Respuesta :
As per the formula of bulk modulus we know that
[tex]B = \frac{\Delta P}{\Delta V/V}[/tex]
as we know that
[tex]V = \frac{4}{3}\pi R^3[/tex]
[tex]\Delta V = 4 \pi R^2 \Delta R[/tex]
now we will have
[tex]\Delta V/ V = \frac{3\Delta R}{R}[/tex]
now we will have from above equation
[tex]\Delta V/ V = \frac{\Delta P}{B}[/tex]
[tex]\frac{3\Delta R}{R} = \frac{\rho g h}{B}[/tex]
[tex]\frac{3 \Delta R }{3} = \frac{10^3 \times 9.8 \times 1000}{14 \times 10^{10}}[/tex]
[tex]\Delta R = 7 \times 10^{-5} m[/tex]
Answer:
The diameter of the sphere decrease is 0.124 m.
Explanation:
Given that,
Bulk modulus [tex]B= 14.03\times10^{10}\ N/m^2[/tex]
Diameter d = 3 m
Depth = 1.00 km
Pressure [tex]\Delta P=\rho gh=1000\times9.81\times1\times10^3[/tex]
[tex]\Delta P= 9.81\times10^{6}\ N/m^2[/tex]
Volume [tex]V =\dfrac{4}{3}\pi r^3[/tex]
[tex]\dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}[/tex]
Formula of the bulk modulus is define as,
[tex]B =- \dfrac{\Delta P}{\dfrac{\Delta V}{V}}[/tex]
Where, [tex]\Delta P[/tex] = Change in pressure
V = volume
Put the value into the formula
[tex]B=-\dfrac{9.81\times10^{6}\times(r)^3}{(\Delta r)^3}[/tex]
[tex](\Delta r)^3=-\dfrac{9.81\times 10^6\times(1.5)^3}{14.03\times10^{10}}[/tex]
[tex](\Delta r)^3=-\dfrac{9.81\times 10^6\times(1.5)^3}{14.03\times10^{10}}[/tex]
[tex](\Delta r)=-(\dfrac{9.81\times 10^6\times(1.5)^3}{14.03\times10^{10}})^{\frac{1}{3}}[/tex]
[tex]\Delta r=-0.062\ m[/tex]
Change in diameter
[tex]\Delta d=-2\times0.062\ m[/tex]
[tex]\Delta d=-0.124\ m[/tex]
Hence, The diameter of the sphere decrease is 0.124 m.
