When a rubber ball is taken to a depth of ...... m in deep sea, its volume decreases by 0.5%. (The b

Q: Why is there a negative sign in the formula for Bulk Modulus?

The negative sign ensures that the Bulk Modulus ($B$) is a positive quantity. When pressure increases (positive $\Delta p$), the volume decreases (negative $\Delta V$). The negative sign cancels out the negative volume change.

Q: Why do we use $\rho g h$ for the change in pressure?

At the surface of the sea, the ball only experiences atmospheric pressure. When it is taken to a depth $h$, the water above it exerts an additional pressure called hydrostatic pressure, which is calculated as $\rho g h$. This is the exact extra pressure causing the compression.

Q: What does a $0.5\%$ decrease in volume mean mathematically?

A percentage decrease means the change in volume relative to the original volume. Mathematically, $\frac{\Delta V}{V} = -0.5\%$. To use this in equations, we convert the percentage to a fraction by dividing by 100, giving $-\frac{0.5}{100}$ or $-0.005$.

Q: Why doesn't the original size or volume ($V$) of the ball matter?

The Bulk Modulus relates pressure to the *fractional* change in volume ($\frac{\Delta V}{V}$), not the absolute change ($\Delta V$). Because the problem gives us the percentage decrease, the original volume $V$ cancels out of the equations entirely.

Q: Can we use this exact formula for any depth in the ocean?

This formula assumes that the density of sea water ($\rho$) remains constant. In reality, water is slightly compressible, so at extreme depths (like the Mariana Trench), the density increases slightly. However, for most calculations, assuming a constant density is a very accurate approximation.

Question

When a rubber ball is taken to a depth of ...... m in deep sea, its volume decreases by 0.5%. (The bulk modulus of rubber $= 9.8 	imes 10^8 	ext{ Nm}^{-2}$. Density of sea water $= 10^3 	ext{ kg m}^{-3}, g = 9.8 	ext{ m/s}^2$)

Sigma Percentile · Accepted Answer

### The Physical Setup: A Journey to the Deep

Imagine standing on the deck of a ship, holding a solid rubber ball. You drop it into the ocean, and it begins its descent into the abyss.

As the ball sinks deeper and deeper, the weight of the water above it increases. This creates a crushing force from all directions, known as **hydrostatic pressure**.

Because rubber is an elastic material, this immense pressure actually squeezes the ball, forcing its volume to shrink. Our mission is to find the exact depth where the ball's volume decreases by exactly $0.5\%$.

### The Master Equation: Bulk Modulus

To solve this, we need a mathematical bridge between the crushing pressure and the shrinking volume. This bridge is the **Bulk Modulus** ($B$).

The Bulk Modulus is a property of the material that measures its resistance to uniform compression. It is defined as the ratio of the change in pressure ($\Delta p$) to the fractional change in volume ($\frac{\Delta V}{V}$).

Mathematically, we write this as:
$$B = \frac{\Delta p}{-\frac{\Delta V}{V}}$$

The negative sign is crucial here. It simply indicates that an increase in pressure (positive $\Delta p$) results in a decrease in volume (negative $\Delta V$).

### The Hydrostatic Connection

Now, what exactly is this change in pressure? At the surface, the pressure is just atmospheric. But at a depth $h$, the water adds an extra pressure.

This extra pressure is the hydrostatic pressure, given by the formula:
$$\Delta p = ho g h$$

Here, $ho$ is the density of sea water, $g$ is the acceleration due to gravity, and $h$ is the depth we want to find.

Let's substitute this into our Bulk Modulus equation:
$$ho g h = -B \frac{\Delta V}{V}$$

### Rearranging and Substituting

Our goal is to isolate the depth $h$. By rearranging the equation, we get:
$$h = -\frac{B}{ho g} \left(\frac{\Delta V}{V}ight)$$

Now, we carefully plug in the given values. The Bulk Modulus of rubber $B$ is $9.8 	imes 10^8 	ext{ N/m}^2$. The density of sea water $ho$ is $10^3 	ext{ kg/m}^3$, and $g$ is $9.8 	ext{ m/s}^2$.

The fractional change in volume is a decrease of $0.5\%$, which means $\frac{\Delta V}{V} = -\frac{0.5}{100}$.

Substituting these into our isolated equation gives:
$$h = -\frac{9.8 	imes 10^8}{10^3 	imes 9.8} 	imes \left(-\frac{0.5}{100}ight)$$

### The Final Calculation

This expression might look intimidating, but the numbers are designed to simplify beautifully. Notice how the $9.8$ in the numerator and denominator cancel out perfectly.

Furthermore, the two negative signs multiply to give a positive result. This makes perfect physical sense, as depth must be a positive quantity.

After canceling $9.8$, we are left with:
$$h = \frac{10^8}{10^3} 	imes \left(\frac{0.5}{100}ight)$$

Using the laws of exponents, $\frac{10^8}{10^3}$ simplifies to $10^5$. Dividing this by $100$ (which is $10^2$) leaves us with $10^3$.

So, the equation simplifies down to:
$$h = 10^3 	imes 0.5$$

Multiplying $1000$ by $0.5$ gives us our final answer:
$$h = 500 	ext{ m}$$

The rubber ball must be taken to a depth of exactly **$500 	ext{ meters}$** to experience a $0.5\%$ decrease in its volume.

This is a fantastic example of how abstract material properties like Bulk Modulus dictate physical behavior in extreme environments!

Animated Solution for Physics - Properties of Solids and Liquids: When a rubber ball is taken to a depth of ...... m in deep sea, its volume decreases by 0.5%. (The bulk modulus of rubber $= 9.8 \times 1 0^{8} Nm^{- 2}$ . Density of sea water $= 1 0^{3} kg m^{- 3}, g = 9.8 m/s^{2}$ )

Enter Numerical Value:

Visualized Solution

$V_{initial} = V$

$h = Depth$

$\frac{Δ V}{V} = - 0.5%$

$B = \frac{Δ p}{- \frac{Δ V}{V}}$

$Δ p = ρ g h$

$ρ g h = - B \frac{Δ V}{V}$

$h = - \frac{B}{ρ g} (\frac{Δ V}{V})$

$Substitution$

$Simplification$

$Powers of 10$

$h = 500 m$

$Conclusion$

The Sigma Insight: Young's Modulus, Bulk Modulus and Modulus of Rigidity

The Physical Setup

The Master Equation

The Hydrostatic Connection

Rearranging and Substituting

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Vinitial​=V

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h=Depth

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } VΔV​=−0.5%

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B=−VΔV​Δp​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δp=ρgh

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ρgh=−BVΔV​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h=−ρgB​(VΔV​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplification

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Powers of 10

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } h=500 m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion