JEE Mains Physics · Mechanical Properties of Solids

Loaded Wires, Combinations and Breaking Stress

Before using ΔL = TL/(AY), find the tension each wire really carries: the same in wires joined end to end, everything below it in a hanging stack, and the result of a force balance when the load accelerates, swings or hangs by its own weight.

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Why this matters

Nineteen PYQs, nine of them asking for a number, and five from 2026. Seven join wires end to end or stack loads, six find the load at which a wire breaks, and six take the tension from mechanics first: a wire's own weight, a sagging wire, a circle or blocks on a pulley. The elasticity is one line; the marks go on the tension.

Concept 1 of 3: Wires in series and stacked loads

Follow the tension. Wires joined end to end carry the same tension, and their extensions add. In a hanging stack, each wire holds everything below it: the top wire carries every block, the lower one only what hangs from it. Once each tension is known, every wire is a separate ΔL = TL/(AY).

Definition

End to end (series): the same tension T in both, and $\Delta L = T\left(\dfrac{L_1}{A_1Y_1} + \dfrac{L_2}{A_2Y_2}\right)$ .
Two wires of equal length and area joined end to end: $Y_{eq} = \dfrac{2Y_1Y_2}{Y_1 + Y_2}$ .
Equal areas in series with equal extensions: $L \propto Y$ .
Stacked loads: the tension in a wire is the weight of everything hanging below it.
Then strain $\propto T/A$ and extension $\propto TL/(AY)$ , wire by wire.

Series wires

\Delta L = T\left(\frac{L_1}{A_1Y_1} + \frac{L_2}{A_2Y_2}\right) \qquad Y_{eq} = \frac{2Y_1Y_2}{Y_1 + Y_2}

Worked example

A wire 1.5 m long hangs from the ceiling and holds a 4 kg block. A second wire, 1 m long, of the same material and cross-section, hangs from that block and holds a 2 kg block. The wires are light. Find the ratio of the upper wire's extension to the lower wire's.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 2 · Q6Moderate

Example 1 · Mechanical Properties of Solids · Loaded Wires, Combinations and Breaking Stress

A metal string A is suspended from a rigid support and its free end is attached to a block of mass M. Second block having mass 2 M is suspended at the bottom of the first block using a string B . The area of cross sections of strings A and B are same. The ratio of lengths of strings of

A

B

is 2 and the ratio of their Young's moduli

\left( Y_{A}/Y_{B} \right)

is 0.5 . The ratio of elongations in A to B is

\_\_\_\_

The upper wire carries more than its own block

It holds every block and wire below it. Giving it only the block attached to it is the commonest wrong option.

Series wires share tension, not strain

Joined end to end, the wires have the same tension. Their strains and extensions differ unless their areas, lengths and moduli match.

Drill 6 more on wires in series and stacked loads

Concept 2 of 3: Breaking stress: the largest load a wire can hold

A wire breaks when its stress reaches the breaking stress, so the largest tension it can carry is the breaking stress times its area. The breaking stress belongs to the material: a thicker wire holds more only because it has more area. With several wires, test each one; the first to reach its limit sets the answer.

Definition

$T_{max} = \sigma_b A$ , where $\sigma_b$ is the breaking stress of the material.
Capacity grows with area: to lift a load W₂ instead of W₁, $A_2 = A_1 W_2/W_1$ .
A load accelerating upward: $T - mg = ma$ , so $a_{max} = \dfrac{\sigma_b A}{m} - g$ .
Stacked wires: write each wire's tension in terms of the unknown load, apply each wire's own limit, and keep the smallest answer.
Two masses over a light pulley: $T = \dfrac{2m_1m_2g}{m_1 + m_2}$ . A mass whirled in a horizontal circle on a wire (gravity ignored): $T = mv^{2}/l$ , so $v_{max} = \sqrt{\sigma_b A l/m}$ .

