NEET2015PhysicsUnits and DimensionsActual
If dimensions of critical velocity, v c of a liquid flowing through a tube are expressed as η x ρ y r z , where, η ,   ρ and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube, respectively, then, the values of x , y and z are given by
Options
- A- 1 , - 1 , 1
- B- 1 , - 1 , - 1
- C1 , 1 , 1
- D1 , - 1 , - 1
Correct answer
D. 1 , - 1 , - 1
Step-by-step solution
Dimensions of velocity, coefficient of viscosity and density are, v = LT - 1 η = F A d v d x = MLT - 2 L 2 T - 1 = ML - 1 T - 1 ρ = ML - 3 Now, v = η x ρ y r z ⇒ LT - 1 = ML - 1 T - 1 x ML - 3 y L z Now, by using dimensional homogeneity, we will get, M 0 = M x + y ⇒ y = - x For T , x = 1 ⇒ y =- x =- 1 For L , 1 = - x - 3 y + z ⇒ z = - 1