About Properties of Matter in Matric Physics
Properties of Matter is one of the 16 main chapters in our Federal Board Matric Physics question bank (9th class), and about 6% of its real questions come from it. Tayyari Karo has 24 real questions on this chapter from 2023 to 2026; 2 are multiple choice and marked instantly, and the rest are short and long questions with a model answer. By difficulty: 13 easy, 11 medium.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
Try 8 real questions
These are real past-paper questions. Think of your answer first, then open the explanation.
2024 Annual · 9th · Properties of Matter
Density of a substance can be calculated by using:
- Archimedes principle
- Pascal's law
- Hooke's law
- Newton's law
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Answer: A. Archimedes principle
Weighing an object in air and in water (Archimedes' principle) gives its density.
2023 Annual · 9th · Properties of Matter
Which of the following is NOT a unit of pressure?
- Pa
- N m⁻²
- atm
- kg m⁻³
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Answer: D. kg m⁻³
kg m⁻³ is the unit of density. Pa, N m⁻² and atmosphere (atm) are all units of pressure.
2026 Annual · 9th · Properties of Matter
What is meant by density? Also write its formula and unit.
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Density is the mass of a substance per unit volume. It tells how compact the matter is.
Formula: ρ = m/V
SI unit: kg m⁻³ (also g cm⁻³; 1 g cm⁻³ = 1000 kg m⁻³).
2025 1st Annual · 9th · Properties of Matter
Briefly describe any three applications of liquid barometer.
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1. Weather forecasting — a falling barometer reading indicates low pressure (rain/storm), a rising reading indicates fair weather.
2. Measuring altitude — atmospheric pressure decreases with height, so a barometer (altimeter) tells the height of an aircraft or a mountaineer.
3. Laboratory/industrial use — measuring atmospheric pressure precisely for experiments (e.g. boiling‑point corrections) and in scientific stations.
2025 1st Annual · 9th · Properties of Matter
Calculate the force required to lift a car weighing 2000 kg by a hydraulic jack whose small piston has an area of 4 cm² whereas its large piston has an area of 80 cm².
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Weight of car (load on large piston) F₂ = mg = 2000 × 10 = 20,000 N
A₁ = 4 cm², A₂ = 80 cm² (units cancel in the ratio)
By Pascal's principle: F₁/A₁ = F₂/A₂
F₁ = F₂ × A₁/A₂ = 20,000 × 4/80 = 1000 N
So a force of only 1000 N (one‑twentieth of the car's weight) on the small piston lifts the car.
2025 2nd Annual · 9th · Properties of Matter
Derive formula for liquid pressure.
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Consider a column of liquid of density ρ, height h and base area A.
Volume V = Ah; mass m = ρAh; weight w = ρAhg.
Pressure at the base = force/area = w/A = ρAhg/A
P = ρgh — liquid pressure depends on depth and density only.
2025 2nd Annual · 9th · Properties of Matter
Calculate the force required to lift a car weighing 2500 kg by a hydraulic jack whose small piston has an area of 6 cm² whereas its large piston has an area of 90 cm².
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Load on large piston F₂ = mg = 2500 × 10 = 25,000 N
A₁ = 6 cm², A₂ = 90 cm²
Pascal's principle: F₁/A₁ = F₂/A₂ ⇒ F₁ = F₂ × A₁/A₂ = 25,000 × 6/90 = 1667 N ≈ 1.67 × 10³ N
2024 Annual · 9th · Properties of Matter
Explain that liquids exert pressure. Show that liquid pressure depends on depth and density of liquid.
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A liquid has weight, so it presses on the bottom and walls of its container and on any object immersed in it: liquids exert pressure. Water spurts out of holes in a can, and the deeper the hole the faster the jet.
Derivation: Consider a cylindrical column of liquid of density ρ, height h and base area A inside the liquid.
Volume of the column V = A h; mass m = ρV = ρAh; weight w = mg = ρAhg.
This weight acts on the base of area A, so the pressure at depth h is
P = w/A = ρAhg / A ⇒ P = ρgh
Hence liquid pressure is directly proportional to the depth h and to the density ρ of the liquid (g being constant). It does not depend on the shape or area of the container. …
The full model answer is in the app.