A rubber ball has bulk modulus $B=9.5 \times 10^{8} \mathrm{~Pa}$. When submerged to a certain depth in Earth's ocean, its volume decreases by only $0.02 \%$ from its original volume. The ocean's salt water has a density of $1025 \mathrm{~kg} / \mathrm{m}^{3}$. At what depth below the surface of the ocean is the ball?
$19 \mathrm{~m}$
$190 \mathrm{~m}$
$1.9 \mathrm{~m}$
$1900 \mathrm{~m}$

Step 1 ：We are given that a rubber ball has a bulk modulus \(B = 9.5 \times 10^{8} \, \mathrm{Pa}\). When submerged to a certain depth in Earth's ocean, its volume decreases by only \(0.02 \% = 0.0002\) from its original volume. The ocean's salt water has a density of \(1025 \, \mathrm{kg/m^{3}}\). We are asked to find the depth below the surface of the ocean at which the ball is submerged.

Step 2 ：We can use the formula for pressure in a fluid at a certain depth, which is given by \(P = \rho g h\), where \(\rho\) is the density of the fluid, \(g\) is the acceleration due to gravity, and \(h\) is the depth below the surface of the fluid.

Step 3 ：We also know that the pressure is related to the bulk modulus and the change in volume by the formula \(P = B \frac{\Delta V}{V}\), where \(B\) is the bulk modulus, \(\Delta V\) is the change in volume, and \(V\) is the original volume.

Step 4 ：Setting these two expressions for the pressure equal to each other, we get \(B \frac{\Delta V}{V} = \rho g h\). We can solve this equation for \(h\), the depth below the surface of the ocean, to get \(h = \frac{B \frac{\Delta V}{V}}{\rho g}\).

Step 5 ：Substituting the given values into this equation, we get \(h = \frac{(9.5 \times 10^{8} \, \mathrm{Pa}) \times 0.0002}{(1025 \, \mathrm{kg/m^{3}}) \times (9.8 \, \mathrm{m/s^{2}})}\).

Step 6 ：Solving this equation, we find that \(h \approx 18.914883026381283 \, \mathrm{m}\).

Step 7 ：Rounding to the nearest meter, we find that the depth below the surface of the ocean at which the ball is submerged is approximately \(\boxed{19 \, \mathrm{m}}\).