Solved Exercise of Chapter 5: Turning Effect of Force | Physics Class 9th

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Get the solved exercise of Chapter 5, Turning Effect of Force, for Physics Class 9th based on the new syllabus. Tailored for Punjab Board and Lahore Board students, this comprehensive guide will help you ace your exams.

4.1

Statement: A particle is simultaneously acted upon by two forces of 4 and 3 newtons. The net force on the particle is:
Options:
(a) 1 N
(b) Between 1 N and 7 N
(c) 5 N
(d) 7 N
Answer: (c) 5 N
Explanation: The net force is calculated using the Pythagorean theorem since the forces are perpendicular: Fnet=√42+32=√16+9=√25=5 

Tips and Tricks: Use the Pythagorean theorem for forces acting at right angles. The formula is Fnet=√F12+F22

4.2

Statement: A force FF is making an angle of 60∘ with the x-axis. Its y-component is equal to:
Options:
(a) F
(b) Fsin⁡60∘
(c) Fcos⁡60∘
(d) Ftan⁡60∘
Answer: (b) Fsin⁡60∘
Explanation: The y-component of a force is calculated as Fy=Fsin⁡θ. Substituting θ=60∘, the y-component becomes Fsin⁡60∘F .
Tips and Tricks: Memorize the formulas for components of force:

  • Fx=Fcos⁡θ
  • Fy=Fsin⁡θ

4.3

Statement: Moment of force is called:
Options:
(a) Moment arm
(b) Couple
(c) Couple arm
(d) Torque
Answer: (d) Torque
Explanation: The moment of force about a point is known as torque, which is the product of force and perpendicular distance. Formula: τ=F×r
Tips and Tricks: Torque is also referred to as the rotational equivalent of force.

4.4

Statement: If F and r are the forces acting on a body and τ is the torque produced in it, the body will be completely in equilibrium under:
Options:
(a) ΣF=0 and Στ= 0
(b) ΣF=0 or Στ= 0
(c) ΣF≠0 and Στ= 0
(d) ΣF=0 and Στ≠0
Answer: (a) ΣF=0 and Στ= 0
Explanation: For a body to be in equilibrium, both the net force and the net torque acting on it must be zero.
Tips and Tricks: Equilibrium conditions:

  • Translational equilibrium: ΣF= 0
  • Rotational equilibrium: Στ = 0.

4.5

Statement: A shopkeeper sells his articles by a balance having unequal arms of the pans. If he puts the weights in the pan having a shorter arm, then the customer:
Options:
(a) Loses
(b) Gains
(c) Neither loses nor gains
(d) Not certain
Answer: (a) Loses
Explanation: When weights are placed on the shorter arm, the force produced is smaller, leading to an imbalance that favors the shopkeeper.
Tips and Tricks: Analyze the torque balance when arms are unequal: τ=F×r.

4.6

Statement: A man walks on a tight rope. He balances himself by holding a bamboo stick horizontally. It is an application of:
Options:
(a) Law of conservation of momentum
(b) Newton’s second law of motion
(c) Principle of moments
(d) Newton’s third law of motion
Answer: (c) Principle of moments
Explanation: The man uses the bamboo stick to shift his center of gravity and balance the moments on either side of the rope.
Tips and Tricks: Recall that balancing requires the moments about a pivot to be equal.

4.7

Statement: In the stable equilibrium, the center of gravity of the body lies:
Options:
(a) At the highest position
(b) At the lowest position
(c) At any position
(d) Outside the body
Answer: (b) At the lowest position
Explanation: In stable equilibrium, the body’s center of gravity is at the lowest point to maintain maximum stability.
Tips and Tricks: Stability depends on the position of the center of gravity: lower is more stable.

4.8

Statement: The center of mass of a body:
Options:
(a) Lies always inside the body
(b) Lies always outside the body
(c) Lies always on the surface of the body
(d) May lie within, outside, or on the surface
Answer: (d) May lie within, outside, or on the surface
Explanation: The center of mass depends on the distribution of mass. For example:

  • A uniform solid has its center of mass within the body.
  • A hollow sphere can have its center of mass outside the material body.
    Tips and Tricks: Think about examples like rings or irregularly shaped bodies to determine the center of mass.

4.9

Statement: A cylinder resting on its circular base is in:
Options:
(a) Stable equilibrium
(b) Unstable equilibrium
(c) Neutral equilibrium
(d) None of these
Answer: (a) Stable equilibrium
Explanation: When the cylinder is slightly disturbed, it will return to its original position because its center of gravity remains low and stable.
Tips and Tricks: Objects with a broad base and low center of gravity are usually in stable equilibrium.

4.10

Statement: Centripetal force is given by:
Options:
(a) rF
(b) rFcos⁡θ
(c) mv2/r
(d) mv/r
Answer: (c) mv2/r
Explanation: Centripetal force is the force required to keep an object moving in a circular path, given by the formula: Fc=mv2/r

where mm is the mass, vv is the velocity, and rr is the radius of the circular path.
Tips and Tricks: Memorize the formula for centripetal force. It directly depends on the mass and velocity squared and inversely on the radius.

Short Answer Questions

4.1 Define like and unlike parallel forces.
Answer:

  • Like parallel forces: Forces acting in the same direction along parallel lines.
  • Unlike parallel forces: Forces acting in opposite directions along parallel lines.

