9th Class Physics Chapter 4 Turning Effects of Force Notes, MCQs, Numericals, Solved Exercise 

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Here you will get 9th Class Physics solved mcqs of the 4th Chapter Turning Effects of Force MCQs Notes and Question Answers, Numericals from Textbook Exercise.

We have created a separate category under name; "9th Class Physics Notes". In this category, We only shares 9th Class Physics Notes, Class 9 Chapter-wise solved MCQs and Question Answers from Sindh Board Textbook Exercise.

Chapter Four Turning Effects of Force: 9th Class Physics MCQs with Answers

While preparing for the board exams, the students of Matric and Intermediate usually focuses on theory and numericals of Physics. 

But, MCQs plays vital role in improving the grades in board exams.

Therefore, We have shared Chapter-wise MCQs notes along with the theory and numericals here.

Here are the 9th class Physics 4th chapter Turning Effects of Force solved MCQs with Answers. 

This is also helpful for the students who are going to appear in board exams this year. 

01. A pair of opposing parallel forces with distinct lines of force result in _______. (a) equilibrium (b) torque (c) a couple (d) unstable equilibrium

02. The number of perpendicular components of a force is:

(a) 1 (b) 3 (c) 4 (d) 2

03. Racing cars are made stable by:

A. Increasing their speed

B. Decrease their mass

C. Lowering their centre of gravity

D. Decreasing their width

04. Head to tail rule can be used to add ______ forces.

(a) two

(b) three

(c) five

(d) any number of

05. To make a body more stable ________ should be lowered:

A. Axis of rotation

B. Center of gravity

C. Stability

D. None

06. The vector that has the same result as the result of all the vectors combined is:

A. Negative vector

B. The vector

C. Resultant vector

D. Zero vector

07. A 15 N force creates a 60° angle with the horizontal. Its vertical element will consist of:

(a) 15 N (b) 10 N (c) 13 N (d) 7 N

08. Gravitational constant depends on:

(a) Medium

(b) Force

(c) Weight

(d) It does not depend on anything

09. What is the orbital speed of a low-orbit satellite?

(a) Zero

(b) 8 ms-1

(c) 800 ms-1

(d) 8000 ms-1

10. A body is in equilibrium when it has:

(a) Uniform Speed

(b) Uniform Acceleration

(c) Both A and B

(d) Zero Acceleration

11. To form a couple, two forces must be

(a) Equal in magnitude (b) Parallel, but opposite in direction (c) Separated by a distance (d) All of these

12. A __________ consist of two forces that are equal in magnitude, but opposite in direction.

(a) pivot (b) net force (c) couple (d) centre of mass

13. After a minor tilt, a body is in stable balance if its centre of gravity:

(a) remains above the point of contact

(b) remains on one side of the point of contact

(c) passes over the point of contact

(d) is at the lowest position

14. The shortest distance between two couple forces is called:

(a) Moment arm (b) Couple arm (c) Radius (d) Double moment

15. Torque depends upon:

(a) The magnitude of the velocity and fulcrum

(b) The mass & fulcrum

(c) the direction of the force & fulcrum

(d) The magnitude of the force & fulcrum

16. A body's centre of gravity indicates neutral equilibrium when: (a) is at the lowest position (b) remains at the same height (c) is at the highest position (d) is at its base

17. What is the S.I unit of the Torque and moment arm:

(a) Newton (b) Newton per second (c) Newton-second (d) Newton-metre

18. If the body's condition of motion remains unchanged, it is considered to be in:

A. Rest

B. Motion

C. Equilibrium

19. The stability of a racing car is increased by:

(a) increasing its height

(b) raising its centre of gravity

(c) decreasing its width

(d) lowering its centre of gravity

20. The velocity at which a body is travelling upward from the earth's centre.

(a) Increases

(b) Decrease

(c) Remain constant

(d) None of these

21. Conventionally, a clock wise torque is taken _______

(a) Positive (b) Zero (c) Negative (d) Parallel

22. sin θ = : (a) base/hypotenuse (b) perpendicular / hypotenuse (c) base/perpendicular (d) perpendicular s/ base

23. One Earth's radius above the planet's surface, the value of g is:

(a) 2g (b) 1/2 g (c) 1/3 g (d) 1/4 g

24. If we increase the length of the spanner the torque produced will be:

A. Increased B. Decreased C. Constant D. Zero

25. Moment of force about a point is equal to: (a) 𝜏 = d x F (b) 𝜏 = F x d (c) 𝜏 = d. F (d) 𝜏 = F . d

9th Class Physics Chapter Turning Effects of Force Notes

Are you studying in 9th Class and looking for the important notes for the preparation of upcoming board exams?

If so, you'll be glad to hear that we're going to share Class 9 Physics Chapter Four Turning Effects of Force Notes here. 

These notes will be beneficial for all students as we have included two main sections of the board paper in these notes. 

These notes includes, solved theoretical questions and important numericals from the 9th Class Physics Chapter Four Turning Effects of Force.

Students are suggested to study from the handwritten notes of 9th class Physics Chapter 4 Turning Effects of Force provided below.

Physics Class 9 Chapter 4 Turning Effects of Force Exercise Questions with Answers

Here are Sindh Board Textbook Exercise Questions with Answers from 9th class Physics Chapter 4 Turning Effects of Force.

