Get Answers to all your Questions

header-bg qa

A point source of power 50 \pi watts is producing sound waves of frequency 1875 Hz. The velocity of sound is 330 m/s, atmospheric pressure is 1.0 \times 10^{5}\; N/m^{2} and density of air is 1\; kg/m^{3}. Pressure amplitude at a distance r=\sqrt{330}m from the point source is

Option: 1

5 \; Nm^{2}


Option: 2

10 \; Nm^{-2}


Option: 3

15 \; Nm^{-2}

 


Option: 4

20 \; Nm^{2}


Answers (1)

best_answer

Answer (1)

I=\frac{P_{0}^{2}}{2\rho v}\Rightarrow \frac{P}{4\pi r^{2}}=\frac{P_{0}^{2}}{2\rho v}

\therefore P_{0}=\sqrt{\frac{50 \pi \times 1 \times 330}{2 \pi \times 330}}=5\; Nm^{-2}

Posted by

Info Expert 30

View full answer

NEET 2024 Most scoring concepts

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