
There are two liquid drops of different radii. The excess pressure over the outside is:-
A. more in the big drop
B. more in the small drop
C. equal in both drops
D. there is no excess pressure inside the drops
Answer
163.5k+ views
Hint: In this question, we have given two liquids of different radii. We know the excess pressure inside a drop is always more than that of the outside pressure. By using the two different radii and comparing them, we get to know that the air flow from a smaller bubble to a larger bubble.
Formula used:
he pressure outside the soap bubble is given by
$\Delta P=\dfrac{2T}{R}$
Where R is the radius of the bubble and T is the surface tension.
Complete step by step solution:
We know a soap bubble has two surfaces, the outer surface and the inner surface. There exists a pressure difference between these two pressures and the pressure inside the bubbles is always greater than the pressure outside the bubble. The pressure outside the soap bubble is given by,
$\Delta P=\dfrac{2T}{R}$
For the radius ${{R}_{1}}$
$P=\dfrac{2T}{{{R}_{1}}}$
And for the radius ${{R}_{2}}$
$P=\dfrac{2T}{{{R}_{2}}}$
Hence the excess pressure due to liquid drop is,
$\Delta P=({{P}_{2}}-{{P}_{1}})=\dfrac{2T}{R}$
Hence we observe that excess pressure in a liquid drop of radius r is directly proportional to ${{r}^{-1}}$. From the above expression, we see that the $\Delta P\propto {{r}^{-1}}$.
Now when we are given two bubbles , then the larger bubble tries to increase its size by decreasing the pressure. Therefore, the air will flow from the smaller bubble to the larger bubble and the size of the larger bubble increases. As the size increases, pressure decreases in the largest bubble. Similarly the pressure outside the smaller bubble increases. Hence, the drop with a smaller radius will have higher excess pressure outside the drop.
Thus, option B is the correct answer.
Note: To solve these types of questions, we must have the knowledge of surface tension and the pressure due to the air- water interface. We know the derivation of excess pressure outside a soap bubble, we can directly use that answer otherwise we have to first derive the expression
Formula used:
he pressure outside the soap bubble is given by
$\Delta P=\dfrac{2T}{R}$
Where R is the radius of the bubble and T is the surface tension.
Complete step by step solution:
We know a soap bubble has two surfaces, the outer surface and the inner surface. There exists a pressure difference between these two pressures and the pressure inside the bubbles is always greater than the pressure outside the bubble. The pressure outside the soap bubble is given by,
$\Delta P=\dfrac{2T}{R}$
For the radius ${{R}_{1}}$
$P=\dfrac{2T}{{{R}_{1}}}$
And for the radius ${{R}_{2}}$
$P=\dfrac{2T}{{{R}_{2}}}$
Hence the excess pressure due to liquid drop is,
$\Delta P=({{P}_{2}}-{{P}_{1}})=\dfrac{2T}{R}$
Hence we observe that excess pressure in a liquid drop of radius r is directly proportional to ${{r}^{-1}}$. From the above expression, we see that the $\Delta P\propto {{r}^{-1}}$.
Now when we are given two bubbles , then the larger bubble tries to increase its size by decreasing the pressure. Therefore, the air will flow from the smaller bubble to the larger bubble and the size of the larger bubble increases. As the size increases, pressure decreases in the largest bubble. Similarly the pressure outside the smaller bubble increases. Hence, the drop with a smaller radius will have higher excess pressure outside the drop.
Thus, option B is the correct answer.
Note: To solve these types of questions, we must have the knowledge of surface tension and the pressure due to the air- water interface. We know the derivation of excess pressure outside a soap bubble, we can directly use that answer otherwise we have to first derive the expression
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