Question

The excess pressure inside a spherical drop of water is four times that of another drop. Then their mass ratio is:
A) 1:6
B) 8:1
C) 1: 4
D) 1:64

Answer 1

Pressure is the force exerted on an object and the excess pressure inside a spherical drop of a liquid is the difference between the pressure inside the spherical drop and pressure outside the spherical drop. We can find the mass ratio with the help of the formula of excess pressure.

Formula used:

& P=\dfrac{2T}{r} \\\ & \rho =\dfrac{m}{V} \\\ & V=\dfrac{4}{3}\pi {{r}^{3}} \\\ \end{aligned}$$ **Complete step-by-step answer:** The excess pressure inside a liquid drop is given as $$P=\dfrac{2T}{r}$$ Where T is the surface tension of the liquid drop and r is the radius of the liquid drop. Now it is given that the first drop has 4 times excess pressure than the second drop. Let say the first drop has excess pressure $${{P}_{1}}$$ and second drop has excess pressure$${{P}_{2}}$$, then as per question we can write $${{P}_{1}}=4{{P}_{2}}$$ Further from the excess pressure formula we can write $$\left( \dfrac{2T}{{{r}_{1}}} \right)=4\left( \dfrac{2T}{{{r}_{2}}} \right)$$ Where $${{r}_{1}}\text{ and }{{r}_{2}}$$are the radius of first and second drop respectively. The surface tension for each drop will be the same as each experiences the same downward force. Solving the above equation we get, $${{r}_{2}}=4{{r}_{1}}$$ As 2T were the common term on both sides so they got cancelled. Now as we have to find mass ratio, let us consider the density which is given as $$\rho =\dfrac{m}{V}$$ Where m is mass and V is volume. We can rewrite the equation for mass as $$m=\rho V$$ Consider the mass of first drop as $${{m}_{1}}$$and mass of second drop as $${{m}_{2}}$$, then there ratio will can be given as $$\dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{\rho {{V}_{1}}}{\rho {{V}_{2}}}$$ Where $${{V}_{1}}\text{ and }{{V}_{2}}$$are the volume of first and second drop respectively. The density of each drop will be the same. As the drop are spherical in shape and volume of sphere is given as $$V=\dfrac{4}{3}\pi {{r}^{3}}$$ Hence we can write $$\begin{aligned} & \dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{\dfrac{4}{3}\pi {{r}_{1}}^{3}}{\dfrac{4}{3}\pi {{r}_{2}}^{3}} \\\ & \dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{{{r}_{1}}^{3}}{{{r}_{2}}^{3}} \\\ \end{aligned}$$ Substituting $${{r}_{2}}=4{{r}_{1}}$$in the above equation we get $$\begin{aligned} & \dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{{{r}_{1}}^{3}}{{{\left( 4{{r}_{1}} \right)}^{3}}} \\\ & \dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{{{r}_{1}}^{3}}{64{{r}_{1}}^{3}} \\\ & \dfrac{{{m}_{1}}}{{{m}_{2}}}=\dfrac{1}{64} \\\ \end{aligned}$$ **So, the correct answer is “Option D”.** **Note:** Pressure is given as force divided by area and by using this and considering the forces experienced by the drop i.e. we can derive the formula for the excess pressure inside the drop of the liquid. The density of each drop will be the same as the liquid here is the same and the drop will have the density of the liquid.