The viscous force acting on a sphere is directly proportional to the following parameters:\tthe radius of the sphere\tcoefficient of viscosity\tthe velocity of the objectMathematically, this is represented asF∝ηarbvcNow let us evaluate the values of a, b and c.Substituting the proportionality sign with an equality sign, we getF=kηarbvc\xa0(1)\xa0Here, k is the constant of proportionality which is a numerical value and has no dimensions.Writing the dimensions of parameters on either side of equation (1), we get[MLT–2] = [ML–1T–1]a\xa0[L]b\xa0[LT-1]c\xa0Simplifying the above equation, we get[MLT–2] = Ma\xa0⋅ L–a+b+c\xa0⋅ T–a–c\xa0(2)\xa0According to\xa0classical mechanics, mass, length and time are independent entities.Equating the superscripts of mass, length and time respectively from equation (2), we geta = 1 (3)–a + b + c = 1 (4)–a –c = 2 or a + c = 2 (5)Substituting (3) in (5), we get1 + c = 2c = 1 (6)Substituting the value of (3) & (6) in (4), we get–1 + b + 1 = 1b = 1 (7)Substituting the value of (3), (6) and (7) in (1), we getF=kηrvThe value of k for a spherical body was experimentally obtained as\xa06πTherefore, the viscous force on a spherical body falling through a liquid is given by the equationF=6πηrv