A narrow capillary tube is dipped $10 c m$ below water surface and a liquid bubble of radius $2 m m$ formed at the lower end by blowing air through the tube.
a. Calculate the excess pressure due to surface tension.
b. What is the pressure required in the tube in order to blow a hemispherical bubble at its end in water?
The surface tension of water at temperature of the experiment is $7.30 \times 10^{- 2} N / m$ . $1 a t m o s p h e r i c p r e s s u r e = 10^{5} P a$ , density of water $= 1000 k g / m^{3}$ .

Open in App

Solution

Excess pressure $= 2 T / r = 146 P a$
Pressure $h$ below the surface $= ρ_{0} + ρ g h$
$∴ P = (P_{0} + ρ g h) + 2 T / r$
$= 1 \times 10^{5} + 10^{3} \times 10 \times 0.1 + 2 \times \frac{7.3 \times 10^{- 2}}{10^{3}}$
$= 1 \times 10^{5} + 10^{3} + 146$
$= 1 \times 10^{5} + 1146$
$= 1.01145 \times 10^{5}$ .

Suggest Corrections

Similar questions

2.00

8.00

= 10^{3}

m^{- 3}

= 7.3 \times 10^{- 2}

m^{- 1}

= 1.01 \times 10^{5}

2.00 m m

8.00 c m

[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}

1.01146 \times 10^{x}

Join BYJU'S Learning Program