If dimensions of critical velocity of a liquid flowing through a tube are expressed as where and are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of and are given by:

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Solution

Q.
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: [2015]

1 $- 1, - 1, - 1$

2 $1, 1, 1$

3 $1, - 1, - 1$

4 $- 1, - 1, 1$

Ans.
(3)

$[v_{c}] = [η^{x} ρ^{y} r^{z}]$ (given) ...(i)

Writing the dimensions of various quantities in eqn. (i), we get $[M^{0} L T^{- 1}] =$ ${[M L^{- 1} T^{- 1}]}^{x}$ ${[M L^{- 3} T^{0}]}^{y}$ ${[M^{0} L T^{0}]}^{z}$

$= [M^{x + y} L^{- x - 3 y + z} T^{- x}]$

Applying the principle of homogeneity of dimensions, we get $x + y = 0;$ $- x - 3 y + z = 1;$ $- x = - 1$

On solving, we get $x = 1, y = - 1, z = - 1$

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