Class XI · Chapter 08
Mechanical Properties of Solids — Exam Practice
Practice questions in CBSE, NEET, and JEE Main style covering stress, strain, Hooke’s Law, and elastic moduli.
A note on these questions: these are original practice questions, written to match the exact style, format, and difficulty of CBSE Board, NEET, and JEE Main papers — not verbatim reproductions of specific past papers.
1CBSE Board Exam Style
Q1. Which material is more elastic — steel or rubber — and why?
Answer
Steel is more elastic. For the same applied stress, steel shows far smaller strain than rubber — meaning steel has a much higher Young’s modulus, which is exactly what defines greater elasticity.
Steel is more elastic. For the same applied stress, steel shows far smaller strain than rubber — meaning steel has a much higher Young’s modulus, which is exactly what defines greater elasticity.
Derive the relation between Young’s modulus and the elongation of a stretched wire, and use it to define Young’s modulus.
Answer
Longitudinal stress = F/A. Longitudinal strain = ΔL/L. By Hooke’s Law (within the elastic limit), stress ∝ strain, with the constant of proportionality defined as Young’s modulus:
Longitudinal stress = F/A. Longitudinal strain = ΔL/L. By Hooke’s Law (within the elastic limit), stress ∝ strain, with the constant of proportionality defined as Young’s modulus:
Y = (F/A) / (ΔL/L) = FL / (AΔL).
Young’s modulus is thus a material’s specific stiffness against longitudinal (stretching or compressing) deformation.
Q3. (a) Sketch and explain the stress-strain curve for a ductile material, labelling the elastic limit, yield point, and fracture point. (b) A wire of length 3 m and diameter 1 mm stretches by 3 mm under a load of 20 N. Find the Young’s modulus of the material.
Answer
(a) The curve rises linearly (Hooke’s Law region) up to the elastic limit; beyond the yield point, strain increases rapidly for little added stress (plastic region), ending at the fracture point where the material breaks.
(a) The curve rises linearly (Hooke’s Law region) up to the elastic limit; beyond the yield point, strain increases rapidly for little added stress (plastic region), ending at the fracture point where the material breaks.
(b) A = π(0.5×10⁻³)² ≈ 7.854×10⁻⁷ m².
Y = FL/(AΔL) = (20×3) / (7.854×10⁻⁷ × 0.003) = 60 / (2.356×10⁻⁹) ≈ 2.5×10¹⁰ N/m².
2NEET Style (Single-Correct MCQ)
Q4. Bulk modulus is meaningfully defined for:
Solution
Bulk modulus relates volume stress to volumetric strain — any substance with a volume to compress qualifies, unlike Young’s and shear modulus, which apply only to solids.
Bulk modulus relates volume stress to volumetric strain — any substance with a volume to compress qualifies, unlike Young’s and shear modulus, which apply only to solids.
Q5. Shear modulus relates:
Solution
G = shearing stress / shearing strain — describes resistance to layers sliding past each other, distinct from stretching (Young’s modulus) or uniform compression (bulk modulus).
G = shearing stress / shearing strain — describes resistance to layers sliding past each other, distinct from stretching (Young’s modulus) or uniform compression (bulk modulus).
Q6. Within the elastic limit, stress is:
Solution
This is exactly Hooke’s Law — the defining relationship of the elastic region on a stress-strain curve.
This is exactly Hooke’s Law — the defining relationship of the elastic region on a stress-strain curve.
Q7. The SI unit of stress is:
Solution
Stress = force/area, giving N/m², named the pascal — the same unit as pressure.
Stress = force/area, giving N/m², named the pascal — the same unit as pressure.
3JEE Main Style
Q8. A wire’s length increases by 0.1% under a stress of 2×10⁸ Pa. Find the Young’s modulus of the material.
Solution
Strain = 0.1% = 0.001. Y = stress/strain = (2×10⁸) / 0.001 = 2×10¹¹ Pa.
Strain = 0.1% = 0.001. Y = stress/strain = (2×10⁸) / 0.001 = 2×10¹¹ Pa.
Q9. A material shows permanent (non-recoverable) deformation when it is stressed:
Solution
Deformation stays fully recoverable only up to the elastic limit; beyond it, the material enters the plastic region and doesn’t return to its original shape.
Deformation stays fully recoverable only up to the elastic limit; beyond it, the material enters the plastic region and doesn’t return to its original shape.
Q10. Elastic potential energy per unit volume is given by ½ × stress × strain. Find the energy density if stress = 10⁸ Pa and strain = 0.0005.
Solution
u = ½ × 10⁸ × 0.0005 = 2.5×10⁴ J/m³.
u = ½ × 10⁸ × 0.0005 = 2.5×10⁴ J/m³.
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