Class 9 Science Chapter 8: Force and Pressure – 18 Important Questions with Full Solutions

Chapter 8 is mandatory across all CBSE Class 9 Science exam papers. The rationalised 2024–25 syllabus emphasises conceptual clarity over rote learning: examiners test your ability to define force, calculate pressure in different scenarios (solids, liquids, gases), explain atmospheric pressure qualitatively, and apply Archimedes' principle to buoyancy problems. Past board papers (2021–2024) show a consistent pattern: 1–2 marks for MCQ or 1-mark definitions (e.g., 'Define pressure'), 2 marks for short explanations (e.g., 'Why does a needle pierce cloth but a blunt end doesn't?'), 3 marks for pressure numericals (e.g., calculating force given pressure and area), and 5 marks for derivations or multi-step applications (e.g., deriving pressure-depth relationship or explaining why boats float). Mastering these question types ensures you're exam-ready across all weightages. The four core sub-topics—force (push/pull), pressure (definition and calculation), atmospheric pressure (existence and measurement), and buoyancy (Archimedes' principle, flotation)—appear in 60–70% of Class 9 Science papers. This guide isolates the most-asked questions so you don't waste time on low-probability content.

**Question 1:** Pressure is defined as: (a) Force divided by time (b) Force per unit area (c) Force multiplied by distance (d) Force divided by mass **Answer:** (b) Force per unit area **Explanation:** Pressure P = F / A, where F is force (in Newtons) and A is area (in m²). This is the NCERT definition. The SI unit is Pascal (Pa) or N/m². --- **Question 2:** Which of the following has the greatest pressure on the ground? (a) A woman standing on both feet (b) A woman standing on one foot (c) A woman lying flat (d) A woman standing on her hands **Answer:** (d) A woman standing on her hands **Explanation:** Pressure P = F / A. When standing on hands, the contact area A is smallest, so pressure is maximum (same weight, much less area). --- **Question 3:** The SI unit of pressure is: (a) Newton (b) Pascal (c) Joule (d) Bar **Answer:** (b) Pascal **Explanation:** 1 Pascal (Pa) = 1 N/m². Bar and atm are non-SI units. Newton is the unit of force. --- **Question 4:** Atmospheric pressure at sea level is approximately: (a) 76 Pa (b) 760 Pa (c) 101,325 Pa (≈ 1 atm or 1 bar) (d) 1000 Pa **Answer:** (c) 101,325 Pa (≈ 1 atm or 1 bar) **Explanation:** This is the standard atmospheric pressure that supports a 76 cm mercury column in a barometer. --- **Question 5:** According to Archimedes' principle, a body immersed in a fluid experiences: (a) An upward force equal to the volume of fluid displaced (b) An upward force equal to the weight of fluid displaced (c) A downward force equal to the weight of the body (d) No net force **Answer:** (b) An upward force equal to the weight of fluid displaced **Explanation:** This is the definition of buoyancy. The buoyant force = weight of displaced fluid = ρ_fluid × g × V_submerged (from NCERT).

**Question 1:** Distinguish between force and pressure. Give one example each. **Answer:** Force is a push or pull acting on an object; it changes the object's state of motion or shape. Pressure is the force exerted per unit area. *Definitions:* Force (F) is measured in Newtons (N); pressure (P) in Pascals (Pa). Pressure = Force / Area. *Examples:* - Force: A hand pushing a door opens it by applying force. - Pressure: A needle pierces cloth because its sharp tip has a tiny contact area, producing very high pressure despite low force. --- **Question 2:** Why does a camel have broad, soft feet? **Answer:** Camels have broad, soft feet to reduce pressure on sand. Since Pressure = Force / Area, spreading the same weight (force) over a larger area decreases pressure. This prevents the camel from sinking into soft desert sand, allowing it to walk comfortably. --- **Question 3:** What is atmospheric pressure? How was it measured historically? **Answer:** Atmospheric pressure is the weight of the air column above a given area pressing downward due to Earth's gravity. Historically, Torricelli measured it using a barometer (1643). He inverted a mercury-filled tube into a dish of mercury; the mercury column rose to ~76 cm because atmospheric pressure balanced the weight of the mercury column. This height (760 mm Hg) equals 101,325 Pa at sea level. --- **Question 4:** A 50 N force acts on an area of 2 m². Calculate the pressure produced. **Answer:** Given: Force (F) = 50 N, Area (A) = 2 m² Pressure = F / A = 50 / 2 = 25 Pa --- **Question 5:** State Archimedes' principle in one sentence. **Answer:** When a body is wholly or partially immersed in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced by it.

**Question 1:** A cube of side 10 cm and mass 500 g is placed on a horizontal surface. Calculate the pressure exerted by the cube. **Solution:** Given: Side of cube = 10 cm = 0.1 m, Mass = 500 g = 0.5 kg, g = 10 m/s² Force (Weight) = m × g = 0.5 × 10 = 5 N Contact area = side² = (0.1)² = 0.01 m² Pressure = F / A = 5 / 0.01 = 500 Pa --- **Question 2:** Why does pressure increase with depth in a liquid? Explain with a formula. **Solution:** Pressure increases with depth because liquid molecules above press down due to gravity, adding their weight to layers below. At depth h below the surface: Pressure (P) = Atmospheric pressure (P₀) + Pressure due to liquid column P = P₀ + ρgh Where ρ = density of liquid (kg/m³), g = 10 m/s², h = depth (m) As h increases, ρgh increases, so total pressure increases. This is why deep-sea divers experience extreme pressure. --- **Question 3:** An object of mass 2 kg is immersed completely in water (density 1000 kg/m³). If the object's volume is 0.0025 m³, calculate the buoyant force. (g = 10 m/s²) **Solution:** Given: Volume = 0.0025 m³, ρ_water = 1000 kg/m³, g = 10 m/s² Buoyant force = Weight of water displaced = ρ_water × g × V Buoyant force = 1000 × 10 × 0.0025 = 25 N Weight of object = m × g = 2 × 10 = 20 N Since buoyant force (25 N) > weight (20 N), the object will float. --- **Question 4:** Explain why a ship made of steel floats despite steel being denser than water. **Solution:** A ship floats because its overall density is less than water. Although steel is denser than water, the ship is hollow and contains air. The average density of the ship (including air inside) is less than water's density (1000 kg/m³). By Archimedes' principle, the buoyant force = weight of water displaced. When the volume of water displaced is large enough, the weight of displaced water equals the ship's weight, and the ship floats in equilibrium. The hollow design maximises water displacement for a given steel mass.

