The temperature of a wire is doubled. The Young's modulus of elasticity (a) will also double. (b) will become

NCERT Exemplar · Q3

Q. The temperature of a wire is doubled. The Young's modulus of elasticity

(a) will also double.

(b) will become four times.

(d) will decrease.

Punjab PsebMCQ· 1mImportance★★★★★est

48% · 24/50 Questions

✓ Free question

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.

Young’s modulus depends on the material’s atomic bonding, which weakens as temperature rises — so when temperature doubles, Young’s modulus decreases, not increases or stays constant.

What Young’s modulus actually measures. Young’s modulus ( $Y$ ) is the ratio of tensile stress to tensile strain within the elastic limit:

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

It is an intrinsic property of the material — it tells you how stiff the material is. But “intrinsic” does not mean “constant under all conditions.” It changes with temperature because the underlying atomic forces change.

Why temperature affects stiffness.

In a solid, atoms are held together by interatomic forces. At higher temperatures, atoms vibrate more vigorously about their equilibrium positions. This increased thermal motion effectively reduces the average restoring force for a given displacement — the bonds become “softer.”

Important

The physical picture: heat makes the atomic bonds less stiff, so the material stretches more easily under the same load. That means Young’s modulus decreases as temperature rises.
What “temperature is doubled” means.

Doubling the temperature (say from $T$ to $2T$ , measured in kelvin) is a significant increase. For most metals, a rise of a few hundred kelvin can reduce $Y$ by several percent. The relationship is approximately linear over moderate ranges:

Y(T) \approx Y_0 \left[1 - \alpha (T - T_0)\right]

where $\alpha$ is a small positive coefficient (of order $10^{-4}\,\text{K}^{-1}$ for metals). So doubling $T$ definitely lowers $Y$ — it does not double, quadruple, or stay the same.

Eliminating the wrong options.
- (A) will also double — No. Heating softens, not stiffens.
- (B) will become four times — Even more absurd; no material behaves that way.
- (C) will remain same — Only true if temperature change is zero. For a finite change, $Y$ changes.
- (D) will decrease — This matches the physics.

Watch out

A common mistake is to think that because Young’s modulus is a material constant, it never changes. It is constant at a given temperature, but varies with temperature. The question explicitly changes temperature, so the modulus must change.

✓Final answer

The correct option is (D) — Young’s modulus will decrease.

←Previous questionQ2 — The maximum load a wire can withstand without breaking, when its lengt…→Next questionQ4 — A spring is stretched by applying a load to its free end. The strain p…

Q.The temperature of a wire is doubled. The Young's modulus of elasticity

🎓Unlock everything free for 14 days

Unlock everything free for 14 days