JEE Challenger
More from Mechanical Properties of Fluids
PhyMedAdv 2017 • Paper 1

Calculate Exponent for Number of Liquid Droplets Split from Surface Energy Change

Consider a spherical drop of liquid with radius R=10−2 mR = 10^{-2}\text{ m} and surface tension S=0.14π N m−1S = \frac{0.1}{4\pi}\text{ N m}^{-1}. It splits into KK identical smaller droplets. In this process, the overall change in surface energy is ΔU=10−3 J\Delta U = 10^{-3}\text{ J}. If K=10αK = 10^{\alpha}, calculate the value of α\alpha.

Official Numerical Answer6

Topics & Concepts

Mechanical Properties of FluidsSurface Tension

Step-by-Step Solution

To find the value of α\alpha, we analyze the surface energy change when a large spherical drop splits into KK identical smaller droplets.

Step 1: Conservation of Volume

Let the radius of the initial large drop be RR and the radius of each of the KK smaller droplets be rr.

The initial volume of the single drop is: Vi=43πR3V_i = \frac{4}{3}\pi R^3

The total final volume of the KK smaller droplets is: Vf=K⋅(43πr3)V_f = K \cdot \left(\frac{4}{3}\pi r^3\right)

Since volume is conserved (Vi=VfV_i = V_f): 43πR3=K⋅43πr3  ⟹  R3=Kr3  ⟹  r=RK−1/3\frac{4}{3}\pi R^3 = K \cdot \frac{4}{3}\pi r^3 \implies R^3 = K r^3 \implies r = R K^{-1/3}

Step 2: Calculate Initial and Final Surface Areas

  • Initial surface area: Ai=4πR2A_i = 4\pi R^2

  • Total final surface area of KK droplets: Af=K⋅(4πr2)=K⋅4π(RK−1/3)2=4πR2K1/3A_f = K \cdot (4\pi r^2) = K \cdot 4\pi (R K^{-1/3})^2 = 4\pi R^2 K^{1/3}

Step 3: Calculate the Change in Surface Energy

The change in surface energy ΔU\Delta U is given by: ΔU=Uf−Ui=S(Af−Ai)=4πR2S(K1/3−1)\Delta U = U_f - U_i = S(A_f - A_i) = 4\pi R^2 S (K^{1/3} - 1)

Given data:

  • R=10−2 mR = 10^{-2}\text{ m}
  • S=0.14π N m−1S = \frac{0.1}{4\pi}\text{ N m}^{-1}
  • ΔU=10−3 J\Delta U = 10^{-3}\text{ J}
  • K=10αK = 10^{\alpha}

Substituting these values into the expression for ΔU\Delta U: 10−3=4π(10−2)2⋅(0.14π)(K1/3−1)10^{-3} = 4\pi (10^{-2})^2 \cdot \left(\frac{0.1}{4\pi}\right) (K^{1/3} - 1)

Simplifying the equation: 10−3=10−4⋅0.1⋅(K1/3−1)10^{-3} = 10^{-4} \cdot 0.1 \cdot (K^{1/3} - 1) 10−3=10−5(K1/3−1)10^{-3} = 10^{-5} (K^{1/3} - 1) 10−310−5=K1/3−1\frac{10^{-3}}{10^{-5}} = K^{1/3} - 1 102=K1/3−110^2 = K^{1/3} - 1 K1/3=101K^{1/3} = 101

Since K≫1K \gg 1, we can approximate K1/3−1≈K1/3K^{1/3} - 1 \approx K^{1/3}, which gives: K1/3≈100=102K^{1/3} \approx 100 = 10^2

Cubing both sides: K=(102)3=106K = (10^2)^3 = 10^6

Given K=10αK = 10^{\alpha}, we have: 10α=106  ⟹  α=610^{\alpha} = 10^6 \implies \alpha = 6

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