Question

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

पर्याय

• 4

• 2

• 1

• 0.125

Answer 1

Let the excess pressure inside the second bubble be P.
​∴ 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.

\[\text{ Excess pressure inside the 2 nd soap bubble :} \]

\[P = \frac{4S}{R} . . . (1)\]

\[\text{ Excess pressure inside the 1st soap bubble: }\]

\[2P = \frac{4S}{x}\]

\[\text{ From (1), we get :} \]

\[2\left( \frac{4S}{R} \right) = \frac{4S}{x}\]

\[ \Rightarrow x = \frac{R}{2}\]

\[\text{ Volume of the first bubble }= \frac{4}{3} \pi x^3 \]

\[\text{ Volume of the second bubble } = \frac{4}{3} \pi R^3 \]

\[ \Rightarrow \frac{4}{3} \pi x^3 = n\frac{4}{3} \pi R^3 \]

\[ \Rightarrow x^3 = {nR}^3 \]

\[ \Rightarrow \left( \frac{R}{2} \right)^3 = {nR}^3 \]

\[ \Rightarrow n = \frac{1}{8} = 0 . 125\]