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Chapter 5: Work, Energy, and Power – Solved Exercise for 9th Class Physics (New Syllabus)

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Learn Chapter 5: Work, Energy, and Power from the new Physics syllabus for 9th class. Explore solved exercises with detailed explanations and tips to excel in your exams.

Question 5.1

Statement: Work done is maximum when the angle between the force F and the displacement d is:
Options:
(a) 0∘
(b) 30∘
(c) 60∘
(d) 90∘

Answer: (a) 0∘

Explanation:
The work done is given by the formula: W=Fdcos⁡θ

The cosine function is maximum when θ=0∘ because cos⁡0∘=1. Thus, work done is maximum when the force and displacement are in the same direction.

Tip: Memorize that cos⁡θ decreases as the angle increases from 0∘ to 90∘. For θ=90∘, work done becomes zero.

Question 5.2

Statement: A joule can also be written as:
Options:
(a) kg m2s−2
(b) kg m s−1
(c) kg m−2s−2
(d) kg m s−2

Answer: (a) kg m2s−2

Explanation:
A joule is the unit of energy or work, given as: Work=Force×Displacement

Force is measured in kg m s−2, and displacement in meters . Therefore, Joule=(kg m s−2)×m=kg m2s−2.

Tip: Remember that joules involve mass, distance squared, and time squared in their derived units.

Question 5.3

Statement: The SI unit of power is:
Options:
(a) joule
(b) newton
(c) watt
(d) second

Answer: (c) watt

Explanation:
Power is the rate at which work is done or energy is transferred. It is defined as: P=W/t

Where P is power in watts, W is work in joules, and t is time in seconds.

Tip: Memorize key SI units: work (joule), force (newton), and power (watt).

Question 5.4

Statement: The power of a water pump is 2 kW. The amount of water it can raise in one minute to a height of 5 m is:
Options:
(a) 1000 litres
(b) 1200 litres
(c) 2000 litres
(d) 2400 litres

Answer: (d) 2400 litres

Explanation:
The power of the pump is P=2000 WP . Work is given by: W=mgh

Where mm is mass, g=9.8 m/s2, and h=5 m. Power is work done per second: P=mgh/t

Rearranging for mm: m=P⋅t/gh=2000×60/9.8×5=240 kg

Since 1 kg of water equals 1 litre, the answer is 2400 litres.

Tip: Always use consistent units and formulas for work, power, and time.

Question 5.5

Statement: A bullet of mass 0.05 kg has a speed of 300 m/s. Its kinetic energy will be:
Options:
(a) 2250 J
(b) 4500 J
(c) 1500 J
(d) 1125 J

Answer: (b) 4500 J

Explanation:
Kinetic energy is given by: KE=1/2mv2

Substitute m=0.05 kg and v=300 m/s

KE=1/2×0.05×(300)2=4500 J

Tip: Remember to square the velocity in the formula.

Question 5.6

Statement: If a car doubles its speed, its kinetic energy will be:
Options:
(a) the same
(b) increased four times
(c) increased two times
(d) reduced two times

Answer: (b) increased four times

Explanation:
Kinetic energy depends on the square of velocity: KE∝v2

If the speed doubles (v→2v), the kinetic energy becomes: KE∝(2v)2=4v2.

Tip: Always relate proportionality with squared terms in energy calculations.

Question 5.7

Statement: The energy possessed by a body by virtue of its position is:
Options:
(a) kinetic energy
(b) potential energy
(c) chemical energy
(d) solar energy

Answer: (b) potential energy

Explanation:
Potential energy is the energy stored in an object due to its position in a gravitational field, given by: PE=mgh

Tip: Potential energy depends on height, while kinetic energy depends on motion.

Question 5.8

Statement: The magnitude of momentum of an object is doubled; the kinetic energy of the object will:
Options:
(a) double
(b) increase to four times
(c) reduce to one-half
(d) remain the same

Answer: (b) increase to four times

Explanation:
Momentum (pp) is related to kinetic energy (KE) as:

KE=p2/2m

If momentum doubles (p→2p)

KE∝(2p)2=4p2

Tip: Understand the relationship between momentum and kinetic energy.

Question 5.9

Statement: Which of the following is not renewable energy?
Options:
(a) Hydro-electric energy
(b) Fossil fuels
(c) Wind energy
(d) Solar energy

Answer: (b) Fossil fuels

Explanation:
Fossil fuels (coal, oil, natural gas) take millions of years to form and are not renewable. Renewable energy sources like solar, wind, and hydro are replenished naturally.

Tip: Fossil fuels are non-renewable because their reserves are finite.

Here are the solutions to the short answer and constructed response questions with clear and simple explanations:

Short Answer Questions (Part B)

Question 5.1

What is the work done on an object that remains at rest when a force is applied to it?
Answer:
The work done is zero.
Explanation:
Work is given by the formula: W=F⋅d⋅cos⁡θ

Since the object remains at rest, the displacement d=0. Therefore, work W=0.

Question 5.2

A slow-moving car may have more kinetic energy than a fast-moving motorcycle. How is this possible?
Answer:
Kinetic energy depends on both mass and velocity: KE=1/2mv2

If the car is very heavy (large m) and the motorcycle is light (small m), the car’s kinetic energy can be greater even if it moves slowly.

Question 5.3

A force F1 does 5 J of work in 10 s. Another force F2 does 3 J of work in 5 s. Which force delivers greater power?
Answer:
Power is given by: P=W/t

For F1: P1=5/10=0.5 W

For F2: P2=3/5=0.6 W

Thus, F2 delivers greater power.

Question 5.4

A woman runs up a flight of stairs. The gain in her gravitational potential energy is 4500 J. If she runs up the stairs with twice the speed, what will be her gain in potential energy?
Answer:
The gain in potential energy will still be 4500 J.
Explanation:
Potential energy depends only on height (PE=mgh), not on speed. Running faster does not change the height.

Question 5.5

Define work and its SI unit.
Answer:
Work is done when a force moves an object in the direction of the force. Mathematically: W=F⋅d⋅cos⁡θ

The SI unit of work is the joule (J).

Question 5.6

What is the potential energy of a body of mass mm when it is raised through a height hh?
Answer:
Potential energy is given by: PE=mgh

Where:

  • m = mass of the body
  • g = gravitational acceleration (9.8 m/s2)
  • h = height

Question 5.7

Find an expression for the kinetic energy of a moving body.
Answer:
Kinetic energy is given by: KE=1/2mv2

Where:

  • m = mass of the body
  • v = velocity of the body

Question 5.8

Define the efficiency of a working system. Why can a system not have 100% efficiency?
Answer:
Efficiency is the ratio of useful energy output to the total energy input: Efficiency=Useful Energy Output/Total Energy Input ×100

No system can have 100%100\% efficiency because some energy is always lost as heat, sound, or friction.

Question 5.9

What is power? Define the unit used for it.
Answer:
Power is the rate at which work is done or energy is transferred: P=W/t

The SI unit of power is the watt (W).

Question 5.10

Differentiate between renewable and non-renewable energy sources.
Answer:

  • Renewable energy: Can be replenished naturally, e.g., solar, wind, and hydro energy.
  • Non-renewable energy: Exists in limited quantities and cannot be replenished, e.g., fossil fuels (coal, oil).

Constructed Response Questions (Part C)

Question 5.1

Can the kinetic energy of a body ever be negative?
Answer:
No, kinetic energy cannot be negative.
Explanation:
Kinetic energy is given by KE=1/2mv2. Since both mass (m) and the square of velocity (v2) are always positive, KE is always positive.

Question 5.2

Which one has the greater kinetic energy: an object traveling with a velocity vv or an object twice as heavy traveling with a velocity of v2\frac{v}{2}?
Answer:
Let the mass of the first object be mm, so KE1=1/2mv2
The second object has mass 2m and velocity v2

KE2=1/2(2m)(v2)/2=1/2(2m)⋅v2/4=mv2/4

Thus, KE1>KE2.

Question 5.3

A car is moving along a curved road at constant speed. Does its kinetic energy change?
Answer:
No, the kinetic energy does not change.
Explanation:
Kinetic energy depends on speed, not direction. Since the speed is constant, the kinetic energy remains the same.

Question 5.4

Comment on the statement: “An object has one joule of potential energy.”
Answer:
The statement means the object’s potential energy is equal to the work done in lifting it to a certain height where: PE=1 J.

Question 5.5

While driving on a motorway, the tyre of a vehicle sometimes bursts. What may be its cause?
Answer:
A tyre may burst due to:

  1. Overheating from friction.
  2. Excessive air pressure.
  3. Damage from sharp objects.

Question 5.6

While playing cricket, a ball smashes a windowpane. Describe the energy changes in this event.
Answer:

  • The ball has kinetic energy while in motion.
  • Upon impact, the ball transfers energy to the windowpane, causing it to break.
  • Some energy is lost as sound and heat.

Question 5.7

A man rows a boat upstream at rest with respect to the shore. Is he doing work?
Answer:
No, he is not doing work with respect to the shore because there is no displacement.

Question 5.8

A cyclist goes downhill from the top of a steep hill without pedaling and takes it to the top of the next hill. Draw a diagram of what happened.
Answer:
The cyclist’s gravitational potential energy at the top of the first hill converts to kinetic energy as he descends, then converts back to potential energy as he climbs the next hill. A diagram would show a curved path between two hills.

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.

Let me know if further explanations are needed!

Top Teaching and Tutoring Apps to Boost Your Online Earnings

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Top Teaching and Tutoring Apps to Boost Your Online Earnings

In today’s digital age, teaching and tutoring apps provide excellent opportunities for professionals to share their expertise and earn a steady income. Whether you’re a seasoned teacher, a skilled professional, or someone with a passion for sharing knowledge, platforms like Skillshare, Udemy, and Preply allow you to connect with learners worldwide. Here’s a detailed overview of these platforms and how you can use them to start earning today.

1. Skillshare: Create and Sell Online Courses

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  • Upload the course to the Skillshare platform.
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2. Udemy: Design Courses and Earn Per Enrollment

Overview: Udemy is one of the largest online learning marketplaces, offering courses in diverse fields such as technology, business, arts, and more. It’s a perfect platform if you’re looking to build your brand as an educator.

