MCQ
The excess pressure inside a soap bubble is twice the excess pressure inside a second soap bubble. The volume of the first bubble is n times the volume of the second where n is:
- A4
- B2
- C1
- ✓0.125.
✓
Answer
Correct option: D.
0.125.
Let the excess pressure inside the second bubble be P.
$\therefore$ Excess pressure inside the first bubble = 2P
Let the radius of the second bubble be R.
Let the radius of the first bubble be x.
Excess pressure inside the 2nd soap bubble:
$\text{P}=\frac{4\text{s}}{\text{R}}\ \cdots(1)$
Excess pressure inside the 1 st soap bubble:
$2\text{P}=\frac{4\text{s}}{\text{x}}$
From (1), we get:
$2\Big(\frac{2\text{S}}{\text{R}}\Big)=\frac{\text{4S}}{\text{x}}$
$\Rightarrow\text{x}=\frac{\text{R}}{2}$
Volume of the fust bubble $=\frac{4}{3}\pi\text{x}^3$
Volume of the second bubble $=\frac{4}{3}\pi\text{x}^3$
$\Rightarrow\frac{4}{3}\pi\text{x}^3=\text{n}\frac{4}{3}\pi\text{R}^3$
$\Rightarrow\text {x}^3=\text{n}\text{R}^3$
$\Rightarrow\Big(\frac{\text{R}}{2}\Big)^3=\text{n}\text{R}^3$
$\Rightarrow\text{n}=\frac{1}{8}=0.125$
$\therefore$ Excess pressure inside the first bubble = 2P
Let the radius of the second bubble be R.
Let the radius of the first bubble be x.
Excess pressure inside the 2nd soap bubble:
$\text{P}=\frac{4\text{s}}{\text{R}}\ \cdots(1)$
Excess pressure inside the 1 st soap bubble:
$2\text{P}=\frac{4\text{s}}{\text{x}}$
From (1), we get:
$2\Big(\frac{2\text{S}}{\text{R}}\Big)=\frac{\text{4S}}{\text{x}}$
$\Rightarrow\text{x}=\frac{\text{R}}{2}$
Volume of the fust bubble $=\frac{4}{3}\pi\text{x}^3$
Volume of the second bubble $=\frac{4}{3}\pi\text{x}^3$
$\Rightarrow\frac{4}{3}\pi\text{x}^3=\text{n}\frac{4}{3}\pi\text{R}^3$
$\Rightarrow\text {x}^3=\text{n}\text{R}^3$
$\Rightarrow\Big(\frac{\text{R}}{2}\Big)^3=\text{n}\text{R}^3$
$\Rightarrow\text{n}=\frac{1}{8}=0.125$
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