JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Fluid Mechanics

A drop of liquid of radius

R = 1 0^{- 2} m

having surface tension

S = \frac{0.1}{4 π} Nm^{- 1}

divides itself into

K

identical drops. In this process the total change in the surface energy

Δ U = 1 0^{- 3} J

. If

K = 1 0^{α}

then the value of

α

Answer: 6

Solution

When a large drop of radius

R

divides into

K

smaller drops of radius

r

, volume conservation gives

\frac{4}{3} π R^{3} = K \cdot \frac{4}{3} π r^{3}

, so

r = R / K^{1/3}

. The initial surface area is

4 π R^{2}

and the final total surface area is

K \cdot 4 π r^{2} = K \cdot 4 π (R / K^{1/3})^{2} = 4 π R^{2} K^{1/3}

. The change in surface energy is

Δ U = S \cdot Δ A = S \cdot 4 π R^{2} (K^{1/3} - 1)

. Substituting values:

1 0^{- 3} = \frac{0.1}{4 π} \cdot 4 π \cdot (1 0^{- 2})^{2} \cdot (K^{1/3} - 1) = 0.1 \cdot 1 0^{- 4} \cdot (K^{1/3} - 1) = 1 0^{- 5} (K^{1/3} - 1)

. Thus

K^{1/3} - 1 = 100

, giving

K^{1/3} = 101 \approx 100

, so

K \approx 1 0^{6}

and

α = 6

JEE Advanced · 2017 · 21 May 2017 · Paper 1 · Integer answer