One spherical liquid drop of radius R is broken into n equal smaller drops without loss of liquid. The surface tension is S. What minimum work must be supplied to create the additional surface area?
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Question

One spherical liquid drop of radius R is broken into n equal smaller drops without loss of liquid. The surface tension is S. What minimum work must be supplied to create the additional surface area?

G1 Physics · Accepted Answer

Let the radius of each small drop be r. Volume conservation gives R³ = nr³. Thus, r = R/∛n. Final total area: A(final) = n(4πr²) = 4πR²∛n. Initial area: A(initial) = 4πR². Therefore, ΔA = 4πR²(∛n−1). The required work equals the increase in surface energy: W = SΔA = 4πSR²(∛n−1). Correct option: A.