Question

A submerged scuba diver hears the sound of a boat horn directly above her on the surface of the lake. At the same time, a friend on dry land 22.0 m from the boat also hears the horn (fig). The horn is 1.2 m above the surface of the water. What is the distance (labeled by "?” in fig) from the horn to the diver? Both air and water are at 20${}^{\circ }$C.

$B_{water}=2.2\times 109Pa,{\gamma }_{air}=1.4,M_{air}\approx 30g$

• A. 90.5 m
• B. 50 m
• C. 10 m
• D. None of these

Answer 1

The correct option is A

90.5 m

For the friend on the surface,

$t=\frac{22}{v_{air}}$

$t=\frac{22}{\sqrt{\frac{\gamma RT}{M}}}=\frac{22}{\sqrt{\frac{1.4\times 8.3\times 293}{30\times {10}^{-3}}}}=0.065$ seconds

Now time for scuba diver is also t

$t=\frac{1.2}{\sqrt{\frac{1.4\times 8.3\times 293}{30\times {10}^{-3}}}}+\frac{x}{\sqrt{\frac{2.2\times 109}{1000}}}[v_{water}=\sqrt{\frac{B}{\rho }}]$

$0.065=\frac{1.2}{\sqrt{\frac{1.4\times 8.3\times 293}{30\times {10}^{-3}}}}+\frac{21}{\sqrt{\frac{2.2\times 109}{1000}}}$

$0.061=\frac{x}{\sqrt{2.2\times {10}^6}}\Rightarrow x=90.5m$