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Pro Tip: Enhance your profile with a professional photo, a video introduction, and positive reviews from students to attract more learners. Popular topics on Preply include “English for Beginners” and “Preparing for IELTS Exams.”

Why Choose These Platforms?

Each platform offers unique advantages, depending on your expertise and preferred teaching method:

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Final Tips for Success

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Chapter 3 Dynamics – Solved Exercise for 9th Class Physics

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3.1 When we kick a stone, we get hurt. This is due to:

  • Statement: When we apply force to kick a stone, it does not move easily.
  • Options:
    (a) inertia
    (b) velocity
    (c) momentum
    (d) reaction
  • Answer: (a) inertia
  • Explanation: The stone resists a change in its state of motion because of its inertia. Since the stone’s mass is large and it is at rest, we feel pain when force is applied.
  • Tip: Inertia is related to the resistance of an object to change its motion or state.

3.2 An object will continue its motion with constant acceleration until:

  • Statement: The object remains under an unbalanced force.
  • Options:
    (a) the net force on it begins to decrease
    (b) the resultant force on it is zero
    (c) the direction of motion changes
    (d) the resultant force is at a right angle to its tangential velocity
  • Answer: (b) the resultant force on it is zero
  • Explanation: According to Newton’s First Law, an object will remain in motion with a constant velocity unless acted upon by an external force. To change its acceleration, a force must act.
  • Tip: Remember Newton’s First Law and that forces cause changes in acceleration.

3.3 Which of the following is a non-contact force?

  • Statement: Non-contact forces act without direct physical contact.
  • Options:
    (a) Friction
    (b) Air resistance
    (c) Electrostatic force
    (d) Tension in the string
  • Answer: (c) Electrostatic force
  • Explanation: Electrostatic force acts over a distance due to charges, while the others require direct contact.
  • Tip: Non-contact forces include gravitational, magnetic, and electrostatic forces.

3.4 A ball with initial momentum pp hits a solid wall and bounces back with the same velocity. Its momentum after collision will be:

  • Statement: Momentum before and after collision is equal in magnitude but opposite in direction.
  • Options:
    (a) p= p
    (b) p=−p
    (c) p=2p
    (d) p=−2p
  • Answer: (b) p=−p
  • Explanation: The ball rebounds with the same speed but opposite direction, so the momentum becomes −p.
  • Tip: Use the principle of conservation of momentum for such problems.

3.5 A particle of mass mm moving with a velocity vv collides with another particle of the same mass at rest. The velocity of the first particle after the collision is:

  • Options:
    (a) v
    (b) −v
    (c) 0
    (d) −1/2v
  • Answer: (c) 0
  • Explanation: In a perfectly elastic collision where the masses are equal, the moving particle transfers all its velocity to the particle at rest.
  • Tip: For elastic collisions, remember velocity exchange occurs between identical masses.

3.6 Conservation of linear momentum is equivalent to:

  • Statement: The total momentum of a system remains constant if no external force acts.
  • Options:
    (a) Newton’s first law of motion
    (b) Newton’s second law of motion
    (c) Newton’s third law of motion
    (d) None of these
  • Answer: (b) Newton’s second law of motion
  • Explanation: Conservation of momentum follows from Newton’s Second Law when no external force acts on the system.
  • Tip: Link conservation laws to the underlying Newtonian principles.

3.7 An object with a mass of 5 kg moves at a constant velocity of 10 m/s. A constant force acts for 5 seconds on the object, and its velocity increases by 2 m/s in the positive direction. The force acting on the object is:

  • Options:
    (a) 5 N
    (b) 9 N
    (c) 12 N
    (d) 15 N
  • Answer: (a) 5 N
  • Explanation: Use F=ma, where a=Δvt=2/5=0.4 m/s2. Then F=5×0.4=2 N
  • Tip: Apply Newton’s Second Law and calculate acceleration first.

3.8 A large force acts on an object for a very short interval of time. In this case, it is easy to determine:

  • Statement: When force acts for a short duration, impulse is involved.
  • Options:
    (a) average force
    (b) time interval
    (c) product of force and time
    (d) none of these
  • Answer: (c) product of force and time
  • Explanation: The impulse is the product of force and time, and it changes momentum.
  • Tip: Think about the concept of impulse whenever force and time are mentioned together.

3.9 Lubricants are introduced between two surfaces to decrease friction. The lubricant:

  • Statement: Lubricants reduce direct contact and rolling resistance.
  • Options:
    (a) decreases temperature
    (b) acts as ball bearings
    (c) prevents direct contact of the surfaces
    (d) provides rolling friction
  • Answer: (c) prevents direct contact of the surfaces
  • Explanation: Lubricants reduce the roughness of surfaces and prevent contact, minimizing friction.
  • Tip: Know the role of lubricants in reducing friction to solve such questions.

Short Answer Questions (B)

3.1 What kind of changes in motion may be produced by a force?

  • Answer: A force can:
    • Start or stop an object.
    • Increase or decrease the speed of an object.
    • Change the direction of motion.
    • Change the shape of an object.

3.2 Give 5 examples of contact forces.

  • Answer:
    • Frictional force
    • Tension in a string
    • Normal force
    • Applied force (pushing or pulling)
    • Air resistance

3.3 An object moves with constant velocity in free space. How long will the object continue to move with this velocity?

  • Answer: The object will continue to move with the same velocity forever because no external force acts on it in free space (Newton’s First Law).

3.4 Define impulse of force.

  • Answer: Impulse is the product of force and the time duration for which the force acts.
    Impulse=F×t
    It changes the momentum of an object.

3.5 Why has Newton’s first law not been proved on the Earth?

  • Answer: On Earth, external forces like friction and air resistance always act on objects, so they don’t continue moving indefinitely, which makes it difficult to directly observe Newton’s First Law.

3.6 When sitting in a car which suddenly accelerates from rest, you are pushed back into the seat. Why?

  • Answer: Your body tends to stay at rest (due to inertia) while the car moves forward, so it feels like you are being pushed back.

3.7 The force expressed in Newton’s second law is a net force. Why is it so?

  • Answer: Newton’s second law considers all forces acting on an object. The net force is the total force after combining all forces acting in different directions.

3.8 How can you show that rolling friction is lesser than the sliding friction?

  • Answer: Rolling a heavy object (like a cylinder) requires less effort than sliding it because rolling friction is smaller than sliding friction. This is why wheels are used in vehicles.

3.9 Define terminal velocity of an object.

  • Answer: Terminal velocity is the constant speed an object reaches when the force of air resistance becomes equal to the weight of the object, and no more acceleration occurs.

3.10 An astronaut walking in space wants to return to his spaceship by firing a hand rocket. In what direction does he fire the rocket?

  • Answer: The astronaut should fire the rocket in the direction opposite to the spaceship. This creates a force pushing him back toward the spaceship (Newton’s Third Law).

Constructed Response Questions (C)

3.1 Two ice skaters weighing 60 kg and 80 kg push off against each other on a frictionless ice track. The 60 kg skater gains a velocity of 4 m/s. Explain how Newton’s third law applies.

  • Answer:
    • According to Newton’s third law, the force exerted by the 60 kg skater on the 80 kg skater is equal and opposite to the force exerted by the 80 kg skater on the 60 kg skater.
    • Since momentum is conserved:
      m1v1=m2v2
      60×4=80×v2
      v2=3 m/s
      The 80 kg skater moves in the opposite direction with a velocity of 3 m/s.

3.2 Inflatable air bags are installed in vehicles as safety equipment. In terms of momentum, what is the advantage of air bags over seatbelts?

  • Answer: Airbags increase the time over which the passenger’s momentum changes during a collision. This reduces the force acting on the body, minimizing injuries compared to seatbelts.

3.3 A horse refuses to pull a cart. The horse argues, “According to Newton’s third law, whatever force I exert on the cart, the cart will exert an equal and opposite force on me. Since the net force will be zero, therefore, I have no chance of accelerating (pulling) the cart.” What is wrong with this reasoning?

  • Answer:
    • The horse’s reasoning is wrong because the equal and opposite forces act on different objects.
    • The force the horse exerts on the ground pushes the horse forward (action-reaction pair). The cart moves because of the force exerted by the horse on the cart.

3.4 When a cricket ball hits high, a fielder tries to catch it. While holding the ball, he/she draws hands backward. Why?

  • Answer: By drawing hands backward, the fielder increases the time of impact. This reduces the force exerted by the ball on the hands, preventing injury.

3.5 When someone jumps from a small boat onto the river bank, why does the jumper often fall into the water? Explain.

  • Answer: When the jumper pushes the boat backward to jump, the boat moves in the opposite direction due to Newton’s Third Law. The jumper’s forward motion and the boat’s backward motion disturb balance, causing the jumper to fall.

3.6 Imagine that if friction vanishes suddenly from everything, then what could be the scenario of daily life activities?

  • Answer:
    • Walking would become impossible as we need friction to push the ground.
    • Vehicles would not move or stop, causing accidents.
    • Objects would keep sliding and never stay in place.
    • Machines would stop working because friction is needed for belts and gears to function.

Comprehensive Questions (D):

3.1 Explain the concept of force by practical examples.

Answer:
Force is a physical quantity that causes a change in the state of motion or shape of an object. It is a push or pull acting upon an object as a result of its interaction with another object.

Practical Examples of Force:

  1. Pushing a shopping cart: When you push a cart in a supermarket, you apply force to move it forward. The harder you push, the faster it moves.
  2. Kicking a football: When a football is kicked, the applied force changes its motion and direction.
  3. Opening a door: To open or close a door, a force is applied to overcome resistance (friction in the hinges).
  4. Stretching a rubber band: Pulling on a rubber band changes its shape due to the applied force.
  5. Gravity pulling objects downward: If you drop an object, the force of gravity pulls it toward the Earth.

3.2 Describe Newton’s laws of motion.

Answer:
Newton’s three laws of motion explain the relationship between an object and the forces acting upon it:

First Law (Law of Inertia):

  • Statement: An object remains at rest or in uniform motion in a straight line unless acted upon by an external force.
  • Example: A book on a table stays at rest until you push it. Similarly, a moving bicycle slows down due to friction if pedaling stops.

