Some of the most important Physics Topics include energy, motion, and force.

## Why The Density of Solid is Greater than Density of Liquid and Gas?

Sound propagates through solids and liquids as longitudinal elastic wave like in gaseous media. But the velocities of sound in solid and liquid are different. If E is the modulus of elasticity of the medium and ρ is its density, Newton’s formula for the velocity of sound is given by [see section 3.7],

c = \(\sqrt{\frac{\underline{E}}{\rho}}\) ……. (1)

The density of any solid or liquid is greater than that of a gas. But the modulus of elasticity of a solid or a liquid is much greater than that of a gas. So, the velocity of sound in a solid or a liquid is greater than that in a gas. For example the velocities of sound in iron and in water are about 15 times and 4.5 times that in air, respectively.

The velocity of sound in iron is greater than that in air – it can be easily understood from a simple experiment. If one end of a long iron pipe is struck heavily the sound of striking is heard twice at the other end of the pipe. Sound is heard for the first time due to its propagation through the iron and for the second time due to its propagation through the air in the pipe. An iron pipe nearly 100 m in length should be taken to hear the two sounds distinctly.

There is one important difference between solids and liquids as media of propagation of sound. The modulus of elasticity for the solid is its, Young’s modulus (Y) while ills the bulk modulus (k) in case of the liquid. So the equation (1)

for solid, c = \(\sqrt{\frac{Y}{\rho}}\) …… (2)

and for liquid, c = \(\sqrt{\frac{k}{\rho}}\) …… (3)

In case of steel, Y = 2 × 10^{11} N ᐧ m^{-2} ; ρ = 7850 kg ᐧ m^{-3};

From equation (2) we get,

c = \(\sqrt{\frac{2 \times 10^{11}}{7850}}\) ≈ 5048 m ᐧ s^{-1}

In case of water, k = 2.1 × 10^{9} N ᐧ m^{-2} ; ρ = 1000 kg ᐧ m^{-3}

From equation (3) we get,

c = \(\sqrt{\frac{2.1 \times 10^9}{1000}}\) = 1449 m ᐧ s^{-1}

### Numerical Examples

**Example 1.**

Two explosions are made simultaneously from ship, one above the surface of water and another just below it. Sound is heard in a hydrophone placed below water from another ship 5 km apart 11 s earlier than the sound reaching the deck of the ship. What is the velocity of sound ¡n water? Given that velocity of sound in air = 348 m ᐧ s^{-1}.

**Solution:**

Distance between the two ships = 5 km = 5000 m

So time taken by the sound to come through air = \(\frac{5000}{348}\)s

According to the question, time taken by the sound to come through work = \(\frac{5000}{348}\) – 11 = \(\frac{5000-3828}{348}\) = \(\frac{1172}{348}\)s

Therefore, velocity of sound in water

= \(\frac{5000}{\frac{1172}{348}}\) = 1485 ᐧ s^{-1} (approx.).

**Example 2.**

If one end of a long steel pipe is struck, sound is heard twice at an interval of 3 s at the other end. What is the length of the pipe? (The velocities of sound in air = 350 m ᐧ s^{-1}, in steel = 5000 m ᐧ s^{-1})

**Solution:**

Let the length of the pipe be l m.

So, time taken by the sound to come through the air in the pipe = \(\frac{l}{350}\)s

Again, time taken by the sound to come through steel = \(\frac{l}{5000}\)s

According to the question,

\(\frac{l}{350}\) – \(\frac{l}{5000}\) = 3 or, l(\(\frac{1}{350}\) – \(\frac{1}{5000}\)) = 3

or, l\(\left(\frac{100-7}{35000}\right)\) = 3

or, l = \(\frac{35000 \times 3}{93}\) = 1129 m = 1.129 km.

**Example 3.**

Sound moves through a liquid with velocity 1340 m ᐧ s^{-1}. If the density of the liquid is 0.8 g ᐧ cm^{-3}, determine the compressibility of the liquid.

**Solution:**

We know c = \(\sqrt{\frac{k}{\rho}}\); k = bulk modulus of the liquid

∴ k = c^{2}ρ [c = 1340 m ᐧ s^{-1} = 134 × 10^{3}cm ᐧ s^{-1}]

Again, compressibility = \(\frac{1}{\left(134 \times 10^3\right)^2 \times 0.8}\) = \(\frac{1}{143648}\) × 10^{-5}

= 6.96 × 10^{-11} cm^{2} ᐧ dyn^{-1}.