**Question 1:** Derive the relationship between pressure and depth in a liquid. State any assumptions and give a real-world application. **Solution:** *Derivation:* Consider a liquid at rest in a container. Imagine a cylinder of liquid of height h and cross-sectional area A at depth d below the surface. Forces on this cylinder: - Downward: atmospheric pressure (P₀) + weight of liquid above = P₀ × A + (ρ × g × h × A) - Upward: pressure at depth d, say P_d, acting on area A For equilibrium (liquid at rest): P_d × A = P₀ × A + ρ × g × h × A Dividing by A: P_d = P₀ + ρgh This is the pressure-depth relationship. *Assumptions:* 1. Liquid is incompressible (density ρ is constant) 2. Liquid is in equilibrium (no flow) 3. Gravity (g) is uniform *Application:* Deep-sea diving: At 100 m depth in ocean water (ρ = 1025 kg/m³): P = 101,325 + (1025 × 10 × 100) = 101,325 + 1,025,000 ≈ 1,126,325 Pa ≈ 11 atm Divers must use special equipment and decompress slowly to avoid decompression sickness. --- **Question 2:** Define Archimedes' principle. Derive the formula for buoyant force and explain why objects sink or float. **Solution:** *Principle:* When a body is immersed in a fluid, it experiences an upward force equal to the weight of the fluid displaced. *Derivation:* Consider a solid object of volume V submerged in a fluid of density ρ_f. The object displaces a volume V of fluid. Mass of fluid displaced = ρ_f × V Weight of fluid displaced = ρ_f × V × g By Archimedes' principle, buoyant force (F_b) = ρ_f × V × g (upward) *Floating and sinking:* Compare buoyant force with weight of object: - If ρ_object × V × g < ρ_f × V × g (i.e., ρ_object < ρ_f): Object floats (net upward force) - If ρ_object × V × g = ρ_f × V × g (i.e., ρ_object = ρ_f): Object remains suspended - If ρ_object × V × g > ρ_f × V × g (i.e., ρ_object > ρ_f): Object sinks (net downward force) *Example:* Ice (ρ ≈ 917 kg/m³) floats in water (ρ = 1000 kg/m³) because ice is less dense. Wood also floats for the same reason. --- **Question 3:** A rectangular block of wood (mass 5 kg, volume 0.01 m³) is placed in water. Calculate whether it floats or sinks. If it floats, find what fraction is submerged. (ρ_water = 1000 kg/m³, g = 10 m/s²) **Solution:** *Step 1: Find density of wood* ρ_wood = mass / volume = 5 / 0.01 = 500 kg/m³ *Step 2: Compare with water* Since ρ_wood (500) < ρ_water (1000), the wood floats. *Step 3: Calculate fraction submerged* At floating equilibrium: Weight of wood = Buoyant force m × g = ρ_water × V_submerged × g 5 = 1000 × V_submerged V_submerged = 5 / 1000 = 0.005 m³ Fraction submerged = V_submerged / V_total = 0.005 / 0.01 = 0.5 = 1/2 or 50% *Answer:* The wood floats, with half its volume submerged (1/2 or 50%).

**Case Study:** An underwater research station is being built 500 m below sea level. Engineers must design the station to withstand extreme pressure. The station has a cylindrical hull with a diameter of 10 m. (i) Calculate the total pressure acting at 500 m depth. (Use ρ_seawater = 1025 kg/m³, atmospheric pressure = 101,325 Pa, g = 10 m/s²) (ii) Find the total force acting on the flat top surface of the cylinder due to water pressure alone. (iii) Why would a hollow, spherical design be better than a cylindrical one for extreme depths? Explain in terms of pressure distribution. **Solution:** **(i) Total Pressure at 500 m depth:** P = P₀ + ρgh P = 101,325 + (1025 × 10 × 500) P = 101,325 + 5,125,000 P = 5,226,325 Pa ≈ 51.6 atm **(ii) Force on top surface:** Diameter = 10 m, so radius = 5 m Area = πr² = π × (5)² = 78.54 m² Force = Pressure × Area = 5,226,325 × 78.54 ≈ 410,557,500 N ≈ 4.1 × 10⁸ N This is enormous (~41 million kg-force), requiring massive structural engineering. **(iii) Why a spherical design is better:** In a cylinder, pressure acts perpendicular to all surfaces. The flat top and bottom experience high concentrated loads, creating stress points. In a sphere, pressure is distributed uniformly across the curved surface. The stress is evenly spread, making the structure more stable. A sphere also has the minimum surface area for a given volume, reducing total force. Additionally, there are no flat surfaces where stress concentrates. This is why submarines and deep-sea vessels use spherical pressure hulls.

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