Physics · Wave Optics

JEE Advanced 2020 — Paper 1 — Question 16

A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a speed of $2 \mathrm{~ms}^{-1}$ in front of the open end of the pipe and parallel to it, the length of the pipe should be changed for the resonance to occur with the moving tuning fork. If the speed of sound in air is $320 \mathrm{~ms}^{-1}$ , the smallest value of the percentage change required in the length of the pipe is $\_\_\_\_\_\_\_\_$ .

Answer: 0.63

Numerical answer — enter this value.

Step-by-step solution

$f=f_{0} \frac{320}{320-2}=\frac{320}{318} f_{0}$

\begin{gathered} \Rightarrow \Delta \mathfrak{f}=\frac{2}{318} \mathrm{f}_{0} ..(1) \end{gathered}

$\because \mathrm{f}_{0}=\frac{(2 \mathrm{n}+1) \mathrm{v}}{4 \mathrm{~L}}$

$\frac{2}{318}=-\frac{\Delta L}{L}$

$100 \times \frac{\Delta \mathrm{L}}{\mathrm{L}}=\frac{200}{318}=0.62983=0.63$

Answer key and solution verified before publishing.

Practise Wave Optics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Advanced 2020
Paper: Paper 1
Subject: Physics
Chapter: Wave Optics
Topic: Defects in Vision, Resolving Power, Doppler Effect of Light