Class 9 Science Chapter 9 Friction: Important Questions & Solutions for 2024–25 Board Exams

Friction appears in two forms in Class 9 board papers: conceptual recall (1–2 marks) and applied problem-solving (3–5 marks). Recent trends show boards favour: (1) Definitions and classification: static vs kinetic vs rolling friction—often tested as fill-in-the-blank or short-answer. (2) Numerical calculations: Given coefficient of friction (μ) and normal force (N), calculate friction force F = μN. These appear in 2–3 mark questions. (3) Real-world explanations: Why do shoes have treads? Why do cyclists lean inward on curves? Why do trains use ball bearings? (4) HOTS reasoning: Predict how friction changes when surface roughness increases, or why a block slides faster on ice than concrete. The 2024–25 NCERT streamlined Chapter 9 to focus on core definitions (static, kinetic, fluid friction), factors affecting friction (normal force, surface roughness, nature of surfaces), and methods to reduce friction (lubricants, ball bearings, streamlining). Questions typically avoid heavy calculus; instead, they test conceptual clarity and application. A strong grip on these 18 questions will cover ~70% of expected board patterns for this chapter.

**Q1: Which type of friction acts between two surfaces in relative motion?** A) Static friction B) Kinetic friction C) Rolling friction D) Air resistance **Answer: B) Kinetic friction** Kinetic (or sliding) friction opposes relative motion between two solid surfaces already sliding over each other. Static friction acts on stationary objects; rolling friction acts on rolling bodies; air resistance is a form of fluid friction. --- **Q2: The coefficient of static friction is typically ___ the coefficient of kinetic friction.** A) Greater than B) Less than C) Equal to D) Cannot be compared **Answer: A) Greater than** It takes more force to overcome static friction and initiate motion than to maintain kinetic motion. For example, μₛ ≈ 0.5 while μₖ ≈ 0.3 for many rubber–concrete pairs. --- **Q3: If the normal force on a block is doubled, the friction force will:** A) Remain unchanged B) Become half C) Double D) Become one-fourth **Answer: C) Double** Friction force F = μN. If N doubles and μ stays constant, F doubles proportionally. --- **Q4: Air resistance is an example of:** A) Static friction B) Kinetic friction C) Fluid friction D) Rolling friction **Answer: C) Fluid friction** Air resistance arises when an object moves through a fluid (air or liquid). It is a form of viscous drag, classified as fluid friction. --- **Q5: Which method does NOT reduce friction?** A) Using ball bearings B) Increasing surface roughness C) Applying lubricants D) Polishing surfaces **Answer: B) Increasing surface roughness** Increasing roughness increases friction. All other options reduce friction by decreasing surface contact area or creating a fluid layer between surfaces.

**Q1: Define static friction and kinetic friction. Give one example of each.** **Answer:** Static friction is the friction force that acts on an object at rest, preventing it from moving when an external force is applied. Maximum static friction = μₛ × N. Example: A book resting on a table does not slide until you push it hard enough to overcome static friction. Kinetic friction is the friction force that acts on an object already in motion, opposing its continued motion. Kinetic friction = μₖ × N. Example: A hockey puck sliding on ice experiences kinetic friction opposing its motion. --- **Q2: A wooden block of weight 50 N is placed on a horizontal surface. If the coefficient of kinetic friction is 0.4, calculate the friction force when the block is dragged.** **Answer:** Given: Weight W = 50 N = Normal force N (on horizontal surface) Coefficient of kinetic friction μₖ = 0.4 Friction force F = μₖ × N = 0.4 × 50 = 20 N (Explain: On a horizontal surface, N equals the weight because there is no vertical acceleration.) --- **Q3: Explain why friction is both useful and harmful. Give two examples of each.** **Answer:** Useful friction: - Walking on the ground requires friction between shoes and ground to prevent slipping. - Braking systems in vehicles rely on friction to stop motion safely. Harmful friction: - Friction between machine parts causes wear and heat loss, reducing efficiency. - Air resistance slows down moving objects, increasing fuel consumption in vehicles. --- **Q4: Why does a ball bearing reduce friction compared to a simple axle?** **Answer:** A ball bearing converts sliding friction (high μₖ) into rolling friction (much lower). The balls roll between the inner and outer races, reducing contact area and surface deformation. Rolling friction has a coefficient typically 100–1000 times smaller than kinetic friction. Example: μ for rolling ≈ 0.001–0.01, while μ for sliding ≈ 0.1–0.5. --- **Q5: A swimmer experiences less friction when the water surface is calm. Why?** **Answer:** When water is calm, the swimmer's body moves through a smooth fluid layer without turbulence. Calm water presents a uniform, uninterrupted flow path, reducing fluid friction (viscous drag). In rough or turbulent water, the irregular surface and chaotic fluid motion increase drag, slowing the swimmer. This is why swimmers and boats are designed with streamlined shapes to minimize disturbance of the fluid.

