Force and Pressure – Numerical Problems with Stepwise Solutions

CBSE Class 8 Science – Chapter 11: Force and Pressure | Numerical Problems with Solutions

CBSE Class 8 – Science

Chapter 11: Force and Pressure

Board: Central Board of Secondary Education (CBSE)

Pattern: NCERT-based Numerical Problems with Stepwise Solutions

Use: Practice for CBSE Class 8 School & Board Examination Pattern

Content Bank – Important Formulas from “Force and Pressure”

Force (F) = Push or pull on an object (SI unit: newton, N)
Net Force (same direction) = F₁ + F₂ + …
Net Force (opposite direction) = Larger force − Smaller force
Weight (W) = m × g, where g ≈ 10 N/kg (for Class 8 problems)
Pressure (P) = Force (F) ÷ Area (A)
SI unit of Pressure = pascal (Pa) = 1 N/m²
Thrust = Force acting perpendicular to a surface (unit: newton, N)
Pressure in Fluids increases with depth.
Atmospheric Pressure = Pressure exerted by the air around us (approximately 1.01 × 10⁵ Pa at sea level – concept only for Class 8).

Note: In these numericals, where needed, we take g = 10 N/kg for easy calculation unless stated otherwise.

These 20 numerical problems with stepwise solutions are designed strictly as per the NCERT syllabus for Class 8 Science, Chapter 11: Force and Pressure. They are ideal for practice and revision for CBSE Class 8 examination pattern, helping students strengthen their concepts of force, pressure, thrust and their applications.

A. Force and Net Force

Problem 1 Topic: Magnitude of Force

A box is pushed by a boy with a force of 30 N and pulled by another boy in the same direction with a force of 20 N. What is the total force acting on the box?

Stepwise Solution:

Both forces act in the same direction.
Net force = Sum of the magnitudes of both forces.
Given: F₁ = 30 N, F₂ = 20 N.
Net force, F_net = 30 N + 20 N = 50 N.

Final Answer: The net force acting on the box is 50 N in the direction of the pushes.

Problem 2 Topic: Balanced vs Unbalanced Forces

Two boys pull a rope in opposite directions. One pulls with a force of 70 N and the other with 50 N. Calculate the net force and state whether the forces are balanced or unbalanced.

Stepwise Solution:

The forces act in opposite directions.
Net force = Larger force − Smaller force.
Given: F₁ = 70 N, F₂ = 50 N.
Net force, F_net = 70 N − 50 N = 20 N.
Since the net force is not zero, the forces are unbalanced.

Final Answer: Net force = 20 N in the direction of the 70 N pull; the forces are unbalanced.

Problem 3 Topic: Balanced Forces

A block is pulled towards the right with a force of 40 N and towards the left with a force of 40 N. Find the net force on the block and state its state of motion if it was initially at rest.

Stepwise Solution:

Two equal forces act in opposite directions.
Net force, F_net = 40 N − 40 N = 0 N.
When net force is zero, forces are balanced.
If the block was initially at rest, balanced forces will not change its state of rest.

Final Answer: Net force = 0 N; the block remains at rest as the forces are balanced.

Problem 4 Topic: Multiple Forces

Three forces of 10 N, 15 N and 25 N act on a body in the same direction. What is the resultant force on the body?

Stepwise Solution:

All forces act in the same direction.
Resultant force = Sum of all forces.
F_net = 10 N + 15 N + 25 N.
F_net = 50 N.

Final Answer: The resultant force on the body is 50 N.

B. Pressure, Force and Area

Problem 5 Topic: Basic Pressure Calculation

A force of 100 N acts on an area of 2 m². Calculate the pressure exerted on the surface.

Stepwise Solution:

Formula: P = F ÷ A
Given: F = 100 N, A = 2 m².
P = 100 N ÷ 2 m².
P = 50 N/m² = 50 Pa.

Final Answer: The pressure exerted is 50 Pa.

Problem 6 Topic: Pressure with smaller area

A nail is pressed against a wall with a force of 20 N. The area of its tip is 0.0005 m². Calculate the pressure exerted by the nail on the wall.

Stepwise Solution:

Formula: P = F ÷ A
Given: F = 20 N, A = 0.0005 m².
P = 20 ÷ 0.0005.
0.0005 = 5 × 10⁻⁴, so P = 20 ÷ 0.0005 = 40,000 N/m².

Final Answer: The pressure exerted by the nail is 40,000 Pa.

Problem 7 Topic: Effect of Area on Pressure

A block of weight 200 N rests on a table on its surface of area 0.5 m². If the same block is turned to rest on another face of area 0.25 m², calculate the pressure in both cases. Which case produces greater pressure?

