Question

After being deformed, a spherical drop of liquid will execute periodic vibrations about its spherical shape. Using the method of dimensions, obtain an expression for the frequency of these vibrations in terms of the related physical quantities.

a) ( f = √{frac{T}{rho R^3}} )
b) ( f = frac{√{T}}{rho R} )
c) ( f = frac{T}{rho R} )
d) ( f = √{frac{T}{rho R}} )

Answer 1

The frequency of vibrations for a deformed spherical liquid can be expressed in terms of surface tension (T), density (rho), and radius (R) as f = √{\frac{T}{\rho R^3}}, obtained through dimensional analysis.

Step-by-step explanation:

The question involves determining the expression for the frequency of vibrations of a deformed spherical drop of liquid using the method of dimensions. The involved physical quantities for this system are surface tension (T), density of the liquid (rho), and radius of the spherical drop (R). The correct expression for the frequency (f) of vibrations should involve these physical parameters.

By considering the dimensional analysis, the expression for frequency solely in terms of the given quantities, and without any other additional constants, would involve these parameters in a way that yields the dimensions of frequency which is T-1. The correct expression that satisfies this is f = √{\frac{T}{\rho R^3}}, which is option (a). This formula is derived by balancing the dimensions on both sides of the equation such that the dimensions of the surface tension divided by the product of the liquid density and cube of the radius should be equal to the dimensions of the frequency squared. Hence, taking a square root provides the dimensions of frequency.