Unit 10: Mechanical Properties of Solids — Online MCQ Test
PHYSICS · CLASS 11 INTERMEDIATE 1 YEAR · Telangana State Board
Practice Unit 10: Mechanical Properties of Solids with a free chapter-wise online MCQ test.
This chapter covers: This chapter covers stress-strain relationships Hooke Law Young modulus bulk modulus shear modulus of elasticity and elastic behavior of materials..
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Unit 10: Mechanical Properties of Solids — Important Questions & Answers
What is stress in a solid?
- A. Restoring force per unit area
- B. Change in length per unit length
- C. Force per unit volume
- D. Work done per unit area
Answer: A. Restoring force per unit area
Stress is defined as the internal restoring force acting per unit area of the deformed body.
Stress is defined as the internal restoring force acting per unit area of the deformed body.
Which of the following is the SI unit of Young's modulus?
- A. N m
- B. N/m
- C. N/m²
- D. kg/m³
Answer: C. N/m²
Young's modulus is stress divided by strain, and since strain is dimensionless, its unit is the same as stress, N/m² or pascal.
Young's modulus is stress divided by strain, and since strain is dimensionless, its unit is the same as stress, N/m² or pascal.
A wire of length 2 m is stretched by 1 mm under a force. The strain produced is:
- A. 2 × 10⁻³
- B. 5 × 10⁻⁴
- C. 1 × 10⁻³
- D. 2 × 10⁻⁴
Answer: B. 5 × 10⁻⁴
Strain = change in length / original length = 0.001 / 2 = 5 × 10⁻⁴.
Strain = change in length / original length = 0.001 / 2 = 5 × 10⁻⁴.
For a given force, which wire will produce the smallest extension?
- A. Long, thin wire
- B. Short, thin wire
- C. Short, thick wire
- D. Long, thick wire
Answer: C. Short, thick wire
Extension is directly proportional to length and inversely proportional to area. A short, thick wire extends the least under the same force.
Extension is directly proportional to length and inversely proportional to area. A short, thick wire extends the least under the same force.
A wire remains within elastic limit when stretched. On removing the force, it returns to its original length because of:
- A. Plastic deformation
- B. Elastic restoring force
- C. Permanent set
- D. Yield point
Answer: B. Elastic restoring force
Within the elastic limit, the material develops an internal restoring force that brings it back to its original shape when the force is removed.
Within the elastic limit, the material develops an internal restoring force that brings it back to its original shape when the force is removed.