MHT CET · Physics · Mechanical Properties of Fluids
Two rain drops of same radius are falling through air each with a steady speed of \(5 \mathrm{~cm} / \mathrm{s}\). If the drops coalesce, the new steady velocity of big drop will be
- A \(5 \mathrm{~cm} / \mathrm{s}\)
- B \(5 \sqrt{2} \mathrm{~cm} / \mathrm{s}\)
- C \(5 \times 2^{1 / 3} \mathrm{~cm} / \mathrm{s}\)
- D \(5 \times(4)^{1 / 3} \mathrm{~cm} / \mathrm{s}\)
Answer & Solution
Correct Answer
(D) \(5 \times(4)^{1 / 3} \mathrm{~cm} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Terminal speed \(\mathrm{v} \propto \mathrm{r}^2\)
\(\begin{array}{ll}
\therefore & \frac{v_1}{v_2}=\frac{r^2}{R^2}=\frac{r^2}{\left(2^{1 / 3} r\right)^2} \\
\therefore & v_2=v_1 \times \frac{\left(2^{1 / 3} r\right)^2}{r^2}=5 \times 2^{2 / 3}=5 \times 4^{1 / 3} \mathrm{~cm} \mathrm{~s}^{-1}
\end{array}\)
\(\begin{array}{ll}
\therefore & \frac{v_1}{v_2}=\frac{r^2}{R^2}=\frac{r^2}{\left(2^{1 / 3} r\right)^2} \\
\therefore & v_2=v_1 \times \frac{\left(2^{1 / 3} r\right)^2}{r^2}=5 \times 2^{2 / 3}=5 \times 4^{1 / 3} \mathrm{~cm} \mathrm{~s}^{-1}
\end{array}\)
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