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  • A spherical liquid drop of diameter D breaks up into n identical spherical drops. If the surface tension of the liquid is <img alt="\mathrm{\sigma,}" src="https://entrancecorner.oncodecog

A spherical liquid drop of diameter D breaks up into n identical spherical drops. If the surface tension of the liquid is \mathrm{\sigma,}  the change in energy in this process is

Option: 1

\mathrm{\pi \sigma D^2\left(n^{1 / 3}-1\right)}


Option: 2

\mathrm{\pi \sigma D^2\left(n^{2 / 3}-1\right)}


Option: 3

\mathrm{\pi \sigma D^2(n-1)}


Option: 4

\mathrm{\pi \sigma D^2\left(n^{4 / 3}-1\right)}


Answers (1)

best_answer

Refer to the solution of Q. 24. Initial energy \mathrm{=4 \pi R^2 \sigma =\pi D^2 \sigma}  where R is the radius of the big drop. Final energy

\mathrm{ =\left(4 \pi r^2\right) n \sigma=\pi d^2 n \sigma, \text { where } d=2 r, \text { is the } }
radius of each tiny drop.
Now 

\mathrm{R^3=n r^3 \, \, or \, \, D^3=n d^3 \, \, or\, \, D=n^{1 / 3} d \, \, or\, \, d=D / n^{1 / 3}.}

Therefore, final energy \mathrm{=\frac{\pi D^2 n \sigma}{n^{2 / 3}}=\pi D^2 \sigma n^{1 / 3}.}

Hence, change in energy \mathrm{=\pi D^2 \sigma n^{1 / 3}-\pi D^2 \sigma}

                                         \mathrm{=\pi D^2 \sigma\left(n^{1 / 3}-1\right),}

Posted by

Pankaj Sanodiya

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