Calculation of true thickness of the bed

Gate_2020

 5) A sample of Granite is observed to have a P- wave velocity of 5 km/sec and density of 2600 kg/m³. The Bulk modules of the granite, assuming it to be a poissions solid is ____________ kilo-Pascal(Kpa).

(Thanks to Chandrasekhar, ANU)

Solution:

Given that $V_{p}$= 5 km/sec =5000m/sec

Density(ρ) = 2600 kg/m³

Bulk Modules(K)= ?

Poisson solid $(\nu)$ =0.25 

$\frac{V_{p}}{V_{s}} =\sqrt{\frac{2(1-\upsilon)}{(1-2\upsilon)}}$

$\frac{V_{p}}{V_{s}} =\sqrt{\frac{2(1-0.25)}{(1-2*0.25)}}$

$\frac{V_{p}}{V_{s}} =\sqrt{\frac{2(0.75)}{(1-0.5)}}$

$\frac{V_{p}}{V_{s}} =\sqrt{\frac{1.5}{0.5}}$

$\frac{V_{p}}{V_{s}} =\sqrt{\frac{1.5}{0.5}}$

$\frac{V_{p}}{V_{s}} =\sqrt{3}$  

$V_{s}= \frac{V_{p}}{\sqrt{3}}$

$V_{s}= \frac{5000}{1.732}$

$V_{s}= 2886.83 m/sec^2$

We know that velocity of S-wave can be defined as follws

$V_{s}^{2}=\frac{\mu}{\rho}$

$\mu=\rho\times V_{s}^{2}$

$\mu=(2600)\times (2886.83)^{2}$

$\mu = 2.16\times 10^{10}$

Velocity of the P-wave can be defined as follows

$V_{p}^{2}= \frac{K + \frac{4}{3}\mu}{\rho}$

$(5000)^{2}= \frac{K + \frac{4}{3}( 2.16\times 10^{10})}{2600}$

$6.5\times 10^{10} = K + 2.88 \times 10^{10}$

$K = 3.62\times 10^{10} pascal$

$K= 3.62 \times 10^{7} kilo - pascal$

K= 36200000 Kpa