MHT CET · Physics · Mechanical Properties of Fluids
Let \(R_1, R_2\) and \(R_3\) be the radii of three mercury drops. \(A\) big mercury drop is formed from them under isothermal conditions. The radius of the resultant drop is
- A \(\left(R_1^3+R_2^3+R_3^3\right)^{\frac{1}{3}}\)
- B \(\left(R_1^2+R_2^3-R_3^3\right)^{\frac{1}{3}}\)
- C \(\left(R_1^3+R_2^3+R_3^3\right)\)
- D \(\left(R_1+R_2+R_3\right)^3\)
Answer & Solution
Correct Answer
(A) \(\left(R_1^3+R_2^3+R_3^3\right)^{\frac{1}{3}}\)
Step-by-step Solution
Detailed explanation
\(V_{initial} = V_{final}\) \(\frac{4}{3}\pi R_1^3 + \frac{4}{3}\pi R_2^3 + \frac{4}{3}\pi R_3^3 = \frac{4}{3}\pi R^3\)
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