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:
- (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
$x+y=0 ;-x=-1 \therefore x=1$
$1+y=0 \therefore y=-1$
$-x-3 y+z=1$
$-1-3(-1)+z=1$
$-1+3+z=1$
$\therefore z=-1$
