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Elasticity, stress-strain relation and elastic moduli: NEET MCQs with answers

75 questions on this topic of Properties of Solids and Liquids (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A metal wire of length L and uniform cross-sectional area A is stretched by a force F within its elastic limit. If the potential energy stored per unit volume in the wire is u, which of the following expressions correctly gives the Young's modulus Y of the material?

  1. u / (2 × (F/A)²)
  2. (F/A) / (2u)
  3. 2u / (F/A)²
  4. (F/A)² / (2u)
Show answer and explanation

Answer: (D) (F/A)² / (2u)

Elastic energy density is given by u = (1/2) × (stress)² / Y. Rearranging this relation yields Y = (stress)² / (2u) = (F/A)² / (2u).

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2. A wire of length 1 m and cross-sectional area 0.5 mm² stretches by 1 mm under a load of 100 N. The Young's modulus of its material is:

  1. 2 × 10¹⁴ N/m²
  2. 4 × 10¹¹ N/m²
  3. 2 × 10⁸ N/m²
  4. 2 × 10¹¹ N/m²
Show answer and explanation

Answer: (D) 2 × 10¹¹ N/m²

Y = FL/(AΔL) = 100 × 1/(0.5 × 10⁻⁶ × 1 × 10⁻³) = 2 × 10¹¹ N/m².

hard

3. A light horizontal rod AB of length 1.2 m is suspended by two vertical wires of equal natural lengths and cross-sections, one of steel (Young's modulus Yₛ = 2.0 × 10¹¹ N m⁻²) at end A and one of brass (Young's modulus Y_b = 1.0 × 10¹¹ N m⁻²) at end B. At what distance from end A must a weight W be hung so that the rod remains horizontal?

  1. 0.4 m
  2. 0.6 m
  3. 0.8 m
  4. 0.3 m
Show answer and explanation

Answer: (A) 0.4 m

For the rod to remain horizontal, the elongations of both wires must be equal: ΔL_s = ΔL_b. Since lengths and areas are equal, T_s / Y_s = T_b / Y_b, which gives T_s / T_b = Y_s / Y_b = 2 / 1. Taking torque about the point where weight W is suspended at distance x from A: T_s × x = T_b × (1.2 - x) ⇒ 2 T_b × x = T_b (1.2 - x) ⇒ 3x = 1.2 ⇒ x = 0.4 m.

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4. A wire of length L and radius r is suspended from a rigid support and carries a load M at its lower end. If the radius of the wire is halved and its length is doubled while keeping the suspended mass unchanged, the elastic potential energy stored in the wire becomes:

  1. 8 times the original value
  2. 4 times the original value
  3. 2 times the original value
  4. 16 times the original value
Show answer and explanation

Answer: (A) 8 times the original value

Elastic potential energy stored in a stretched wire is U = F²L / (2AY) = (M²g²L) / (2πr²Y). When length becomes 2L and radius becomes r/2, the new energy is U' = (M²g²(2L)) / [2π(r/2)²Y] = 2 / (1/4) × U = 8U.

easy

5. The stress-strain graphs for two materials A and B are plotted on the same scale. The graph for material A has a steeper slope in the linear region, and the distance between its yield point and fracture point is very small. In contrast, for material B, the linear slope is lower and the distance between the yield point and fracture point is large. Which of the following statements is correct?

  1. Material A is brittle and has a larger Young's modulus than B
  2. Material A is ductile and has a smaller Young's modulus than B
  3. Material B is brittle and has a larger Young's modulus than A
  4. Material B is brittle and has a smaller Young's modulus than A
Show answer and explanation

Answer: (A) Material A is brittle and has a larger Young's modulus than B

The slope of the stress-strain graph in the linear region represents Young's modulus, so material A has a larger Young's modulus. A short plastic range between the yield point and the fracture point indicates that material A is brittle, whereas a large plastic deformation range indicates that material B is ductile.

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6. A light rigid bar of length 1.0 m is suspended horizontally by two vertical wires P and Q of equal original lengths attached at its ends. Wire P is made of steel (cross-sectional area A₁ = 1.0 mm², Y₁ = 2.0 × 10¹¹ N m⁻²) and wire Q is made of brass (cross-sectional area A₂ = 2.0 mm², Y₂ = 1.0 × 10¹¹ N m⁻²). At what distance from wire P must a weight W be hung on the bar so that the bar remains horizontal?

