Mechanical Properties of Solids

6 NEET-UG practice questions on Mechanical Properties of Solids, part of Physics. The first 6 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

Question 1 · Medium
Arrange these points on the stress-strain curve of a ductile metal in the order in which they are reached as the strain increases. A. Fracture point B. Yield point C. Proportional limit D. Point of ultimate tensile strength
1. AB, D, C, A
2. BB, C, D, A
3. CC, B, D, A
4. DC, D, B, A
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Answer: C · C, B, D, A
Stress is proportional to strain up to the proportional limit. The yield point comes next, where plastic behaviour begins. Stress then rises to its greatest value, the ultimate tensile strength, and the wire finally breaks at the fracture point.
- A: This puts the proportional limit after the plastic region.
- B: The proportional limit comes before the yield point.
- D: Ultimate strength lies beyond the yield point, not before it.
Remember this
The maximum stress occurs before fracture; between these two points the stress falls as the wire thins.
Question 2 · Easy
In an experiment to find Young's modulus, a wire 2 m long with a cross-section of 1 mm² stretches by 1 mm under a load of 100 N. What value of Young's modulus do these readings give?
1. A2 × 10⁸ Pa
2. B5 × 10¹⁰ Pa
3. C2 × 10⁵ Pa
4. D2 × 10¹¹ Pa
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Answer: D · 2 × 10¹¹ Pa
Y = FL/(A ΔL) = (100 × 2)/(10⁻⁶ × 10⁻³) = 2 × 10¹¹ Pa.
- A: This does not convert the extension from millimetres to metres.
- B: This divides by the length instead of multiplying by it.
- C: This does not convert the area from mm² to m².
Remember this
Convert to SI first: 1 mm² = 10⁻⁶ m² and 1 mm = 10⁻³ m.
Question 3 · Medium
A sealed 2 litre sample of a liquid of bulk modulus 2 × 10⁹ Pa experiences a pressure rise of 4 × 10⁶ Pa. By how much does its volume decrease?
1. A2 cm³
2. B1 cm³
3. C0.4 cm³
4. D4 cm³
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Answer: D · 4 cm³
B = Δp/(ΔV/V), so ΔV/V = 4 × 10⁶/(2 × 10⁹) = 2 × 10⁻³. Then ΔV = 2000 cm³ × 2 × 10⁻³ = 4 cm³.
- A: This takes the original volume as 1 litre.
- B: This divides the strain by the 2 litres instead of multiplying.
- C: This makes a power-of-ten slip in the conversion from litres to cm³.
Remember this
Volumetric strain = Δp/B; multiply it by the original volume to get the volume change.
Question 4 · Medium
Load against extension is plotted for a wire 2 m long with a cross-section of 1 mm². What is the Young's modulus of its material?
Figure: MedExams
1. A2 × 10¹¹ Pa
2. B2 × 10⁵ Pa
3. C1 × 10¹¹ Pa
4. D2 × 10⁸ Pa
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Answer: A · 2 × 10¹¹ Pa
From the graph, 40 N gives 0.4 mm. Stress = 40/10⁻⁶ = 4 × 10⁷ Pa and strain = (0.4 × 10⁻³)/2 = 2 × 10⁻⁴, so Y = 2 × 10¹¹ Pa. Direct use of Y = FL/(AΔL) gives 80/(4 × 10⁻¹⁰), the same value.
- B: Leaves the area in mm², not converting it to m².
- C: Leaves out the wire's length of 2 m.
- D: Treats the extension in mm without converting it to metres.
Remember this
Y = FL/(AΔL) = (slope of load-extension) × L/A, in SI units.
Question 5 · Easy
Match List-I with List-II for the stress-strain curve of a metal wire shown. List-I (Point on the curve) A. P B. Q C. R D. S List-II (What it marks) I. Fracture point II. Proportional limit III. Ultimate tensile strength IV. Yield point
Figure: MedExams
1. AA-IV, B-II, C-III, D-I
2. BA-II, B-IV, C-III, D-I
3. CA-II, B-III, C-IV, D-I
4. DA-II, B-IV, C-I, D-III
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Answer: B · A-II, B-IV, C-III, D-I
Hooke's law holds on the straight part up to P, the proportional limit. Q, just beyond, is the yield point. The peak R is the ultimate tensile strength, and the wire breaks at S.
- A: Interchanges the proportional limit and the yield point.
- C: Interchanges the yield point and the peak stress.
- D: Places fracture at the peak, though the wire breaks later.
Remember this
Going up the curve: proportional limit, yield point, ultimate strength, then fracture.
Question 6 · Hard
A light rigid rod hangs from wires A and B of equal length and equal cross-section, and load W is hung at distance x from wire A (drawing not to scale). For what value of x does the rod stay horizontal?
Figure: MedExams
1. A45 cm
2. B18 cm
3. C30 cm
4. D60 cm
Answer and explanation in LearnFree account

Mechanical Properties of Solids

Arrange these points on the stress-strain curve of a ductile metal in the order in which they are reached as the strain increases. A. Fracture point B. Yield point C. Proportional limit D. Point of ultimate tensile strength

In an experiment to find Young's modulus, a wire 2 m long with a cross-section of 1 mm² stretches by 1 mm under a load of 100 N. What value of Young's modulus do these readings give?

A sealed 2 litre sample of a liquid of bulk modulus 2 × 10⁹ Pa experiences a pressure rise of 4 × 10⁶ Pa. By how much does its volume decrease?

Load against extension is plotted for a wire 2 m long with a cross-section of 1 mm². What is the Young's modulus of its material?

Match List-I with List-II for the stress-strain curve of a metal wire shown. List-I (Point on the curve) A. P B. Q C. R D. S List-II (What it marks) I. Fracture point II. Proportional limit III. Ultimate tensile strength IV. Yield point

A light rigid rod hangs from wires A and B of equal length and equal cross-section, and load W is hung at distance x from wire A (drawing not to scale). For what value of x does the rod stay horizontal?