Problem 922 - JEE Physics

Question

To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by \(0.1 \%\) [Take : density of sea water \(=1000\left(\mathrm{~kg} / \mathrm{m}^{3}\right)\) Bulk modulus of rubber \(=9 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\), acceleration due to gravity \(\left.=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]\) (A) \(9 \mathrm{~m}\) (B) \(18 \mathrm{~m}\) (C) \(180 \mathrm{~m}\) (D) \(90 \mathrm{~m}\)

Step-by-Step Solution

Answer

The depth below the surface of the sea required to decrease the volume of the rubber ball by 0.1% is \(h = 90 \ m\). The correct answer is (D).

Step 1: Understand the given information

We are given the following information: - Density of sea water (ρ) = 1000 kg/m³ - Bulk modulus of rubber (B) = 9 × 10⁸ N/m² - Acceleration due to gravity (g) = 10 m/s² Our goal is to find the depth (h) below the sea surface so that the volume of the rubber ball decreases by 0.1%.

Step 2: Formulate the equation for the volume decrease

To find the depth, we use the formula for the decrease in volume as a result of a pressure increase: ΔV/V = -ΔP/B Here, ΔV/V is the percentage change in volume, which is 0.1% (or 0.001 as a decimal), ΔP is the change in pressure, and B is the bulk modulus of the rubber.

Step 3: Calculate the change in pressure

To calculate the change in pressure (ΔP) due to gravity, we can use the formula: ΔP = ρgh Where: ρ is the density of sea water, g is the acceleration due to gravity, h is the depth below the sea surface.

Step 4: Substitute the pressure change and volume ratio in the equation

Now, we can substitute ΔP = ρgh and the ratio \(ΔV/V = -0.001 \) in the equation from step 2: -0.001 = -ρgh/B

Step 5: Solve for depth h

Rearrange the equation from step 4 to solve for the depth h: h = (0.001 * B) / (ρg) We can substitute the given values: h = (0.001 * 9 × 10⁸ N/m²) / (1000 kg/m³ × 10 m/s²) h = 90 m

Step 6: Choose the correct answer

According to our calculations, the depth h must be 90 meters (m) for the volume of the rubber ball to decrease by 0.1%. Therefore, the correct answer is: (D) 90 m

Key Concepts

Bulk Modulus

The bulk modulus is a fundamental concept in the study of how substances respond to pressure changes. It measures a material's resistance to uniform compression. When you compress something, its volume tends to decrease. The bulk modulus, represented by B, quantifies how much pressure is needed to compress a material by a certain amount. Mathematically, the bulk modulus is expressed by the equation:\[B = - \frac{\Delta P}{\Delta V/V}\]where:

• \(\Delta P\) is the change in pressure.
• \(\Delta V/V\) is the fractional volume change (volume decrease).

A larger bulk modulus indicates a material that is harder to compress. In the example, the rubber ball's bulk modulus was given as \(9 \times 10^8\, \text{N/m}^2\), meaning significant pressure is needed to cause even a small volume change.

Pressure Change in Fluids

In the context of fluids, such as water, pressure change is a critical factor affecting submerged objects. When an object is submerged underwater, the pressure it experiences increases with depth due to the weight of the water above it. This pressure change is computed using the formula:\[\Delta P = \rho g h\]where:

• \(\rho\) is the fluid density (\(1000\, \text{kg/m}^3\) for seawater).
• \(g\) is the acceleration due to gravity (\(10\, \text{m/s}^2\)).
• \(h\) is the depth below the fluid's surface.

Different materials will react differently to pressure changes based on their intrinsic properties, like the bulk modulus. Understanding how pressure variation with depth works is vital for applications ranging from oceanography to engineering.

Volume Decrease in Fluids

Volume decrease in fluids occurs when an object is subjected to increased pressure, leading to compression. This phenomenon is particularly noticeable in elastic materials like rubber. The relation explaining volume decrease due to pressure increase is derived from the bulk modulus concept, where:\[\frac{\Delta V}{V} = -\frac{\Delta P}{B}\]Here, a negative sign indicates that the volume reduces when pressure increases.In practical terms, determining how much the volume of an object decreases involves connecting this relation with the depth underwater via pressure change. Using values from the sea depth exercise, we notice:

• A 0.1% decrease in volume is equivalent to a \(0.001\) fractional volume change.
• This connects back to how much pressure increase is needed, weighing in the object's bulk modulus.

Hence, calculations like these help predict how deep an elastic object must be submerged before it compresses to a specified extent.