Second Law (Force and Acceleration):

  • Statement: The force acting on an object is equal to the product of its mass and acceleration.
    F=m⋅a
  • Example: A heavier object requires more force to accelerate than a lighter object. For example, pushing a truck requires more force than pushing a bicycle.

Third Law (Action and Reaction):

  • Statement: For every action, there is an equal and opposite reaction.
  • Example: When a swimmer pushes water backward, the water exerts an equal force forward, propelling the swimmer.

3.3 Define momentum and express Newton’s second law of motion in terms of change in momentum.

Answer:
Momentum: Momentum (pp) is the product of the mass of an object and its velocity. It measures the quantity of motion in an object.
p=m⋅v
Where:

  • pp = momentum,
  • mm = mass,
  • vv = velocity.

Newton’s Second Law in Terms of Momentum:

  • Newton’s second law can also be written as:
    F=Δp/Δt
    Where:
    Δp = change in momentum,
    Δt = time interval.
  • Explanation: Force is equal to the rate of change of momentum of an object.
  • Example: When a cricketer catches a fast ball and pulls his hands backward, he increases the time to change the ball’s momentum, which reduces the force exerted on his hands.

3.4 State and explain the principle of conservation of momentum.

Answer:
Principle of Conservation of Momentum:

  • Statement: The total momentum of an isolated system remains constant if no external forces act on it.
  • Mathematically,
    m1u1+m2u2=m1v1+m2v2
    where:
    m1,m2= masses of two objects,
    u1,u2 = initial velocities,
    v1,v2 = final velocities.

Explanation:

  • During a collision or interaction, the momentum lost by one object is gained by the other, keeping the total momentum constant.

Example:

  • When a gun is fired, the bullet moves forward while the gun recoils backward. The forward momentum of the bullet is equal to the backward momentum of the gun, conserving the total momentum.

3.5 Describe the motion of a block on a table taking into account the friction between the two surfaces. What is the static friction and kinetic friction?

Answer:
When a block is placed on a table and you try to push it, friction acts between the block and the surface.

Friction Types:

  1. Static Friction (fs):
    • Static friction acts when the object is at rest. It prevents the block from moving until a certain threshold force is applied.
    • Static friction is higher than kinetic friction.
    • Formula: fs≤μs⋅N, where μs = coefficient of static friction, N= normal force.
  2. Kinetic Friction (fk):
    • Kinetic friction acts when the object is sliding. It resists the motion of the block while it is in motion.
    • Formula: fk=μk⋅N, where μk= coefficient of kinetic friction, N = normal force.

Example:

  • When you try to push a heavy box, it initially resists (static friction). Once the force exceeds static friction, the box begins to move, and kinetic friction acts.

3.6 Explain the effect of friction on the motion of vehicles in the context of tire surface and braking force.

Answer:
Friction plays a crucial role in the motion of vehicles, both in terms of tire grip and braking.

1. Role of Tire Surface:

  • The grooves on the tire surface increase the friction between the tire and the road. This prevents the vehicle from slipping and allows better control while driving.
  • On wet or icy roads, friction reduces, causing tires to slip. Special tires with deeper grooves or chains are used in such conditions to increase friction.

2. Braking Force:

  • When brakes are applied, friction between the brake pads and the wheels slows the rotation of the tires, reducing the vehicle’s speed.
  • In the absence of friction, the vehicle would not stop.
  • Overuse of brakes may reduce friction due to overheating of the brake pads, which can lead to brake failure.

Importance of Friction in Safety:

  • Friction ensures grip and prevents skidding during turns or sudden stops.
  • Anti-lock Braking Systems (ABS) are designed to maintain optimal friction between the tires and the road, preventing skidding.

Solved Exercise of Chapter 2 Kinematics: 9th Class Physics

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Get step-by-step solutions for Chapter 2 “Kinematics” from the 9th class physics new syllabus. Specifically designed for Lahore Board and all Punjab Boards, this guide helps students excel in their exams.

MCQs

2.1 The numerical ratio of displacement to distance is:
Options:
(a) always less than one
(b) always equal to one
(c) always greater than one
(d) equal to or less than one

Answer: (d) equal to or less than one
Explanation: Displacement is the shortest distance between two points and can be equal to or less than the actual distance traveled. It cannot exceed the distance.

2.2 If a body does not change its position with respect to some fixed point, then it will be in a state of:
Options:
(a) rest
(b) motion
(c) uniform motion
(d) variable motion

Answer: (a) rest
Explanation: A body is said to be at rest when it does not change its position relative to a reference point.
Tip: Relate to the definition of rest and motion.

2.3 A ball is dropped from the top of a tower; the distance covered by it in the first second is:
Options:
(a) 5 m
(b) 10 m
(c) 50 m
(d) 100 m

Answer: (a) 5 m
Explanation: The distance covered in free fall is given by s=1/2gt2
s=1/2×10×(1)2=5 m
Tip: Memorize the formula s=1/2gt2

2.4 A body accelerates from rest to a velocity of 144 km/h in 20 seconds. Then the distance covered by it is:
Options:
(a) 100 m
(b) 400 m
(c) 1400 m
(d) 1440 m

Answer: (c) 1400 m
Explanation: Convert 144 km/h
v=144×1000/3600=40 m/s
Using the formula s=1/2at2
First, calculate acceleration: a=vt=40/20=2 m/s2
Then, s=1/2×2×202=1400 m
Tip: Convert units before calculations.

2.5 A body is moving with constant acceleration starting from rest. It covers a distance S in 4 seconds. How much time does it take to cover one-fourth of this distance?
Options:
(a) 1 s
(b) 2 s
(c) 4 s
(d) 16 s

Answer: (b) 2 s
Explanation: For constant acceleration, distance is proportional to the square of time:
S∝t2
If the total time is t=4 s, one-fourth of the distance is covered in t/2=2 
Tip: Remember the proportionality S∝t2

2.6 The displacement-time graphs of two objects A and B are shown in the figure. Point out the true statement from the following:
Options:
(a) The velocity of A is greater than B.
(b) The velocity of A is less than B.
(c) The velocity of A is equal to that of B.
(d) The graph gives no information in this regard.

Answer: (a) The velocity of A is greater than B.
Explanation: The slope of a displacement-time graph represents velocity. Since the slope of A’s graph is steeper than B’s, A has a greater velocity.
Tip: Compare slopes for velocity on such graphs.

2.7 The area under the speed-time graph is numerically equal to:
Options:
(a) velocity
(b) uniform velocity
(c) acceleration
(d) distance covered

Answer: (d) distance covered
Explanation: The area under a speed-time graph represents the distance traveled by the object.
Tip: Always associate “area under the curve” with specific physical quantities based on the graph type.

2.8 Gradient of the speed-time graph is equal to:
Options:
(a) speed
(b) velocity
(c) acceleration
(d) distance covered

Answer: (c) acceleration
Explanation: The gradient (slope) of a speed-time graph gives the rate of change of speed, which is acceleration.
Tip: For speed-time graphs:

  • Slope → Acceleration
  • Area under the curve → Distance.

2.9 Gradient of the distance-time graph is equal to:
Options:
(a) speed
(b) velocity
(c) distance covered
(d) acceleration

Answer: (b) velocity
Explanation: The gradient of a distance-time graph represents the rate of change of distance with time, which is velocity.
Tip: Remember, distance-time graph slope indicates motion speed or velocity.

2.10 A car accelerates uniformly from 80.5 km/h at t=0 to 113 km/h at t=9 s. Which graph best describes the motion of the car?
Answer: (a)
Explanation: For uniform acceleration, the velocity-time graph is a straight line with a positive slope, as shown in option (a).
Tip: Uniform acceleration always produces a straight, inclined line in velocity-time graphs.

B: Short Answer Questions

2.1 Define scalar and vector quantities.
Answer:

  • Scalar quantities: Physical quantities that have magnitude only (e.g., mass, temperature).
  • Vector quantities: Physical quantities that have both magnitude and direction (e.g., force, velocity).

2.2 Give 5 examples each for scalar and vector quantities.
Answer:

  • Scalars: Speed, mass, temperature, time, energy.
  • Vectors: Velocity, force, acceleration, displacement, momentum.

2.3 State head-to-tail rule for addition of vectors.
Answer: Place the tail of the second vector at the head of the first vector. The resultant vector is drawn from the tail of the first vector to the head of the second vector.

2.4 What are distance-time graph and speed-time graph?
Answer:

  • Distance-time graph: Represents the motion of an object by plotting distance against time. Slope indicates speed.
  • Speed-time graph: Represents the variation of speed with time. Slope gives acceleration, and the area under the curve gives distance.

2.5 Falling objects near the Earth have the same constant acceleration. Does this imply that a heavier object will fall faster than a lighter object?
Answer: No, all objects fall with the same acceleration (9.8 m/s²) near the Earth, regardless of mass, due to gravity (neglecting air resistance).

2.6 The vector quantities are sometimes written in scalar notation (not bold face). How is the direction indicated?
Answer: Direction is indicated using angles, signs (+/-), or directional symbols (e.g., North, South, East, West).

2.7 A body is moving with uniform speed. Will its velocity be uniform? Give reason.
Answer: Not necessarily. If the body changes direction, the velocity will not remain uniform even if the speed is constant because velocity is a vector quantity (depends on both magnitude and direction).

2.8 Is it possible for a body to have acceleration when moving with:
(i) Constant velocity?
Answer: No, because acceleration is the rate of change of velocity, and with constant velocity, there is no change.
(ii) Constant speed?
Answer: Yes, if the direction changes (e.g., circular motion), there can be centripetal acceleration.

C: Constructed Response Questions

2.1 Distance and displacement may or may not be equal in magnitude. Explain this statement.
Answer:

  • Equal: When the motion is in a straight line without changing direction. For example, walking 5 meters straight.
  • Not Equal: When the motion involves a change in direction, displacement (shortest path) will be less than the distance (total path). For example, walking in a circular path.

