In dimension of circal velocity `v_(0)` liquid following through a take are expressed as `(eta^(x) rho^(y) r^(z))` where `eta, rhoand r `are the coefficient of viscosity of liquid density of liquid and radius of the tube respectively then the value of `x,y` and `z` are given by

Options

A. `1,-1,-1`
B. `-1,-1,1`
C. `-1,-1,-1`
D. `1,1,1`

Answer

A

Explanation

Key concept According to principle of homogeneity of dimension states that, a physical quantity equation will be dimensionally correct, if the dimensions of all the terms occuring on both sides of the equations are same. Given critical velocity of liquid flowing through a tube are expressed as
`V_(c) prop eta^(n) rho^(y) r^(z)`
Coefficient of viscosity of liquid, `eta = [ML^(-1)T^(-1)]`
Density of liquid, `rho = [ML^(-3)]`
Radius of tube `r = [L]`
Critical velocity of liquid, `v_(c) = [ML^(0)T^(-1)]`
`rArr [M^(0)L^(1)T^(-1)] = [ML^(-1)T^(-1)]^(x) [ML^(-3)]^(y) [L]^(z)`
`[M^(0)L^(1)T^(-1)] = [M^(x+y)L^(-x-3y+z)T^(-x)]`
Comparing exponents of M,L and L, we get
`x + y =0, - x - 3y +z = 1, - x =- 1`
`rArr z =- 1,x = 1, y =-1`