Fluid Mechanics

JEE Main level, numerical answer, medium

99 of 192

Question 99 of 192

A thin transparent film with refractive index 1.4 , is held on circular ring of radius 1.8 cm . The fluid in the film evaporates such that transmission through the film at wavelength 560 nm goes to a minimum every 12 seconds. Assuming that the film is flat on its two sides, the rate of evaporation is ____ π×10−13 m3/s\pi \times 10^{-13} \mathrm{~m}^3 / \mathrm{s}.

Show answer and solution

Answer: 54

For a thin film interference, a fringe for transmission is formed.
When
2μx=nλ2 \mu x=n \lambda
⇒dxdt=(dndt)λ2μ=(112)560×10−92×1.4=53×10−8 m/s\begin{aligned} \Rightarrow & \frac{d x}{d t}=\left(\frac{d n}{d t}\right) \frac{\lambda}{2 \mu} \\ & =\left(\frac{1}{12}\right) \frac{560 \times 10^{-9}}{2 \times 1.4}=\frac{5}{3} \times 10^{-8} \mathrm{~m} / \mathrm{s}\end{aligned}
V= Volume of film =πR2xdVdt=πR2dxdt=π(1.8×10−2)253×10−8 m3/s=54π×10−13 m3/s\begin{aligned} & V=\text { Volume of film }=\pi R^2 x \\ & \frac{d V}{d t}=\pi R^2 \frac{d x}{d t} \\ & =\pi\left(1.8 \times 10^{-2}\right)^2 \frac{5}{3} \times 10^{-8} \mathrm{~m}^3 / \mathrm{s} \\ & =54 \pi \times 10^{-13} \mathrm{~m}^3 / \mathrm{s}\end{aligned}

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.