2.2 When a bullet is fired, its velocity with which it leaves the barrel is called the muzzle velocity of the gun. The muzzle velocity of one gun with a longer barrel is less than that of another gun with a shorter barrel. In which gun is the acceleration of the bullet larger? Explain your answer.
Answer:
The gun with the shorter barrel has larger acceleration because the same change in velocity (muzzle velocity) occurs over a shorter distance, leading to greater acceleration (since a=v2−u2/2s, where ss is the distance).

2.3 For a car moving at uniform speed, the area under the speed-time graph is calculated. Its value came out to be positive. Is it possible that its instantaneous velocity at any time during the trip had the negative sign? Give justification of your answer.
Answer:
No, because the speed-time graph shows the magnitude of velocity, which is always positive. If the graph is used to compute displacement (not speed), the instantaneous velocity could be negative if the car changes direction.

Comprehensive questions

2.1 How can a vector be represented graphically? Explain.

  • A vector is represented graphically as a directed line segment.
  • The length of the line represents the magnitude of the vector, and the arrowhead shows its direction.
  • For example, if a vector shows a displacement of 5 meters to the right, draw a 5 cm arrow pointing to the right (scale: 1 cm = 1 m).

2.2 Differentiate between:
(i) Rest and Motion:

  • Rest: An object is at rest when it does not change its position relative to a reference point.
    Example: A book lying on a table is at rest.
  • Motion: An object is in motion when it changes its position relative to a reference point.
    Example: A car moving on a road is in motion.

(ii) Speed and Velocity:

  • Speed: It is the rate of change of distance and has no direction (scalar quantity).
    Example: A car moving at 60 km/h.
  • Velocity: It is the rate of change of displacement and includes direction (vector quantity).
    Example: A car moving 60 km/h east.

2.3 Describe different types of motion. Also give examples.

  1. Translational Motion: Movement in a straight or curved path.
    Example: A car driving on a straight road or a ball rolling downhill.
  2. Rotational Motion: Movement around a fixed axis.
    Example: The spinning of a fan.
  3. Oscillatory Motion: Repeated to-and-fro motion.
    Example: The swinging of a pendulum.
  4. Random Motion: Unpredictable movement in any direction.
    Example: The movement of dust particles in the air.

2.4 Explain the difference between distance and displacement.

  • Distance:
    • The total path covered by an object.
    • It is a scalar quantity (only magnitude).
    • Example: If a person walks 4 m north and then 3 m south, the distance is 4+3=7 m
  • Displacement:
    • The shortest straight-line distance between the initial and final position of an object.
    • It is a vector quantity (magnitude and direction).
    • Example: For the same movement above, displacement = 4−3=1 m north.

2.5 What do gradients of distance-time graph and speed-time graph represent? Explain it by drawing diagrams.

  • Distance-Time Graph:
    • The gradient (slope) represents the speed. A steeper slope means higher speed.
    • Example: A straight, slanted line shows uniform speed, while a curved line shows acceleration or deceleration.
  • Speed-Time Graph:
    • The gradient represents acceleration. A straight, inclined line shows uniform acceleration.
    • Example: If the slope is zero (horizontal line), the speed is constant.

2.6 Prove that the area under speed-time graph is equal to the distance covered by an object.

  • The area under a speed-time graph represents the product of speed and time, which gives distance.
  • Proof:
    • Speed = Distance ÷ Time → Distance = Speed × Time
    • For a speed-time graph, the area of a rectangle (or triangle for acceleration) gives the distance:
      • Area = Base × Height = Time × Speed = Distance.
    • Example: For a car moving at 10 m/s for 5 seconds, the graph’s area = 10×5=50 m

2.7 How equations of motion can be applied to bodies moving under the action of gravity?

  • Equations of motion are:
    1. v=u+at
    2. s=ut+1/2at2
    3. v2=u2+2as
  • For objects in free fall:
    • Initial velocity u=0u = 0 (if dropped).
    • Acceleration a=g=9.8 m/s2 (gravity).
  • Example: If a ball is dropped from a height of 20 m:
    • Use s=1/2gt2
      20=1/2(9.8)t2 → t=2.02 s
      The equations help determine time, velocity, or height for objects under gravity.

Chapter 1 Physical Quantities and Measurements Solved Exercise

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MCQ 1

Statement: The instrument most suitable for measuring the thickness of a few sheets of cardboard is:
Options:
(a) Metre rule
(b) Measuring tape
(c) Vernier Callipers
(d) Micrometer screw gauge
Answer: (d) Micrometer screw gauge
Explanation: A micrometer screw gauge is specifically designed to measure very small thicknesses, such as the thickness of thin materials like sheets of cardboard, with high precision.

MCQ 2

Statement: One femtometre is equal to:
Options:
(a) 10−9 m
(b) 1015 m
(c) 10−15 m
(d) 105 m
Answer: (c) 10−15 m
Explanation: A femtometre (fm) is a unit of length equal to 10−15 metres, commonly used in nuclear physics to measure distances at the subatomic level.

MCQ 3

Statement: A light year is a unit of:
Options:
(a) Light
(b) Time
(c) Distance
(d) Speed
Answer: (c) Distance
Explanation: A light year is the distance that light travels in one year in a vacuum, which is approximately 9.46×1012 kilometers.

MCQ 4

Statement: Which one is a non-physical quantity?
Options:
(a) Distance
(b) Density
(c) Colour
(d) Temperature
Answer: (c) Colour
Explanation: Colour is a perceptual property and not a measurable physical quantity like distance, density, or temperature.

MCQ 5

Statement: When using a measuring cylinder, one precaution to take is to:
Options:
(a) Check for the zero error
(b) Look at the meniscus from below the level of the water surface
(c) Take several readings by looking from more than one direction
(d) Position the eye in line with the bottom of the meniscus
Answer: (d) Position the eye in line with the bottom of the meniscus
Explanation: To ensure accurate readings, the observer must position their eye level with the bottom of the meniscus, which is the curved surface of the liquid.

MCQ 6

Statement: Volume of water consumed by you per day is estimated in:
Options:
(a) Millilitre
(b) Litre
(c) Kilogram
(d) Cubic metre
Answer: (b) Litre
Explanation: The volume of water consumption is typically measured in litres, which is a convenient unit for daily use.

MCQ 7

Statement: A displacement can is used to measure:
Options:
(a) Mass of a liquid
(b) Mass of a solid
(c) Volume of a liquid
(d) Volume of a solid
Answer: (d) Volume of a solid
Explanation: A displacement can is used to measure the volume of an irregularly shaped solid by observing the amount of liquid it displaces.

MCQ 8

Statement: Two rods with lengths 12.321 cm and 10.3 cm are placed side by side, the difference in their lengths is:
Options:
(a) 2.02 cm
(b) 2.0 cm
(c) 2.021 cm
(d) 2.021 cm
Answer: (b) 2.0 cm
Explanation: The difference in length is calculated as 12.321−10.3=2.021, but the result is rounded off to 2.02.0 cm based on the significant figures.

MCQ 9

Statement: Which of the following measures are likely to represent the thickness of a sheet of this book?
Options:
(a) 6×10−56 m
(b) 1×10−41 m
(c) 1.2×10−15 m
(d) 4×10−24 m
Answer: (b) 1×10−41 m
Explanation: The thickness of a sheet of paper in a book is typically in the range of 10−4 meters, equivalent to 0.1 mm.

1.1 Can a non-physical quantity be measured? If yes, then how?

No, a non-physical quantity, such as emotions, feelings, or color, cannot be measured directly because they are not tangible. However, we can assess them indirectly through surveys, psychological methods, or other qualitative approaches.

1.2 What is measurement? Name its two parts.

Measurement is the process of comparing an unknown quantity with a standard quantity of the same kind. The two parts of a measurement are:

  1. Numerical value (indicates the magnitude).
  2. Unit (specifies the standard of measurement, e.g., meters, kilograms).

1.3 Why do we need a standard unit for measurements?

We need standard units to ensure consistency, reliability, and uniformity in measurements. Without standard units, comparing and sharing results across different places or systems would become difficult and confusing.

1.4 Write the names of three base quantities and three derived quantities.

Base quantities:

  1. Length
  2. Mass
  3. Time

Derived quantities:

  1. Speed (derived from length/time)
  2. Volume (derived from length³)
  3. Force (derived from mass × acceleration).

1.5 Which SI unit will you use to express the height of your desk?

The height of a desk is typically expressed in meters (m) or centimeters (cm), depending on its size.

1.6 Write the names and symbols of all SI base units.

  1. Length: Meter (m)
  2. Mass: Kilogram (kg)
  3. Time: Second (s)
  4. Electric current: Ampere (A)
  5. Temperature: Kelvin (K)
  6. Amount of substance: Mole (mol)
  7. Luminous intensity: Candela (cd)

1.7 Why is a prefix used? Name three sub-multiples and three multiples with their symbols.

Why prefixes are used: Prefixes are added to SI units to express very large or very small quantities in a convenient way, avoiding the need for many zeros.

Sub-multiples:

  1. Milli (m) = 10−3
  2. Micro (µ) = 10−6
  3. Nano (n) = 10−9

Multiples:

  1. Kilo (k) = 103
  2. Mega (M) = 106
  3. Giga (G) = 109

1.8 What is meant by:

(a) 55 pm = 5×10−12 meters (picometers, used to measure atomic distances).
(b) 1515 ns = 15×10−9 seconds (nanoseconds, used for time intervals in electronics).
(c) 66 µm = 6×10−6 meters (micrometers, used for measuring microscopic distances).
(d) 55 fs = 5×10−15 seconds (femtoseconds, used in ultrafast phenomena).

1.9 For what purpose is a Vernier Callipers used?

A Vernier Callipers is used to measure:

  1. The external dimensions of an object (e.g., diameter of a cylinder).
  2. The internal dimensions of an object (e.g., diameter of a hole).
  3. The depth of an object.

Main parts:

  • Main scale
  • Vernier scale

How least count is found:
The least count is calculated as: Least count=Smallest division on main scale/Total number of divisions on Vernier scale

1.10 State least count and Vernier scale reading as shown in the figure and hence find the length.

Least count: Assume the smallest division on the main scale is 1 mm and there are 10 divisions on the Vernier scale. Least count=110=0.1 mm

Vernier scale reading: Check the alignment of the Vernier and main scale; the reading will be calculated as: Length=Main scale reading+(Vernier division×Least count).

