MHT CET · Physics · Mechanical Properties of Fluids
Under isothermal conditions, two soap bubbles of radii ' \(r_{1}\) ' and ' \(r_{2}\) ' coalesce to form a big drop. The radius of the big drop is
- A \(\left(r_{1}-r_{2}\right)^{\frac{1}{2}}\)
- B \(\left(r_{1}+r_{2}\right)^{\frac{1}{2}}\)
- C \(\left(r_{1}^{2}+r_{2}^{2}\right)^{\frac{1}{2}}\)
- D \(\left(r_{1}^{2}-r_{2}^{2}\right)^{\frac{1}{2}}\)
Answer & Solution
Correct Answer
(C) \(\left(r_{1}^{2}+r_{2}^{2}\right)^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
Under isothermal conditions the surface tension remains constant. If \(r\) is the radius of the digger drop, then Final surface energy \(=\) Initial surface energy \(\therefore 8 \pi r^{2} T=8 \pi r_{1}^{2} T+8 \pi r_{2}^{2} T\)
\(\therefore r^{2}=r_{1}^{2}+r_{2}^{2}\)
\(\therefore r=\left(r_{1}^{2}+r_{2}^{2}\right)^{\frac{1}{2}}\)
\(\therefore r^{2}=r_{1}^{2}+r_{2}^{2}\)
\(\therefore r=\left(r_{1}^{2}+r_{2}^{2}\right)^{\frac{1}{2}}\)
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