A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a

Q: Why does the pressure from the mass on top affect the sphere at the bottom equally?

This is the magic of Pascal's Law! In an enclosed, incompressible (or even compressible) fluid at rest, any change in pressure applied at one point is transmitted undiminished to every other point in the fluid. So, the extra pressure $\frac{mg}{a}$ is felt exactly the same way by the sphere, regardless of its depth.

Q: Why did we use $3 \frac{\Delta r}{r}$ for the volumetric strain instead of calculating the exact new volume?

Great question! When deformations are very small (which is the case for solid objects under normal pressures), we can use calculus approximations. Since $V \propto r^3$, taking the derivative gives $dV = 3r^2 dr$. Dividing by $V$ yields $\frac{dV}{V} = 3 \frac{dr}{r}$. It's a powerful shortcut!

Q: Does the initial atmospheric pressure or the weight of the liquid matter here?

No, they don't! The Bulk Modulus relates the *change* in pressure ($\Delta p$) to the *change* in volume ($\Delta V$). The initial pressure determines the initial state, but the compression we are calculating is purely due to the *additional* pressure from the mass $m$.

Q: What if the liquid was highly compressible, like a gas?

The beauty of this problem is that the liquid's compressibility doesn't change the answer for the sphere! If the fluid were a gas, the piston would just sink much further down until it reached equilibrium. However, the extra pressure it exerts on the fluid (and thus on the sphere) would still be exactly $\frac{mg}{a}$.

Q: Why is the negative sign often used in the Bulk Modulus formula missing here?

The negative sign in $K = - \frac{\Delta p}{\Delta V / V}$ simply indicates that an increase in pressure (positive $\Delta p$) causes a decrease in volume (negative $\Delta V$). Since the question asks for the 'fractional decrement' (which implies the magnitude of the decrease), we just use the absolute values.

Question

A solid sphere of radius $r$ made of a soft material of bulk modulus $K$ is surrounded by a liquid in a cylindrical container. A massless piston of area $a$ floats on the surface of the liquid, covering entire cross-section of cylindrical container. When a mass $m$ is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, $\left(\frac{dr}{r}\right)$ is

Options:

(A) $\frac{Ka}{mg}$

(B) $\frac{Ka}{3mg}$

(C) $\frac{mg}{3Ka}$

(D) $\frac{mg}{Ka}$

Sigma Percentile · Accepted Answer

### The Setup: A Sphere in a Bath

Imagine a serene, perfectly cylindrical container filled to the brim with a liquid. Suspended within this liquid bath is a solid sphere of radius $r$. Resting gently on the surface of the liquid is a massless piston, perfectly sealing the container.

This system is in perfect equilibrium until we introduce a disturbance: a block of mass $m$ is placed squarely on the piston. The weight of this mass, $mg$, pushes down, attempting to compress the liquid. But how does this affect the solid sphere hidden beneath the surface?

### Pascal's Principle in Action

When the mass $m$ is placed on the piston of area $a$, it exerts a downward force. This creates an additional pressure on the liquid surface. We can calculate this extra pressure, let's call it $\Delta p$, using the fundamental definition of pressure: force divided by area.

$$\Delta p = \frac{mg}{a}$$

Here is where the magic of fluid mechanics comes into play. According to **Pascal's Law**, any change in pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of its container. This means that our solid sphere, regardless of how deep it is submerged, experiences this exact same additional pressure $\Delta p$ uniformly from all directions. It is being squeezed!

### The Geometry of Compression

To understand how the sphere responds to this squeezing, we must look at its material properties, specifically its **Bulk Modulus ($K$)**. The Bulk Modulus is a measure of a substance's resistance to uniform compression and is defined as the ratio of volumetric stress to volumetric strain.

$$K = \frac{	ext{Volumetric Stress}}{	ext{Volumetric Strain}} = \frac{\Delta p}{\frac{\Delta V}{V}}$$

We already know the volumetric stress is our extra pressure $\Delta p$. But the question asks for the fractional decrement in the *radius* ($\frac{\Delta r}{r}$), not the volume. We need a mathematical bridge between volume and radius.

The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$. For very small deformations, we can use calculus to relate the changes. Taking the natural logarithm of both sides and differentiating, or simply applying the power rule for small errors, we find that the fractional change in volume is exactly three times the fractional change in radius.

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

### The Master Equation

Now we have all the pieces of the puzzle. We substitute our expressions for the volumetric stress ($\Delta p$) and the volumetric strain ($\frac{\Delta V}{V}$) back into the Bulk Modulus equation.

$$K = \frac{\frac{mg}{a}}{3\frac{\Delta r}{r}}$$

This equation beautifully links the macroscopic force applied at the top of the container to the microscopic deformation of the sphere at the bottom.

### The Final Revelation

Our final goal is to isolate the fractional decrement in the radius, $\frac{\Delta r}{r}$. By simply rearranging the terms in our master equation, we arrive at the solution.

$$\frac{\Delta r}{r} = \frac{mg}{3aK}$$

This elegant result tells us exactly how much the sphere shrinks. Notice a fascinating detail: the result depends only on the applied mass, the piston area, and the sphere's material properties. It is completely independent of the liquid's own compressibility or the depth of the sphere! Even if the liquid were highly compressible, the extra pressure transmitted to the sphere would remain exactly the same.

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Visualized Solution

$Visualizing the Setup$

$Applying the Load$

$Calculating Additional Pressure$

$Defining Bulk Modulus$

$Relating Volume to Radius$

$Substituting Values$

$Final Rearrangement$

$Conceptual Reflection$

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

The Setup

Pascal's Principle in Action

The Geometry of Compression

The Master Equation

The Final Revelation

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Visualized Solution

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

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

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

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

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

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

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

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