(Values can be estimated based on the image provided.)

1.11 Which reading out of A, B, and C shows the correct length and why?

The correct length is the one where the zero of the Vernier scale aligns perfectly with the reading on the main scale. (Specific answer depends on analyzing the given figure in detail.)

C.1.1 In what unit will you express each of the following?

(a) Thickness of a five-rupee coin:
The thickness of a coin is small, so it is best measured in millimeters (mm) or micrometers (µm) for greater precision.

(b) Length of a book:
The length of a book can be expressed in centimeters (cm) or millimeters (mm), depending on the level of detail required.

(c) Length of a football field:
A football field is large, so its length is expressed in meters (m) or sometimes in yards (if using non-metric units).

(d) The distance between two cities:
The distance between two cities is usually measured in kilometers (km) because the distance is large.

(e) Mass of a five-rupee coin:
The mass of a coin is small, so it is measured in grams (g) or milligrams (mg) for high precision.

(f) Mass of your school bag:
The mass of a school bag is measured in kilograms (kg) because it is heavier than smaller objects like a coin.

(g) Duration of your class period:
The duration of a class period is expressed in minutes (min) or hours (h).

(h) Volume of petrol filled in the tank of a car:
The volume of petrol is expressed in litres (L), which is the standard unit for liquid volumes.

(i) Time to boil one litre of milk:
The time to boil milk is usually measured in minutes (min) or seconds (s), depending on how precise the timing is.

Chapter 11 Biostatistics – 9th Class New Syllabus for Punjab & Federal Boards

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Explore Chapter 11: Biostatistics from the 9th class new syllabus. Comprehensive guide for Lahore Board, all Punjab Boards, and Federal Board students, tailored to meet the updated syllabus requirements.

1. What is the primary purpose of biostatistics?

  • Options:
    a) To analyze financial data
    b) To apply statistical methods to biological sciences
    c) To design engineering models
    d) To study historical events
  • Answer: b) To apply statistical methods to biological sciences
  • Explanation:
    Biostatistics is the branch of statistics that focuses on applying statistical methods to biological and medical research.
  • Tip:
    Remember, “bio” refers to life sciences and “statistics” involves data analysis. Combine both for the purpose.

2. In biostatistics, which method is used to predict future outcomes based on current data?

  • Options:
    a) Designing experiments
    b) Interpreting results
    c) Predicting outcomes
    d) Analyzing data
  • Answer: c) Predicting outcomes
  • Explanation:
    In biostatistics, prediction involves using models and analysis of current data to make future estimations, like disease trends.
  • Tip:
    Focus on the keyword “future outcomes,” which points to prediction.

3. Which of the following best describes the mean of a data set?

  • Options:
    a) The most frequently occurring value
    b) The middle value when data is ordered
    c) The sum of all values divided by the number of values
    d) The difference between the highest and lowest values
  • Answer: c) The sum of all values divided by the number of values
  • Explanation:
    The mean is calculated by adding all values and dividing by how many values there are. It’s often referred to as the “average.”
  • Tip:
    Mean = Total Sum ÷ Number of Values. This formula always helps!

4. If the data set is 5, 8, 12, 15, 20, what is the median?

  • Options:
    a) 8
    b) 12
    c) 15
    d) 20
  • Answer: b) 12
  • Explanation:
    The median is the middle value when data is arranged in order. Here, 12 is the middle value.
  • Tip:
    For odd-numbered data, directly find the middle number after sorting.

5. What is the mean of the data set: 7, 8, 9, 10, 11?

  • Options:
    a) 7
    b) 8
    c) 9
    d) 10
  • Answer: c) 9
  • Explanation:
    Mean = (7 + 8 + 9 + 10 + 11) ÷ 5 = 45 ÷ 5 = 9.
  • Tip:
    Use the formula and divide step by step to avoid calculation mistakes.

6. When the number of values in a data set is even, how is the median calculated?

  • Options:
    a) By choosing the middle value
    b) By taking the average of the two middle values
    c) By selecting the most frequent value
    d) By adding all values and dividing by the total number of values
  • Answer: b) By taking the average of the two middle values
  • Explanation:
    For an even number of values, there’s no single middle value. You take the average of the two central numbers.
  • Tip:
    If there are two middle values, calculate their mean to get the median.

7. In a data set with values 3, 3, 6, 7, 8, 9, 9, what is the mode?

  • Options:
    a) 3
    b) 9
    c) Both 3 and 9
    d) 7
  • Answer: c) Both 3 and 9
  • Explanation:
    The mode is the value(s) that appear most frequently. Here, both 3 and 9 occur twice.
  • Tip:
    A data set can have more than one mode (bimodal), or no mode at all if no number repeats.

8. If a data set has no repeated values, what is the mode?

  • Options:
    a) The highest value
    b) The average of the data set
    c) There is no mode
    d) The median value
  • Answer: c) There is no mode
  • Explanation:
    The mode refers to the most frequent value. If no value repeats, there is no mode.
  • Tip:
    When solving, check if any numbers are repeated. If none, the mode does not exist.

A. Select the correct answers

9. In a bar chart, what does the height or length of each bar represent?

  • Options:
    a) The total number of categories
    b) The value of the corresponding category
    c) The average of all values
    d) The difference between the highest and lowest values
  • Answer: b) The value of the corresponding category
  • Explanation:
    The height or length of each bar in a bar chart shows the value of the specific category it represents.
  • Tip:
    Look at the y-axis (vertical axis) for the values.

10. When constructing a bar chart, which axis usually represents the categories?

  • Options:
    a) Vertical axis (y-axis)
    b) Horizontal axis (x-axis)
    c) Both axes equally represent the categories
    d) Neither axis represents the categories
  • Answer: b) Horizontal axis (x-axis)
  • Explanation:
    In a bar chart, the x-axis typically represents the categories (e.g., types of items), while the y-axis represents the values (e.g., quantities).
  • Tip:
    Remember: categories = x-axis, values = y-axis.

B. Write short answers

1. Define biostatistics.
Biostatistics is the application of statistical methods to biological and medical research. It helps analyze data to make informed decisions in health and science.

2. What is the median of a data set?
The median is the middle value of a data set when arranged in ascending or descending order.

3. How is the mean calculated?
The mean is calculated by dividing the sum of all values by the total number of values. Formula:
Mean = (Sum of all values) ÷ (Number of values).

4. What does the height of a bar in a bar chart represent?
The height of a bar represents the value of the corresponding category in the data.

5. What is the mode of a data set?
The mode is the value that appears most frequently in a data set.

C. Write answers in detail

1. Explain the importance of biostatistics in the field of public health.
Biostatistics plays a vital role in public health by:

  • Analyzing disease patterns to understand their spread and prevention.
  • Guiding health policies and resource allocation.
  • Evaluating the effectiveness of treatments and interventions through statistical studies.
  • Helping in predicting future health trends based on current data.
    Example: Tracking COVID-19 cases and designing vaccination strategies.

2. Discuss the differences between mean, median, and mode. Include examples where each measure is most appropriate to use.

  • Mean: The average of all values. Used when data is evenly distributed.
    Example: Average score of students in a test.
  • Median: The middle value in a sorted data set. Useful for skewed data.
    Example: Median income in a neighborhood with varying incomes.
  • Mode: The most frequent value. Used in categorical data.
    Example: The most sold product in a store.

3. Describe the steps involved in creating a bar chart using Excel. Include a discussion on how to customize the chart for better visualization.
Steps to create a bar chart in Excel:

  1. Enter data into two columns: categories and their values.
  2. Select the data and click on the “Insert” tab.
  3. Choose the “Bar Chart” option.
  4. Customize the chart:
    • Add titles for the chart and axes.
    • Adjust colors to improve readability.
    • Add data labels to show exact values.
  5. Save or export the chart for use.

4. Provide a detailed example of how to calculate the mean, median, and mode of a data set.
Data set: 12, 22, 8, 19, 25, 15

  • Mean:
    Mean=12+22+8+19+25+156=1016=16.83\text{Mean} = \frac{12 + 22 + 8 + 19 + 25 + 15}{6} = \frac{101}{6} = 16.83
  • Median:
    Arrange in order: 8, 12, 15, 19, 22, 25.
    The middle values are 15 and 19. Median = 15+192=17\frac{15 + 19}{2} = 17.
  • Mode:
    No value repeats, so there is no mode.

5. You are given the following data set. Create a bar chart to represent the number of different types of fruits sold at a market in one week:

  • Apples: 30
  • Bananas: 45
  • Oranges: 25
  • Grapes: 20

To create the chart:

  1. List fruits and their numbers.
  2. Create a bar chart (using Excel or by hand) with fruits on the x-axis and their quantities on the y-axis.
  3. Label the chart with a title: “Fruits Sold in One Week.”

Chapter 10 Reproduction: Solved Exercise for 9th Class

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Looking for solved exercises for Chapter 10: Reproduction? Our comprehensive solutions are specifically designed for 9th Class students based on the updated 2025 syllabus. Simplified answers to key topics like binary fission, vegetative propagation, and the life cycle of flowering plants are included to make your preparation easier and more efficient.

MCQs with Solutions

1. Which of the following organisms commonly reproduce by binary fission?

  • Options:
    a) Yeast
    b) Bacteria
    c) Rhizopus
    d) Plants
  • Answer: b) Bacteria
  • Explanation: Binary fission is a simple asexual reproduction method where a single cell divides into two identical daughter cells. This process is typical in prokaryotic organisms like bacteria.
  • Tip/Trick: Remember that binary fission is exclusive to unicellular organisms like bacteria.

2. What is the primary method of reproduction in yeast?

  • Options:
    a) Binary fission
    b) Spore formation
    c) Budding
    d) Fragmentation
  • Answer: c) Budding
  • Explanation: Yeast, a unicellular fungus, reproduces primarily by budding, where a new cell forms as an outgrowth of the parent cell.
  • Tip/Trick: Associate yeast with budding since it involves a “bud” growing on the parent cell.

