Fluid Mechanics

JEE Main level, single correct, medium

90 of 192

Question 90 of 192

A liquid drop of diameter 22 mm breaks into 512512 droplets. The change in surface energy is α×10−6\alpha \times 10^{-6} J. The value of α\alpha is _______. (Take surface tension of liquid =0.08= 0.08 N/m)

Show answer and solution

Answer: (B) 77

Let the radius of the initial drop be RR and the radius of each small droplet be rr.

Given diameter D=2D = 2 mm, so R=1R = 1 mm =10−3= 10^{-3} m.

By conservation of volume:

43πR3=512×43πr3\dfrac{4}{3}\pi R^3 = 512 \times \dfrac{4}{3}\pi r^3

R3=512r3⇒R=8r⇒r=R8R^3 = 512 r^3 \Rightarrow R = 8r \Rightarrow r = \dfrac{R}{8}

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

Final surface area, Af=512×4πr2=512×4π(R8)2=32πR2A_f = 512 \times 4\pi r^2 = 512 \times 4\pi \left(\dfrac{R}{8}\right)^2 = 32\pi R^2

Change in surface area, ΔA=Af−Ai=32πR2−4πR2=28πR2\Delta A = A_f - A_i = 32\pi R^2 - 4\pi R^2 = 28\pi R^2

Change in surface energy, ΔU=TΔA\Delta U = T \Delta A

ΔU=0.08×28πR2\Delta U = 0.08 \times 28\pi R^2

Substituting R=10−3R = 10^{-3} m and π≈227\pi \approx \dfrac{22}{7}:

ΔU=0.08×28×227×(10−3)2\Delta U = 0.08 \times 28 \times \dfrac{22}{7} \times (10^{-3})^2

ΔU=0.08×4×22×10−6\Delta U = 0.08 \times 4 \times 22 \times 10^{-6}

ΔU=7.04×10−6\Delta U = 7.04 \times 10^{-6} J

Comparing with α×10−6\alpha \times 10^{-6} J, we get α≈7\alpha \approx 7.

Answer: 77

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Fluid mechanics simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Fluid Mechanics (Surface tension), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.