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PART - 2 CH - 9 Mechanical Properties of Fluids — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 9 Mechanical Properties of Fluids2 Marks
Question
Blood is flowing 0.1 cm long and $2 \times 10^{-4}$ cm radius in the tube. The pressure difference at the ends of this tube is 20 mm mercury. Coefficient of viscosity of blood is $0.5 \times 10^3 kg m ^{-1} sec ^{-1}$. Find the rate of blood flow in the tube.
✓

Answer

Given :
$l=0.1 cm=10^{-3} m,$
radius $r=2 \times 10^{-4} cm=2 \times 10^{-6} m$
Pressure difference $P =20 mm$ mercury
$\begin{aligned}& =20 \times 10^{-3} m \text { mercury } \\& =20 \times 10^{-3} \times\left(13.6 \times 10^3\right) \times 9.8 N / m^2 \\\eta & =0.5 \times 10^{-3} kg / m-sec\end{aligned}$
Hence, rate of blood flow in the tube
$\begin{array}{l}Q=\frac{\pi Pr^4}{8 \eta l} \\Q=\frac{3.14 \times\left(20 \times 10^{-3} \times 13.6 \times10^3 \times 9.8\right) \times\left(2 \times 10^{-6}\right)^4}{8 \times 0.5 \times 10^{-3} \times10^{-3}} \\Q=3.42 \times 10^{-15} m^3 / sec .\end{array}$

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