Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Medical
  • The maximum pressure variation that the human ear can tolerate in loud sound is about 30 N/m2. The corresponding maximum displacement for a sound wave in air having a frequency of 10<sup

The maximum pressure variation that the human ear can tolerate in loud sound is about 30 N/m2. The corresponding maximum displacement for a sound wave in air having a frequency of 103 Hz is (take velocity of sound in air as 300 m/s and density of air 1.5 kq/ m3)

Option: 1

\frac{2 \pi}{3} \times 10^{-2} \mathrm{~m}


Option: 2

\frac{2 \times 10^{-4}}{\pi} \mathrm{m}


Option: 3

\frac{\pi}{3} \times 10^{-2} \mathrm{~m}


Option: 4

\frac{10^{-4}}{3 \pi} \mathrm{m}


Answers (1)

best_answer

\begin{aligned} & (\Delta P)_{\text {max }}=B A K \\ \therefore \quad & A=\frac{(\Delta P)_{\text {max }}}{B K}-\left ( 1 \right ) \end{aligned}\\\\ Here, \: v=\sqrt{\frac{B}{\rho}}=\frac{\omega}{k}\\\\\ \therefore \quad k=\omega \sqrt{\frac{\rho}{B}}=2 \pi f \sqrt{\frac{\rho}{B}}\\\\ \text{further,} B=\rho r^2\\\\ \text{substituting in equation (1) we get}\\\\\ A=\frac{(\Delta P)_{\text {max }}}{2 \pi f \sqrt{B \rho}}=\frac{(\Delta P)_{\text {max }}}{2 \pi f \rho V}\\\\ \text {substituting the values, we here}\\\\ A=\frac{30}{2 \pi \times 10^3 \times 1.5 \times 300}=\frac{10^{-4}}{3 \pi} \mathrm{m}

Posted by

Divya Prakash Singh

View full answer

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

Similar Questions

Trending Articles/News

anam-khan

Uttarakhand NEET PG 2025 round 3 counselling registration begins; merit list on January 6
anam-khan

Tamil Nadu opens BDS special stray round 2 with NEET cut-off as low as 113, sets Rs 2 lakh security deposit
anam-khan

Karnataka NEET UG Counselling 2025: KEA to hold special stray vacancy round for dental seats on December 31

Latest Question