UPSC Prelims 2007 Q108: Science & Technology – Physics
Four wires of the same material and of the dimensions given below are stretched by equal
loads separately. Which one of them will be elongated the maximum?
- (a) Wire of 1 m length and 2 mm diameter
- (b) Wire of 2 m length and 2 mm diameter
- (c) Wire of 3 m length and 1.5 mm diameter
- (d) Wire of 1 m length and 1 mm diameter
Answer: (c) Wire of 3 m length and 1.5 mm diameter
Explanation
Which wire stretches most? — Hooke's law / Young's modulus
Formula: Elongation ΔL = F·L / (A·Y).
Same material → same Y. Same load → same F.
Area A = πd²/4, so ΔL ∝ L / d².
| Wire | Length L (m) | Diameter d (mm) | L / d² |
|---|---|---|---|
| a | 1 | 2 | 1/4 = 0.25 |
| b | 2 | 2 | 2/4 = 0.50 |
| c | 3 | 1.5 | 3/2.25 = 1.33 |
| d | 1 | 1 | 1/1 = 1.00 |
Wire (c) has the largest L/d², so it stretches the most.
Answer: (c) Wire of 3 m length and 1.5 mm diameter.
Why each option is right or wrong
- (a) Incorrect. L/d² = 0.25 — the smallest stretch.
- (b) Incorrect. L/d² = 0.5, so this wire stretches less than wire (c).
- (c) CORRECT. L/d² = 1.33 — the highest, so maximum elongation.
- (d) Incorrect. L/d² = 1.0 — close, but less than wire (c).
Key point
Elongation ∝ L/d² (same material, same load). Long + thin = most stretch.
Remember
Stretch ∝ length ÷ (diameter)².
Common mistakes
- Comparing only length or only thickness — you must use L/d².
- Using d instead of d² — area depends on the square of the diameter.
- Picking (d) because it is the thinnest — thinness helps, but (c) is three times longer.
Exam tip
Shortcut: compute L/d² for each option. It takes 20 seconds and avoids traps.
Sources
- NCERT Class XI — Physics Part II, Ch 9 (Mechanical Properties of Solids)
- HC Verma — Concepts of Physics, Vol 1, Ch 14 (Some Mechanical Properties of Matter)