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NVAEasyJEE 2023Surface Tension & Capillarity

JEE Physics 2023 Question with Solution

A spherical drop of liquid splits into 10001000 identical spherical drops. If uiu_i is the surface energy of the original drop and ufu_f is the total surface energy of the resulting drops, the ratio uf/ui=10/xu_f/u_i = 10/x. Then value of xx is _____:

Answer

Correct answer:1

Step-by-step solution

Standard Method

Given: A spherical drop splits into 10001000 identical smaller spherical drops.

Find: The value of xx if

ufui=10x\frac{u_f}{u_i} = \frac{10}{x}

Surface energy is proportional to surface area, so for surface tension TT:

ui=T⋅4πR2u_i = T \cdot 4\pi R^2

where RR is the radius of the bigger drop. Let rr be the radius of each smaller drop.

Since volume remains the same,

43πR3=1000×43πr3\frac{4}{3}\pi R^3 = 1000 \times \frac{4}{3}\pi r^3

So,

R=10rR = 10r

Now total final surface energy is proportional to total final surface area. Hence,

ufui=1000r2R2\frac{u_f}{u_i} = \frac{1000r^2}{R^2}

Substituting R=10rR = 10r,

ufui=1000r2(10r)2=1000r2100r2=101\frac{u_f}{u_i} = \frac{1000r^2}{(10r)^2} = \frac{1000r^2}{100r^2} = \frac{10}{1}

Comparing with

ufui=10x\frac{u_f}{u_i} = \frac{10}{x}

we get x=1x = 1.

Therefore, the required value is 11.

Common mistakes

  • Using surface area directly without first conserving volume. The bigger drop and the smaller drops must satisfy equal total volume before comparing areas. Always apply volume conservation first to relate RR and rr.

  • Assuming surface energy is proportional to volume instead of surface area. Surface energy for a liquid drop is proportional to T×4πr2T \times 4\pi r^2, not to r3r^3.

  • Calculating the area ratio for one small drop only. There are 10001000 identical small drops, so the final surface area must include the factor 10001000.

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