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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 \[W = - T\left( {{A_f} - {A_i}} \right)\].

**Complete step by step answer:**

As in the question we are given with radius and surface tension as \[R\]and \[T\]respectively and we know the formula of volume of spherical drop is

\[V = \dfrac{{4\pi {R^3}}}{3}\], 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

\[{V_i} = 8{V_f}\], where \[{V_i}\] is initial and\[{V_f}\] 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 \[V = \dfrac{{4\pi {R^3}}}{3}\]so\[V \propto {R^3}\].

So for initial state we can write, \[{V_i} \propto {R^3}\]

And for the final state we can write, \[{V_f} \propto {R_1}^3\],where \[{R_1}\] is the final radius of small droplets.

And substituting these values in \[{V_i} = 8{V_f}\],we get

\[{R^3} = 8{R_1}^3\]

\[R = {\left( {8{R_1}^3} \right)^{1/3}}\]

\[R = 2{R_1}\], so the final radius will be half of initial and now to calculate work done we know the formula of work done as

\[W = - T\left( {{A_f} - {A_i}} \right)\]

\[W = - T\left( {4\pi {R_1}^2 - 4\pi {R^2}} \right)\]

\[W = - 4\pi T\left( {\dfrac{{{R^2}}}{4} - {R^2}} \right)\]

\[W = - 4\pi T\left( { - 3\dfrac{{{R^2}}}{4}} \right)\]

\[W = 3\pi T{R^2}\]

**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.

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