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Pascal's Law

The lower end...

Question

The lower end of a capillary tube of diameter $2.00$ mm is dipped $8.00$ below the water surface in a beaker. What is the pressure required in the tube in order to blow a hemisphere bubble at its end in water? Given that density of water $= 10^{3}$ kg $m^{- 3}$ , surface tension of water $= 7.3 \times 10^{- 2}$ N $m^{- 1}$ and atmospheric pressure $= 1.01 \times 10^{5}$ Pa.

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Solution

The excess pressure in a bubble of gas in a liquid is given by 2S/r

There is only one liquid surface in this case. Pressure outside the bubble

wy = Atmospheric pressure + pressure due to 8cm of water column
$w y = 1.01 \times 10 g w + 0.08 \times 1000 \times 9.80 = 1.01784 \times 10 g w$

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2.00 m m

8.00 c m

10 c m

2 m m

7.30 \times 10^{- 2} N / m

1 a t m o s p h e r i c p r e s s u r e = 10^{5} P a

= 1000 k g / m^{3}

[T_{w a t e r} = 7.3 \times 10^{- 3} N m^{- 1}, ρ_{w a t e r} = 1000 k g m^{- 3} a n d 1 a t m = 1.01 \times 10^{5} P a a n d g = 9.8 m s^{- 2}]

1

2

13600 k g m^{- 3}

35 \times 10^{- 3} N m^{- 1}

0.28 mm

({in N/m}^{2})

= 0.07 N/m

= 10^{5} {N/m}^{2})

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