MHT CET · Physics · Mechanical Properties of Fluids
The energy needed for breaking a liquid drop of radius ' R ' into ' n ' droplets each of radius ' \(r\) ' is
\(\text { [ } \mathrm{T}=\text { surface tension of the liquid] }\)
- A \(4 \pi \mathrm{~T} \mathrm{R}^2\left[\frac{\mathrm{R}}{\mathrm{r}}+1\right]\)
- B \(4 \pi \mathrm{~T} \mathrm{R}^2\left[\frac{\mathrm{R}}{\mathrm{r}}-1\right]\)
- C \(4 \pi \mathrm{~T} \mathrm{R}^2\left[\frac{\mathrm{r}}{\mathrm{R}}+1\right]\)
- D \(4 \pi \mathrm{Tr}^2\left[\frac{\mathrm{R}}{\mathrm{r}}-1\right]\)
Answer & Solution
Correct Answer
(B) \(4 \pi \mathrm{~T} \mathrm{R}^2\left[\frac{\mathrm{R}}{\mathrm{r}}-1\right]\)
Step-by-step Solution
Detailed explanation
\(R^3 = n r^3 \implies n = \frac{R^3}{r^3}\) \(E = T (\text{Final Surface Area} - \text{Initial Surface Area})\)
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