A solid brass sphere is initially surrounded by air, and the air pressure exerted on it is 1.0 × 10^5 N/m^2 (normal atmospheric pressure). The sphere is lowered into the ocean to a depth where the pressure is 2.0 × 10^7 N/m^2. The volume of the sphere in air is 0.50 m^3. By how much does this

Question

A solid brass sphere is initially surrounded by air, and the air pressure exerted on it is [latex] 1.0 imes 10^{5} N / m ^{2} [/latex] (normal atmospheric pressure). The sphere is lowered into the ocean to a depth where the pressure is [latex] 2.0 imes 10^{7} N / m ^{2} [/latex]. The volume of the sphere in air is 0.50 m³. By how much does this volume change once the sphere is submerged?

Accepted Answer

From the definition of bulk modulus, we have

[latex] B=-\frac{\Delta P}{\Delta V / V_{i}} [/latex].

[latex] \Delta V=-\frac{V_{i} \Delta P}{B} [/latex].

Substituting the numerical values, we obtain

[latex] \Delta V=-\frac{\left(0.50 m ^{3} ight)\left(2.0 imes 10^{7} N / m ^{2}-1.0 imes 10^{5} N / m ^{2} ight)}{6.1 imes 10^{10} N / m ^{2}} [/latex].

[latex] =-1.6 imes 10^{-4} m ^{3} [/latex].

The negative sign indicates that the volume of the sphere decreases.

Question 12.7: A solid brass sphere is initially surrounded by air, and the...

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