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
• B. 19 erg
• C. 46 erg
• D. 74 erg

Answer 1

The correct option is D

74 erg

Step1: Given data

A 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

$$\begin{gathered} Volum{e}_{big-drop}=nVolum{e}_{small-drop}[n=numberofthesmalldrop] \\ Changeinenergy=Surfacetenison\times Increasedarea \end{gathered}$$

Step3: Calculating the change in energy

The 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.

$$\begin{gathered} Volum{e}_{big-drop}=nVolum{e}_{small-drop} \\ \frac{4}{3}\pi R^3=n\times \frac{4}{3}\pi r^3[R=raidusofbigdrop,r=radiusofsmalldrop] \\ \frac{4}{3}\pi {(0.14)}^3=125\times \frac{4}{3}\pi r^3 \\ r=\frac{R}{5}=\frac{0.14}{5}=0.028cm \\ Changeinenergy=Surfacetension\times increasedarea \\ E=75\times [125\times 4\pi r^2-4\pi R^2]erg \\ E=75\times 4\pi [125\times {(0.028)}^2-{(0.14)}^2]erg \\ E=73.89erg\simeq 74erg \end{gathered}$$

The energy is nearly 74 erg.

Hence, option D is the correct option.