JEE Mains Physics · Mechanical Properties of Solids

Hooke's Law and Elastic Potential Energy

Within the elastic limit the tension is proportional to the extension, T = k(l − l₀), and a stretched wire stores energy ½FΔL, which is ½ × stress × strain in every cubic metre.

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Why this matters

Thirteen PYQs, seven of them asking for a number, and one from 2026. Seven find a natural length, or a tension, from two loaded lengths, and six find the energy stored in a stretched wire. Both are short once the right relation is written; the slips are proportional to length instead of extension, and the half in ½FΔL.

Concept 1 of 2: Natural length from two loaded lengths

Within the elastic limit the tension is proportional to the extension, not to the length: T = k(l − l₀). Two readings of length under two tensions are two equations in k and l₀. Subtracting them gives k; either one then gives l₀.

Definition

$T = k(l - l_0)$ , so $l = l_0 + T/k$ .
From $(T_1, l_1)$ and $(T_2, l_2)$ : $k = \dfrac{T_2 - T_1}{l_2 - l_1}$ and $l_0 = \dfrac{T_2l_1 - T_1l_2}{T_2 - T_1}$ .
A length $pl_1 - ql_2$ with $p - q = 1$ goes with the tension $pT_1 - qT_2$ , because the $l_0$ parts leave exactly one $l_0$ .
For extensions there is no $l_0$ to cancel: $px_1 - qx_2$ goes with $pT_1 - qT_2$ for any p and q.
Masses hung on a wire: use their weights as the tensions; g cancels in $l_0$ .

Natural length

T = k(l - l_0) \qquad l_0 = \frac{T_2l_1 - T_1l_2}{T_2 - T_1}

Worked example

A wire is 2.04 m long under a tension of 4 N and 2.10 m long under a tension of 10 N. Find its natural length.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q28Moderate

Example 1 · Mechanical Properties of Solids · Hooke's Law and Elastic Potential Energy

An elastic spring under tension of

3\text{ }N

has a length a. Its length is

b

under tension

2\text{ }N

. For its length

(3a - 2b)

, the value of tension will be ____

N

Tension proportional to length

T₁/T₂ = l₁/l₂ is wrong: the tension follows the extension l − l₀. Setting up T = k(l − l₀) for both readings avoids it.

Swapping the pairs in the l₀ formula

Check the formula by putting T₁ = 0: it must give l₀ = l₁. The other arrangement fails this test.

Drill 6 more on natural length from two loaded lengths

Concept 2 of 2: Energy stored in a stretched wire

Stretching a wire takes work, and the wire stores it as elastic energy. The force grows from zero to F as the wire stretches, so the work is the average force times the stretch, ½FΔL. Shared over the wire's volume, it is half the stress times the strain in each cubic metre.

Definition

$U = \tfrac{1}{2}F\,\Delta L = \dfrac{YA\,\Delta L^{2}}{2L}$ .
Energy per unit volume: $u = \tfrac{1}{2}\,\text{stress} \times \text{strain} = \tfrac{1}{2}Y\varepsilon^{2} = \dfrac{\text{stress}^{2}}{2Y}$ ; total $U = u \times AL$ .
Given Poisson's ratio and the lateral strain, the longitudinal strain is lateral strain ÷ σ; use that in u.
From a stress–strain graph: Y is the slope, and u is the area under the line up to the strain asked.
Stored energy handed to a mass becomes kinetic energy: $U = \tfrac{1}{2}mv^{2}$ .

Elastic energy

u = \tfrac{1}{2}\,(\text{stress})(\text{strain}) = \tfrac{1}{2}Y\varepsilon^{2} \qquad U = \tfrac{1}{2}F\,\Delta L

Worked example

A steel wire 2 m long with a cross-section of

1\ \text{mm}^{2}

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

is stretched by 1 mm. Find the energy stored.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 5 Apr 2026 Shift 2 · Q21Moderate

Example 2 · Mechanical Properties of Solids · Hooke's Law and Elastic Potential Energy

A copper wire of length 3 m is stretched by 3 mm by applying an external force. The volume of the wire is

600 \times 10^{- 6}{\text{ }m}^{3}

. The elastic potential energy stored in the wire in stretched condition would be

\_\_\_\_

J. (Given Young modulus of copper

\left. \ = 1.1 \times 10^{11}\text{ }N/m^{2} \right)

The load's work is not the energy stored

A load that stretches a wire by ΔL loses mgΔL of potential energy, but the wire stores only ½mgΔL. The other half leaves as heat or oscillation.

Using the lateral strain as the strain

u = ½Yε² needs the lengthwise strain. When the question gives the sideways strain, divide it by Poisson's ratio first.

Drill 5 more on energy stored in a stretched wire

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 (2)

Natural length from two loaded lengths
Natural length
$T = k(l - l_0) \qquad l_0 = \frac{T_2l_1 - T_1l_2}{T_2 - T_1}$
Energy stored in a stretched wire
Elastic energy
$u = \tfrac{1}{2}\,(\text{stress})(\text{strain}) = \tfrac{1}{2}Y\varepsilon^{2} \qquad U = \tfrac{1}{2}F\,\Delta L$

Watch out for (4)

Tension proportional to length
→ Natural length from two loaded lengths
Swapping the pairs in the l₀ formula
→ Natural length from two loaded lengths
The load's work is not the energy stored
→ Energy stored in a stretched wire
Using the lateral strain as the strain
→ Energy stored in a stretched wire

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