Question

A space of elastic material is taken to bottom of sea (1 km deep). If radius changes by 0.02%, then the bulk modulus of elasticity of material is (Excess pressure = $\rho$gh)

• A. $2.4\times {10}^{10}$
• B. $6\times {10}^{10}$
• C. $1.7\times {10}^{10}$
• D. $3.1\times {10}^{10}$

Answer 1

Answer: (C)

. $1.7\times {10}^{10}$.

We know that, the volume of the sphere is given by, $V=\frac{4}{3}\pi r^3$
Therefore, the fractional change in volume is given by,

$\frac{\Delta v}{v}=\frac{3\Delta r}{r}=3\times \frac{0.02}{100}=\frac{0.06}{100}$
Pressure at the bottom of the sea is,

$\Delta P=\rho gh={10}^3\times 10\times {10}^3={10}^7$ Pa.
We know that,

Bulk modulus, $B=\frac{\Delta P}{\frac{\Delta v}{v}}$

Therefore, $B=\frac{\Delta P}{\frac{\Delta v}{v}}=\frac{{10}^7\times 100}{0.06}=\frac{{10}^9}{6\times {10}^{-2}}=\frac{{10}^{11}}{6}$

$B=\frac{10}{6}\times {10}^{10}$

$B=1.7\times {10}^{10}N/m^2$.
B=ΔPΔvvB=\frac{\Delta P}{\frac{\Delta v}{v}