Verified Solution • 2025–26 Session moderate

QJEEAdv20Ph3598: JEE Advanced 2020 Physics - Class 11 JEEAdv20Ph3598 Solution: A stationary tuning fork is in resonance with an air column in…

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 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 ms−1, the smallest value of the percentage change required in the length of the pipe is ____________.

Step-by-Step Verified Solution

f∝1l1⇒f=kl1f \propto {1 \over {{l_1}}} \Rightarrow f = {k \over {{l_1}}} ....(i)

(where, l1 ⇒\Rightarrow initial length of pipe)

(vv−vT)f=kl1\left( {{v \over {v - {v_T}}}} \right)f = {k \over {{l_1}}} ....(ii)

(where, vT = speed of tuning fork, l2 = new length of pipe)

Dividing Eq. (i) by Eq. (ii), we get

v−vTc=l2l1⇒l2l1−1=v−vTv−1{{v - {v_T}} \over c} = {{{l_2}} \over {{l_1}}} \Rightarrow {{{l_2}} \over {{l_1}}} - 1 = {{v - {v_T}} \over v} - 1

l2−l1l1=−vTv{{{l_2} - {l_1}} \over {{l_1}}} = {{ - {v_T}} \over v}

l2−l1l1×100=−2320×100=−0.625{{{l_2} - {l_1}} \over {{l_1}}} \times 100 = {{ - 2} \over {320}} \times 100 = - 0.625

Therefore, smallest value of percentage change required in the length of pipe is 0.625.

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring