📘 Chapter 6: Mechanical Properties of Matter – Constructed Response Questions
Prepared by Muhammad Tayyab, Subject Specialist Physics, Govt Christian High School Daska. Based on PECTAA 2026 syllabus (National Curriculum 2023).
📖 What's Inside: This section covers constructed response questions from Chapter 6 Mechanical Properties of Matter including Spring Combinations, Elasticity, Pressure, Pascal's Law, Barometer, Density, and Load Estimation. Each question is presented with a detailed answer as per the official PECTAA 2026 Physics curriculum. Perfect for Punjab Boards (Lahore, Gujranwala, Multan, etc.) and all BISE boards across Pakistan.
📚 Related Resources – Chapter 6: Mechanical Properties of Matter
📋 Exercise Short Qs
📝 Short Qs
📖 Comprehensive Qs
🔢 Numerical Problems
✅ MCQs
✏️ Constructed Response
Mechanical Properties of Matter covers elasticity, pressure, Hooke's law, Pascal's law, and atmospheric pressure.
📖 Constructed Response Questions & Answers (PECTAA 2026)
6.1 A spring having spring constant \(k\) hangs vertically from a fixed point. A load of weight \(L\) when hung from the spring, causes an extension \(x\). The elastic limit of the spring is not exceeded. Some identical springs, each with spring constant \(k\) are arranged as shown below. For each arrangement, complete the table by determining: (i) the total extension in terms of \(x\) (ii) the spring constant in terms of \(k\).
Completed Table:
| Arrangement | Total Extension | Spring Constant |
|---|---|---|
| Series (Top Row) | \(2x\) | \(\frac{k}{2}\) |
| Parallel (Bottom Row) | \(\frac{x}{2}\) | \(2k\) |
Step-by-Step Calculation / Derivation:
1. Series Arrangement (Top Row)
- Extension: In a series combination, the full load \(L\) acts on both springs.
- Extension in top spring = \(x\)
- Extension in bottom spring = \(x\)
- Total Extension = \(x + x = 2x\)
- Equivalent Spring Constant: Using Hooke's law \(\left(k = \frac{\text{Force}}{\text{Extension}} = \frac{L}{x}\right)\)
- \(k_{\text{new}} = \frac{L}{2x}\)
- \(k_{\text{new}} = \frac{1}{2}\left(\frac{L}{x}\right)\)
- \(k_{\text{new}} = \frac{1}{2}k\)
- \(k_{\text{new}} = \frac{k}{2}\)
2. Parallel Arrangement (Bottom Row)
- Extension: In a parallel combination, the load \(L\) is shared equally between the two springs. Each spring carries a load of \(\frac{L}{2}\). Since extension is proportional to load \((x \propto L)\):
- Total Extension = \(\frac{x}{2}\)
- Equivalent Spring Constant: Using Hooke's law \(\left(k = \frac{\text{Force}}{\text{Extension}} = \frac{L}{x}\right)\):
- \(k_{\text{new}} = \frac{L}{x/2}\)
- \(k_{\text{new}} = \frac{2L}{x}\)
- \(k_{\text{new}} = 2\left(\frac{L}{x}\right)\)
- \(k_{\text{new}} = 2k\)
✅ Series: \(2x\), \(\frac{k}{2}\) | Parallel: \(\frac{x}{2}\), \(2k\)
6.2 Springs are made of steel instead of iron. Why?
Answer:
Springs are made of steel instead of iron because steel is more elastic than iron and has a higher value of spring constant \((k)\). It requires a larger force to produce permanent deformation and regains its original shape effectively within the elastic limit.
✅ Steel is more elastic than iron → higher spring constant
6.3 Which of the following material is more elastic? (a) Iron or rubber (b) Air or water
Answer:
(a) Iron: Iron is more elastic than rubber because a larger deforming force is required to produce the same change in shape or length in iron than in rubber.
(b) Water: Water is more elastic than air because water offers much higher resistance to compression than air.