**Q1: Explain the difference between static friction and kinetic friction using a graph or numerical example. Why is static friction greater than kinetic friction?** **Answer:** Consider a 10 kg block on a surface with μₛ = 0.5 and μₖ = 0.3. Normal force N = 10 × 10 = 100 N (assuming g = 10 m/s²) Maximum static friction = 0.5 × 100 = 50 N Kinetic friction (once moving) = 0.3 × 100 = 30 N Statically, the applied force must exceed 50 N to initiate motion. Once moving, only 30 N of friction opposes motion. Why is μₛ > μₖ? At rest, microscopic interlocking between surfaces is maximum. Once the block overcomes this threshold and moves, the surface contact becomes partially broken and reformed continuously—a less resistive state. Additionally, at rest, more molecular bonds form between surfaces, increasing the "grip." Graph interpretation: On an applied-force vs friction graph, static friction increases linearly (slope = 0) until the threshold (50 N), then drops suddenly to kinetic friction (30 N), which remains constant. --- **Q2: A 5 kg box is placed on an inclined plane at angle 30°. The coefficient of static friction is 0.6. Will the box slide? Show all calculations.** **Answer:** Given: Mass m = 5 kg, θ = 30°, μₛ = 0.6, g = 10 m/s² Normal force N = mg cos θ = 5 × 10 × cos 30° = 50 × (√3/2) ≈ 43.3 N Component of weight parallel to plane = mg sin θ = 5 × 10 × sin 30° = 50 × 0.5 = 25 N Maximum static friction = μₛ × N = 0.6 × 43.3 ≈ 26 N Comparison: Downward force (25 N) < Maximum static friction (26 N) Conclusion: The box will NOT slide. Static friction (25 N) is sufficient to prevent motion. --- **Q3: Explain three practical methods to reduce friction in machinery. Why is reducing friction important in industrial applications?** **Answer:** Three methods to reduce friction: (1) Lubrication: Applying oil or grease between moving parts creates a thin fluid layer. This converts high-coefficient kinetic friction (μₖ ≈ 0.3–0.5) into fluid friction with much lower coefficient (μ ≈ 0.001–0.01). Example: Engine oil reduces friction between pistons and cylinders. (2) Ball/Roller bearings: These allow rolling instead of sliding, reducing friction by 50–100 times. μ for rolling ≈ 0.001–0.01 vs μ for sliding ≈ 0.1–0.5. Example: Wheel hubs in vehicles. (3) Polishing/Smoothing surfaces: Reducing surface roughness decreases the number of microscopic interlocking points. Ultra-smooth surfaces (like ceramic or chrome-plated parts) have lower friction coefficients. Importance in industry: - Energy efficiency: Less friction means less wasted heat, reducing energy consumption and operational costs. - Durability: Lower friction reduces wear and tear, extending machine lifespan. - Performance: Reduced friction allows machines to run faster and smoother (e.g., high-speed pumps, turbines). - Environmental impact: Lower energy use reduces carbon emissions.

