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Two spherical soap bubbles formed in vacuum have diameters 3.0 \mathrm{~mm} and 4.0 \mathrm{~mm}. They coalesce to form a single spherical bubble. If the temperature remains unchanged, the diameter of the bubble so formed will be

Option: 1

5.0 mm


Option: 2

5.8 mm


Option: 3

6.2 mm


Option: 4

7.0 mm


Answers (1)

best_answer

Since the bubbles are in vacuum, the pressure of air inside them are \mathrm{P_1=\frac{4 \sigma}{r_1} \, \, and \, \, P_2=\frac{4 \sigma}{r_2}, where \, \, r_1=3.0 \mathrm{~mm} \, \, and\, \, r_2=4.0 \mathrm{~mm}}

. Since the temperature remains unchanged, we have from Boyle's law

\mathrm{ \begin{gathered} P_1 V_1+P_2 V_2=P V \\\\ \text { or } \frac{4 \sigma}{r_1} \cdot \frac{4}{3} \pi r_1^3+\frac{4 \sigma}{r_2} \cdot \frac{4}{3} \pi r_2^3=\frac{4 \sigma}{r} \cdot \frac{4}{3} \pi r^3 \end{gathered} }

where r is the radius of the single bubble formed.

From (i), we get 

\mathrm{r^2=r_1^2+r_2^2 \, \, or \, \, r=\sqrt{r_1^2+r_2^2}}

\mathrm{=\sqrt{(3.0)^2+(4.0)^2}=5.0 \mathrm{~mm}},

which is choice (a).

Posted by

Ritika Harsh

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