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Chapter-8 Mechanical Properties of Solids — Online MCQ Test

PHYSICS · CLASS 11 INTER I YEAR · Andhra State Board
Practice Chapter-8 Mechanical Properties of Solids with a free chapter-wise online MCQ test. This chapter covers: Elastic behavior - Stress-strain relationship - Hooke's law - Young's modulus - Bulk modulus - Shear modulus - Poisson's ratio - Elastic potential energy. AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter-8 Mechanical Properties of Solids — Important Questions & Answers

What is stress defined as?
  • A. Force applied per unit area
  • B. Change in length per unit original length
  • C. Deformation produced in a body
  • D. Energy stored in a deformed body
Answer: A. Force applied per unit area
Stress is the restoring force per unit area acting on a body when it is deformed.
Strain is a dimensionless quantity because it represents:
  • A. Force per unit area
  • B. Ratio of change in dimension to original dimension
  • C. Energy per unit volume
  • D. Deformation in a body
Answer: B. Ratio of change in dimension to original dimension
Strain is the ratio of change in dimension to the original dimension, making it a dimensionless quantity.
A wire of length L and cross-sectional area A is stretched by a force F. Its Young's modulus can be calculated using:
  • A. Y = (F × L) / (A × ΔL)
  • B. Y = (F × ΔL) / (A × L)
  • C. Y = F / (A × L)
  • D. Y = (A × L) / (F × ΔL)
Answer: A. Y = (F × L) / (A × ΔL)
Young's modulus Y = Stress/Strain = (F/A)/(ΔL/L) = (F × L)/(A × ΔL).
The relationship between Young's modulus (Y), bulk modulus (K), and shear modulus (G) is given by:
  • A. Y = 3KG / (3K + G)
  • B. Y = 9KG / (3K + G)
  • C. K = YG / (3Y - 6G)
  • D. G = 3YK / (9K + Y)
Answer: B. Y = 9KG / (3K + G)
The relationship Y = 9KG/(3K + G) connects the three elastic moduli for isotropic materials.
A steel wire of diameter 2 mm and length 5 m is stretched by applying a force of 5000 N. If Young's modulus of steel is 2 × 10¹¹ Pa, the extension produced is:
  • A. 2.5 mm
  • B. 3.98 mm
  • C. 5 mm
  • D. 7.96 mm
Answer: B. 3.98 mm
Y = (F/A) × (L/ΔL); ΔL = FL/(AY) = (5000 × 5)/(π × 10⁻⁶ × 2 × 10¹¹) ≈ 3.98 × 10⁻³ m = 3.98 mm.