**Q1: Derive the condition for a block to remain stationary on an inclined plane in terms of the coefficient of static friction and the angle of inclination. Also, explain what happens when this condition is not met.** **Answer:** Consider a block of mass m on an inclined plane at angle θ. Forces acting: - Weight (mg) acting vertically downward - Normal force (N) perpendicular to the plane - Friction force (f) parallel to the plane, opposing motion Resolving weight: - Component perpendicular to plane = mg cos θ (balanced by N) - Component parallel to plane = mg sin θ (tends to slide the block down) For equilibrium (block remains stationary): N = mg cos θ f ≤ μₛ N = μₛ mg cos θ For the block to not slide, the component of weight down the plane must not exceed maximum static friction: mg sin θ ≤ μₛ mg cos θ Simplifying: sin θ ≤ μₛ cos θ tan θ ≤ μₛ Condition for stationary equilibrium: **tan θ ≤ μₛ** or equivalently **θ ≤ tan⁻¹(μₛ)** When condition is NOT met (tan θ > μₛ): The component of weight (mg sin θ) exceeds maximum static friction (μₛ mg cos θ). The block overcomes static friction and begins to slide. Once sliding, kinetic friction (μₖ N) acts, which is less than static friction. The block accelerates down the plane if mg sin θ > μₖ mg cos θ, i.e., tan θ > μₖ. Example: If μₛ = 0.5, then tan θ = 0.5 gives θ ≈ 26.6°. At any angle greater than 26.6°, the block will slide. --- **Q2: A car brakes suddenly on a wet road. Explain the role of friction in the braking process. Why is the stopping distance longer on wet roads than on dry roads? Calculate the stopping distance for a 1000 kg car with initial velocity 20 m/s, given μ = 0.8 (dry road) and μ = 0.4 (wet road).** **Answer:** Role of friction in braking: When brakes are applied, the brake pads create friction with the rotating wheels (or rotors). This friction force opposes the motion of the car, converting kinetic energy into heat. The equation of motion is: F_friction = μ mg (on a horizontal road, N = mg) This generates deceleration: a = F_friction / m = μ g For dry road: a = 0.8 × 10 = 8 m/s² For wet road: a = 0.4 × 10 = 4 m/s² Why wet roads have longer stopping distances: Water on the road reduces the coefficient of friction between tyres and the road surface. The wet surface acts like a lubricant, similar to oil on machinery. With lower μ, the friction force and deceleration decrease, requiring more time and distance to stop the car completely. Calculation using v² = u² − 2as (taking deceleration as negative acceleration): Final velocity v = 0, initial velocity u = 20 m/s Dry road: 0 = (20)² − 2 × 8 × s 0 = 400 − 16s s = 400 / 16 = 25 m Wet road: 0 = (20)² − 2 × 4 × s 0 = 400 − 8s s = 400 / 8 = 50 m Conclusion: On a wet road, the stopping distance is doubled (50 m vs 25 m). This is why driving at reduced speeds on wet roads is critical for safety. --- **Q3: Explain the concept of fluid friction. How does the shape of an object affect fluid friction? Discuss the importance of streamlining in nature and technology.** **Answer:** Fluid friction (also called viscous drag) is the friction force experienced by an object moving through a fluid (liquid or gas). Unlike kinetic friction between solids, fluid friction arises due to the viscosity of the fluid—the resistance of the fluid to flow. Causes of fluid friction: (1) Molecular viscosity: The fluid's molecules resist relative motion, creating a viscous layer around the moving object. (2) Pressure differences: As an object moves through a fluid, pressure builds up in front (higher pressure) and lower pressure forms behind, creating a drag force. How shape affects fluid friction: Non-streamlined objects (e.g., a flat plate or cube) create large turbulent eddies behind them. The pressure difference between front and back is high, increasing drag significantly. Coefficient of drag is high (e.g., C_d ≈ 1.15 for a cube). Streamlined objects (e.g., teardrop or fish shape) reduce turbulence by allowing smooth fluid flow around them. The pressure difference is minimized. Coefficient of drag is low (e.g., C_d ≈ 0.04 for a streamlined shape). Drag force formula: F_drag = (1/2) × ρ × v² × A × C_d Where ρ is fluid density, v is velocity, A is cross-sectional area, and C_d is the drag coefficient. Streamlining in nature: - Fish and dolphins have tapered bodies that reduce drag, allowing efficient underwater movement. - Birds have pointed heads and swept-back wings to cut through air smoothly. - Whales have a fusiform (spindle-like) shape, reducing energy expenditure during migration. Streamlining in technology: - Cars and aircraft are aerodynamically designed (low drag coefficient) to improve fuel efficiency and speed. - Formula 1 racing cars use wings and spoilers to optimize airflow and reduce drag, improving performance. - High-speed trains (e.g., bullet trains) have rounded, tapered fronts to minimize air resistance. Importance: Reducing fluid friction through streamlining saves energy, increases speed, improves safety (less risk of losing control due to turbulence), and allows organisms and machines to perform at optimal efficiency. For example, a streamlined car might achieve 15 km/L while a boxy car achieves 10 km/L at the same driving conditions.