  1. 0.25 m
  2. 0.33 m
  3. 0.50 m
  4. 0.67 m
Show answer and explanation

Answer: (C) 0.50 m

For the bar to remain horizontal, the elongation in both wires must be equal: ΔL₁ = ΔL₂. Thus, (T₁ L) / (A₁ Y₁) = (T₂ L) / (A₂ Y₂), which means T₁ / (A₁ Y₁) = T₂ / (A₂ Y₂). We calculate A₁ Y₁ = (1.0 × 10⁻⁶)(2.0 × 10¹¹) = 2.0 × 10⁵ N and A₂ Y₂ = (2.0 × 10⁻⁶)(1.0 × 10¹¹) = 2.0 × 10⁵ N. Since A₁ Y₁ = A₂ Y₂, the tensions must be equal: T₁ = T₂. For rotational equilibrium about the center of mass or point of suspension of W, T₁ x = T₂ (1 - x), which yields x = 0.50 m.

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7. A wire of length 1 m and cross-sectional area 0.5 mm² stretches by 1 mm under a load of 200 N. The Young's modulus of its material is:

  1. 4 × 10¹⁴ N/m²
  2. 4 × 10¹¹ N/m²
  3. 8 × 10¹¹ N/m²
  4. 4 × 10⁸ N/m²
Show answer and explanation

Answer: (B) 4 × 10¹¹ N/m²

Y = FL/(AΔL) = 200 × 1/(0.5 × 10⁻⁶ × 1 × 10⁻³) = 4 × 10¹¹ N/m².

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8. A wire of length 3 m and cross-sectional area 2 mm² stretches by 2 mm under a load of 100 N. The Young's modulus of its material is:

  1. 2.5 × 10¹⁰ N/m²
  2. 7.5 × 10¹³ N/m²
  3. 7.5 × 10¹⁰ N/m²
  4. 7.5 × 10⁷ N/m²
Show answer and explanation

Answer: (C) 7.5 × 10¹⁰ N/m²

Y = FL/(AΔL) = 100 × 3/(2 × 10⁻⁶ × 2 × 10⁻³) = 7.5 × 10¹⁰ N/m².

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9. A wire of length 1.5 m and cross-sectional area 1 mm² stretches by 1 mm under a load of 200 N. The Young's modulus of its material is:

  1. 2 × 10¹¹ N/m²
  2. 3 × 10¹¹ N/m²
  3. 3 × 10⁸ N/m²
  4. 3 × 10¹⁴ N/m²
Show answer and explanation

Answer: (B) 3 × 10¹¹ N/m²

Y = FL/(AΔL) = 200 × 1.5/(1 × 10⁻⁶ × 1 × 10⁻³) = 3 × 10¹¹ N/m².

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10. A wire of length 2 m and cross-sectional area 0.5 mm² stretches by 0.5 mm under a load of 150 N. The Young's modulus of its material is:

  1. 1.2 × 10¹² N/m²
  2. 1.2 × 10⁹ N/m²
  3. 6 × 10¹¹ N/m²
  4. 1.2 × 10¹⁵ N/m²
Show answer and explanation

Answer: (A) 1.2 × 10¹² N/m²

Y = FL/(AΔL) = 150 × 2/(0.5 × 10⁻⁶ × 0.5 × 10⁻³) = 1.2 × 10¹² N/m².

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11. A lead block of volume 0.5 m³ is taken from the surface of an ocean to a depth where the water pressure is 4.0 × 10⁷ N m⁻² greater than atmospheric pressure. If the bulk modulus of lead is 8.0 × 10¹⁰ N m⁻², the change in volume of the block is:

  1. 2.50 × 10⁻⁴ m³
  2. 1.25 × 10⁻⁴ m³
  3. 5.00 × 10⁻⁴ m³
  4. 1.00 × 10⁻³ m³
Show answer and explanation

Answer: (A) 2.50 × 10⁻⁴ m³

Bulk modulus B = ΔP / (|ΔV| / V). Hence, |ΔV| = (ΔP × V) / B = (4.0 × 10⁷ N m⁻² × 0.5 m³) / (8.0 × 10¹⁰ N m⁻²) = 2.0 × 10⁷ / 8.0 × 10¹⁰ = 2.50 × 10⁻⁴ m³.

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12. A wire of length 2 m and cross-sectional area 1 mm² stretches by 1 mm under a load of 50 N. The Young's modulus of its material is:

  1. 5 × 10¹⁰ N/m²
  2. 1 × 10⁸ N/m²
  3. 1 × 10¹⁴ N/m²
  4. 1 × 10¹¹ N/m²
Show answer and explanation

Answer: (D) 1 × 10¹¹ N/m²

Y = FL/(AΔL) = 50 × 2/(1 × 10⁻⁶ × 1 × 10⁻³) = 1 × 10¹¹ N/m².

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