Volume of a liquid flowing out per second of a pipe of length l and radius r is wr

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1.Units, Dimensions and Measurement

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The volume of a liquid flowing out per second of a pipe of length $l$ and radius $r$ is written by a student as $V\, = \,\frac{{\pi p{r^4}}}{{8\eta l}}$ where $p$ is the pressure difference between the two ends of the pipe and $\eta $ is coefficent of viscosity of the liquid having dimensional formula $[M^1L^{-1}T^{-1}] $. Check whether the equation is dimensionally correct.

Option A

Option B

Option C

Option D

Solution

The volume of a liquid flowing out per second of a pipe is given by $\mathrm{V}=\frac{\pi}{8} \frac{p r^{4}}{\eta l}$

${[\mathrm{V}]=\frac{[\text { Volume }]}{[\mathrm{Time}]}=\frac{\left[\mathrm{L}^{3}\right]}{[\mathrm{T}]}=\left[\mathrm{L}^{3} \mathrm{~T}^{-1}\right]}$

${[p]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}$

${[\eta]=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]}$

${[l]=[\mathrm{L}]}$

${[r]=[\mathrm{L}]}$

${[\mathrm{LHS}]=[\mathrm{V}]=\frac{\left[\mathrm{L}^{3}\right]}{[\mathrm{T}]}=\left[\mathrm{L}^{3} \mathrm{~T}^{-1}\right]}$

${[\mathrm{RHS}]=\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \times\left[\mathrm{L}^{4}\right]}{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right] \times[\mathrm{L}]}=\left[\mathrm{L}^{3} \mathrm{~T}^{-1}\right]}$

${[\mathrm{LHS}]=[\mathrm{RHS}]}$

Thus, equation is dimensionally correct.

Standard 11

Physics