Modulus of rigidity of ideal liquids is (a) infinity. (b) zero. (c) unity. (d) some finite small non-zero cons

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Q. Modulus of rigidity of ideal liquids is

(a) infinity.

(b) zero.

(d) some finite small non-zero constant value.

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Concept understanding — Youngs Modulus

Young’s Modulus: The Stretchiness of a Solid

When you pull on a rubber band, it stretches easily. When you pull on a steel rod of the same size, it barely moves. Both are elastic — they return to their original shape when you let go — but they resist stretching very differently. Young’s modulus is the number that tells you exactly how much a material resists being stretched or compressed lengthwise.

The Intuition: Stiffness per Unit Size

Think of a spring. A stiff spring requires a large force to stretch it a little. A soft spring stretches a lot with a small force. Young’s modulus is like the “stiffness” of a material, but it’s cleverly designed to be independent of the object’s shape and size.

If you take a thick steel rod and a thin steel wire of the same length, the rod is harder to stretch. That’s because you’re pulling on more material. Young’s modulus removes this size effect — it tells you the stiffness of the material itself, not the particular piece you’re holding.

The Precise Definition

Young’s modulus ( $E$ or $Y$ ) is defined as the ratio of tensile stress to tensile strain, as long as the material obeys Hooke’s law (the deformation is reversible and proportional to the force).

Y = \frac{\text{Tensile Stress}}{\text{Tensile Strain}}

Let’s break down the two parts.

Tensile Stress ( $\sigma$ ) is the force per unit area. If you pull with a force $F$ on a rod of cross-sectional area $A$ , the stress is:

\sigma = \frac{F}{A}

Stress has units of pressure — pascals (Pa) or $\text{N/m}^2$ . It tells you how “intense” the pulling is, regardless of the rod’s thickness.

Tensile Strain ( $\varepsilon$ ) is the fractional change in length. If the original length is $L_0$ and it stretches by $\Delta L$ , the strain is:

\varepsilon = \frac{\Delta L}{L_0}

Strain is a pure number — it has no units. A strain of 0.01 means the rod stretched by 1% of its original length.

Putting it together:

Y = \frac{F/A}{\Delta L / L_0} = \frac{F L_0}{A \Delta L}

What the Number Tells You

A high Young’s modulus means the material is very stiff — it takes a huge stress to produce even a tiny strain. Steel has $Y \approx 200 \times 10^9$ Pa. A low Young’s modulus means the material is easily stretched. Rubber has $Y \approx 0.01 \times 10^9$ Pa — about 20,000 times smaller than steel.

Note

Young’s modulus is only valid in the elastic region — where the material returns to its original shape after the force is removed. If you stretch too far (past the elastic limit), the material deforms permanently or breaks, and Young’s modulus no longer applies.

A Worked Example

A steel wire of length 2.0 m and cross-sectional area $1.0 \times 10^{-6} \text{ m}^2$ is pulled by a force of 100 N. How much does it stretch? (Young’s modulus of steel = $2.0 \times 10^{11}$ Pa)

From $Y = \frac{F L_0}{A \Delta L}$ , rearrange:

\Delta L = \frac{F L_0}{A Y} = \frac{100 \times 2.0}{(1.0 \times 10^{-6}) \times (2.0 \times 10^{11})} = \frac{200}{2.0 \times 10^5} = 1.0 \times 10^{-3} \text{ m} = 1.0 \text{ mm}

The wire stretches by just 1 mm. If you tried the same with a rubber band of the same dimensions ( $Y \approx 10^7$ Pa), the stretch would be about 20,000 times larger — 20 metres! (Of course, a real rubber band would break long before that.)

Key Points for Exams

Young’s modulus is a material property — it doesn’t depend on the object’s length or thickness.
It applies only to axial (lengthwise) tension or compression, not to bending or twisting.
The units are the same as pressure: pascals (Pa) or $\text{N/m}^2$ .
For most materials, Young’s modulus is the same in tension and compression (within the elastic limit).

Watch out

Do not confuse Young’s modulus with stiffness ( $k = F/\Delta L$ ). Stiffness depends on the object’s dimensions ( $k = YA/L_0$ ). Young’s modulus is the intrinsic material property; stiffness is the property of a particular object.

