If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as [eta ^x ρ

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1.Units, Dimensions and Measurement

hard

If dimensions of critical velocity $v_c$ of a liquid flowing through a tube are expressed as$ [\eta ^x \rho ^yr^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

$1,1,1$

$1,-1,-1$

$-1,-1,1$

$-1,-1,-1$

(AIPMT-2015)

Solution

$[ {{v_c}} ] = [ {{\eta ^x}{\rho ^y}{r^z}}] $( {given}
Writing the dimensions of various quantities in
eqn. ($i$), we get
$\left[ {{M^0}L{T^{ – 1}}} \right] = {\left[ {M{L^{ – 1}}{T^{ – 1}}} \right]^x}{\left[ {M{L^{ – 3}}{T^0}} \right]^y}{\left[ {{M^0}L{T^0}} \right]^z}$
$ = \left[ {{M^{x + y}}{L^{ – x – 3y + z}}{T^{ – x}}} \right]$
Applying the principle of homogeneity of

dimensions we get

$,x + y = 0; – x – 3y + z = 1; – x = – 1$
On solving we get
$x = 1,y = – 1,z = – 1$

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If dimensions of critical velocity $v_c$ of a liquid flowing through a tube are expressed as$ [\eta ^x \rho ^yr^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

Solution

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Let us consider an equation

$\frac{1}{2} m v^{2}=m g h$

where $m$ is the mass of the body. velocity, $g$ is the acceleration do gravity and $h$ is the height. whether this equation is dimensionally correct.

The equation of stationary wave is

$\mathrm{y}=2 \mathrm{a} \sin \left(\frac{2 \pi \mathrm{nt}}{\lambda}\right) \cos \left(\frac{2 \pi \mathrm{x}}{\lambda}\right)$

Which of the following is NOT correct