The angle \( \theta \) of the string is a result of the balance between these forces. From the force balance, the angle \( \theta \) can be derived as: \[ \tan \theta = \frac{F_{\text{gravity}}}{F_{\text{drag}}} \] Substituting the expressions for gravitational force \( F_{\text{gravity}} = \rho_s V g = \rho_s \frac{\pi D^3}{6} g \) and drag force \( F_{\text{drag}} = 6 \pi \mu \frac{D}{2} U_\infty \), we can simplify the expression for \( \theta \) to: \[ \tan \theta = \frac{\rho_s D^2 g}{18 \mu U_\infty} \] Thus, the correct expression for the angle \( \theta \) is: \[ \theta = \tan^{-1} \left[ \frac{D^2 \rho_s g}{18 \mu U_\infty} \right] \] Therefore, the correct answer is (D).
Final Answer: (D) \( \tan^{-1} \left[ \frac{D^2 \rho_s g}{18 \mu U_\infty} \right] \)