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JEE Main 202528 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2025 · 28 January · Evening Shift · Physics · Mechanical Properties of Fluids

A thin transparent film with refractive index 1.4 , is held on...

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}.

Correct answer

0

Step-by-step solution

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=53×10−8m/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)253×10−8m3/s=54π×10−13m3/s\begin{aligned} & V=\text{ Volume\text{ 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}$$
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