If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as [ x y rz] where , 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
- -1, -1, -1
- 1, 1, 1
- 1, -1, -1
- -1, -1, 1
Answer
(3) 1, -1, -1
Solution
[vc] = [ M0 L T-1], [ ] = [ M L-1 T-1], [ ] = [ M L-3], [r] = [L]
[ M0 L T-1] = [ M L-1 T-1]x [ M L-3]y [L]z
Comparing powers:
T: -x = -1 ⇒ x = 1
M: x + y = 0 ⇒ y = -1
L: -x - 3y + z = 1 ⇒ -1 + 3 + z = 1 ⇒ z = -1
∴ x, y, z = 1, -1, -1, i.e. vc r
Answer: (3)