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ICSE Class 8 Physics Question 11 of 24

Force and Pressure — Question 13

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Question 13

Figure below shows a brick of weight 2 kgf and dimensions 20 cm x 10 cm x 5 cm placed in three different positions on the ground. Find the pressure exerted by the brick in each case.

Figure below shows a brick of weight 2 kgf and dimensions 20 cm x 10 cm x 5 cm placed in three different positions on the ground. Find the pressure exerted by the brick in each case. Force and Pressure, Concise Physics Solutions ICSE Class 8.
Answer

First case :

Thrust = Weight = 2 kgf
Area = 20 cm x 10 cm = 200 cm2
Pressure = ?

Pressure=ThrustArea=2200=0.01 kgf cm−2\text{Pressure}= \dfrac{\text{Thrust}}{\text{Area}} \\[1em] = \dfrac{\text{2}}{\text{200}} \\[1em] = 0.01 \text{ kgf cm}^{-2}

So Pressure exerted by brick in first figure is 0.01 kgf cm-2.

Second case :

Thrust = Weight = 2 kgf
Area = 5 cm x 10 cm = 50 cm2
Pressure = ?

Pressure=ThrustArea=250=0.04 kgf cm−2\text{Pressure}= \dfrac{\text{Thrust}}{\text{Area}} \\[1em] = \dfrac{\text{2}}{\text{50}} \\[1em] = 0.04 \text{ kgf cm}^{-2}

So Pressure exerted by brick in second figure is 0.04 kgf cm-2.

Third case :

Thrust = Weight = 2 kgf
Area = 20 cm x 5 cm = 100 cm2
Pressure = ?

Pressure=ThrustArea=2100=0.02 kgf cm−2\text{Pressure}= \dfrac{\text{Thrust}}{\text{Area}} \\[1em] = \dfrac{\text{2}}{\text{100}} \\[1em] = 0.02 \text{ kgf cm}^{-2}

So, Pressure exerted by brick in third figure is 0.02 kgf cm-2.

Chapter Overview: Force and Pressure

This chapter covers the fundamental concepts of force and pressure for ICSE Class VIII. A force is a push or pull that can change the speed, direction, or shape of an object. Forces are classified as contact forces (muscular, friction, normal) and non-contact forces (gravitational, magnetic, electrostatic). Pressure is defined as force per unit area (P = F/A), measured in pascals (Pa). Students learn that smaller area produces greater pressure (sharp knife, drawing pin) and larger area reduces pressure (wide tyres, camel feet). The chapter also covers liquid pressure (P = hρg), atmospheric pressure measured by a barometer (76 cm Hg = 1 atm = 101,325 Pa), and Pascal’s law which states that pressure in an enclosed liquid is transmitted equally. Hydraulic machines (press, brakes, jack) work on Pascal’s law to multiply force.

Key Definitions & Concepts

Term Definition / Details
ForceA push or pull that changes the state of motion or shape of an object. Unit: newton (N)
PressureForce per unit area. P = F/A. Unit: pascal (Pa) = N/m²
Contact ForceForce requiring physical contact: muscular, friction, normal reaction
Non-Contact ForceForce acting at a distance: gravitational, magnetic, electrostatic
Atmospheric PressurePressure exerted by the atmosphere. Standard: 760 mm Hg = 1 atm
Pascal’s LawPressure in an enclosed liquid is transmitted equally and undiminished in all directions
Hydraulic MachineDevice using Pascal’s law to multiply force. F1/A1 = F2/A2

Must-Know Points for Exams

  • Force is measured in newtons (N). 1 N is the force needed to accelerate 1 kg by 1 m/s².
  • Pressure = Force / Area. More area means less pressure for the same force.
  • Liquid pressure increases with depth: P = h × ρ × g.
  • Atmospheric pressure = 76 cm Hg at sea level.
  • Pascal’s law allows small forces to produce large forces in hydraulic machines.
  • Examples of high pressure: sharp knife, needle, nail. Low pressure: wide tyres, snowshoes, camel feet.

Quick Self-Test

  1. Define the main concept of this chapter in one sentence.
  2. List the key types or categories discussed in this chapter.
  3. Give three real-life examples related to the main concept.
  4. Draw and label the key diagram of this chapter from memory.
  5. State the main law or principle covered in this chapter.