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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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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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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Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Set ANNUAL1 mark
$Q. State Hooke's law.$
›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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$Q. How are we able to break a wire by repeated bending ?$
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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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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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