Physical World Units and Measurements - Result Question 10

10. If dimensions of critical velocity $v_c$ of a liquid flowing through a tube are expressed as $[\eta^{x} \rho^{y} r^{z}]$, where $\eta, \rho$ 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 RS]

(a) $-1,-1,1$

(b) $-1,-1,-1$

(c) $1,1,1$

(d) $1,-1,-1$

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Answer:

Correct Answer: 10. (d)

Solution:

  1. (d) Applying dimensional method: $v_c=\eta^{x} \rho^{y_r z}$

here,

dimension of critical velocity, $V_0=[LT^{-1}]$

co-efficient of viscosity, $\eta=\frac{F}{6 \pi r v}$

so dimension of $\eta=\frac{[MLT^{-2}]}{[L][L T^{-1}]}=[ML^{-1} T^{-1}]$

dimension of density, $\rho=\frac{[M]}{[L^{3}]}=[ML^{-3}]$

dimension of radius, $r=[L]$

Put these values in equation (i), $[M^{0} LT^{-1}]=[ML^{-1} T^{-1}]^{x}[ML^{-3} T^{0}]^{y}[M^{0} LT^{0}]^{z}$

Equating powers both sides

$x+y=0 ;-x=-1 \therefore x=1$

$1+y=0 \therefore y=-1$

$-x-3 y+z=1$

$-1-3(-1)+z=1$

$-1+3+z=1$

$\therefore z=-1$