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Question

# 'n' droplets of equal size of radius r coalesce to form a bigger drop of radius R. The energy liberated is equal to _________.(T= Surface tension of water)

A
4πR2T[n1/31]
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B
4πr2T[n1/31]
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C
4πR2T[n2/31]
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D
4πr2T[n2/31]
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Solution

## The correct option is A 4πR2T[n1/3−1]As the mass of the water is conserved, thus mass of bigger drop is equal to the mass of n droplets.∴ ρ(4π3R3)=ρ(n4π3r3) where ρ is the density of water⟹ r=n−1/3R .......(1)Surface area of bigger drop A=4πR2Total Surface area of n droplets A′=n(4πr2)∴ A′=n4π(n−2/3R2)=4πR2n1/3Change in surface area ΔA=A′−A=4πR2[n1/3−1]∴ Energy liberated E=TΔA=4πR2T[n1/3−1]

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