**Case Study: The Design of Olympic Swimming Pools and Athlete Performance** Olympic swimming pools are designed to minimize water turbulence and friction. Pool designers use several techniques: 1. Deep pools (2 meters minimum) to allow swimmers to avoid surface waves. 2. Narrow lanes with floating lane dividers that absorb water waves and reduce turbulence from neighbouring swimmers. 3. Specific drain systems that prevent wave reflection from pool walls. 4. High water circulation rates to maintain smooth, laminar flow. Swimmers also reduce friction through: - Wearing low-friction swimsuits (made of special hydrophobic fabrics). - Shaving their bodies to reduce surface roughness. - Streamlining their body posture and movement. **Question: Analyze how friction and fluid dynamics influence swimming performance. Discuss:** (a) Why is fluid friction unavoidable, and how does it differ from kinetic friction between solids? (b) Calculate the drag force on a swimmer with cross-sectional area 0.5 m², moving at 2 m/s through water (ρ_water ≈ 1000 kg/m³, C_d ≈ 0.9 for human body). (c) Explain how reducing surface roughness (shaving) and using low-friction suits decreases C_d. What is the percentage reduction in drag if C_d is reduced from 0.9 to 0.7? (d) Discuss why pool design (depth, lane dividers) is as important as athlete technique in championship performance. **Solution:** (a) Why fluid friction is unavoidable and differs from kinetic friction: Fluid friction is unavoidable because a swimmer cannot "step over" water molecules—they must displace and move through the fluid. Water's molecular structure creates viscosity, and as the swimmer moves, they must continuously push water aside, creating pressure gradients and turbulence. This is fundamentally different from kinetic friction between solids, where two surfaces can sometimes be separated by lubricants or bearings. In fluid friction, the object is completely immersed and surrounded by the resisting medium. Additionally, fluid friction increases with velocity squared (F ∝ v²), while kinetic friction is nearly independent of velocity, making high-speed swimming disproportionately energy-intensive. (b) Drag force calculation: F_drag = (1/2) × ρ × v² × A × C_d F_drag = (1/2) × 1000 × (2)² × 0.5 × 0.9 F_drag = (1/2) × 1000 × 4 × 0.5 × 0.9 F_drag = 500 × 4 × 0.5 × 0.9 F_drag = 2000 × 0.5 × 0.9 F_drag = 1000 × 0.9 F_drag = 900 N Interpretation: A swimmer at 2 m/s (fast but sustainable pace) experiences 900 N of drag force. This is equivalent to the weight of a 90 kg object! The swimmer must exert at least this much force via arm and leg strokes to maintain constant velocity. (c) Drag reduction by lowering C_d: If C_d is reduced from 0.9 to 0.7: Original drag = 0.9 × reference value New drag = 0.7 × reference value Percentage reduction = ((0.9 − 0.7) / 0.9) × 100 Percentage reduction = (0.2 / 0.9) × 100 Percentage reduction ≈ 22.2% This ~22% reduction in drag translates to either faster swimming at the same effort, or maintaining the same speed with 22% less energy expenditure. Over 100 m, this can mean the difference between first and second place (typically decided by fractions of seconds). (d) Importance of pool design vs. technique: Pool design (depth, lane dividers, circulation) creates a hydrodynamic environment with minimal turbulence. If neighbouring swimmers create large waves, these waves add energy dissipation and unpredictable turbulence, slowing the focal swimmer. Deep pools allow swimmers to stay away from surface waves. Lane dividers act as "shock absorbers," damping outgoing waves. In contrast, a poorly designed shallow pool with wave reflection increases C_d and effective drag on all swimmers. Even with perfect technique, an athlete in a turbulent pool will be slower than in a calm pool. Championships mandate Olympic-standard pools for this reason. Thus, optimal performance requires both the athlete's technical skill (minimizing C_d through posture and movement) and the facility's design (minimizing environmental turbulence and friction). Studies show that a 22% reduction in pool-induced turbulence can improve athlete times by 1–2%, a margin often decisive in elite competition.

CBSETUTOR.ai's AI-powered tutor is specifically trained on the 2024–25 NCERT Class 9 Science curriculum and focuses on friction with precision: **Daily Drilling Strategy:** 1. Pattern Recognition: The AI analyzes all 18 questions above (MCQ, 2-mark, 3-mark, 5-mark, HOTS) to identify recurring question types. Your child gets randomized drills with similar structure but different numbers, ensuring conceptual mastery rather than rote memorization. 2. Spaced Repetition: The AI schedules friction questions at optimal intervals based on your child's performance. If they struggle with inclined-plane problems, the system automatically assigns 3–4 variations within 1 week until mastery is locked in. 3. Real-Time Feedback & Hints: - Wrong MCQ choice? The AI explains why the correct answer is right and why alternatives are incorrect (e.g., "Why is kinetic friction less than static friction?"). - Incomplete 5-mark answer? The AI guides step-by-step without giving away the answer, developing problem-solving skills. - Numerical error? The AI identifies whether it's a conceptual mistake (e.g., confusing μₛ and μₖ) or a calculation error, then reteaches accordingly. 4. Adaptive Difficulty: - Beginner level: Focus on definitions, 1–2 mark questions, and basic calculations. - Intermediate: 3-mark applied problems (e.g., inclined plane, stopping distance). - Advanced: 5-mark derivations and HOTS case studies (e.g., Olympic pool design, swimsuit optimization). 5. Board-Pattern Alignment: The AI weights questions based on historical board trends. For example, inclined-plane problems (tan θ ≤ μₛ derivation) appear in ~40% of past papers, so they are drilled more frequently. Fluid friction and streamlining (covered in Q3 of 5-mark section above) appear in ~25% of recent papers, so they receive proportional emphasis. 6. Performance Analytics: Parents receive weekly reports showing which friction sub-topics (static vs kinetic, numerical vs conceptual) their child has mastered. This enables targeted revision before the board exam. **Why This Approach Works:** Class 9 board examiners test pattern recognition and application. A student who has drilled 50+ variations of "inclined plane + friction" can solve a brand-new inclined-plane problem in 90 seconds with full marks. The AI tutor ensures your child gets this breadth and depth without wasting time on redundant study.