MHT CET · Physics · Mechanical Properties of Fluids
Two identical drops of water are falling through air with steady velocity ' \(V\) '. If the two drops come together to form a single drop. The new velocity of the single drop is
- A \((2)^{1 / 3} \mathrm{~V}\)
- B \((2)^{3 / 2} \mathrm{~V}\)
- C \((2)^{2 / 3} \mathrm{~V}\)
- D \((2)^{1 / 4} \mathrm{~V}\)
Answer & Solution
Correct Answer
(C) \((2)^{2 / 3} \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Let the radius of the 2 rain droplets be r each.
They coalesce to form a drop of radius R .
As volume is conserved, \(\mathrm{R}^3=2 \mathrm{r}^3\)
\(\therefore \quad R=2^{\frac{i}{3}} r\)
Terminal velocity \(\mathrm{V}=\frac{2}{9} \frac{\mathrm{r}^2(\rho-\sigma) g}{\eta}\)
\(\therefore \quad \mathrm{V} \propto \mathrm{r}^2\)
\(\therefore \quad \mathrm{V}^{\prime}=\mathrm{V} \frac{\mathrm{R}^2}{\mathrm{r}^2}=\mathrm{V} \times \frac{2^{\frac{2}{3}} \mathrm{r}^2}{\mathrm{r}^2}=\mathrm{V} \cdot 2^{2 / 3}\)
They coalesce to form a drop of radius R .
As volume is conserved, \(\mathrm{R}^3=2 \mathrm{r}^3\)
\(\therefore \quad R=2^{\frac{i}{3}} r\)
Terminal velocity \(\mathrm{V}=\frac{2}{9} \frac{\mathrm{r}^2(\rho-\sigma) g}{\eta}\)
\(\therefore \quad \mathrm{V} \propto \mathrm{r}^2\)
\(\therefore \quad \mathrm{V}^{\prime}=\mathrm{V} \frac{\mathrm{R}^2}{\mathrm{r}^2}=\mathrm{V} \times \frac{2^{\frac{2}{3}} \mathrm{r}^2}{\mathrm{r}^2}=\mathrm{V} \cdot 2^{2 / 3}\)
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