3. Which of the following statements is true about spore formation in fungi?

  • Options:
    a) They produce spores during sexual reproduction.
    b) They produce two kinds of spores.
    c) Spores can only grow into new fungi in dry environments.
    d) Spores are produced to withstand harsh conditions.
  • Answer: d) Spores are produced to withstand harsh conditions.
  • Explanation: Spores in fungi are designed to survive extreme environmental conditions, including dryness, heat, or lack of nutrients.
  • Tip/Trick: Focus on “harsh conditions” when considering the purpose of fungal spores.

4. What happens in some bacteria during harsh conditions?

  • Options:
    a) Creation of a bud that detaches from the cell.
    b) Formation of thick-walled endospores.
    c) Splitting the cell into two identical daughter cells.
    d) Fusion of two bacterial cells.
  • Answer: b) Formation of thick-walled endospores.
  • Explanation: Endospores are highly resistant structures formed by bacteria to protect their genetic material during unfavorable conditions.
  • Tip/Trick: Recall that endospores = bacterial survival strategy.

5. Which of the following is an example of vegetative propagation through runners?

  • Options:
    a) Potato
    b) Strawberry
    c) Onion
    d) Ginger
  • Answer: b) Strawberry
  • Explanation: Strawberries propagate vegetatively through runners, which are horizontal stems that grow along the soil’s surface and develop new plants.
  • Tip/Trick: Think of “running strawberries” for runners.

6. Which plant propagates through tubers?

  • Options:
    a) Onion
    b) Potato
    c) Ginger
    d) Garlic
  • Answer: b) Potato
  • Explanation: Tubers are thickened underground stems, such as potatoes, that store food and enable vegetative propagation.
  • Tip/Trick: Visualize a potato’s “eyes” as propagation points.

7. The horizontal aboveground stem, which produces leaves and roots at its nodes:

  • Options:
    a) Stolon
    b) Bulb
    c) Rhizome
    d) Corm
  • Answer: a) Stolon
  • Explanation: Stolons are horizontal stems that grow above the ground and help plants like strawberries propagate.
  • Tip/Trick: Stolons are “above-ground runners.”

8. Which of these does NOT help a plant in vegetative propagation?

  • Options:
    a) Rhizome
    b) Corm
    c) Runner
    d) Flower
  • Answer: d) Flower
  • Explanation: Flowers are reproductive structures for sexual reproduction, not vegetative propagation.
  • Tip/Trick: Vegetative propagation relies on structures like stems, roots, and leaves, not flowers.

9. Which part of the flower is responsible for producing pollen?

  • Options:
    a) Stigma
    b) Anther
    c) Ovary
    d) Petal
  • Answer: b) Anther
  • Explanation: The anther is a part of the stamen in flowers and is responsible for producing and releasing pollen grains.
  • Tip/Trick: Associate “anther” with “pollen production.”

10. Which of the following is NOT a part of the carpel?

  • Options:
    a) Filament
    b) Style
    c) Stigma
    d) Ovary
  • Answer: a) Filament
  • Explanation: The carpel (or pistil) is the female reproductive part of the flower and consists of the stigma, style, and ovary. The filament is a part of the stamen, the male reproductive structure.
  • Tip/Trick: Remember the carpel components as “SOS” — Stigma, Ovary, Style.

11. Which structure forms the female gametophyte in flowering plants?

  • Options:
    a) Pollen grain
    b) Ovule
    c) Anther
    d) Sepal
  • Answer: b) Ovule
  • Explanation: The ovule in flowering plants develops into the female gametophyte, which contains the egg cell and is involved in reproduction.
  • Tip/Trick: The ovule = female gametophyte, while pollen grain = male gametophyte.

12. The male gametophyte in flowering plants is known as:

  • Options:
    a) Pollen grain
    b) Embryo sac
    c) Ovary
    d) Carpel
  • Answer: a) Pollen grain
  • Explanation: The pollen grain is the male gametophyte, carrying the male reproductive cells (sperm). It fertilizes the ovule during pollination.
  • Tip/Trick: Associate “pollen” with “male” and “grain” with small particles.

13. In the life cycle of flowering plants, which structure is triploid (3n)?

  • Options:
    a) Egg
    b) Fusion nucleus
    c) Endosperm nucleus
    d) Sperm
  • Answer: c) Endosperm nucleus
  • Explanation: The endosperm nucleus is triploid (3n) as it forms after the fusion of one sperm with two polar nuclei during double fertilization.
  • Tip/Trick: Triploid = three sets of chromosomes, found in endosperm for nutrient storage.

14. Embryo sac is formed inside:

  • Options:
    a) Filament
    b) Anther
    c) Style
    d) Ovule
  • Answer: d) Ovule
  • Explanation: The embryo sac is the female gametophyte and is located inside the ovule of flowering plants.
  • Tip/Trick: Embryo sac = female gametophyte = ovule.

15. Double fertilization involves:

  • Options:
    a) Fertilization of the egg by two male gametes.
    b) Fertilization of two eggs in the same embryo sac by two sperms.
    c) Fertilization of the egg and the fusion nucleus by two sperms.
    d) Fertilization of the egg and the tube cell by two sperms.
  • Answer: c) Fertilization of the egg and the fusion nucleus by two sperms.
  • Explanation: In double fertilization, one sperm fertilizes the egg to form a zygote (2n), while the other sperm fuses with two polar nuclei to form the endosperm nucleus (3n).
  • Tip/Trick: Think “double duty” — one sperm for the egg, the other for the polar nuclei.

Short Questions with Detailed Answers

1. Write a short note on budding in yeast.

  • Answer: Budding in yeast is a form of asexual reproduction. A small bud grows on the parent cell, enlarges, and eventually detaches to form a new yeast cell. This process ensures rapid multiplication under favorable conditions.

2. Write a short note on spore formation in fungi.

  • Answer: Spore formation in fungi is an asexual reproduction method where spores are produced inside a sporangium. These spores are resistant to harsh environmental conditions and germinate to form new fungi when conditions are favorable.

3. What are the advantages of spore formation in fungi and bacteria?

  • Answer:
    • Advantages:
      1. Allows survival in unfavorable conditions due to spore resistance.
      2. Enables rapid reproduction and dispersal over large areas.
      3. Requires minimal resources for reproduction.

4. Describe how vegetative propagation occurs through runners.

  • Answer: Runners are horizontal stems that grow above the ground. Nodes on these stems produce roots and shoots, which develop into new plants. This is commonly seen in plants like strawberries.

5. State how potatoes reproduce through tubers.

  • Answer: Potatoes reproduce vegetatively through tubers, which are swollen underground stems. “Eyes” on the tubers sprout into new shoots and roots, growing into new potato plants.

6. Describe the advantages and disadvantages of vegetative propagation.

  • Answer:
    • Advantages:
      1. Produces genetically identical offspring.
      2. Quick and efficient reproduction.
      3. No need for seeds or pollination.
    • Disadvantages:
      1. No genetic diversity, making plants susceptible to diseases.
      2. Overcrowding can occur, leading to competition for resources.

7. Name the four whorls present in a flower and their components.

  • Answer:
    1. Calyx: Made of sepals, protects the flower bud.
    2. Corolla: Made of petals, attracts pollinators.
    3. Androecium: Made of stamens, produces pollen.
    4. Gynoecium: Made of carpels, contains ovules.

8. Briefly describe the formation of the egg cell and polar nuclei within the embryo sac of a flower.

  • Answer: The embryo sac forms inside the ovule during megasporogenesis. One megaspore divides to form eight nuclei, which organize into cells, including the egg cell (near the micropylar end) and two polar nuclei (in the center of the sac).

9. Differentiate between:

  • i. Asexual and sexual reproduction:
    • Asexual: Single parent, no gametes, offspring identical.
    • Sexual: Two parents, gametes involved, offspring genetically diverse.
  • ii. Binary fission in bacteria and amoeba:
    • Bacteria: Simple splitting without a nucleus.
    • Amoeba: Nucleus divides first, followed by cytoplasm.
  • iii. Stolon and rhizome:
    • Stolon: Horizontal above-ground stem.
    • Rhizome: Horizontal underground stem.
  • iv. Bulb and corm:
    • Bulb: Modified stem with fleshy leaves (e.g., onion).
    • Corm: Solid, fleshy underground stem (e.g., taro).

C. Write answers in detail

1. Explain the process of binary fission in bacteria and describe how it leads to the formation of two daughter bacteria.
Binary fission is the process by which bacteria reproduce. In this process:

  • The bacterial cell grows and its DNA (genetic material) is copied.
  • The cell divides into two parts, with each part getting one copy of the DNA.
  • This results in two identical daughter bacteria.

2. What do you mean by vegetative propagation? Differentiate among different plant structures modified for vegetative propagation.
Vegetative propagation is a way plants reproduce without seeds. New plants grow from parts like roots, stems, or leaves of the parent plant.

  • Modified roots: Examples are sweet potato and carrot.
  • Modified stems: Examples are potato (tuber) and ginger (rhizome).
  • Modified leaves: An example is Bryophyllum, where buds grow on leaves.

3. Describe the ways by which humans can grow new plants by using the vegetative parts of the parent plants.
Humans grow new plants using methods like:

  • Cutting: A part of a plant like a stem or leaf is cut and planted to grow into a new plant. Example: rose.
  • Grafting: Two plants are joined together so they grow as one. Example: mango.
  • Layering: A branch of a plant is bent to the ground and covered with soil. It grows roots and becomes a new plant. Example: jasmine.

4. Define sporophyte and gametophyte. State their roles in the life cycle of plants.

  • Sporophyte: The part of the plant that produces spores. It is usually larger, like the main body of a plant.
  • Gametophyte: The part that produces gametes (sperm and egg cells). It is smaller and often hidden.
    These two alternate in the life cycle of plants. Spores from the sporophyte grow into gametophytes, and gametophytes create gametes that form a new sporophyte.

5. Explain the life cycle of flowering plants, focusing on the alternation between the gametophyte and sporophyte generations.
The life cycle of flowering plants has two stages:

  • Sporophyte stage: The main plant that produces flowers and seeds. It creates spores.
  • Gametophyte stage: The spores grow into tiny structures inside the flower, which produce male and female gametes.
    When the gametes fuse (fertilization), a seed forms, and the cycle starts again.

