Answer:
Diameter decreases by the diameter of 0.0312 m.
Explanation:
Given that,
Bulk modulus = 14.0 × 10¹⁰ N/m²
Diameter d = 2.20 m
Depth = 2.40 km
Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000
= 24.25 × 10⁶ N/m²
Volume = [tex]\dfrac{4}{3} \pi r^3[/tex]
[tex]\dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}[/tex]
Bulk modulus is equal to
[tex]B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }[/tex]
now
[tex]B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }[/tex]
[tex]B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{1.1^3} }[/tex]
[tex](\Delta r)^3 = \dfrac{24.25 \times 10^6 \times 1.1^3}{14.0 \times 10^{10}}[/tex]
Δ r = -0.0156 m
change in diameter
Δ d = -2 × 0.0156
Δ d = -0.0312 m
Diameter decreases by the diameter of 0.0312 m.
