Question

An air bubble of radius 0.02 mm is at a depth of 25 cm in oil of density 0.8 gm/cm. If the surface tension of
oil is 25 x 10 N/m find the pressure inside the bubble
(atmospheric pressure = 10 N/m?)?

Answer 1

Answer:

Pressure inside the oil drop is given as 104462 Pa

Explanation:

As we know that excess pressure on the surface of oil drop is given as

$P_i - P_o = \frac{2S}{R}$

here we know that

$P_o$ = pressure just outside the oil drop

now we know that oil drop is inside the oil surface at depth of 25 cm

So pressure is given as

$P = P_{atm} + \rho g h$

$P = 10^5 + (800)(9.81)(0.25)$

$P = 101962 Pa$

Now we can use above formula to find the pressure inside the oil

$P_i = P_o + \frac{2S}{r}$

here we have

$S = 25 \times 10^{-3} N/m$

$r = 0.02 \times 10^{-3} m$

$P_i = 101962 + \frac{2\times 25 \times 10^{-3}}{0.02 \times 10^{-3}}$

$P_i = 104462 Pa$

#Learn

Topic : Excess pressure inside the drop

https://brainly.in/question/6570243

Answer 2

Answer:

10 raise 5 into (800)(9.8)(0.25)