Question

If a rubber ball is taken at the depth of \(200 \mathrm{~m}\) in a pool, Its volume decreases by \(0.1 \%\). If the density of the water is \(1 \times 10^{3}\left(\mathrm{~kg} / \mathrm{m}^{3}\right) \& \mathrm{~g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\). Then what will be the volume elasticity in ? (A) \(10^{8}\) (B) \(2 \times 10^{8}\) (C) \(10^{9}\) (D) \(2 \times 10^{9}\)

Short answer

The short answer is: The volume elasticity of the rubber ball is (D) \(2 \times 10^9 ~\mathrm{Pa}\).

Step-by-step solution

Calculate the change in pressure (dP)

First, let's calculate the change in pressure at a depth of 200 meters. The formula to calculate the pressure at a depth h is: \(P = ρgh\), where P is the pressure, ρ is the density of the fluid (water), g is the acceleration due to gravity, and h is the depth. In this case, ρ = \(1 \times 10^3 ~\mathrm{kg/m^3}\), g = \(10 ~\mathrm{m/s^2}\), and h = 200 m. Plugging these values, we get: \(P = (1 \times 10^3) \times (10) \times (200) = 2 \times 10^6 ~\mathrm{Pa}\).

Determine the relative change in volume (dV/V)

The ball's volume decreases by 0.1%. To express this as a fraction, divide by 100: \( \frac{dV}{V} = -\frac{0.1}{100} = -1 \times 10^{-3} \).

Calculate the volume elasticity (K)

Now, we can plug in the values we found in Steps 1 and 2 into the formula for volume elasticity: \(K = -V\frac{dP}{dV} = -V \frac{2 \times 10^6}{-1 \times 10^{-3}} \). Cancel out the negative sign and we get: \(K = V \times 2\times 10^9\). Since we are only looking for the ratio of volume elasticity (K), we do not need the exact volume of the ball, and therefore, the answer is: \(K = 2 \times 10^9 ~\mathrm{Pa}\) The correct answer is (D) \(2 \times 10^9\).

Key concepts

Pressure Calculation

Calculating pressure in a liquid environment involves understanding how fluid dynamics work under different depths. Pressure at a particular depth in a liquid such as water can be found using the formula:

• \(P = \rho gh\)

Here, \(P\) stands for pressure, \(\rho\) for the fluid's density, \(g\) for acceleration due to gravity, and \(h\) for the depth. When we plug in the given values: \(\rho = 1 \times 10^3 ~\mathrm{kg/m^3}\), \(g = 10 ~\mathrm{m/s^2}\), and \(h = 200 ~\mathrm{m}\), we determine that the pressure is \(P = 2 \times 10^6 ~\mathrm{Pa}\). This means that as you descend deeper into the pool, the pressure exerted by the water increases linearly with depth.

Change in Volume

Changes in volume, especially under pressure, signify how a material reacts to external forces. In the exercise, the rubber ball's volume decreases by a fractional amount under the pressure of water. A decrease of \(0.1\%\) of its original volume can be turned into a decimal to work with:

• \(\frac{dV}{V} = -\frac{0.1}{100}\)
• \(\frac{dV}{V} = -1 \times 10^{-3}\)

This decrement is critical for figuring out how much "squeeze" or compressibility the object is undergoing. The negative sign indicates a reduction rather than an expansion in volume.

Density of Water

Density is a key factor in calculating fluid pressure and describes how much mass is contained in a given volume. For water, the density is typically taken as \(1 \times 10^3 ~\mathrm{kg/m^3}\) under standard conditions. This means that each cubic meter of water weighs about 1000 kilograms.
This density is crucial when calculating pressure, as seen in our previous section. By knowing the density, other physical properties, such as buoyancy and potential energy in water, can be calculated too. In this problem, density plays an indispensable role in finding the pressure that affects the ball's volume at a specific depth.

Acceleration Due to Gravity

Gravity or the acceleration due to gravity, usually denoted by \(g\), is a constant that dictates how strongly a planet pulls objects toward its center. On Earth, \(g\) is generally approximated to \(9.81 \mathrm{~m/s^2}\), but in many calculations, a round figure of \(10 \mathrm{~m/s^2}\) is used for simplicity.
In the context of fluid pressure, gravity is the driving force behind the pressure exerted by water or any fluid with a certain depth. In our case, using \(g = 10 \mathrm{~m/s^2}\), the formula \(P = \rho gh\) showed how gravity, combined with depth and density, dictates how much pressure is put on the rubber ball. Understanding this concept helps in comprehending phenomena like buoyancy and hydrostatic pressure.