Answer
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Hint:
Total pressure inside the drop of mercury is the sum total of excess pressure inside the mercury drop and the atmospheric pressure. We can calculate the excess pressure using the formula$\dfrac{2S}{r}$, where S is the surface tension of fluid and r is the radius of mercury drop. Once we get the excess pressure, we can find the total pressure inside the mercury drop by adding the atmospheric pressure to it.
Formulae used:
Excess pressure inside a liquid drop =$\dfrac{2S}{r}$, where S is the surface tension of the liquid and r is the radius of drop.
Total pressure inside the liquid drop = excess pressure inside the drop + Atmospheric pressure
Complete step by step solution:
We have been given that the radius of mercury drop,$r=3.00\;mm=3\times 10^{-3}$ m, surface tension of the mercury, $S=4.65\times 10^{-1}$ N/m and the atmospheric pressure, ${P}_{0}=1.01\times 10^5$ Pa
So, now we will find the excess pressure inside the drop using the formula $\dfrac{2S}{r}$, where S is the surface tension of the liquid and r is the radius of drop
On putting the values, we will get the excess pressure$=\dfrac{2\times 4.65\times 10^{-1}}{3\times 10^{-3}}=3.1\times 10^{2}=310$ Pa
So, total pressure inside the drop = excess pressure + atmospheric pressure $({P}_{0})= 3.1\times 10^{2}+1.01\times 10^5=1.0131\times 10^{5}$ Pa
Hence the pressure inside the drop of the mercury is $1.0131\times 10^{5}\;Pa$ and the excess pressure inside the drop is 310 Pa.
Note:
We know that because of the surface tension, the molecules on the surface film experience the net force normal to the surface and towards inward direction. Thus, we can say that there is more pressure inside the bubble than outside. The difference between these two pressures gives us the excess pressure drop. So, we should note that the net pressure on the film of the bubble is just the excess pressure, but inside the bubble is both excess pressure and atmospheric pressure.
Total pressure inside the drop of mercury is the sum total of excess pressure inside the mercury drop and the atmospheric pressure. We can calculate the excess pressure using the formula$\dfrac{2S}{r}$, where S is the surface tension of fluid and r is the radius of mercury drop. Once we get the excess pressure, we can find the total pressure inside the mercury drop by adding the atmospheric pressure to it.
Formulae used:
Excess pressure inside a liquid drop =$\dfrac{2S}{r}$, where S is the surface tension of the liquid and r is the radius of drop.
Total pressure inside the liquid drop = excess pressure inside the drop + Atmospheric pressure
Complete step by step solution:
We have been given that the radius of mercury drop,$r=3.00\;mm=3\times 10^{-3}$ m, surface tension of the mercury, $S=4.65\times 10^{-1}$ N/m and the atmospheric pressure, ${P}_{0}=1.01\times 10^5$ Pa
So, now we will find the excess pressure inside the drop using the formula $\dfrac{2S}{r}$, where S is the surface tension of the liquid and r is the radius of drop
On putting the values, we will get the excess pressure$=\dfrac{2\times 4.65\times 10^{-1}}{3\times 10^{-3}}=3.1\times 10^{2}=310$ Pa
So, total pressure inside the drop = excess pressure + atmospheric pressure $({P}_{0})= 3.1\times 10^{2}+1.01\times 10^5=1.0131\times 10^{5}$ Pa
Hence the pressure inside the drop of the mercury is $1.0131\times 10^{5}\;Pa$ and the excess pressure inside the drop is 310 Pa.
Note:
We know that because of the surface tension, the molecules on the surface film experience the net force normal to the surface and towards inward direction. Thus, we can say that there is more pressure inside the bubble than outside. The difference between these two pressures gives us the excess pressure drop. So, we should note that the net pressure on the film of the bubble is just the excess pressure, but inside the bubble is both excess pressure and atmospheric pressure.
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