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Question: A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity \(\eta\...

A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity η\eta flowing per second through a tube of radius rr and length l and having a pressure difference p across its end, is

A

V=πpr48ηlV = \frac{\pi pr^{4}}{8\eta l}

B

V=πηl8pr4V = \frac{\pi\eta l}{8pr^{4}}

C

V=8pηlπr4V = \frac{8p\eta l}{\pi r^{4}}

D

V=πpη8lr4V = \frac{\pi p\eta}{8lr^{4}}

Answer

V=πpr48ηlV = \frac{\pi pr^{4}}{8\eta l}

Explanation

Solution

Given V = Rate of flow =Volume sec=[L3T−1]\frac{\text{Volume }}{\text{sec}} = \lbrack L^{3}T^{- 1}\rbrack,

P = Pressure = [ML−1T−2]\lbrack ML^{- 1}T^{- 2}\rbrack, r = Radius = [L]

η\eta = Coefficient of viscosity = [ML−1T−1]\lbrack ML^{- 1}T^{- 1}\rbrack, l = Length = [L]

By substituting the dimension of each quantity we can check the accuracy of the formula V=πPr48ηlV = \frac{\pi Pr^{4}}{8\eta l}

∴\therefore [L3T−1]=[ML−1T−2][L4][ML−1T−1][L]\lbrack L^{3}T^{- 1}\rbrack = \frac{\lbrack ML^{- 1}T^{- 2}\rbrack\lbrack L^{4}\rbrack}{\lbrack ML^{- 1}T^{- 1}\rbrack\lbrack L\rbrack} = [L3T−1]\lbrack L^{3}T^{- 1}\rbrack

L.H.S. = R.H.S. i.e., the above formula is Correct**.**