6. Describe how the female gametophyte (embryo sac) develops within the ovule of a flower.

  • Inside the ovule of a flower, a cell divides to form a structure with 8 nuclei.
  • This structure becomes the embryo sac, which is the female gametophyte.
  • The embryo sac has one egg cell, which combines with sperm during fertilization to form a seed.

Chapter 9: Physiology – Class 9th Biology Solved

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Delve into Chapter 9: Physiology from the Class 9th Biology New Syllabus, tailored for Lahore and Punjab Boards. This detailed post covers the human organ systems, their physiological functions, and related concepts, including the nervous system, circulatory system, and respiration. Includes solved MCQs, short questions, and exam tips to ensure success in your exams. A must-read for Class 9 students preparing for Punjab board exams.

1. Which of the following plant nutrients is required in large amounts?

Options:
a) Iron
b) Potassium
c) Phosphorus
d) Boron
Answer: b) Potassium
Explanation: Potassium is a macronutrient essential for plant growth. It regulates water uptake, photosynthesis, and enzyme activation.
Tip: Macronutrients like potassium, nitrogen, and phosphorus are needed in large amounts, while micronutrients like boron and iron are required in smaller amounts.

2. Which element is required by plants for the formation of chlorophyll?

Options:
a) Phosphorus
b) Calcium
c) Magnesium
d) Sulphur
Answer: c) Magnesium
Explanation: Magnesium is a core element in chlorophyll, enabling photosynthesis by capturing sunlight.
Tip: Link magnesium to chlorophyll by remembering that green leafy vegetables are rich in magnesium.

3. The primary function of root hairs is:

Options:
a) Transport of nutrients
b) Storage of food
c) Increase surface area for absorption
d) Synthesis of proteins
Answer: c) Increase surface area for absorption
Explanation: Root hairs increase the surface area, making water and nutrient uptake from the soil more efficient.
Tip: Think of root hairs as “extensions” that maximize absorption.

4. Root hairs absorb salts from soil by:

Options:
a) Diffusion
b) Active transport
c) Filtration
d) Osmosis
Answer: b) Active transport
Explanation: Active transport uses energy (ATP) to move salts and nutrients from lower to higher concentrations in the roots.
Tip: Remember, “active” means energy is required.

5. Water moves from the soil into root cells by:

Options:
a) Osmosis
b) Facilitated diffusion
c) Active transport
d) Bulk flow
Answer: a) Osmosis
Explanation: Osmosis is the movement of water from high to low water potential across a semi-permeable membrane.
Tip: Water = Osmosis; nutrients = Active transport.

6. The transpiration is regulated by:

Options:
a) Mesophyll
b) Guard cells
c) Xylem
d) Phloem
Answer: b) Guard cells
Explanation: Guard cells control the opening and closing of stomata, regulating water loss through transpiration.
Tip: Link “guard” cells with “guarding” the stomata.

7. Under which condition will there be high rate of transpiration?

Options:
a) High humidity
b) Low light intensity
c) Wind
d) Waterlogged soil
Answer: c) Wind
Explanation: Wind removes water vapor around leaves, increasing the rate of transpiration.
Tip: Think of wind as a “dryer” increasing water loss.

8. Which ion plays a role in the opening of stomata?

Options:
a) Sodium (Na⁺)
b) Potassium (K⁺)
c) Calcium (Ca²⁺)
d) Magnesium (Mg²⁺)
Answer: b) Potassium (K⁺)
Explanation: Potassium ions regulate the turgor pressure in guard cells, causing stomata to open or close.
Tip: Associate potassium (K⁺) with “Key” for stomata.

9. In most plants, the food is transported in the form of:

Options:
a) Glucose
b) Sucrose
c) Starch
d) Maltose
Answer: b) Sucrose
Explanation: Sucrose is water-soluble and easily transported through the phloem.
Tip: Glucose is stored, sucrose is transported.

10. What is TRUE according to the pressure flow mechanism of food transport?

Options:
a) Water enters the source, creating pressure.
b) Water is pulled from the sink.
c) Movement of food in phloem is due to gravity.
d) Solutes move from low to high concentration.
Answer: a) Water enters the source, creating pressure.
Explanation: In the source (e.g., leaves), sugar concentration draws water in, creating pressure that pushes food through the phloem to the sink (e.g., roots).
Tip: Think of the source as the “pump” for food transport.

MCQs:

11. Succulent organs are present in:
Options:
a) Xerophytes
b) Hydrophytes
c) Mesophytes
d) Halophytes
Answer: a) Xerophytes
Explanation: Xerophytes are plants adapted to dry environments, and their succulent organs store water to survive droughts.
Tip: “Xero” means dry, so remember xerophytes store water.

Short Answers (3 lines each):

  1. Define mineral nutrition in plants.
    Mineral nutrition involves the uptake of essential nutrients like nitrogen, phosphorus, and potassium from the soil, which plants use for growth and development.
  2. Define macronutrients and micronutrients and give examples.
    Macronutrients are needed in large amounts (e.g., nitrogen, phosphorus), while micronutrients are needed in small amounts (e.g., iron, zinc).
  3. State the roles of nitrogen and magnesium in plants.
    Nitrogen helps in protein synthesis and plant growth. Magnesium is essential for chlorophyll production and photosynthesis.
  4. Define transpiration and its types.
    Transpiration is the loss of water vapor from plant surfaces, mainly through stomata. Types include stomatal, cuticular, and lenticular transpiration.
  5. How is the transpiration pull important in plants?
    Transpiration pull helps transport water and minerals from roots to leaves, supporting photosynthesis and maintaining plant structure.
  6. Transpiration is the loss of water from plants. Is it harmful?
    Transpiration can be harmful during droughts, but it also cools the plant, supports nutrient transport, and maintains water flow.
  7. Differentiate between Xylem and Phloem:
  • Xylem: Transports water and minerals.
  • Phloem: Transports food (sucrose).
  1. How do the plants of rubber and keekar excrete their wastes?
    Rubber trees excrete waste as latex, while keekar trees excrete waste through leaf shedding and bark.

Detailed Answers:

1. Describe the events involved in the opening and closing of stomata.
Stomata are tiny openings on the surface of leaves, controlled by guard cells. Their opening and closing depend on water movement and ion concentration:

  • Opening of stomata:
    • During the day, guard cells actively absorb potassium ions (K⁺) from surrounding cells.
    • This increases the solute concentration inside guard cells, causing water to enter by osmosis.
    • The guard cells swell and become turgid, bending outward to open the stomatal pore.
    • This allows gases like carbon dioxide to enter for photosynthesis and oxygen to exit.
  • Closing of stomata:
    • At night or in dry conditions, potassium ions leave the guard cells, reducing their solute concentration.
    • Water moves out of the guard cells, making them flaccid and closing the stomatal pore.
    • This prevents water loss through transpiration.

Importance:
The stomata help regulate water loss, maintain plant hydration, and allow essential gas exchange for photosynthesis and respiration.

2. Explain the internal structure of roots and describe the uptake of salt and water.
Roots are specially adapted for water and mineral uptake. The key parts are:

  • Root hairs: Tiny hair-like structures increase surface area for water and nutrient absorption.
  • Cortex: A layer of loosely packed cells that allows easy movement of water and nutrients.
  • Endodermis: A barrier that ensures selective absorption of minerals.
  • Xylem and phloem: Xylem transports water and minerals, while phloem transports food.

Water uptake:

  • Water enters the root hairs from the soil by osmosis (movement from high to low water potential).
  • It travels through the root cortex to the xylem, either through cell walls (apoplast pathway) or through the cytoplasm (symplast pathway).

Salt uptake:

  • Minerals are absorbed by active transport, a process that requires energy in the form of ATP. This allows plants to take up nutrients even when their concentration in the soil is low.

Importance:
This mechanism ensures the plant receives water and nutrients for growth, photosynthesis, and development.

3. Describe temperature, wind, and humidity as factors affecting transpiration.
Transpiration is the loss of water vapor from plant leaves, mainly through stomata. It is influenced by environmental factors:

  • Temperature:
    • High temperatures increase evaporation of water from leaf surfaces, raising the rate of transpiration.
    • At low temperatures, transpiration slows down because evaporation is reduced.
  • Wind:
    • Wind blows away water vapor around the leaves, creating a low-humidity environment.
    • This increases the water concentration gradient between the leaf and the surrounding air, speeding up transpiration.
  • Humidity:
    • High humidity (moist air) reduces the rate of transpiration because the air already has a high water content.
    • Low humidity (dry air) increases transpiration as water vapor diffuses more quickly.

Importance:
These factors affect a plant’s water balance, cooling, and nutrient transport.

4. Describe the mechanism of transport of water and salt in plants.
Plants transport water and salts from the roots to other parts through the xylem:

  • Root pressure:
    • Minerals actively absorbed by roots create a pressure that pushes water upward in the xylem.
  • Capillary action:
    • Water rises in the narrow xylem vessels due to adhesion (water sticking to the vessel walls) and cohesion (water molecules sticking to each other).
  • Transpiration pull:
    • As water evaporates from leaves during transpiration, it creates a negative pressure in the xylem that pulls water upward from the roots.

Salt transport:
Salts are absorbed by active transport and move with water through the xylem. This ensures the plant receives essential nutrients for growth.

5. Explain the mechanism of food translocation by Pressure Flow Mechanism.
Food (mainly sucrose) is transported through the phloem using the pressure flow mechanism:

  • At the source (e.g., leaves):
    • Sucrose is actively loaded into the phloem sieve tubes.
    • This increases solute concentration, causing water from the xylem to enter by osmosis.
    • The pressure builds up, pushing the sucrose solution (sap) toward the sink.
  • At the sink (e.g., roots or fruits):
    • Sucrose is actively unloaded and used for energy or storage.
    • Water exits the phloem, reducing pressure and maintaining flow from source to sink.

Importance:
This mechanism efficiently transports food to growing parts, storage organs, and roots.

6. How do the plants excrete extra water and salts from their bodies?
Plants excrete waste through:

  • Transpiration: Excess water is lost as vapor through stomata.
  • Guttation: In some plants, water droplets are expelled from leaf edges (hydathodes) during the night or early morning.
  • Salt excretion: Halophytes excrete salts through special salt glands. Other plants store excess salts in leaves, which are later shed.

