Question

The radius of a sphere of lead (B = 4.2 x 1010 N/m2) is 1.000 m on the surface of the earth where the pressure is 1.01 x 105 N/m2. The sphere is taken by submarine tothe deepest part of the ocean to a depth of 1.10 x 104 m where it is exposed to a pressure is 1.25 x 108 N/m2. What is the volume of the sphere at the bottom of theocean?

Answer 1

Bulk modulaus of lead B=4.2 x 10^10 N/m^2
Radius of sphere on surface of earth r = 1.000 m
The pressure on surface P = 1.01 x 10^5 N/m^2
Depth h = 1.10 x 10^4 m
Pressure at depth P ' = 1.25 x 10^8 N/m^2
The volume of the sphere at the bottom of the ocean V ' = ?
Initial volume of the sphere V = ( 4/ 3) p r 3
= 4.188 m 3
We know B = ( P '-P ) / (dV / V )
= ( P ' - P ) V / dV
From this Change in volume dV = ( P ' - P ) V / B
= 0.012456 m 3
? V ' = V - dV
= 4.175 m 3


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