Question

calculate the pressure inside the water droplets of 0.03mm , if the surface of water is 0.075 n/m.​

Answer 1

Answer:

1.05 atm

Explanation:

The excess pressure in an water droplet is given by

$p = \frac{2t}{r}$

Where t= surface tension=0.075N/m

r=radius of droplet=0.03mm= 0.03×10^(-3) m

$p = \frac{2 \times 0.075}{0.03 \times {10}^{ - 3} } = 5000 pascal$

Now 10^(5) pascal=1atm, so

5000 pascal= 5000/10^5 = 0.05

So excess pressure is 0.05 atm

Now excess pressure=pressure inside liquid -pressure of atmosphere

so, 0.05=liquid pressure-1

Pressure of liquid droplet=1.05 atm