A cubical solid aluminium (bulk modulus = -V dP/dV = 70 GPa) block has an edge length of 1 m on the

Q: Why do we use gauge pressure instead of absolute pressure?

Because the bulk modulus is defined for the *change* in pressure ($\Delta P$) from the surface to the depth. The atmospheric pressure is already acting on the block at the surface, so the additional pressure causing the compression is just the gauge pressure $\rho g h$.

Q: Why is $\frac{\Delta V}{V} = 3 \frac{\Delta a}{a}$?

This is derived from the volume formula of a cube, $V = a^3$. Using calculus for small changes, we differentiate to get $dV = 3a^2 da$. Dividing by $V = a^3$ yields $\frac{dV}{V} = 3 \frac{da}{a}$. It's a standard approximation for small percentage changes.

Q: Does the shape of the object affect the volume change?

No, the bulk modulus applies to the overall volume change regardless of the object's shape. However, how that volume change translates to linear dimensions (like radius or edge length) depends entirely on the geometry.

Q: Why did we ignore the atmospheric pressure?

Atmospheric pressure acts on the surface and is already present before the block is submerged. We only care about the *additional* pressure causing the *change* in volume when it is taken to the bottom of the ocean.

Q: Is the negative sign in the bulk modulus formula important?

The negative sign simply indicates that as pressure increases (positive $\Delta P$), the volume decreases (negative $\Delta V$). Since we are asked for the magnitude of the change in edge length, we can safely work with absolute values.

Question

A cubical solid aluminium (bulk modulus = $-V \frac{dP}{dV} = 70 	ext{ GPa}$) block has an edge length of $1 	ext{ m}$ on the surface of the earth. It is kept on the floor of a $5 	ext{ km}$ deep ocean. Taking the average density of water and the acceleration due to gravity to be $10^3 	ext{ kg m}^{-3}$ and $10 	ext{ ms}^{-2}$, respectively, the change in the edge length of the block in mm is _________.

Sigma Percentile · Accepted Answer

### The Deep Ocean Setup

Imagine a solid aluminium cube resting at the bottom of a $5 	ext{ km}$ deep ocean. At this immense depth, the water exerts a crushing hydrostatic pressure from all directions, trying to squeeze the cube into a smaller volume.

This is a classic problem of elasticity, specifically dealing with the **Bulk Modulus**, which measures a substance's resistance to uniform compression.

### Calculating the Hydrostatic Pressure

First, let's figure out exactly how much extra pressure this cube is experiencing. We know from fluid mechanics that the gauge pressure at a depth $h$ is simply given by:

$$ \Delta P = ho g h $$

Now, let's plug in the numbers. The density of water $ho$ is $10^3 	ext{ kg/m}^3$, gravity $g$ is $10 	ext{ m/s}^2$, and the depth $h$ is $5000 	ext{ m}$. Multiplying these gives us a massive pressure:

$$ \Delta P = (10^3) 	imes (10) 	imes (5000) = 50 	imes 10^6 	ext{ Pa} = 50 	ext{ MPa} $$

### Bulk Modulus & Volume Change

How does the cube respond to this pressure? That's where the Bulk Modulus $B$ comes in. It tells us how much a material resists compression. The fractional change in volume, $\frac{\Delta V}{V}$, is equal to the pressure change divided by the Bulk Modulus:

$$ \frac{\Delta V}{V} = \frac{\Delta P}{B} $$

Let's substitute our pressure and the given Bulk Modulus of $70 	ext{ GPa}$ ($70 	imes 10^9 	ext{ Pa}$).

$$ \frac{\Delta V}{V} = \frac{50 	imes 10^6}{70 	imes 10^9} = \frac{5}{7000} = \frac{1}{1400} $$

This is the fractional decrease in the cube's volume.

### Relating Volume to Edge Length

But the question asks for the change in edge length, not volume. Since the volume of a cube is $V = a^3$, we can use a neat calculus trick. Differentiating both sides gives $dV = 3a^2 da$. Dividing this by the original volume $V = a^3$ gives us:

$$ \frac{dV}{V} = 3 \frac{da}{a} $$

This means the fractional change in volume is exactly three times the fractional change in edge length.

### Final Calculation

Equating the two expressions, we get:

$$ 3 \frac{\Delta a}{a} = \frac{1}{1400} $$

Solving for $\Delta a$, we find it is $\frac{a}{4200}$. Since the original length is $1 	ext{ m}$, or $1000 	ext{ mm}$, we just plug that in:

$$ \Delta a = \frac{1000}{4200} 	ext{ mm} = \frac{10}{42} 	ext{ mm} = \frac{5}{21} 	ext{ mm} $$

Calculating this gives approximately **$0.238 	ext{ mm}$**. So, the edge length decreases by about $0.24 	ext{ mm}$. What a beautiful application of elasticity!

Enter Numerical Value:

Visualized Solution

$T h e D ee p O ce an S e t u p$

$H y d r os t a t i c P r ess u r e$

$C a l c u l a t in g Δ P$

$B u l k M o d u l u s & V o l u m e C han g e$

$F r a c t i o na l V o l u m e C han g e$

$R e l a t in g V o l u m e t o E d g e L e n g t h$

$C a l c u l a t in g C han g e in E d g e L e n g t h$

$F ina l R es u l t$

$T h e W a y F or w a r d$

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

The Deep Ocean Setup

Calculating the Hydrostatic Pressure

Bulk Modulus & Volume Change

Relating Volume to Edge Length

Final Calculation

Similar Questions

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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; } The Deep Ocean Setup

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bulk Modulus & Volume Change

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relating Volume to Edge Length

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Change in Edge Length

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

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