JEE Mains · Physics · STD 11 - 9.2 surface tension
The radii of two soap bubbles are \(r_1\) and \(r_2\). In isothermal conditions, two meet together in vaccum. Then the radius of the resultant bubble is given by
- A \(\frac{r_{1} r_{2}}{r_{1}+r_{2}}\)
- B \(\sqrt{r_{1} r_{2}}\)
- C \(\sqrt{r_{1}^{2}+r_{2}^{2}}\)
- D \(\frac{r_{1}+r_{2}}{2}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{r_{1}^{2}+r_{2}^{2}}\)
Step-by-step Solution
Detailed explanation
(c) \(8\pi r_1^2T + 8\pi r_2^2T = 8\pi {R^2}T\) \(⇒\) \({R^2} = r_1^2 + r_2^2\)
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