Physics · Sound Waves

JEE Main 2024 — 30 January, Shift 1 — Question 54

In a closed organ pipe, the frequency of fundamental note is 30 Hz . A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 Hz . If the organ pipe has a cross-sectional area of $2 \mathrm{~cm}^{2}$ , the amount of water poured in the organ tube is $\_\_\_\_\_\_$ g. (Take speed of sound in air is $330 \mathrm{~m} / \mathrm{s}$ )

Answer: 400

Numerical answer — enter this value.

Step-by-step solution

$\frac{V}{4 \ell_{1}}=30 \Rightarrow \ell_{1}=\frac{11}{4} \mathrm{~m}$

$\frac{V}{4 \ell_{2}}=110 \Rightarrow \ell_{2}=\frac{3}{4} m$ $\Delta \ell=2 m$ , Change in volume

$=A \Delta \ell=400 \mathrm{~cm}^{3}$

$M=400 \mathrm{~g} ;\left(\because \rho=1 \mathrm{~g} / \mathrm{cm}^{3}\right)$

Answer key and solution verified before publishing.

Practise Sound Waves

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 Main 2024
Paper: 30 January, Shift 1
Subject: Physics
Chapter: Sound Waves
Topic: Vibrations in rod and Air Columns - Organ pipes