A rigid bar of mass $M$ is supported symmetrically by three wires each of length $l$. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to (a) $Y_{copper}/Y_{iron}$ (b) $\sqrt{\dfrac{Y_{iron}}{Y_{copper}}}$ (c) $\dfrac{Y_{iron}^{2}}{Y_{copper}^{2}}$ (d) $\dfrac{Y_{iron}}{Y_{copper}}$.

A rigid bar of mass M is supported symmetrically by three wires each of length l. Those at each end are of cop

NCERT Exemplar · Q5

M

(a)

Y_{copper}/Y_{iron}

(b)

\sqrt{\dfrac{Y_{iron}}{Y_{copper}}}

(c)

\dfrac{Y_{iron}^{2}}{Y_{copper}^{2}}

(d)

\dfrac{Y_{iron}}{Y_{copper}}

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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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For equal tension in wires of different materials supporting the same load, the extensions must be equal. Since extension depends on stress (tension/area) and Young's modulus, wires with higher modulus need larger diameter. The diameter ratio is $\boxed{\sqrt{\frac{Y_{\text{iron}}}{Y_{\text{copper}}}}}$ .

The physical picture is straightforward: a horizontal rigid bar hangs from three vertical wires—copper at the ends, iron in the middle—and because the bar is rigid and the support is symmetric, all three wires stretch by the same amount when the bar is loaded. The question asks for the diameter ratio that ensures equal tension in each wire.

Young's modulus $Y$ measures a material's stiffness: how much stress is needed to produce a given strain. For a wire under tension $T$ ,

Y = \frac{\text{stress}}{\text{strain}} = \frac{T/A}{\Delta l / l}

where $A$ is the cross-sectional area and $\Delta l$ is the extension. Rearranging,

\Delta l = \frac{T l}{A Y}.

Because the bar is rigid and supported symmetrically, all three wires must extend by the same amount—otherwise the bar would tilt. This geometric constraint is the key.

Step-by-step reasoning:

Equal extension condition.

Let the copper wires have diameter $d_c$ , area $A_c = \frac{\pi d_c^2}{4}$ , and Young's modulus $Y_c$ . The iron wire has diameter $d_i$ , area $A_i = \frac{\pi d_i^2}{4}$ , and modulus $Y_i$ . All wires have the same length $l$ and, by the problem statement, the same tension $T$ .
Extension of copper wire:

\Delta l_c = \frac{T l}{A_c Y_c} = \frac{T l}{\frac{\pi d_c^2}{4} Y_c} = \frac{4 T l}{\pi d_c^2 Y_c}.

Extension of iron wire:

\Delta l_i = \frac{T l}{A_i Y_i} = \frac{4 T l}{\pi d_i^2 Y_i}.

Equate the extensions: Since the bar remains horizontal,

\Delta l_c = \Delta l_i \implies \frac{4 T l}{\pi d_c^2 Y_c} = \frac{4 T l}{\pi d_i^2 Y_i}.

Simplify: Cancel common factors ( $4Tl/\pi$ ):

\frac{1}{d_c^2 Y_c} = \frac{1}{d_i^2 Y_i} \implies d_c^2 Y_c = d_i^2 Y_i.

Solve for the diameter ratio: …

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Q.A rigid bar of mass MMM is supported symmetrically by three wires each of length lll. Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to

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Q.A rigid bar of mass $M$ is supported symmetrically by three wires each of length $l$ . Those at each end are of copper and the middle one is of iron. The ratio of their diameters, if each is to have the same tension, is equal to

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