A cube of metal is subjected to a hydrostatic pressure of 4 GPa. The percentage change in the length

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

The negative sign is a mathematical convention to ensure that the Bulk Modulus ($B$) is a positive quantity. Since an increase in pressure (positive $\Delta p$) always causes a decrease in volume (negative $\Delta V$), the extra negative sign cancels it out.

Q: Can we use $\frac{\Delta V}{V} = 3 \frac{\Delta l}{l}$ for any amount of compression?

No, this is an approximation derived from binomial expansion or calculus, and it is only highly accurate for very small fractional changes (typically under $5\%$). For massive deformations, you would need to use exact algebraic expansion: $(l - \Delta l)^3$.

Q: What if the object was a sphere instead of a cube?

The relationship would be exactly the same! For a sphere, $V = \frac{4}{3}\pi r^3$. When you take fractional changes, the constants drop out, leaving $\frac{\Delta V}{V} = 3 \frac{\Delta r}{r}$. This is true for any isotropic 3D shape.

Q: Why did we convert GPa to Pa?

In physics, it is critical to maintain dimensional consistency. Since the Bulk Modulus was provided in Pascals (Pa), the pressure must also be in Pascals. $1 \text{ GPa} = 10^9 \text{ Pa}$.

Q: Does the material of the cube matter?

Absolutely. The material dictates the value of the Bulk Modulus ($B$). A material with a higher Bulk Modulus (like steel) is stiffer and will compress less under the same pressure compared to a material with a lower Bulk Modulus (like rubber).

Question

A cube of metal is subjected to a hydrostatic pressure of $4	ext{ GPa}$. The percentage change in the length of the side of the cube is close to
(Take bulk modulus of metal, $B = 8 	imes 10^{10}	ext{ Pa}$)

Options:

(A) $1.67$

(B) $0.6$

(C) $20$

(D) $5$

Sigma Percentile · Accepted Answer

### The Deep Sea Thought Experiment

Imagine a solid metal cube submerged deep in the Mariana Trench, or placed inside a high-pressure industrial chamber. It experiences a uniform squeezing force from all directions. This is what physicists call **hydrostatic pressure**.

When a material is subjected to such pressure, it doesn't just sit there; it compresses. The extent to which it resists this compression is governed by a property known as the **Bulk Modulus** ($B$).

### The Master Equation: Bulk Modulus

To understand how much the cube shrinks, we rely on the definition of Bulk Modulus. It is defined as the ratio of volumetric stress (the change in pressure, $\Delta p$) to the volumetric strain (the fractional change in volume, $\frac{\Delta V}{V}$).

$$B = \frac{\Delta p}{-\left(\frac{\Delta V}{V}ight)}$$

The negative sign is crucial here. It mathematically represents a physical reality: an *increase* in pressure results in a *decrease* in volume. By rearranging this formula, we can isolate the fractional change in volume:

$$\frac{\Delta V}{V} = -\frac{\Delta p}{B}$$

### Crunching the Numbers

Now, let's bring in the numbers provided in the problem. The applied pressure is $4 	ext{ GPa}$, which we must convert to standard SI units: $4 	imes 10^9 	ext{ Pa}$. The Bulk Modulus is given as $8 	imes 10^{10} 	ext{ Pa}$.

We substitute these values into our rearranged equation and multiply by $100$ to find the percentage change:

$$\frac{\Delta V}{V} 	imes 100\% = -\frac{4 	imes 10^9}{8 	imes 10^{10}} 	imes 100\%$$

The powers of ten cancel out beautifully, leaving us with:

$$\frac{\Delta V}{V} 	imes 100\% = -\frac{1}{20} 	imes 100\% = -5\%$$

So, the overall volume of the cube decreases by exactly $5\%$.

### From Volume to Length: The Geometric Link

But the question doesn't ask for the change in volume; it asks for the percentage change in the *side length* of the cube. How do we bridge this gap?

We know that for a cube, the volume $V$ is the cube of its side length $l$:

$$V = l^3$$

For very small fractional changes (typically less than $5\%$), we can use the rules of error analysis or logarithmic differentiation. Taking the natural logarithm of both sides and differentiating yields:

$$\frac{\Delta V}{V} = 3 \frac{\Delta l}{l}$$

This tells us that the fractional change in volume is three times the fractional change in length. Rearranging this to solve for the length change gives:

$$\frac{\Delta l}{l} = \frac{1}{3} \left(\frac{\Delta V}{V}ight)$$

### The Final Verdict

Finally, we substitute our previously calculated percentage change in volume ($-5\%$) into this new relation:

$$\frac{\Delta l}{l} 	imes 100\% = \frac{1}{3} (-5\%)$$

$$\frac{\Delta l}{l} 	imes 100\% \approx -1.67\%$$

The side length of the cube decreases by approximately **$1.67\%$**. This elegant interplay between macroscopic geometry and material elasticity leads us directly to the correct answer.

Animated Solution for Physics - Properties of Solids and Liquids: A cube of metal is subjected to a hydrostatic pressure of $4 GPa$ . The percentage change in the length of the side of the cube is close to (Take bulk modulus of metal, $B = 8 \times 1 0^{10} Pa$ )

Select Answer:

Visualized Solution

$Hydrostatic Compression$

$Bulk Modulus (B)$

$Volumetric Strain Setup$

$Calculating %Δ V$

$Volume and Side Length$

$Fractional Change Relation$

$Final Calculation$

$Conclusion$

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

The Deep Sea Thought Experiment

The Master Equation

Crunching the Numbers

From Volume to Length

The Final Verdict

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Select Answer:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bulk Modulus (B)

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

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

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

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

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

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