Question

# A drop of liquid of diameter 2.8 mm breaks up into 125 identical drops. The change in energy is nearly (S.T. of liquid = 75 dynes/cm)

A

zero

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B

19 erg

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C

46 erg

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D

74 erg

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Solution

## The correct option is D 74 ergStep1: Given dataA drop of liquid of diameter 2.8 mm breaks up into 125 identical drops, $R=\frac{2.8}{2}mm=1.4mm=0.14cm,n=125$Surface tension of liquid = 75 dynes/cm.Step2: Formula used$Volum{e}_{big-drop}=nVolum{e}_{small-drop}\left[n=numberofthesmalldrop\right]\phantom{\rule{0ex}{0ex}}Changeinenergy=Surfacetenison×Increasedarea$Step3: Calculating the change in energyThe volume of the big drop is equal to the total volume of all the small drops. From this concept, the radius of each small drop can be calculated. And, the radius will help to measure the area. $Volum{e}_{big-drop}=nVolum{e}_{small-drop}\phantom{\rule{0ex}{0ex}}\frac{4}{3}\pi {R}^{3}=n×\frac{4}{3}\pi {r}^{3}\left[R=raidusofbigdrop,r=radiusofsmalldrop\right]\phantom{\rule{0ex}{0ex}}\frac{4}{3}\pi {\left(0.14\right)}^{3}=125×\frac{4}{3}\pi {r}^{3}\phantom{\rule{0ex}{0ex}}r=\frac{R}{5}=\frac{0.14}{5}=0.028cm\phantom{\rule{0ex}{0ex}}Changeinenergy=Surfacetension×increasedarea\phantom{\rule{0ex}{0ex}}E=75×\left[125×4\pi {r}^{2}-4\pi {R}^{2}\right]erg\phantom{\rule{0ex}{0ex}}E=75×4\pi \left[125×{\left(0.028\right)}^{2}-{\left(0.14\right)}^{2}\right]erg\phantom{\rule{0ex}{0ex}}E=73.89erg\simeq 74erg$The energy is nearly 74 erg.Hence, option D is the correct option.

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