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प्रश्न
The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as `v = π/8 (pr^4)/(ηl)` where P is the pressure difference between the two ends of the pipe and η is coefficient of viscosity of the liquid having dimensional formula ML–1 T–1. Check whether the equation is dimensionally correct.
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उत्तर
If the dimensions of LHS of an equation are equal to the dimensions of RHS, then the equation is said to be dimensionally correct.
According to the problem, the volume of a liquid flowing out per second of a pipe is given by V = `pi/8 (pr^4)/(ηl)`
(where V = rate of volume of liquid per unit time)
Dimension of given physical quantities,
`[V] = "Dimension of Volume"/"Dimension of time" = ([L^3])/([T]) = [L^3T^-1], [p] = [ML^-1T^-2]`
`[η] = [ML^-1T^-1], [l] = [L], [r] = [L]`
LHS = `[V] = ([L^3])/([T]) = [L^3T^-1]`
RHS = `([ML^-1T^-2] xx [L^4])/([ML^-1 T^-1] xx [L]) = [L^3T^-1]`
Dimensionally, L.H.S = R.H.S.
Therefore, the equation is correct dimensionally.
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