Which one of these is more elastic, steel or rubber? Why?

II. Short Answer Questions · Q5

Q. Which one of these is more elastic, steel or rubber? Why?

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★

21% · 18/84 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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:

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

Start your 14-day free trial to unlock the full solution →

Step 1. 'More elastic', in the strict physics sense, means the material that more strongly resists deformation for a given applied stress -- i.e. the material with the HIGHER Young's modulus.

Step 2. If an equal stress is applied to both steel and rubber, steel produces far LESS strain than rubber does.

Step 3. Since $Y=\text{stress}/\text{strain}$ , a smaller strain for the same stress means a LARGER Young's modulus for steel than for rubber. …

Start your 14-day free trial to unlock the full solution →

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

Showing the 12 most recent of 55 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Among the following, the material with the highest elasticity is :$
(1) Copper
(2) Aluminium
(3) Brass
(4) Steel
›Reveal solutionSolution

Elasticity is measured by the modulus of elasticity (e.g. Young's modulus): the higher the modulus, the more elastic the material (the smaller the strain for a given stress, and the more completely it returns to its original shape).

Typical Young's modulus values (approximate, standard NCERT comparison): steel is markedly stiffer/more elastic than copper, brass, or aluminium — steel deforms the least under a given stress and recovers its shape most completely, so among Copper, Aluminium, Brass and Steel, steel has the highe …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following materials is most elastic?$
(a) Steel
(b) Rubber
(c) Copper
(d) Glass
›Reveal solutionSolution

"More elastic" means better at returning to the original shape/size after a deforming force is removed -- steel does this more effectively (higher Young's modulus, closer to ideal elastic behaviour under normal loads) than rubber, which is often mistakenly assumed to be "more elastic" just because it stretches a lot.

Elasticity is the property of a material to regain its original shape and size after the removal of a deforming force -- it is NOT the same as flexibility or how much a material can stretch.

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The restoring force per unit area is known as$
(a) strain
(b) elasticity
(c) stress
(d) plasticity
›Reveal solutionSolution

Stress = restoring force / area, by definition.

When a deforming force is applied to a body, internal restoring forces develop within it that try to bring it back to its original shape. The magnitude of this internal restoring force, per unit cross-sectional area, is called STRESS:

Stress = Restoring force / Area

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. ............. law states that stress is directly proportional to strain.$
›Reveal solutionSolution

The proportionality between stress and strain, within the elastic limit, is Hooke's Law.

When a solid material is deformed by an external force, the internal restoring force per unit area is called stress, and the fractional change in dimension (length, volume, or shape) is called strain. For small deformations — within the material's elastic limit — experiments show stress is directly proportional to strain:

$\text{stress} \propto \text{strain} \quad\Rightarrow\quad \text{stress} = k\times\text{strain}$ …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. What is Young's Modulus of a rigid body?$
›Reveal solutionSolution

Since $Y=\text{stress}/\text{strain}$ and a rigid body has zero strain for any stress, $Y\to\infty$ .

Young's modulus is defined as:

$Y = \frac{\text{longitudinal stress}}{\text{longitudinal strain}}$

A 'rigid body' is, by definition, a body whose shape and size do not change no matter how large a force (stress) is applied — its strain is always exactly zero. Substituting strain $=0$ into the formula makes the ratio undefined in the finite sense, i.e. it tends to i …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The Young's Modulus for a perfect rigid body is:$
(a) 0.5
(b) zero
(c) infinity
(d) 1
›Reveal solutionSolution

Young's modulus is stress divided by strain; a perfectly rigid body has zero strain for any stress (it never deforms), so its Young's modulus is infinite.

Young's modulus is defined as

Y = (longitudinal stress)/(longitudinal strain) = (F/A)/(deltaL/L)

A 'perfectly rigid body' is an idealization in which the body does not change its shape or size at all, no matter how large a force (stress) is applied — that is, deltaL = 0 always, so strain = deltaL/L = 0.

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. Substances like tissue of aorta, rubber etc. which can be stretched to cause large strain are called -$
(a) elastic substances
(b) Inelastic substances
(c) elastomer substances
(d) brittle substances
›Reveal solutionSolution

Materials that show a large elastic (recoverable) strain for a given stress, without obeying Hooke's law over most of that range, are called elastomers.

Most elastic materials, like steel, obey Hooke's law and can only sustain small strains (typically under 1%) before permanent deformation. Elastomers — such as rubber and the elastic tissue found in blood vessels like the aorta — are a special class of elastic substances that can be stretched to very large strains (several hundred percent) while still recovering their original shape when the deforming force is removed. Their stress-strain curve is markedly non-linear, unlike Hookean solids.

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: The ratio of lateral strain to the longitudinal strain in a stretched wire is called ____.$
›Reveal solutionSolution

The ratio of lateral strain to longitudinal strain in a stretched wire is called Poisson's ratio.

When a wire of original length L and diameter D is stretched by a longitudinal force, it becomes longer (longitudinal strain = ΔL/L) and simultaneously its diameter decreases slightly (lateral strain = ΔD/D). For a given material, within the elastic limit, the ratio of these two strains is constant: …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The unit of Young's modulus of elasticity is$
(a) N/m^2
(b) N-m^2
(c) joule (J)
(d) unitless
›Reveal solutionSolution

Young's modulus has the unit N/m^2 (pascal). Answer (A).

Young's modulus Y = (longitudinal stress)/(longitudinal strain).

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The Young's modulus of a perfectly rigid body is$
(a) zero
(b) 1
(c) infinite
(d) between zero and one
›Reveal solutionSolution

A perfectly rigid body has zero strain, giving infinite Young's modulus. Answer (C).

Y = stress/strain. A perfectly rigid body cannot be deformed, so no matter how much stress is applied, the strain remains zero.

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. Elastic potential energy in a stretched wire is$
(a) 1/2 × stress × strain
(b) 1/2 × stress × strain^2
(c) 1/2 × stress × strain × volume of the wire
(d) 1/2 × stress × strain^2 × volume of the wire
›Reveal solutionSolution

Elastic PE = ½ x stress x strain x volume. Answer (C).

The energy stored per unit volume (energy density) in a stretched wire is u = ½ x stress x strain.

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANN1 mark
$Q. Steel is more elastic than rubber. (True / False)$
›Reveal solutionSolution

True: steel is more elastic than rubber because it regains its shape with a much larger restoring stress (higher Young's modulus).

In physics, a material is more elastic if it develops a greater restoring force (stress) for a given deformation (strain) and returns more exactly to its original shape when the load is removed. Elasticity is measured by the modulus of elasticity (Young's modulus): the larger the modulus, the more elastic the material.

…

Start your 14-day free trial to see the full solution →

←Previous question · 7.8Q4 — Explain elasticity using intermolecular forces.→Next question · 7.8Q6 — A spring balance shows wrong readings after using for a long time. Why…

Q.Which one of these is more elastic, steel or rubber? Why?

🎓Unlock everything free for 14 days

Unlock everything free for 14 days