MHT CET · Physics · Mechanical Properties of Fluids
A liquid drop of density ' \(Q\) ' is floating half immersed in a liquid of density ' \(d\) '. Diameter of the liquid drop is
( \(\mathrm{Q}>\mathrm{d}, \mathrm{g}=\) acceleration due to gravity, \(\mathrm{T}=\) surface tension)
- A \(\left[\frac{3 T}{g(2 Q-d)}\right]^{\frac{1}{2}}\)
- B \(\left[\frac{6 T}{g(Q-d)}\right]^{\frac{1}{2}}\)
- C \(\left[\frac{12 \mathrm{~T}}{\mathrm{~g}(2 \mathrm{Q}-\mathrm{d})}\right]^{\frac{1}{2}}\)
- D \(\left[\frac{9 T}{g(Q-d)}\right]^{\frac{1}{2}}\)
Answer & Solution
Correct Answer
(B) \(\left[\frac{6 T}{g(Q-d)}\right]^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
The force balance invovles, surface-tension force and boyancy force balance the weight of the droplet, so the equation is:
\((2 \pi r) T+\left(\frac{4}{3} \pi r^3\right) \mathrm{dg}=\left(\frac{4}{3} \pi r^3\right) \mathrm{Qg}\)
Where, \(\mathrm{r}\) is the radius of the drop, \(\mathrm{T}\) the interfacial tension between the liquids, \(g\) is acceleration due to gravity.
Therefore,
\(\mathrm{T}=\left(\frac{2}{3} \mathrm{r}^2\right)(\mathrm{Q}-\mathrm{d}) \mathrm{g}\)
Diameter \(=2 r=\left[\frac{6 T}{g(Q-d)}\right]^{\frac{1}{2}}\).
\((2 \pi r) T+\left(\frac{4}{3} \pi r^3\right) \mathrm{dg}=\left(\frac{4}{3} \pi r^3\right) \mathrm{Qg}\)
Where, \(\mathrm{r}\) is the radius of the drop, \(\mathrm{T}\) the interfacial tension between the liquids, \(g\) is acceleration due to gravity.
Therefore,
\(\mathrm{T}=\left(\frac{2}{3} \mathrm{r}^2\right)(\mathrm{Q}-\mathrm{d}) \mathrm{g}\)
Diameter \(=2 r=\left[\frac{6 T}{g(Q-d)}\right]^{\frac{1}{2}}\).
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