Question

If dimensions of critical velocity \[{V_C}\] of a liquid flowing through a tube are expressed as \[\left[ {{{{\eta }}^{{x}}}{{{\rho }}^{{y}}}{{{r}}^{{z}}}} \right]\], 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
A. \[1, - 1, - 1\]
B. \[ - 1, - 1,1\]
C. \[ - 1, - 1, - 1\]
D. \[1,1,1\]

Answer 1

Hint: Recall the units of coefficient of viscosity, density and radius. Then determine the dimensions of these units. Substitute the dimensions in the given formula for critical velocity and equate the powers on the both sides.

Complete step by step answer:
To answer this question, we have to recall the units of coefficient of viscosity, density and radius. We know that the unit of coefficient of viscosity is poise or \[kg\,{s^{ - 1}}\,{m^{ - 1}}\]. Therefore, the dimension of coefficient of viscosity is,
\[\eta = \left[ {{{{M}}^1}\,{{{L}}^{ - 1}}\,{{{T}}^{ - 1}}} \right]\]
We have the unit of density is \[kg/{m^3}\]. Therefore, the dimension of density is,
\[\rho = \left[ {{{{M}}^1}{{{L}}^{ - 3}}\,{{{T}}^0}} \right]\]
We have the unit of radius m. Therefore, the dimension of radius is,
\[r = \left[ {{{{L}}^1}} \right]\]
Now, the unit of critical velocity is m/s. Therefore, the dimension of critical velocity is,
\[{V_C} = \left[ {{{{M}}^0}{{{L}}^{{1}}}{{{T}}^{ - 1}}} \right]\]
We have given that the critical velocity is given as,
\[{V_C} = \left[ {{{{\eta }}^{{x}}}{{{\rho }}^{{y}}}{{{r}}^{{z}}}} \right]\]
Therefore,
\[\left[ {{{{M}}^0}{{{L}}^{{1}}}{{{T}}^{ - 1}}} \right] = {\left[ {{{{M}}^1}\,{{{L}}^{ - 1}}\,{{{T}}^{ - 1}}} \right]^x}{\left[ {{{{M}}^1}{{{L}}^{ - 3}}\,{{{T}}^0}} \right]^y}{\left[ {{{{L}}^1}} \right]^z}\]
\[ \Rightarrow \left[ {{{{M}}^0}{{{L}}^{{1}}}{{{T}}^{ - 1}}} \right] = \left[ {{{{M}}^{x + y}}\,{{{L}}^{ - x - 3y + z}}\,{{{T}}^{ - x}}} \right]\]
Now, comparing the powers of M, L and T on the both sides, we get,
\[x + y = 0\] …… (1)
\[\Rightarrow - x - 3y + z = 1\] …… (2)
\[\Rightarrow - x = - 1\] …… (3)
From equation (3), we get,
\[x = 1\]
Substituting \[x = 1\] in equation (1), we get,
\[1 + y = 0\]
\[ \Rightarrow y = - 1\]
Substituting \[x = 1\] and \[y = - 1\] in equation (2), we get,
\[ - 1 - 3\left( { - 1} \right) + z = 1\]
\[ \Rightarrow - 1 + 3 + z = 1\]
\[ \therefore z = - 1\]
Then the value of x, y and z is $1,-1$ and $-1$ respectively.

So, the correct answer is “Option A”.

Note:
To solve such type questions, one thing that must be on your fingertips is the units of all the quantities asked in the question. To calculate the dimensions of the quantities, you should refer to the unit of the quantity. Since the value of x, y and z is \[1, - 1,\,\,{{and }} - 1\] respectively, the critical velocity of the fluid can be expressed as, \[{V_C} = \dfrac{\eta }{{\rho r}}\].