Water flows through a capillary tube of length $20 c m$ and radius of $1 m m$ under a constant pressure difference of $14 c m$ of water level.
If $η_{w a t e r} = 1$ x $10^{- 3} P a s; g = 10 m s^{- 2}$ , the rate of flow of water through the tube in $m^{3} s^{- 1}$ is:

27.5 \times 10^{- 7}

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275 \times 10^{- 7}

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275

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2.75 \times 10^{- 7}

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Solution

The correct option is A $27.5 \times 10^{- 7}$
Applying poiseuillis eqn,
$Q = \frac{π \times 10^{- 12} \times 1000 \times 10 \times 0.14}{8 \times 10^{- 3} \times 0.2} m^{3} / s$
$Q = 2.7488 \times 10^{- 3} m^{3} / s$
$Q \approx 27.5 \times 10^{- 7} m^{3} / s$

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Water flows through a capillary tube of length 20 cm and radius of 1mm under a constant pressure difference of 14 cm of water level. If ηwater=1 x10−3Pa s; g=10ms−2 , the rate of flow of water through the tube in m3s−1 is:

The correct option is A 27.5×10−7Applying poiseuillis eqn,Q=π×10−12×1000×10×0.148×10−3×0.2m3/sQ=2.7488×10−3m3/sQ≈27.5×10−7m3/s

Water flows through a capillary tube of length $20 c m$ and radius of $1 m m$ under a constant pressure difference of $14 c m$ of water level.
If $η_{w a t e r} = 1$ x $10^{- 3} P a s; g = 10 m s^{- 2}$ , the rate of flow of water through the tube in $m^{3} s^{- 1}$ is:

The correct option is A $27.5 \times 10^{- 7}$
Applying poiseuillis eqn,
$Q = \frac{π \times 10^{- 12} \times 1000 \times 10 \times 0.14}{8 \times 10^{- 3} \times 0.2} m^{3} / s$
$Q = 2.7488 \times 10^{- 3} m^{3} / s$
$Q \approx 27.5 \times 10^{- 7} m^{3} / s$