Question

Example 2.6: What should be the diameter
of a water drop so that the excess pressure
inside it is 80 N/m2? (Surface tension of
water = 7.27 x 10-2 N/m)​

Answer 1

Answer:

Explanation:

In down u will know the answer image

Answer 2

Answer:

The diameter of a water drop so that the excess pressure

inside it is 80 N/m² is  $d=1.81 \times 10^{-3}m$.

Explanation:

Excess pressure inside water droplet, ΔP= 80 N/m²

Surface tension of water = $7.27 \times10^{-2}$ N/m

To find: diameter of water droplet, d=?

We know that, excess pressure inside a water droplet is given by

$deltaP = \frac{2T}{d}$

d=\frac{2T}{ΔP}

Substituting all the values,

$d =\frac{2(7.27 \times 10^{-2})}{80}$

$=\frac{ 2\times 7.27}{80 \times 10^2}$

$= \frac{14.54}{ 8 \times 10^3}$

$d=1.81 \times 10^{-3}m$