Stepwise Solution:

Pressure formula: P = F ÷ A.
Here, F = 200 N (weight).
Case 1: A₁ = 0.5 m² → P₁ = 200 ÷ 0.5 = 400 Pa.
Case 2: A₂ = 0.25 m² → P₂ = 200 ÷ 0.25 = 800 Pa.
Since P₂ > P₁, pressure is greater when area is smaller.

Final Answer: P₁ = 400 Pa, P₂ = 800 Pa; pressure is greater when the block rests on the 0.25 m² face.

Problem 8 Topic: Comparing Pressure

A girl wearing pencil heels exerts a force of 400 N on the ground. The total area of contact of the heels with the ground is 0.002 m². Calculate the pressure on the ground. How will the pressure change if she wears flat shoes with an area of 0.02 m² (force same)?

Stepwise Solution:

Given: F = 400 N.
Pencil heels: A₁ = 0.002 m².
P₁ = F ÷ A₁ = 400 ÷ 0.002 = 200,000 Pa.
Flat shoes: A₂ = 0.02 m².
P₂ = F ÷ A₂ = 400 ÷ 0.02 = 20,000 Pa.
Pressure decreases when area of contact increases.

Final Answer: With pencil heels, pressure = 200,000 Pa; with flat shoes, pressure = 20,000 Pa. Flat shoes exert much less pressure on the ground.

Problem 9 Topic: Finding Force from Pressure

A pressure of 500 Pa is acting on a surface of area 0.8 m². Calculate the force exerted on the surface.

Stepwise Solution:

Formula: P = F ÷ A ⇒ F = P × A.
Given: P = 500 Pa, A = 0.8 m².
F = 500 × 0.8.
F = 400 N.

Final Answer: The force exerted is 400 N.

C. Thrust and Weight

Problem 10 Topic: Thrust and Pressure

A rectangular wooden block has a weight of 250 N and its base area is 0.5 m². (a) What is the thrust on the table? (b) What pressure does it exert on the table?

Stepwise Solution:

(a) Thrust is the force acting perpendicular to a surface. Here it equals the weight of the block.
Thrust, F = 250 N.
(b) Pressure, P = F ÷ A.
Given A = 0.5 m² ⇒ P = 250 ÷ 0.5 = 500 Pa.

Final Answer: (a) Thrust on the table = 250 N. (b) Pressure on the table = 500 Pa.

Problem 11 Topic: Weight (F = m×g)

The mass of a stone is 4 kg. Taking g = 10 N/kg, calculate the weight of the stone and the pressure it exerts on the ground if the area of contact is 0.02 m².

Stepwise Solution:

Weight formula: W = m × g.
Given: m = 4 kg, g = 10 N/kg.
W = 4 × 10 = 40 N.
Pressure P = W ÷ A.
Given A = 0.02 m², so P = 40 ÷ 0.02 = 2,000 Pa.

Final Answer: Weight of the stone = 40 N; pressure exerted on the ground = 2,000 Pa.

Problem 12 Topic: Comparing Weight and Pressure

Two blocks A and B have equal weights of 300 N each. Block A has a base area of 0.6 m², while block B has a base area of 0.2 m². Calculate the pressure exerted by each block on the floor. Which block will sink more into soft mud?

Stepwise Solution:

Weight is same: F = 300 N for both.
For block A: A₁ = 0.6 m² ⇒ P₁ = F ÷ A₁ = 300 ÷ 0.6 = 500 Pa.
For block B: A₂ = 0.2 m² ⇒ P₂ = F ÷ A₂ = 300 ÷ 0.2 = 1,500 Pa.
Block B exerts higher pressure as its area is smaller.
Higher pressure means it will sink more into soft mud.

Final Answer: Block A pressure = 500 Pa, Block B pressure = 1,500 Pa. Block B will sink more into soft mud.

D. Pressure in Liquids and Applications

Problem 13 Topic: Liquid Pressure with Depth (Conceptual Numeric)

In a water tank, the pressure at a depth of 2 m is 2,000 Pa. What will be the approximate pressure at a depth of 4 m, assuming pressure is directly proportional to depth?

Stepwise Solution:

Given: At depth h₁ = 2 m, pressure P₁ = 2,000 Pa.
If pressure ∝ depth, then P₂ / P₁ = h₂ / h₁.
Let h₂ = 4 m. Then P₂ / 2,000 = 4 / 2 = 2.
P₂ = 2 × 2,000 = 4,000 Pa.

Final Answer: The pressure at a depth of 4 m will be approximately 4,000 Pa.

Problem 14 Topic: Dam Walls and Pressure

The pressure at the bottom of a dam is 50,000 Pa. If the area of the bottom surface in contact with water is 200 m², calculate the total thrust exerted by water on the bottom surface of the dam.