4.2 What are rectangular components of a vector and their values?
Answer:
Rectangular components of a vector are the projections of the vector along mutually perpendicular axes (usually x and y).

  • Vx=Vcos⁡θ
  • Vy=Vsin⁡θ

4.3 What is the line of action of a force?
Answer:
The line of action of a force is an imaginary line that extends along the direction of the force. It determines the point of application and the torque produced by the force.

4.4 Define moment of force. Prove that τ=Fsin⁡θ, where θ\theta is the angle between r and F.
Answer:

  • Definition: The moment of a force (or torque) is the measure of its ability to rotate an object about an axis or a point.
  • Proof:
    Torque (τ\tau) is given by the formula:

τ=r×F=rFsin⁡θ

where r is the perpendicular distance from the axis of rotation, F is the applied force, and θ is the angle between r and F.

4.5 With the help of a diagram, show that the resultant force is zero but the resultant torque is not zero.
Answer:
Draw a diagram where two equal and opposite forces are acting on a body (e.g., a rectangular body). The forces cancel each other out, resulting in ΣF=0. However, since these forces do not share the same line of action, they produce a couple, resulting in τ≠0.

4.6 Identify the state of equilibrium in the given figure.
Answer:
(a) Stable equilibrium (the cone on its base)
(b) Neutral equilibrium (sphere on a flat surface)
(c) Unstable equilibrium (cylinder on its curved surface).

Constructed Response Questions

4.1 A car travels at the same speed around two curves with different radii. For which radius does the car experience more centripetal force? Prove your answer.
Answer:

  • Formula: Centripetal force Fc=mv2/r
  • When v (velocity) and m (mass) are constant, Fc is inversely proportional to rr.
  • For the smaller radius, Fc is larger because r is smaller.
  • Proof: Assume two radii r1<r2r_1 < r_2, then:

Fc1=mv2/r1>Fc2=mv2/r2

4.2 A ripe mango does not normally fall from the tree. But when the branch of the tree is shaken, the mango falls down easily. Can you tell the reason?
Answer:
The mango remains stationary due to inertia. When the branch is shaken, the inertia of the mango resists the motion of the branch, causing the mango to lose its support and fall due to gravity.

4.3 Discuss the concepts of stability and center of gravity in the context of objects toppling over. Provide an example where an object’s center of gravity affects its stability.
Answer:

  • Stability: An object is stable if its center of gravity lies within its base of support.
  • Toppling: If the line of action of weight (passing through the center of gravity) falls outside the base of support, the object topples.
  • Example: A tall, narrow vase is less stable compared to a wide, flat vase because its center of gravity is higher and more likely to shift outside the base.

Short Questions

4.4 Why an accelerated body cannot be considered in equilibrium?
Answer:
An accelerated body is not in equilibrium because:

  1. In equilibrium, the net force (ΣF) and net torque (Στ) acting on the body must be zero.
  2. An accelerated body has a non-zero net force (F=ma), which violates the equilibrium condition.

4.5 Two boxes of the same weight but different heights are lying on the floor of a truck. If the truck makes a sudden stop, which box is more likely to tumble over? Why?
Answer:
The taller box is more likely to tumble over because it has a higher center of gravity.

  • When the truck stops suddenly, inertia causes the box to tilt.
  • A higher center of gravity increases the chance of the line of action of weight moving outside the base of support, leading to toppling.

Comprehensive Questions

4.1 Explain the principle of moments with an example.
Answer:

  • Principle of Moments: For a body to be in rotational equilibrium, the sum of clockwise moments must equal the sum of anticlockwise moments about a pivot point:

ΣClockwise moments=ΣAnticlockwise moments

  • Example: A seesaw in balance:
    • If a child of weight W1 sits at a distance d1d_1 from the pivot on one side, and another child of weight W2 sits at d2 on the other side, equilibrium is achieved when:

W1×d1=W2×d2

4.2 Describe how you could determine the center of gravity of an irregular-shaped lamina experimentally.
Answer:

  1. Suspend the irregular lamina freely from one point.
  2. Use a plumb line to draw a vertical line along the lamina from the suspension point.
  3. Suspend the lamina from another point and repeat the process.
  4. The intersection of the lines is the center of gravity.

4.3 State and explain two conditions of equilibrium.
Answer:

  • Translational Equilibrium:
    The net force acting on the body is zero (ΣF=0).
    • Example: A book resting on a table where gravitational force is balanced by the normal force.
  • Rotational Equilibrium:
    The net torque acting on the body is zero (Στ=0).
    • Example: A balanced beam on a fulcrum where clockwise and anticlockwise torques are equal.

4.4 How the stability of an object can be improved? Give a few examples to support your answer.
Answer:
Ways to Improve Stability:

  1. Lower the center of gravity: The closer the center of gravity is to the base, the more stable the object.
    • Example: Racing cars have low centers of gravity for stability at high speeds.
  2. Widen the base of support: A broader base increases stability by making it harder for the center of gravity to shift outside the base.
    • Example: Tripods have wide bases to prevent tipping.
  3. Add weight near the base: Heavier bases prevent toppling.
    • Example: Tall structures like towers have heavy foundations.
  4. Align weight along the base: Ensure the line of action of weight remains within the base of support.
    • Example: A gymnast balances carefully to maintain stability.

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