Question

The excess pressure inside a spherical drop of water is four times that of another drop .the their respective mass ratio is ??????????????????????????

Answer 1

The excess pressure inside a spherical drop of water is four times that of another drop. Then their respective mass ratio is?

The excess pressure inside a spherical drop of water equals:

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

where $\gamma$ – surface tension for water, $r$ – radius of the drop

So, if we have 2 drops with excess pressures $4p_{1} = p_{2}$ then:

$$4 \frac {2 \gamma}{r _ {1}} = \frac {2 \gamma}{r _ {2}}$$

or:

$$\frac {r _ {1}}{r _ {2}} = 4$$

But on other hand, mass of spherical drop of water equals:

$$m = \frac {4}{3} \pi r ^ {3} \rho$$

where $\rho$ – density of water, $r$ – radius of the drop

Therefore,

$$\frac {m _ {1}}{m _ {2}} = \frac {r _ {1} ^ {3}}{r _ {2} ^ {3}} = \left(\frac {r _ {1}}{r _ {2}}\right) ^ {3} = 4 ^ {3} = 64$$

Answer: 64