(a) A steel wire of mass $\mu$ per unit length with a circular cross section has a radius of 0.1 cm. The wire is of length 10 m when measured lying horizontal, and hangs from a hook on the wall. A mass of 25 kg is hung from the free end of the wire. Assuming the wire to be uniform and lateral strains $\ll$ longitudinal strains, find the extension in the length of the wire. The density of steel is 7860 kg m$^{-3}$ (Young's modules $Y = 2 \times 10^{11}$ Nm$^{-2}$). (b) If the yield strength of steel is $2.5 \times 10^{8}$ Nm$^{-2}$, what is the maximum weight that can be hung at the lower end of the wire?

(a) A steel wire of mass μ per unit length with a circular cross section has a radius of 0.1 cm. The wire is o

NCERT Exemplar · Q25

\mu

(b) If the yield strength of steel is

2.5 \times 10^{8}

^{-2}

, what is the maximum weight that can be hung at the lower end of the wire?

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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:

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

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The $25$ kg load stretches the wire by $\Delta L\approx 3.9\times10^{-3}$ m ( $\approx 3.9$ mm); the yield strength limits the hangable load to $W_{\max}\approx 785$ N (about $80$ kg).

Part (a): Extension under the load

1. Set up Hooke's law for the wire

\Delta L=\frac{FL}{AY},

with $F=Mg=25\times 9.8=245$ N, $L=10$ m, $Y=2\times10^{11}$ N m $^{-2}$ .

2. Cross-sectional area (radius $r=0.1$ cm $=1\times10^{-3}$ m):

A=\pi r^2=\pi(1\times10^{-3})^2=3.14\times10^{-6}\ \text{m}^2.

3. Compute the extension

\Delta L=\frac{245\times 10}{(3.14\times10^{-6})(2\times10^{11})}=\frac{2450}{6.28\times10^{5}}=3.9\times10^{-3}\ \text{m}.

(The wire's own weight, $\approx\rho A L g\approx 2.4$ N against the $245$ N load, is negligible.)

Part (b): Maximum hangable weight

4. Apply the yield-strength limit …

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

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The shear modulus (modulus of rigidity) of an ideal liquid is zero.

Shear modulus G is defined as G = shear stress / shear strain. A liquid (unlike a solid) has no fixed shape; when a tangential (shear) force is applied to a liquid layer, it does not develop a restoring elastic shear stress — it simply flows/deforms continuously. Since a liquid offers essentially zero resistance to shape-changing (shear) deformation whil …

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$Q. Draw the Force-Displacement graph for a body obeying the relation (F = kx), where (k) is a constant and$
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›Reveal solutionSolution
Figure — Force-Displacement graph for F=kx

The Force–Displacement graph for F = kx is a straight line passing through the origin, with slope equal to k.

Given F = kx, F is directly proportional to x (k is a constant, e.g., a spring/elastic constant, as in Hooke's Law).

To draw the graph:

Put displacement x on the horizontal axis and force F on the vertical axis.

Since F = kx is a linear equation in x with zero intercept, the graph is a straight line that starts at the origin (0,0) and rises with a constant slope equal to k, i.e., for every unit increase in x, F increases by k. …
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$Q. State Hooke's law.$
›Reveal solutionSolution

Within the elastic limit, the stress produced in a body is directly proportional to the strain it undergoes.

When a deforming force is applied to an elastic body, it develops an internal restoring force per unit area called stress, and undergoes a fractional change in dimension called strain. Hooke's law states that as long as the deformation stays within the elastic limit of the material (i.e., the body can still fully recover its original shape when the force is removed), the stress developed is directly proportional to the strain produced:

$\text{Stress} \propto \text{Strain} \quad \Rightarrow \quad \text{Stress} = k \times \text{Strain}$

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$Q. How are we able to break a wire by repeated bending ?$
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Bending a wire repeatedly at the same point strains it beyond its elastic limit again and again, causing progressive internal damage (elastic fatigue) until it fractures, even though one such bend alone would not break it.

When a wire is bent, the material on the outer side of the bend is stretched (tensile strain) while the material on the inner side is compressed. If the bend is sharp, the strain at that point can exceed the elastic limit of the material, so the material does not fully return to its original state when straightened — some permanent (plastic) deformation and internal microscopic damage remains.

When this bending is repeated again and again at the same point, each cycle adds a little more damage: microscopic cracks form and grow within the material, and the local heating produced by the repeated internal friction/deformation further reduces the wire's elastic strength at that spot. This cumulative loss of strength through repeated stress cycles is called elastic fatigue.

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$Q. How are we able to break a wire by repeated bending ?$
›Reveal solutionSolution

Bending a wire back and forth repeatedly causes elastic fatigue — the metal at the bend loses its elasticity and develops cracks, so it eventually breaks even without a large single force.

When a wire is bent, the layer on the outside of the bend is stretched (put under tensile stress) and the layer on the inside is compressed. If the bend is sharp enough, the local stress at the fold can exceed the material's elastic limit even though the wire as a whole isn't under much load.

Each time we bend the wire back and forth, the material at that point undergoes this stress reversal again. Every cycle produces tiny, permanent internal deformations (dislocations) that the material cannot fully recover from — this cumulative loss of elastic strength with repeated stress cycles is called elastic fatigue. Over many cycles, microscopic cracks form at the bend and grow with each further flex, until the cross-section remaining is too weak to bear even the wire's own handling force, and it snaps.

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Part (a): Extension under the load

Part (b): Maximum hangable weight

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