MCQ
Two soap bubbles of radii $2 \,cm$ and $4 \,cm$ join to form a double bubble in air, then radius of curvature of interface is .......... $cm$
- A$2 \sqrt{5}$
- B$2$
- ✓$4$
- D$2 \sqrt{3}$
✓
Answer
Correct option: C.
$4$
c
(c)
(c)
$P_1-P_0=\frac{4 T}{R_1} ; P_2-P_0=\frac{4 T}{R_2}$
$P_1-P_2=4 T\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{4 T}{R}$
$\frac{1}{R}=\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{2}-\frac{1}{4}$
$\frac{1}{R}=\frac{2-1}{4}=\frac{1}{4}$
$R=4\,Cm$
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