What will be the change in surface energy when a drop of liquid (S = 0.08N/m) is divided into 10 equal droplets? Radius of initial single drop (R)= 5cm.(a) 0.1259 J(b) 0.1141 J(c) 0.2356 J(d) 0.1765 JThe question was posed to me in unit test.My question comes from Fluids Mechanical Properties in portion Mechanical Properties of Fluids of Physics – Class 11

What will be the change in surface energy when a drop of liquid (S = 0

1.	What will be the change in surface energy when a drop of liquid (S = 0.08N/m) is divided into 10 equal droplets? Radius of initial single drop (R)= 5cm.(a) 0.1259 J(b) 0.1141 J(c) 0.2356 J(d) 0.1765 JThe question was posed to me in unit test.My question comes from Fluids Mechanical Properties in portion Mechanical Properties of Fluids of Physics – Class 11
Answer» Right option is (B) 0.1141 J Explanation: Volume of DROP = 4/3πR^2.Let the RADIUS of droplets be ‘r’. Volume of big drop will be equal to volume of 10 droplets. 104/3πr^3=4/3πR^3 ⇒ r = 10^1/3 R. SURFACE area of big drop = 4πR^2& surface area of 10 droplets = 10 * 4πr^2 = 40πR^2(10^2/3). Surface energy of big drop = 4πR^2 S = 4π(0.0025)0.08 = 0.0025 Surface energy of droplets = 40πR^2(10^2/3) S = 40π(0.0025)(10^2/3)0.08 = 0.1166. Change in surface energy = 0.1166 – 0.0025 = 0.1141 J.

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