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Chapter 9: Problem 36

A longitudinal wave sent by a ship to the bottom of the sea returns after a lapse of \(2.64 \mathrm{~s}\). Elasticity of water is \(220 \mathrm{~kg} / \mathrm{mm}^{2}\) and density of sea water is \(1.1 \mathrm{gm} / \mathrm{cc}\). The depth of the sea is (in metres) (A) 1400 (B) 1848 (C) 924 (D) 700

Short Answer

Expert verified
The depth of the sea is approximately 18664.05 meters, so none of the given answer choices are correct.

Step by step solution

01

Convert Units

First, convert all given units into the SI system. The elasticity is given in kg/mm虏, convert it to N/m虏. So, \(E = 220 \times 10^{6} N/m虏\). The density is given in gm/cc, convert it to kg/m鲁. So, \(\rho = 1.1 \times 10^{3} kg/m鲁\).
02

Calculate Wave Speed

Next, use the formula for wave speed in a medium. That is \(v = \sqrt{\frac{E}{\rho}}\). Substitute the appropriate values to get the wave speed \(v = \sqrt{\frac{220 \times 10^{6}}{1.1 \times 10^{3}}}= \sqrt{200000000} \approx 14142.14 m/s\).
03

Calculate Depth of the Sea

Finally, use the wave speed in the formula for speed in terms of distance and time to find the depth of the sea. The formula is \(v = \frac{2d}{t}\), solving for d gives us \(d = \frac{vt}{2}\). With v as 14142.14 m/s and t as 2.64 s, we have \(d = \frac{14142.14 \times 2.64}{2} \approx 18664.05 m\).

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