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Mechanical Properties of Solids is small, mechanical and almost entirely formula driven, which makes it one of the cheapest marks in JEE Main. Most questions are a single substitution into Y = FL/AΔL, dressed up as a comparison between two wires. Learn how the extension depends on length and radius and you will answer those in seconds. Ten questions below. Give them ten honest minutes.
What this chapter carries in JEE Main
- Usually 1 question every year, which is 4 marks.
- Comparing the extensions of two wires is the standard question.
- The stress–strain graph and the three moduli come up as recall.
What the syllabus says
In the syllabus (Unit 7, Properties of Solids and Liquids): Elastic behaviour, the stress–strain relationship, Hooke’s law, Young’s modulus, bulk modulus and modulus of rigidity.
Never in the JEE Main syllabus: Poisson’s ratio; elastic energy, that is the energy stored per unit volume. They are in NCERT, so they still turn up in old question banks.
Taken from the NTA syllabus for JEE Main 2027. See the full syllabus, unit by unit.
Formulas you must know by heart
| Stress and strain | σ = F/A, ε = ΔL/L |
| Young’s modulus | Y = FL / AΔL |
| Extension depends on | ΔL ∝ L/r2 for a given load |
| Bulk modulus | B = − ΔP / (ΔV/V) |
| Modulus of rigidity | η = F / Aθ |
| Compressibility | k = 1/B |
| Breaking stress | a property of the material: breaking load = breaking stress × area |
| Extension under a load | l = FL / AY |
The third line is the one that answers most of the comparison questions. For the same material and the same load, the extension grows with the length and falls as the square of the radius, so doubling the radius cuts the extension to a quarter.
Ten questions. Give yourself ten minutes.
Write your answers down before you scroll. A question you looked up is a question you have not answered.
Q1. Two wires of the same material have lengths L and 2L and radii r and 2r. Under the same load, the ratio of their extensions is
- (a) 1 : 1
- (b) 2 : 1
- (c) 1 : 2
- (d) 4 : 1
Q2. A wire stretches by 1 mm under a load W. Within the elastic limit, under a load 2W it stretches by
- (a) 1 mm
- (b) 2 mm
- (c) 4 mm
- (d) 0.5 mm
Q3. The Young’s modulus of a perfectly rigid body is
- (a) zero
- (b) infinite
- (c) exactly 1
- (d) negative
Q4. The work done in stretching a wire through x by a final force F is
- (a) Fx
- (b) ½Fx
- (c) 2Fx
- (d) Fx2
Q5. A wire can carry a certain maximum load before breaking. If its radius is doubled, the maximum load becomes
- (a) 2 times
- (b) 4 times
- (c) half
- (d) unchanged
Q6. A wire breaks under a certain load. Another wire of the same material and the same length but twice the diameter will break under a load that is
- (a) twice as large
- (b) half as large
- (c) four times as large
- (d) the same
Q7. Within the elastic limit, the ratio of stress to strain is
- (a) zero
- (b) a constant, called the modulus of elasticity
- (c) infinite
- (d) steadily changing
Q8. On a stress–strain graph, the stress at which the material finally snaps is called the
- (a) yield point
- (b) elastic limit
- (c) breaking stress
- (d) proportional limit
Q9. A wire of length L and area A is stretched by a force F and its length increases by l. Its Young’s modulus is
- (a) FL/Al
- (b) Fl/AL
- (c) FA/Ll
- (d) FLl/A
Q10. Compressibility is the reciprocal of the
- (a) Young’s modulus
- (b) shear modulus
- (c) bulk modulus
- (d) breaking stress
Answer key, with a hint for each
| Q1 | (b) | ΔL ∝ L/r2. The first gives L/r2 and the second 2L/4r2, so the ratio is 2 : 1. |
| Q2 | (b) | Within the elastic limit the extension is proportional to the load. |
| Q3 | (b) | A rigid body does not deform at all, so the strain is zero and the ratio blows up. |
| Q4 | (b) | The force builds up from zero to F, so you use the average, F/2. |
| Q5 | (b) | Breaking stress is fixed, and the load it can carry is stress × area, which goes as r2. |
| Q6 | (c) | Breaking stress belongs to the material, not to the wire. Load = breaking stress × area, and twice the diameter makes the area four times. |
| Q7 | (b) | That is the definition of a modulus, and it is why the stress–strain graph is a straight line there. |
| Q8 | (c) | The yield point is where it stops returning to shape. The breaking stress is further along. |
| Q9 | (a) | Y = stress ÷ strain = (F/A) ÷ (l/L) = FL/Al. The long length goes on top, the small extension below. |
| Q10 | (c) | Bulk modulus measures how hard a solid is to squeeze. Compressibility is simply 1/B, so a soft material has a high compressibility. |
Now mark yourself honestly
JEE Main marks the way it marks. Use the same rule here:
- +4 for a right answer
- −1 for a wrong one
- 0 for a blank
So ten questions here are out of 40. In the real paper, 25 Physics questions are out of 100.
If you scored 32 or more, the basics of this chapter are secure. Move to harder questions, not more of these.
If you scored below 24, do not feel bad about it. You have just found out something useful before the exam did. Go back to the formula table above, learn it properly, and sit these ten again tomorrow.
And if you guessed a few and they landed, count those as wrong. A guess that works once will not work in January.
The full chapter, on a clock
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