Breaking load

T_{max} = \sigma_b A \qquad a_{max} = \frac{\sigma_b A}{m} - g

Worked example

An upper wire of area

2\ \text{mm}^{2}

holds a 10 kg block. A lower wire of area

1\ \text{mm}^{2}

hangs from the block and holds a light pan. Both wires break at

5 \times 10^{8}\ \text{N/m}^{2}

(g = 10)

. What is the largest mass that can go in the pan?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q5Moderate

Example 2 · Mechanical Properties of Solids · Loaded Wires, Combinations and Breaking Stress

A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is

4 \times10^{8}\text{ }N/m^{2}

and its radius is 4 mm , then maximum acceleration the lift can take is

\_\_\_\_

m/s^{2}

. (take

g = 10\text{ }m/s^{2}

and

\pi= 3.14

)

Testing only the upper wire

The upper wire carries more, but it is often thicker. Test every wire against its own limit; the thinner lower wire often breaks first.

An accelerating lift needs more than mg

Going up with acceleration a, the tension is m(g + a). Setting σ_b A = ma forgets the weight.

Drill 5 more on breaking stress: the largest load a wire can hold

Concept 3 of 3: Tension from mechanics: own weight, sag, circles and pulleys

Here the tension is not a hanging weight you can read off. Find it first with mechanics, then use ΔL = TL/(AY). A wire hanging under its own weight has zero tension at the bottom and its full weight at the top, so its stretch uses the average, half the weight.

Definition

Own weight: $\Delta L = \dfrac{MgL}{2AY}$ . The largest stress is at the top, $\rho g L$ , so the longest wire that can hang without breaking is $L_{max} = \dfrac{\sigma_b}{\rho g}$ , whatever its area.
A wire held between two supports, with a mass m at its middle sagging by a small angle θ: $2T\theta = mg$ , and the strain is $\theta^{2}/2$ .
Vertical circle, at the lowest point: $T = mg + mv^{2}/r$ .
Blocks on a smooth table pulled by a hanging block: find the common acceleration, then each wire's tension from the blocks it pulls.

Tensions to use

\Delta L_{own} = \frac{MgL}{2AY} \qquad L_{max} = \frac{\sigma_b}{\rho g} \qquad 2T\theta = mg

Worked example

A uniform rod of mass 50 kg, length 2 m and cross-section

1\ \text{cm}^{2}

hangs from one end.

Y = 10^{11}\ \text{N/m}^{2}

g = 10\ \text{m/s}^{2}

. How much does it stretch under its own weight?

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The same idea in a real exam question:

JEE Mains · 2022 · 26 July 2022 · Q115Moderate

Example 3 · Mechanical Properties of Solids · Loaded Wires, Combinations and Breaking Stress

A uniform heavy rod of mass

20\text{ }kg

, cross-sectional area

0.4{\text{ }m}^{2}

and length

20\text{ }m

is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is

x \times10^{- 9}\text{ }m

. The value of

x

is :(Given. Young's modulus

Y = 2 \times10^{11}Nm^{- 2}

and

\left. \ g = 10{\text{ }ms}^{- 2} \right)

The full weight for an own-weight stretch

Only the top of the rod carries the full weight; the bottom carries none. The stretch uses Mg/2, so the full weight doubles the answer.

Leaving out mg at the bottom of a vertical circle

At the lowest point the string supports the weight and supplies the centripetal force: T = mg + mv²/r, not mv²/r.

Drill 5 more on tension from mechanics: own weight, sag, circles and pulleys

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Wires in series and stacked loads
Series wires
$\Delta L = T\left(\frac{L_1}{A_1Y_1} + \frac{L_2}{A_2Y_2}\right) \qquad Y_{eq} = \frac{2Y_1Y_2}{Y_1 + Y_2}$
Breaking stress: the largest load a wire can hold
Breaking load
$T_{max} = \sigma_b A \qquad a_{max} = \frac{\sigma_b A}{m} - g$
Tension from mechanics: own weight, sag, circles and pulleys
Tensions to use
$\Delta L_{own} = \frac{MgL}{2AY} \qquad L_{max} = \frac{\sigma_b}{\rho g} \qquad 2T\theta = mg$

Watch out for (6)

The upper wire carries more than its own block
→ Wires in series and stacked loads
Series wires share tension, not strain
→ Wires in series and stacked loads
Testing only the upper wire
→ Breaking stress: the largest load a wire can hold
An accelerating lift needs more than mg
→ Breaking stress: the largest load a wire can hold
The full weight for an own-weight stretch
→ Tension from mechanics: own weight, sag, circles and pulleys
Leaving out mg at the bottom of a vertical circle
→ Tension from mechanics: own weight, sag, circles and pulleys

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