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Question

Physics Question on Surface tension

Eight drops of water of 0.6 mm radius each merge to form one big drop. If the surface tension of water is 0.072 Nm−1,0.072\text{ }N{{m}^{-1}}, the energy dissipated in the process is

A

16 π×10−7J16~\pi \times {{10}^{-7}}J

B

4.15 π×10−7J4.15~\pi \times {{10}^{-7}}J

C

2.0.75 π×10−7J2.0.75~\pi \times {{10}^{-7}}J

D

8 π×10−7J8~\pi \times {{10}^{-7}}J

Answer

4.15 π×10−7J4.15~\pi \times {{10}^{-7}}J

Explanation

Solution

Volume of 8 drops of water = Volume of big drop of water Given that, Radius of small drop r = 0.6 mm ∴\therefore 8×43π(0.6)3=43πR38\times \frac{4}{3}\pi {{(0.6)}^{3}}=\frac{4}{3}\pi {{R}^{3}} or 2×0.6=R2\times 0.6=R or R=1.2 mmR=1.2\,mm where R is the radius of big drop. Change in surface area ΔA=4πr2×8−4πR2\Delta A=4\pi {{r}^{2}}\times 8-4\pi {{R}^{2}} =4π[(0.6)2×8−(1.2)2]×10−6=4\pi [{{(0.6)}^{2}}\times 8-{{(1.2)}^{2}}]\times {{10}^{-6}} =4π(0.36×8−1.44)×10−6=4\pi (0.36\times 8-1.44)\times {{10}^{-6}} =4π(2.88−1.44)×10−6=4\pi (2.88-1.44)\times {{10}^{-6}} ΔA=4π(1.44)×10−6m2\Delta A=4\pi (1.44)\times {{10}^{-6}}{{m}^{2}} Energy dissipated in this process E=T×ΔAE=T\times \Delta A =0.072×4π×1.44×10−6=0.072\times 4\pi \times 1.44\times {{10}^{-6}} =28.8×10−2×π×1.44×10−6=28.8\times {{10}^{-2}}\times \pi \times 1.44\times {{10}^{-6}} E=4.15π×10−7JE=4.15\pi \times {{10}^{-7}}J