The maximum load a wire can withstand without breaking, when its length is reduced to half of its original len

NCERT Exemplar · Q2

Q. The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will

(a) be double.

(b) be half.

(d) remain same.

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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.

The breaking load depends only on the material's tensile strength and the wire's cross-sectional area, both of which are unchanged when you cut the length. The maximum load remains the same.

When we talk about the maximum load a wire can withstand, we're asking: at what applied force does the wire snap? This breaking point is governed by the tensile strength of the material—the maximum stress (force per unit area) the material can tolerate before fracturing.

The key insight is to separate what changes from what doesn't when you shorten a wire. Young's modulus tells us how stiff a material is (how much it stretches under load), but the breaking condition is about stress reaching a critical value, not about total extension.

Why length doesn't matter for breaking load

The stress in a wire under tension is defined as:

\sigma = \frac{F}{A}

where $F$ is the applied force and $A$ is the cross-sectional area. The wire breaks when this stress reaches the material's ultimate tensile strength $\sigma_{\text{max}}$ , a property of the material itself (like steel, copper, etc.).

So the breaking force is:

F_{\text{max}} = \sigma_{\text{max}} \cdot A

Notice that length $L$ doesn't appear in this equation. The breaking load depends only on:

The material (through $\sigma_{\text{max}}$ )
The cross-sectional area $A$

When you reduce the wire's length to half, you're not changing the material or the thickness of the wire—you're just making it shorter. The cross-sectional area $A$ stays exactly the same.

Step-by-step reasoning

Original wire: Length $L$ , area $A$ , breaking stress $\sigma_{\text{max}}$

Maximum load before breaking: $F_{\text{max}} = \sigma_{\text{max}} \cdot A$
Shortened wire: Length $\frac{L}{2}$ , area still $A$ (you haven't stretched or compressed it radially), same material so $\sigma_{\text{max}}$ unchanged
Breaking condition: The wire still breaks when stress reaches $\sigma_{\text{max}}$ , which happens at force:

F'_{\text{max}} = \sigma_{\text{max}} \cdot A

Comparison: $F'_{\text{max}} = F_{\text{max}}$

The maximum load is identical.

Watch out

A common confusion: "A shorter wire is stiffer (extends less for the same force), so shouldn't it hold more?" Stiffness and strength are different. A shorter wire does have a larger effective spring constant ( $k = \frac{YA}{L}$ increases as $L$ decreases), meaning it stretches less—but it still breaks at the same stress, hence the same force.

Tip

Think of it this way: if you have a rope and cut it in half, each piece can still hold the same weight before snapping. The length you're pulling doesn't change how much force the material's atomic bonds can resist.

✓Final answer

The correct option is (D) — the maximum load remains the same.

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Q.The maximum load a wire can withstand without breaking, when its length is reduced to half of its original length, will

Why length doesn't matter for breaking load

Step-by-step reasoning

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