Stepwise Solution:

Thrust, F = P × A.
Given: P = 50,000 Pa, A = 200 m².
F = 50,000 × 200.
F = 10,000,000 N (1.0 × 10⁷ N).

Final Answer: Thrust on the bottom of the dam = 1.0 × 10⁷ N.

Problem 15 Topic: Pressure at Different Depths

At a certain place, the pressure at 1 m depth in a liquid is 900 Pa. (a) What will be the pressure at 3 m depth? (b) By what factor has the pressure increased?

Stepwise Solution:

Pressure ∝ depth in a given liquid.
(a) h₁ = 1 m, P₁ = 900 Pa; h₂ = 3 m.
P₂ / P₁ = h₂ / h₁ = 3 / 1 = 3.
P₂ = 3 × P₁ = 3 × 900 = 2,700 Pa.
(b) Pressure has increased by a factor of 3.

Final Answer: (a) Pressure at 3 m depth = 2,700 Pa. (b) Pressure has increased 3 times.

Problem 16 Topic: Thrust of Liquid on a Surface

A horizontal surface at the bottom of a tank has an area of 3 m². The water above it exerts a pressure of 2,500 Pa. Calculate the thrust exerted by the water on this surface.

Stepwise Solution:

Formula: Thrust F = P × A.
Given: P = 2,500 Pa, A = 3 m².
F = 2,500 × 3.
F = 7,500 N.

Final Answer: Thrust exerted by water on the surface = 7,500 N.

E. Atmospheric Pressure and Applications

Problem 17 Topic: Drinking with a Straw (Conceptual Numeric)

Atmospheric pressure at a place is taken as 1.0 × 10⁵ Pa. If a straw has a cross-sectional area of 1.0 × 10⁻⁴ m², calculate the maximum theoretical force that atmospheric pressure can exert on the liquid inside the straw to push it up (assume vacuum is created inside the straw).

Stepwise Solution:

Force due to pressure: F = P × A.
Given: P = 1.0 × 10⁵ Pa, A = 1.0 × 10⁻⁴ m².
F = (1.0 × 10⁵) × (1.0 × 10⁻⁴).
F = 1.0 × 10¹ N = 10 N.

Final Answer: Maximum theoretical force = 10 N (due to atmospheric pressure on the liquid).

Problem 18 Topic: Force on Surface due to Atmospheric Pressure

A notebook lying on a table has a surface area of 0.06 m² exposed to air. If atmospheric pressure is 1.0 × 10⁵ Pa, calculate the force exerted by air on this surface of the notebook.

Stepwise Solution:

Force due to pressure: F = P × A.
Given: P = 1.0 × 10⁵ Pa, A = 0.06 m².
F = 1.0 × 10⁵ × 0.06.
F = 6,000 N.

Final Answer: Air exerts a force of about 6,000 N on the notebook’s surface.

F. Mixed Concept Numerical Problems

Problem 19 Topic: Pressure and Area – Comparison

A wooden plank of weight 600 N is placed on sand. Case 1: It rests on a face of area 1.5 m². Case 2: It rests on a face of area 0.75 m². Calculate the pressure in each case and state in which case the plank will sink deeper into the sand.

Stepwise Solution:

Given: Weight (force) F = 600 N.
Case 1: A₁ = 1.5 m².
P₁ = F ÷ A₁ = 600 ÷ 1.5 = 400 Pa.
Case 2: A₂ = 0.75 m².
P₂ = F ÷ A₂ = 600 ÷ 0.75 = 800 Pa.
Higher pressure occurs in Case 2, so plank will sink deeper when area is 0.75 m².

Final Answer: Case 1 pressure = 400 Pa, Case 2 pressure = 800 Pa. The plank sinks deeper in Case 2.

Problem 20 Topic: Changing Area and Pressure

A force of 300 N acts on a surface and produces a pressure of 600 Pa. (a) Calculate the area of contact. (b) If the same force now acts on half of this area, what will be the new pressure?

Stepwise Solution:

(a) Use P = F ÷ A ⇒ A = F ÷ P.
Given: F = 300 N, P = 600 Pa.
A = 300 ÷ 600 = 0.5 m².
(b) New area A′ = ½ × 0.5 = 0.25 m².
New pressure P′ = F ÷ A′ = 300 ÷ 0.25 = 1,200 Pa.
By reducing area to half, pressure becomes double.

Final Answer: (a) Area of contact = 0.5 m². (b) New pressure when area is halved = 1,200 Pa.

Force and Pressure – Numerical Problems with Stepwise Solutions

Content Bank – Important Formulas from “Force and Pressure”

A. Force and Net Force

B. Pressure, Force and Area

C. Thrust and Weight

D. Pressure in Liquids and Applications

E. Atmospheric Pressure and Applications

F. Mixed Concept Numerical Problems

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