✅ Iron > Rubber | Water > Air
6.4 How does water pressure one metre below the surface of a swimming pool compare to water pressure one metre below the surface of a very large and deep lake?
Answer:
The water pressure at one metre below the surface is the same in both cases because pressure depends only on the depth and density of the liquid.
\[ P = \rho gh \]
Since the depth and density of water are the same in both cases, the pressure will be the same.
✅ Pressure is the same at the same depth in any body of water
6.5 What will happen to the pressure in all parts of a confined liquid if pressure is increased in one part? Give an example from your daily life where such principle is applied.
Answer:
By Pascal's law, any pressure applied at one point in an enclosed liquid is transmitted equally in all directions.
Example: Hydraulic brakes in vehicles. The pressure applied to the brake fluid is transmitted equally to the pistons at the wheels.
✅ Pressure is transmitted equally throughout the liquid (Pascal's Law)
6.6 If some air remains trapped within the top of the mercury column of the barometer which is supposed to be vacuum, how would it affect the height of the mercury column?
Answer:
The trapped air would exert its own pressure on the mercury column, pushing it down. This would cause the height of the mercury column to be less than the true atmospheric pressure reading.
✅ Trapped air reduces the mercury column height → lower reading
6.7 How does the long neck is not a problem to a giraffe while raising its neck suddenly?
Answer:
The giraffe's circulatory system has special adaptations. Valves in the blood vessels of the neck prevent blood from rushing to the head when the neck is raised suddenly, maintaining proper blood pressure.
✅ Valves prevent blood from rushing to the head
6.8 The end of glass tube used in a simple barometer is not properly sealed, some leak is present. What will be its effect?
Answer:
The leak would allow air to enter the tube. The air pressure inside the tube would push the mercury down, resulting in a lower reading than the actual atmospheric pressure.
✅ Leak allows air in → mercury pushed down → lower reading
6.9 Comment on the statement: "Density is a property of a material not the property of an object made of that material."
Answer:
Density is a property of the material itself (mass per unit volume) and does not depend on the size or shape of the object. An object made of a material will have the same density as the material regardless of its size.
\[ \rho = \frac{m}{V} \]
✅ Density is an intensive property of the material
6.10 How the load of a large structure is estimated by an engineer?
Answer:
Engineers estimate the load of a large structure by determining the density of the construction materials. Using the density and volume of each material, its mass and weight can be calculated. This helps engineers estimate the strength required in foundations and supporting pillars.
\[ \text{Mass} = \text{Density} \times \text{Volume} \]
\[ \text{Weight} = \text{Mass} \times g \]
✅ Load = Density × Volume × g
📐 Key Concepts – Mechanical Properties of Matter (Constructed Response Questions)
Hooke's Law: \(F = kx\)
Pressure: \(P = \frac{F}{A}\)
Liquid Pressure: \(P = \rho gh\)
Density: \(\rho = \frac{m}{V}\)
Series Spring: \(k_{\text{eq}} = \frac{k}{2}\), \(x_{\text{total}} = 2x\)
Parallel Spring: \(k_{\text{eq}} = 2k\), \(x_{\text{total}} = \frac{x}{2}\)
💡 Exam Tip:
For constructed response questions, write detailed answers with clear reasoning, derivations, and calculations. Show all steps in your working. These questions test your analytical and problem-solving skills. These questions follow the PECTAA 2026 pattern and are prepared by Subject Specialist Muhammad Tayyab.
Chapter 1Physical Quantities and Measurements
Chapter 2Kinematics
Chapter 3Dynamics
Chapter 4Turning Effect of Forces
Chapter 5Work and Energy
Chapter 6✨ Mechanical Properties of Matter
Chapter 7Thermal Properties of Matter
Chapter 8Magnetism
Chapter 9Nature of Science
📖 Complete syllabus coverage for Class 9 Physics (PECTAA 2026) – Chapters 1 to 9
Created by Hira Science Academy | Aligned with PECTAA 2026 Syllabus