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A spherical liquid drop of radius R is divided into eight equal droplets. If the surface tension is T, then the work done in this process will be
A) 2πR2T
B) 3πR2T
C) 4πR2T
D) 2πRT2

Answer
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Hint: As we know if anything is divided into any number of parts it can be any solid container in which we can store anything then total volume will remain the same but its size like in this case radius will decrease. And then calculating work done from W=T(AfAi).

Complete step by step answer:
As in the question we are given with radius and surface tension as Rand Trespectively and we know the formula of volume of spherical drop is
V=4πR33, and as given in question that spherical drop has divided into eight equal drops so we will equate the volume of both initial and final state of drops as
Vi=8Vf, where Vi is initial andVf is the final state of droplets. As it has divided into eight droplets we have written it above and as from the above formula we can see that V=4πR33soVR3.
So for initial state we can write, ViR3
And for the final state we can write, VfR13,where R1 is the final radius of small droplets.
And substituting these values in Vi=8Vf,we get
R3=8R13
R=(8R13)1/3
R=2R1, so the final radius will be half of initial and now to calculate work done we know the formula of work done as
W=T(AfAi)
W=T(4πR124πR2)
W=4πT(R24R2)
W=4πT(3R24)
W=3πTR2

So, The correct option is B option.

Note: From the above question we have seen that its along process in some questions you can also work from options that as in this question we are have D. option where dimension doesn’t match to work done so we can neglect that option and take care of negative sign in work done.