"Youngs Modulus important questions" is a common search among CBSE and competitive-exam aspirants alike, since Youngs Modulus sits squarely within the Mechanical Properties of Solids coverage of NCERT Class 11 Physics, so it is fair game for both CBSE board questions and competitive-exam numericals. Pairing this explanation with NCERT Physics textbook practice and previous years' questions is the surest way to lock the concept in before an exam.

Ideal liquids cannot sustain a static shear stress, meaning they offer no resistance to changes in shape. Therefore, their modulus of rigidity is zero.

The modulus of rigidity, also known as the shear modulus, is a measure of a material's resistance to deformation when subjected to a tangential (shear) force. It quantifies how much a material will deform in shape, rather than volume, under stress.

Concept and Intuition

Imagine pushing the top surface of a block while keeping its bottom surface fixed.

For a solid, the block will deform by a certain amount, changing its shape, but it will resist this change and try to return to its original shape once the force is removed (if the deformation is within its elastic limit). This resistance to shape change is what the modulus of rigidity measures. A higher modulus means greater resistance to shape change.
For a liquid, if you apply a tangential force to its surface, the liquid will not just deform by a fixed amount and stop; it will continuously flow as long as the force is applied. It offers no static resistance to a change in shape. This fundamental difference in behavior between solids and liquids is key to understanding their respective moduli of rigidity.

Step-by-Step Explanation

Understanding Modulus of Rigidity ( $\eta$ or $G$ ):

The modulus of rigidity is defined as the ratio of shear stress to shear strain.

$\eta = \frac{\text{Shear Stress}}{\text{Shear Strain}}$
- Shear Stress ( $\tau$ ): This is the tangential force ( $F$ ) applied per unit area ( $A$ ) of the surface. So, $\tau = F/A$ .
- Shear Strain ( $\phi$ ): This is the ratio of the relative displacement ( $x$ ) of any layer with respect to a fixed layer, to the perpendicular distance ( $h$ ) between the layers. So, $\phi = x/h$ . For small deformations, $\phi \approx \tan \theta$ , where $\theta$ is the angle of shear.
Behavior of Solids under Shear Stress:

When a shear stress is applied to a solid, it undergoes a finite shear strain. The solid resists this deformation, and if the stress is removed, it returns to its original shape (within the elastic limit). Since a finite shear stress produces a finite shear strain, the modulus of rigidity for solids is a finite, non-zero value.
Behavior of Ideal Liquids under Shear Stress:

Ideal liquids, by definition, are incompressible and have zero viscosity. More generally, liquids (even real ones) cannot sustain a static shear stress. If a tangential force is applied to a liquid, it will start to flow. This flow means that the layers of the liquid continuously slide past each other.
- As long as the tangential force is applied, the liquid continues to deform.
- This continuous deformation implies that the relative displacement ( $x$ ) between layers keeps increasing indefinitely with time.
- Consequently, the shear strain ( $\phi = x/h$ ) tends towards infinity for any non-zero applied shear stress.
Watch out

Do not confuse the modulus of rigidity with viscosity. Viscosity describes a liquid's resistance to flow (dynamic resistance to shear), while the modulus of rigidity describes its resistance to static deformation (static resistance to shear). Even real liquids, which have viscosity, cannot sustain a static shear stress and will flow indefinitely if such a stress is applied.
Calculating Modulus of Rigidity for Ideal Liquids:

Using the formula $\eta = \frac{\text{Shear Stress}}{\text{Shear Strain}}$ :
- For any finite (even very small) shear stress applied to an ideal liquid, the shear strain becomes infinitely large because the liquid flows continuously.
- Therefore, $\eta = \frac{\text{finite value}}{\text{infinite value}} = 0$ .
Conclusion:

Since ideal liquids offer no resistance to a change in shape and deform continuously under any applied shear stress, their modulus of rigidity is zero.

The correct option is (B).

✓Final answer

\boxed{\text{zero}}

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Q.Modulus of rigidity of ideal liquids is

Concept and Intuition

Step-by-Step Explanation

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