Q.1: What is Force? Ans: Definition: Force is a push or pull. It moves and stops the objects. It also changes the shape to the objects. "Force is an agent which tends to change the state of an object." OR "Force is the agent that changes the state of rest or uniform motion of a body."

Force is a vector quantity. Therefore, it has both magnitude (size) and specific direction. It is denoted by 'F'.

Formula of Force: F=ma Where, F is force applied on a body m is mass of a body and a is acceleration of a body Unit of Force:

In SI system, unit of force is Newton (N).

Q.2: What do you mean by parallel forces? Define like and unlike parallel forces? 

Ans: PARALLEL FORCES: The parallel forces can be define as: "When multiple forces occur in parallel directions on a body, they are referred to be parallel forces.." Thus lines of action of parallel forces are parallel to each other.

LIKE PARALLEL FORCES: "The forces that act along the same direction are called like parallel forces."

They can be substituted with a single force as like parallel forces add up to a single resultant force. It is found that most of the time, some or all of the forces are acting in the same direction.

UNLIKE PARALLEL FORCES: "The forces that act along opposite directions are called unlike parallel forces."

Q.3: How can a force be represented? Ans: Representation Of A Force: One vector quantity is force. It has a defined direction in addition to magnitude (size). It is shown in diagrams as a segment of a line with an arrowhead (⟶) at one end to indicate the direction of action. On a suitable scale, the length of the line segment indicates the force's magnitude.

Q.4: Define resultant of forces? Ans: Resultant Of Force: Anywhere multiple forces act on an item, we must sum them up to obtain the single resulting force. Thus it can be define as: "Single force that has the same effect as the combined effect of the forces to be added is called resultant force." OR

"The sum of the two or more forces is called the resultant of forces."

Q.5: Define Head to Tail rule used for addition of forces?

Ans: Head to tail rule of vector addition consist of following steps:

• Step 1: Choose a suitable scale
• Step 2: Draw all the force vectors according to scale. Vectors A and B in this case.
• Step 3: Now, using any vector as the starting point, construct the next vector so that its tail and the preceding vector's head coincide. If there are more than two vectors, go through the process until you reach the last vector.
• Step 4: To connect the tail of the first vector to the head of the last vector, draw a straight line with an arrow pointing towards the last vector. This is the final vector.

Q.6: Define Define moment of force or torque. Write down its formula and units? List the factors on which moment of force depends?

Ans: TORQUE OR MOMENT OF FORCE: Definition: "The turning effect of force is called Torque or moment of force."

In other words, "The product of the force and the moment arm of the force is equal to the torque." Formula:

Moment of force about a point = Force × Perpendicular distance from point. or τ = F × d

Unit of Torque: Newton-meter (Nm) is the SI unit for torque or moment of force, depending on its direction. Factors On Which Moment Of Force Depends: It depends upon:

• The magnitude of force.
• The perpendicular distance measured in degrees from the pivot or fulcrum to the point where force is applied.

Class 9 Physics Chapter 4 Turning Effects of Force Solved Numericals

According to the students, Physics Numericals section in the board paper is considered as the most difficult part of the exam.

Therefore, We have prepared notes of Class 9 Physics Chapter 4 Turning Effects of Force Solved Numericals from the past papers.

These numericals will help students of Class 9 in getting good marks in the board exams.

Here are some important numericals from 9th class Sindh Board Textbook Exercise of Physics Chapter 4 Turning Effects of Force in PDF.

Q. At a distance of 10 cm from a nut, a force of 100N is applied perpendicularly to the spanner. Find the torque.

Solution: F = 100 N r = 10 cm = 10/100 m = 0.1m τ = r x F τ = 0.1 x 100 = 10 Nm

Q. A man is pushing a wheelbarrow on a horizontal ground with a force of 300 N making an angle of 60° with ground. Find the horizontal and vertical components of the force. Solution: Step 1: Write data from the given statement. F = 200 N θ = 60° with horizontal. Fx = ? Fy =? Step 2: Write the formula and rearrange if necessary. Fx = F cos θ Fy = F sin θ Step 3: Put the values in the formula and calculate. Fx = F cos 60° Fx = 300 N x cos 60° Fx = 300 N x 0.5 Fx = 150 N

Fy = F sin 60° Fy = 300 N x sin 60° Fy = 300 N x 0.8660 Fy = 259.8N Ans: Therefore, the pushing force's horizontal and vertical components are 150 N and 259.8 N, respectively.

Q. A car driver tightens the nut of wheel using 20 cm long spanner by exerting a force of 300N. Find the torque. Solution: Step 1: Write data from the given statement. F = 300 N L = 20 cm = 0.20 m τ =? Step 2: Write the formula and rearrange it if necessary. τ = F × L Step 3: Put the values in the formula and calculate τ = 300 N × 0.20 m = 60 Nm

Ans: Thus, the torque of 60 Nm is used to tighten the nut.

Q. A force of 150 N can loosen a nut when applied at the end of a spanner 10cm long.

Solution: F = 150 N L =10 cm = 0.1 m So t = F x L t = 150 x 0.1 t = 15 Nm

Q. What should be the length of the spanner to loosen the same nut with a 60 N force? Solution: As; t = F x L 15 = 60 x L L = 15 / 60 L = 0.25 m L = 25 cm