What is the Bulk modulus for a perfect rigid body?

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Q. What is the Bulk modulus for a perfect rigid body?

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Concept understanding — Bulk Modulus Application

Bulk Modulus: The Resistance to Squeezing

Imagine you have a sponge. When you squeeze it from all sides — say, by pushing it into a smaller space — it gets compressed. Now imagine a block of steel. If you try to squeeze it from all sides, it barely changes size. The bulk modulus is the number that tells you how hard it is to compress a material when you apply pressure evenly from every direction.

This is different from stretching or bending. Here, the force is uniform all around — like the pressure deep underwater, where water pushes on every surface of an object.

The Intuition: Pressure vs. Volume Change

Take a cube of material. If you increase the pressure on it (push harder from all sides), its volume decreases. The bulk modulus $K$ is defined as:

Bulk modulus = $\frac{\text{pressure applied}}{\text{fractional change in volume}}$

In symbols:

K = -\frac{\Delta P}{\Delta V / V_0}

Where:

$\Delta P$ = change in pressure (force per area)
$\Delta V$ = change in volume (final minus initial)
$V_0$ = original volume

The minus sign is there because when pressure increases ( $\Delta P > 0$ ), volume decreases ( $\Delta V < 0$ ), so the ratio comes out positive.

Note

A large $K$ means the material is hard to compress (like diamond or steel). A small $K$ means it's easy to compress (like air or a sponge).

The Precise Statement

The bulk modulus is a material property. It tells you how much the volume of a substance changes when you apply a uniform pressure. The reciprocal of bulk modulus is called compressibility ( $\beta = 1/K$ ), which is often used for gases and liquids.

For a solid, $K$ is usually very large — a few hundred gigapascals for metals. For water, $K \approx 2.2 \times 10^9$ Pa (about 2.2 GPa). For air at room temperature, $K \approx 1.4 \times 10^5$ Pa — much smaller, which is why you can easily squeeze a balloon.

K = -\frac{\Delta P}{\Delta V / V_0}

Where It Shows Up in Exams

You'll typically see three types of problems:

Direct calculation: Given $\Delta P$ and $\Delta V/V_0$ , find $K$ (or vice versa).
Comparing materials: Which has higher bulk modulus? (Steel > water > air)
Applications: Why does a submarine's hull need to be strong? Because at depth, $\Delta P$ is huge, and a small $K$ would mean dangerous compression.

Tip

For solids, the volume change is tiny — often given in scientific notation. For gases, the volume change can be large, so always check units carefully.

A Common Mistake

Students often forget the negative sign in the formula. Remember: when pressure goes up, volume goes down. The ratio $\Delta V / V_0$ is negative, so $-\Delta P / (\Delta V/V_0)$ gives a positive $K$ . If you drop the minus sign, you'll get a negative bulk modulus — which is physically meaningless.

Real-World Example …

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A perfectly rigid body cannot be compressed at all, so its bulk modulus—which measures resistance to volume change—is infinite.

Why bulk modulus captures rigidity

The bulk modulus $K$ quantifies how much pressure you need to apply to produce a given fractional change in volume. It is defined as

K = -V \frac{dP}{dV} = -\frac{dP}{dV/V}.

The negative sign ensures $K$ is positive, since an increase in pressure ( $dP > 0$ ) causes a decrease in volume ( $dV < 0$ ). Rearranging, the volumetric strain is

\frac{\Delta V}{V} = -\frac{\Delta P}{K}.

A material with a large bulk modulus resists compression strongly: you need enormous pressure to squeeze out even a tiny volume change. Conversely, a small bulk modulus means the material is easily compressible.

What "perfectly rigid" means

A perfect rigid body is an idealization in which the material does not deform under any applied stress. No matter how much pressure you apply, the volume (and shape) remain absolutely unchanged:

\Delta V = 0 \quad \text{for any finite } \Delta P.

Finding the bulk modulus

From the definition, if $\Delta V = 0$ while $\Delta P$ can be finite (or even arbitrarily large), then …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2025Set 2025-E1 markMCQ
$Q. Select the correct statement from the following: (A) Shear modulus of an ideal liquid is infinite (B) Rubber is more elastic than steel (C) The bulk modulus of a perfect rigid body is infinite (D) The bulk modulus of an ideal liquid is zero$
›Reveal solutionSolution
The key idea is that elastic moduli measure resistance to deformation: shear modulus resists shape change (zero for fluids), bulk modulus resists volume change (infinite for incompressible bodies). The correct statement is that the bulk modulus of a perfect rigid body is infinite.

Concept & Intuition

Elastic moduli describe how a material responds to different types of stress.

Shear modulus ( $G$ ) measures resistance to shape change (shear stress). Fluids cannot sustain shear stress — they flow — so their shear modulus is zero, not infinite.

Bulk modulus ( $K$ ) measures resistance to volume change under uniform pressure. A perfect rigid body cannot change volume at all, so its bulk modulus is infinite. An ideal liquid is nearly incompressible, so its bulk modulus is large but finite — definitely not zero.

Elasticity refers to the ability to return to original shape after deformation. Steel returns to shape more perfectly than rubber under small stresses, so steel is more elastic, even though rubber stretches more.

Now evaluate each option step by step.

Option (A): "Shear modulus of an ideal liquid is infinite"

An ideal liquid (inviscid, incompressible) cannot resist shear stress — it deforms continuously under any shear.

Shear modulus is defined as shear stress / shear strain. Since shear stress is zero at equilibrium, $G = 0$ .

Conclusion: False. (Infinite shear modulus would mean it’s a rigid solid.)

Option (B): "Rubber is more elastic than steel"

"More elastic" means a greater ability to recover original shape after deformation.

