
Excess pressure of one soap bubble is four times more than that of another. Then the ratio of volume of first bubble to second one is :
A. 1 : 64
B. 64 : 1
C. 4 : 1
D. 1 : 2
Answer
161.1k+ views
Hint:We are given that the excess pressure is four times that of another and we have to find the ratio of volume of two bubbles. First we use the formula for excess pressure and then we use the formula for volume of spherical objects. First find the ratio of radii of the soap bubbles and then the volume of the soap bubbles.
Formula Used:
The excess pressure $\Delta P$ inside a soap bubble is,
$\Delta P=\dfrac{4T}{R}$………………………………. (1)
Where T is the surface tension on the soap and R is the radius of the soap bubble.
and the volume V of the object is given by
$V=\dfrac{4}{3}\pi {{R}^{3}}$…………………………….. (2)
Complete step by step solution:
Consider that the excess pressure inside the first and second soap bubble is $\Delta {{P}_{1}}$ and $\Delta {{P}_{2}}$ respectively. Let the radii and the volume of first and second soap bubbles are ${{R}_{1}}$ and ${{R}_{2}}$ and the volumes be ${{V}_{1}}$ and ${{V}_{2}}$ respectively.
From equation (1), we conclude that the excess pressure is inversely proportional to the radius. That is
$\Delta P\propto \dfrac{1}{R}$
Given in the question ${{P}_{1}}=4{{P}_{2}}$ ,we get
$\dfrac{4T}{{{R}_{1}}}=4\times \dfrac{4T}{{{R}_{2}}}$
Solving the above equation, we get
$\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{1}{4}$
Hence the bubbles are spherical in shape.
From equation (2), we conclude that Volume is directly proportional to the cube of radius of the soap bubble
$V\propto {{R}^{3}}$
For two soap bubbles
$\dfrac{{{V}_{1}}}{{{V}_{2}}}={{\left( \dfrac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3}} \\ $
By substituting the value, we get
$\dfrac{{{V}_{1}}}{{{V}_{2}}}={{\left( \dfrac{1}{4} \right)}^{3}} \\ $
$\therefore \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{1}{64}$
Thus, the ratio of volume of the soap bubble is 1 : 64.
Thus, option A is the correct answer.
Note:We can find the shape of the object by considering that the surface tension and the excess pressure acting on the soap bubble balances such that the shape of the soap bubble in steady state is spherical. So we consider the volume of the sphere.
Formula Used:
The excess pressure $\Delta P$ inside a soap bubble is,
$\Delta P=\dfrac{4T}{R}$………………………………. (1)
Where T is the surface tension on the soap and R is the radius of the soap bubble.
and the volume V of the object is given by
$V=\dfrac{4}{3}\pi {{R}^{3}}$…………………………….. (2)
Complete step by step solution:
Consider that the excess pressure inside the first and second soap bubble is $\Delta {{P}_{1}}$ and $\Delta {{P}_{2}}$ respectively. Let the radii and the volume of first and second soap bubbles are ${{R}_{1}}$ and ${{R}_{2}}$ and the volumes be ${{V}_{1}}$ and ${{V}_{2}}$ respectively.
From equation (1), we conclude that the excess pressure is inversely proportional to the radius. That is
$\Delta P\propto \dfrac{1}{R}$
Given in the question ${{P}_{1}}=4{{P}_{2}}$ ,we get
$\dfrac{4T}{{{R}_{1}}}=4\times \dfrac{4T}{{{R}_{2}}}$
Solving the above equation, we get
$\dfrac{{{R}_{1}}}{{{R}_{2}}}=\dfrac{1}{4}$
Hence the bubbles are spherical in shape.
From equation (2), we conclude that Volume is directly proportional to the cube of radius of the soap bubble
$V\propto {{R}^{3}}$
For two soap bubbles
$\dfrac{{{V}_{1}}}{{{V}_{2}}}={{\left( \dfrac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3}} \\ $
By substituting the value, we get
$\dfrac{{{V}_{1}}}{{{V}_{2}}}={{\left( \dfrac{1}{4} \right)}^{3}} \\ $
$\therefore \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{1}{64}$
Thus, the ratio of volume of the soap bubble is 1 : 64.
Thus, option A is the correct answer.
Note:We can find the shape of the object by considering that the surface tension and the excess pressure acting on the soap bubble balances such that the shape of the soap bubble in steady state is spherical. So we consider the volume of the sphere.
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