NCERT Exercise - Q8
Textbook Exercise Problem
A force of 100 N is applied to an object of an area 2 m². Calculate the pressure exerted on the object.
Step-by-Step Verified Solution:
- Step 1: Identify given parameters: Force F = 100 N, Area A = 2 m².
- Step 2: Recall pressure formula: P = F / A.
- Step 3: Substitute values: P = 100 / 2 = 50 N/m² (or 50 Pa).
Final Answer: 50 Pascal (Pa)
Key Concept / Exam Strategy: Pressure is force acting perpendicularly per unit surface area.
Generated Exercise #1
Foundation
Question: Name two contact forces and two non-contact forces acting in nature.
Core Concept Tested: Physical categorization of force interactions
Mentor Hint: Contact forces require physical touch.
Step-by-Step Solution:
- Step 1: Contact Forces: (i) Muscular force, (ii) Frictional force.
- Step 2: Non-contact Forces: (i) Gravitational force, (ii) Magnetic force (or Electrostatic force).
Verified Answer: Contact: Muscular, Friction; Non-contact: Gravitational, Magnetic
Generated Exercise #2
Intermediate
Question: Why do porters place a round piece of cloth on their heads when carrying heavy loads?
Core Concept Tested: Pressure minimization via contact area expansion
Mentor Hint: P = F / A. Increasing A decreases P.
Step-by-Step Solution:
- Step 1: The round cloth increases the contact area between the heavy load and the porter head.
- Step 2: Since Pressure P = Force / Area, an increase in area significantly reduces the pressure exerted on the skull.
- Step 3: This makes carrying the load easier and painless.
Verified Answer: Increases contact surface area, thereby reducing the pressure exerted on the head.
Generated Exercise #3
Intermediate
Question: A rectangular brick of dimensions 20 cm × 10 cm × 5 cm has weight 40 N. Calculate the maximum and minimum pressure it can exert on a tabletop.
Core Concept Tested: Geometric surface area variation in pressure generation
Mentor Hint: Max pressure on smallest face (10×5 cm); Min pressure on largest face (20×10 cm).
Step-by-Step Solution:
- Step 1: Smallest area = 0.10 m × 0.05 m = 0.005 m².
- Step 2: P_max = 40 / 0.005 = 8,000 Pa.
- Step 3: Largest area = 0.20 m × 0.10 m = 0.02 m².
- Step 4: P_min = 40 / 0.02 = 2,000 Pa.
Verified Answer: Maximum Pressure = 8,000 Pa; Minimum Pressure = 2,000 Pa
Generated Exercise #4
JEE-NEET Foundation
Question: Explain why water spurts out with greater speed from a hole near the bottom of a container compared to a hole near the top.
Core Concept Tested: Hydrostatic pressure dependency on fluid column depth
Mentor Hint: P = ρgh. Depth h is measured from the liquid surface.
Step-by-Step Solution:
- Step 1: Liquid pressure increases directly with column depth (h): P = ρgh.
- Step 2: The bottom hole experiences the greatest hydrostatic head and pressure.
- Step 3: Greater pressure drives water out with higher velocity (Torricelli theorem).
Verified Answer: Because fluid hydrostatic pressure is directly proportional to depth (P = ρgh).
Generated Exercise #5
Advanced HOTS
Question: Why are our bodies not crushed under the immense weight of atmospheric pressure (~101,325 Pa)?
Core Concept Tested: Equilibrium between internal body fluid pressure and external atmospheric pressure
Mentor Hint: Blood pressure and internal fluid pressures balance atmospheric force.
Step-by-Step Solution:
- Step 1: The pressure of fluids (blood, intracellular fluids) inside our body is approximately equal to atmospheric pressure.
- Step 2: This internal pressure pushes outward with the exact same magnitude as atmospheric pressure pushing inward.
- Step 3: Net unbalanced force is zero, preventing collapse.
Verified Answer: Internal blood and body fluid pressure equals and balances atmospheric pressure from the outside.