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Question: In the relation \[V = \dfrac{\pi }{8}\dfrac{{{{\Pr }^4}}}{{\eta l}}\], where the letter have their u...

In the relation V=π8Pr⁡4ηlV = \dfrac{\pi }{8}\dfrac{{{{\Pr }^4}}}{{\eta l}}, where the letter have their usual meanings, the dimension of V is
A. M0L3T0{M^0}{L^3}{T^0}
B. M0L3T−1{M^0}{L^3}{T^{ - 1}}
C. M0L−3T0{M^0}{L^{ - 3}}{T^0}
D. M1L3T0{M^1}{L^3}{T^0}

Explanation

Solution

Learn about the different dimensions of the quantities that are given in the expression of the V here. The physical quantities given in the expression are the pressure, radius, coefficient of viscosity and length respectively. The dimension of any physical quantities are written in terms of mass length and time.

Complete step by step answer:
In the given expression the physical quantities given are P that is the pressure, r is the radius η\eta is the coefficient of viscosity l is the length. Now, we know that the dimension of any physical quantities are written in terms of mass, length and time by the form [MxLyTz][{M^x}{L^y}{T^z}] where, x,y,zx,y,z are some constant for that particular quantity.

Now, we know that the dimension of pressure is, [M1L−1T−2][{M^1}{L^{ - 1}}{T^{ - 2}}].
Also the dimension of radius is [L][L].
The dimension of coefficient of viscosity is [M1L−1T−1][{M^1}{L^{ - 1}}{T^{ - 1}}]
And the dimension of length is [L][L].
We know that the constants are dimensionless quantities.

Hence, putting this dimensions in the given expression we can find the dimension of V as,
[V]=[M1L−1T−2][L]4[M1L−1T−1][L][V] = \dfrac{{[{M^1}{L^{ - 1}}{T^{ - 2}}]{{[L]}^4}}}{{[{M^1}{L^{ - 1}}{T^{ - 1}}][L]}}
Upon simplifying we will have,
[V]=[M1−1L−1+4−1+1T−2+1][V] = [{M^{1 - 1}}{L^{ - 1 + 4 - 1 + 1}}{T^{ - 2 + 1}}]
∴[V]=[M0L3T−1]\therefore [V] = [{M^0}{L^3}{T^{ - 1}}]
Hence the dimension of the term VV is [M0L3T−1][{M^0}{L^3}{T^{ - 1}}].

Hence, option B is the correct answer.

Note: We can also determine the dimension of each quantity from analyzing the unit of the individual quantities. Especially for the coefficient of viscosity. We know that the unit of coefficient of viscosity is nothing but the Nsm−2Ns{m^{ - 2}}. In the unit newton is the unit of force and force is mass times acceleration like this. We can easily find the dimension of compound physical quantities if we can’t recall the dimension of each quantity.