Question

The average density of Earth's crust $10\,km$ beneath the surface is $2.7\,gm/c{m^3}$ The speed of longitudinal seismic waves at that depth is $5.4\,km/s$ The bulk modulus of earth's crust Its considering its behavior as fluid at that depth Is:
A) $7.9\, \times {10^{10}}\,Pa$
B) $5.6\, \times {10^{10}}\,Pa$
C) $7.9\, \times {10^7}\,Pa$
D) $1.46\, \times {10^7}\,Pa$

Answer 1

The question has given us the values of speed of the longitudinal seismic waves, the density of the medium, the medium is told to be considered as a fluid instead of solid, thus we can use the formula of velocity of sound or longitudinal mechanical waves which relates the velocity with the ratio of bulk modulus of the fluid and the density of the fluid raised to a power.

Complete step by step answer:
We will approach the solution to the asked question exactly the same way as explained in the hint section of the solution. We will use the formula which gives us the velocity of longitudinal mechanical waves through relating it with the bulk modulus of the fluid and the density of the fluid.
The formula about which we are talking is:
$v = \sqrt {\dfrac{B}{\rho }}$
Where, $B$ is the bulk modulus of the given fluid medium
$\rho$ is the density of the given fluid medium and,
$v$ is the velocity of the sound or the longitudinal mechanical wave in that particular medium whose bulk modulus and density we are mentioning and using in the relation.
Now, let’s have a look at what’s given to us in the question.
The density of the crust, which can be considered as fluid is given to be:
$\rho = 2.7\,gm/c{m^3}$
If we try to convert the density from $gm/c{m^3}$ to $kg/{m^3}$ , we need to multiply the numerator by a thousand. Then, the density becomes:
$\rho = 2.7 \times {10^3}\,kg/{m^3}$
The question has also told us that the speed of the longitudinal seismic waves is:
$v = 5.4\,km/s$
Or, $v = 5.4 \times {10^3}\,m/s$
The bulk modulus of the crust is asked in the question, for that, we need to modify the formula a little bit.
If we transpose in the formula, we can write:
$B = \rho {v^2}$
Substituting the values that we just found out, we get:
$B = \left( {2.7 \times {{10}^3}} \right) \times {\left( {5.4 \times {{10}^3}} \right)^2}$
Upon solving this, we get:
$B = 7.87 \times {10^{10}}\,Pa \\\ B \approx 7.9 \times {10^{10}}\,Pa \\\$
We can see that the value matches with the one in option (A).
Hence, option (A) is the correct answer.

Note: Many students do not read the full question leaving the most important piece of information out, that the crust can be considered as a fluid. This is the most important piece of information as it enables us to use the above-mentioned formula, thus finding out the bulk modulus of the earth’s crust.