Steel obeys Hooke’s law up to a high stress and returns exactly to its original shape; rubber undergoes large deformation but has more internal friction and hysteresis.

Steel has a much higher Young’s modulus and is considered more elastic in the technical sense.

Conclusion: False.

Option (C): "The bulk modulus of a perfect rigid body is infinite" …
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KCET 2024Set D-21 markMCQ
$\rho$
›Reveal solutionSolution

Integrate the extension of each element: the tension at a point is only the weight hanging below it, so the answer is half of what a load $Mg$ at the free end would give.

1. Why we cannot just use $\Delta L = FL/AY$

That formula assumes a uniform tension $F$ throughout the wire. Here the stretching force is the wire's own weight, which is not uniform: the top of the wire supports the entire wire below it, while the bottom end supports nothing at all. So we must integrate.

2. Set up an element

Let the wire have length $L$ , cross-sectional area $A$ , density $\rho$ , hung from the top. Take an element of thickness $dx$ at a distance $x$ below the support.

The weight hanging below this element is the weight of the remaining length $(L - x)$ :
$T(x) = \rho\,A\,(L-x)\,g$
3. Extension of the element

From Hooke's law, $Y = \dfrac{\text{stress}}{\text{strain}} = \dfrac{T/A}{d(\delta)/dx}$ , so
$d(\delta) = \frac{T(x)\,dx}{AY} = \frac{\rho A (L-x) g\,dx}{AY} = \frac{\rho g (L-x)}{Y}\,dx$
4. Integrate over the whole wire
$\Delta L = \int_{0}^{L} \frac{\rho g (L-x)}{Y}\,dx = \frac{\rho g}{Y}\left[ Lx - \frac{x^{2}}{2}\right]_{0}^{L} = \frac{\rho g}{Y}\left(L^{2} - \frac{L^{2}}{2}\right)$ $\boxed{\;\Delta L = \frac{\rho g L^{2}}{2Y}\;}$
5. The physical shortcut …
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KCET 2023Set A-31 markMCQ
$Y = 2 \times 10^{11}\,\mathrm{Nm}^{-2}$
›Reveal solutionSolution

Convert lateral strain to longitudinal strain via Poisson's ratio, then use $u=\frac12 Y\varepsilon^2$ .

Step 1 — Get the longitudinal strain.

Poisson's ratio is defined as
$\sigma = \frac{\text{lateral strain}}{\text{longitudinal strain}} = \frac{\varepsilon_l}{\varepsilon}.$
So
$\varepsilon = \frac{\varepsilon_l}{\sigma} = \frac{10^{-3}}{0.25} = 4\times10^{-3}.$
This step is essential: the elastic energy stored in a stretched wire is governed by the longitudinal strain, not the lateral one.

Step 2 — Elastic energy density.

Stretching a wire stores energy; per unit volume it is the area under the stress–strain line:
$u = \frac{1}{2}\times\text{stress}\times\text{strain} = \frac{1}{2}(Y\varepsilon)(\varepsilon) = \frac{1}{2}Y\varepsilon^{2},$
using Hooke's law, stress $=Y\varepsilon$ .

Step 3 — Substitute. …
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KCET 2022Set B-31 markMCQ
$0.2\%$
›Reveal solutionSolution

Convert the breaking strain into a breaking stress via $\sigma = Y\varepsilon$ , then get the area from $A = F/\sigma$ .

Step 1 — The governing law.

Within the elastic limit, Hooke's law in modulus form states that stress is proportional to strain, the constant being Young's modulus $Y$ :
$Y = \frac{\text{stress}}{\text{strain}} = \frac{\sigma}{\varepsilon} \qquad\Longrightarrow\qquad \sigma = Y\varepsilon$
Step 2 — Convert the breaking strain.

The rod breaks at a strain of $0.2\%$ . A percentage strain must be turned into a fraction:
$\varepsilon = \frac{0.2}{100} = 2 \times 10^{-3}$
(Forgetting this conversion is the single commonest error in this question — it costs a factor of 100.)

Step 3 — Compute the breaking stress.

This is the maximum stress the material can bear before it fractures:
$\sigma_{\max} = Y\varepsilon = (7 \times 10^9\ \mathrm{N/m^2})(2 \times 10^{-3})$ $\sigma_{\max} = 14 \times 10^{6} = 1.4 \times 10^{7}\ \mathrm{N/m^2}$
Step 4 — Relate stress to the load and area.

By definition, $\sigma = F/A$ , so the cross-sectional area must be large enough that the applied load $F$ does not push the stress past $\sigma_{\max}$ :

$A = \frac{F}{\sigma_{\max}}$ …
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KCET 2019Set A-11 markMCQ
$Q. A wire is stretched such that its volume remains constant. The Poission's ratio of the material of the wire is (A) 0.50 (B) -0.50 (C) 0.25 (D) -0.25$
›Reveal solutionSolution

Constant volume during stretching is exactly the condition that gives Poisson's ratio its maximum value of +0.50.

Step 1 — Volume of a wire.

$V = \pi r^2 L$

Step 2 — Constant volume condition.

Differentiating and setting $dV = 0$ : $\dfrac{2\,dr}{r} + \dfrac{dL}{L} = 0 \;\Rightarrow\; \dfrac{dr}{r} = -\dfrac{1}{2}\dfrac{dL}{L}$

Step 3 — Definition of Poisson's ratio.

$\sigma = -\dfrac{dr/r}{dL/L} = -\left(-\dfrac{1}{2}\right) = 0.50$ …
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Q.What is the Bulk modulus for a perfect rigid body?

Why bulk modulus captures rigidity

What "perfectly rigid" means

Finding the bulk modulus

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