A) \[2\pi \,{{R}^{2}}T\]
B) \[8\pi \,{{R}^{2}}T\]
C) \[4\pi \,{{R}^{2}}T\]
D) \[2\pi \,R{{T}^{2}}\]
Correct Answer: C
Solution :
Here, volume of 8 small droplets each of radius r = volume of big drop of radius R, \[\Rightarrow \] \[8\times \frac{4}{3}\pi {{r}^{3}}=\frac{4}{3}\pi {{R}^{3}}\] \[\Rightarrow \] \[{{r}^{3}}=\frac{{{R}^{3}}}{8}\] \[\Rightarrow \] \[r=\frac{R}{2}\] ?..(i) Work done (W) = surface tension (T) \[\times \] increase in surface area \[(\Delta A)=T[8\times 4\pi {{r}^{2}}-4\pi {{R}^{2}}]\] But \[r=\frac{R}{2}\] So, \[W=T\left[ 32\times \pi {{\left( \frac{R}{2} \right)}^{2}}-4\pi {{R}^{2}} \right]\] \[=4\pi {{R}^{2}}T.\]
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