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The work done to get n smaller equal size spherical drops from a bigger size spherical drop of water is proportional to :

\[\begin {array} {1 1} (a)\;\bigg[\frac{1}{n^{2/3}}-1\bigg] & \quad  (b)\;\bigg[1-\frac{1}{n^{1/3}}\bigg] \\  (c)\;\bigg[n^{2/3}-1\bigg] & \quad (d)\;\bigg[n^{4/3}-1\bigg] \end {array}\]

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$(b)\;\bigg[1-\frac{1}{n^{1/3}}\bigg] $
answered Nov 7, 2013 by pady_1
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