Importance:
These processes help plants maintain a balance of water and salts, preventing toxicity and dehydration.

7. Describe the process of gaseous exchange in plants.
Gaseous exchange in plants occurs through stomata and lenticels:

  • Daytime:
    • Carbon dioxide enters through stomata for photosynthesis.
    • Oxygen, a byproduct of photosynthesis, exits through stomata.
  • Nighttime:
    • Plants take in oxygen for respiration and release carbon dioxide as a waste product.

Importance:
This exchange is crucial for photosynthesis and cellular respiration, providing energy and maintaining life processes.

8. Describe the mechanisms/adaptations in plants for excretion of wastes.
Plants manage waste through:

  • Storage in vacuoles: Toxic substances are stored in vacuoles or in old tissues like bark and leaves.
  • Excretion through leaves: Some waste products are expelled when leaves shed.
  • Salt excretion: Halophytes (e.g., mangroves) excrete salts through specialized salt glands.

Importance:
These mechanisms help plants survive in challenging environments and prevent waste accumulation.

9. Explain osmotic adjustments in hydrophytes, xerophytes, and halophytes.

  • Hydrophytes:
    • Adapted to water-rich environments.
    • Excess water is stored or lost through transpiration.
  • Xerophytes:
    • Adapted to dry environments.
    • Store water in succulent tissues, have thick waxy cuticles, and small leaves to reduce water loss.
  • Halophytes:
    • Adapted to saline conditions.
    • Excrete excess salts through salt glands or store salts in vacuoles.

Importance:
These adjustments allow plants to survive and grow in their specific environments1. Describe the events involved in the opening and closing of stomata.
Stomata are tiny openings on the surface of leaves, controlled by guard cells. Their opening and closing depend on water movement and ion concentration:

  • Opening of stomata:
    • During the day, guard cells actively absorb potassium ions (K⁺) from surrounding cells.
    • This increases the solute concentration inside guard cells, causing water to enter by osmosis.
    • The guard cells swell and become turgid, bending outward to open the stomatal pore.
    • This allows gases like carbon dioxide to enter for photosynthesis and oxygen to exit.
  • Closing of stomata:
    • At night or in dry conditions, potassium ions leave the guard cells, reducing their solute concentration.
    • Water moves out of the guard cells, making them flaccid and closing the stomatal pore.
    • This prevents water loss through transpiration.

Importance:
The stomata help regulate water loss, maintain plant hydration, and allow essential gas exchange for photosynthesis and respiration.

2. Explain the internal structure of roots and describe the uptake of salt and water.
Roots are specially adapted for water and mineral uptake. The key parts are:

  • Root hairs: Tiny hair-like structures increase surface area for water and nutrient absorption.
  • Cortex: A layer of loosely packed cells that allows easy movement of water and nutrients.
  • Endodermis: A barrier that ensures selective absorption of minerals.
  • Xylem and phloem: Xylem transports water and minerals, while phloem transports food.

Water uptake:

  • Water enters the root hairs from the soil by osmosis (movement from high to low water potential).
  • It travels through the root cortex to the xylem, either through cell walls (apoplast pathway) or through the cytoplasm (symplast pathway).

Salt uptake:

  • Minerals are absorbed by active transport, a process that requires energy in the form of ATP. This allows plants to take up nutrients even when their concentration in the soil is low.

Importance:
This mechanism ensures the plant receives water and nutrients for growth, photosynthesis, and development.

3. Describe temperature, wind, and humidity as factors affecting transpiration.
Transpiration is the loss of water vapor from plant leaves, mainly through stomata. It is influenced by environmental factors:

  • Temperature:
    • High temperatures increase evaporation of water from leaf surfaces, raising the rate of transpiration.
    • At low temperatures, transpiration slows down because evaporation is reduced.
  • Wind:
    • Wind blows away water vapor around the leaves, creating a low-humidity environment.
    • This increases the water concentration gradient between the leaf and the surrounding air, speeding up transpiration.
  • Humidity:
    • High humidity (moist air) reduces the rate of transpiration because the air already has a high water content.
    • Low humidity (dry air) increases transpiration as water vapor diffuses more quickly.

Importance:
These factors affect a plant’s water balance, cooling, and nutrient transport.

4. Describe the mechanism of transport of water and salt in plants.
Plants transport water and salts from the roots to other parts through the xylem:

  • Root pressure:
    • Minerals actively absorbed by roots create a pressure that pushes water upward in the xylem.
  • Capillary action:
    • Water rises in the narrow xylem vessels due to adhesion (water sticking to the vessel walls) and cohesion (water molecules sticking to each other).
  • Transpiration pull:
    • As water evaporates from leaves during transpiration, it creates a negative pressure in the xylem that pulls water upward from the roots.

Salt transport:
Salts are absorbed by active transport and move with water through the xylem. This ensures the plant receives essential nutrients for growth.

5. Explain the mechanism of food translocation by Pressure Flow Mechanism.
Food (mainly sucrose) is transported through the phloem using the pressure flow mechanism:

  • At the source (e.g., leaves):
    • Sucrose is actively loaded into the phloem sieve tubes.
    • This increases solute concentration, causing water from the xylem to enter by osmosis.
    • The pressure builds up, pushing the sucrose solution (sap) toward the sink.
  • At the sink (e.g., roots or fruits):
    • Sucrose is actively unloaded and used for energy or storage.
    • Water exits the phloem, reducing pressure and maintaining flow from source to sink.

Importance:
This mechanism efficiently transports food to growing parts, storage organs, and roots.

6. How do the plants excrete extra water and salts from their bodies?
Plants excrete waste through:

  • Transpiration: Excess water is lost as vapor through stomata.
  • Guttation: In some plants, water droplets are expelled from leaf edges (hydathodes) during the night or early morning.
  • Salt excretion: Halophytes excrete salts through special salt glands. Other plants store excess salts in leaves, which are later shed.

Importance:
These processes help plants maintain a balance of water and salts, preventing toxicity and dehydration.

7. Describe the process of gaseous exchange in plants.
Gaseous exchange in plants occurs through stomata and lenticels:

  • Daytime:
    • Carbon dioxide enters through stomata for photosynthesis.
    • Oxygen, a byproduct of photosynthesis, exits through stomata.
  • Nighttime:
    • Plants take in oxygen for respiration and release carbon dioxide as a waste product.

Importance:
This exchange is crucial for photosynthesis and cellular respiration, providing energy and maintaining life processes.

8. Describe the mechanisms/adaptations in plants for excretion of wastes.
Plants manage waste through:

  • Storage in vacuoles: Toxic substances are stored in vacuoles or in old tissues like bark and leaves.
  • Excretion through leaves: Some waste products are expelled when leaves shed.
  • Salt excretion: Halophytes (e.g., mangroves) excrete salts through specialized salt glands.

Importance:
These mechanisms help plants survive in challenging environments and prevent waste accumulation.

9. Explain osmotic adjustments in hydrophytes, xerophytes, and halophytes.

  • Hydrophytes:
    • Adapted to water-rich environments.
    • Excess water is stored or lost through transpiration.
  • Xerophytes:
    • Adapted to dry environments.
    • Store water in succulent tissues, have thick waxy cuticles, and small leaves to reduce water loss.
  • Halophytes:
    • Adapted to saline conditions.
    • Excrete excess salts through salt glands or store salts in vacuoles.

Importance:
These adjustments allow plants to survive and grow in their specific environments

Chapter 8: Bioenergetics – Class 9th Biology

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Explore Chapter 8: Bioenergetics from the Class 9th Biology New Syllabus for Punjab Boards. This post provides detailed insights into photosynthesis, cellular respiration, ATP production, and factors affecting energy metabolism. It includes solved MCQs, short questions, and key points for exam preparation. Ideal for students studying under Punjab boards who want a comprehensive understanding of bioenergetics for better exam performance.

A. Select the correct answers for the following questions:

1. When we get energy from ATP, which bonds are broken?

  • Options:
    a) P-P bonds
    b) C-H bonds
    c) C-N bonds
    d) C=O bonds
  • Answer: a) P-P bonds
  • Explanation: Energy is released from ATP when the high-energy phosphate bonds (P-P bonds) are hydrolyzed. Typically, the terminal phosphate group is removed, converting ATP to ADP.
  • Tip: Remember “ATP = Adenosine Tri-Phosphate,” where the energy is stored in the phosphate bonds.

2. Light reactions of photosynthesis occur in:

  • Options:
    a) Plasma membrane of cell
    b) Cytoplasm of cell
    c) Stroma of chloroplasts
    d) Thylakoids of chloroplasts
  • Answer: d) Thylakoids of chloroplasts
  • Explanation: The light-dependent reactions of photosynthesis occur in the thylakoid membranes, where chlorophyll absorbs light and converts it into chemical energy (ATP and NADPH).
  • Tip: Think of “thylakoid” as the “power plant” of the chloroplast for light reactions.

3. Which type of chlorophyll is most common in plants?

  • Options:
    a) Chlorophyll a
    b) Chlorophyll b
    c) Chlorophyll c
    d) Chlorophyll d
  • Answer: a) Chlorophyll a
  • Explanation: Chlorophyll a is the primary pigment responsible for photosynthesis in plants. It absorbs light most efficiently in the blue-violet and red regions of the spectrum.
  • Tip: Chlorophyll a is “essential,” while others are “accessory pigments.”

4. Where does the reaction of photosynthesis take place?

  • Options:
    a) Chloroplast
    b) Mitochondria
    c) Cytoplasm
    d) Ribosomes
  • Answer: a) Chloroplast
  • Explanation: Photosynthesis occurs in chloroplasts, with light reactions in the thylakoids and the Calvin cycle in the stroma.
  • Tip: Remember that chloroplasts are exclusive to plants and are the “photosynthesis factory.”

5. When yeast ferments glucose, the products are:

  • Options:
    a) Alcohol and CO₂
    b) Alcohol and water
    c) Lactic acid and H₂O
    d) Alcohol and H₂O
  • Answer: a) Alcohol and CO₂
  • Explanation: In anaerobic conditions, yeast ferments glucose to produce ethanol (alcohol) and carbon dioxide as by-products.
  • Tip: Associate “yeast” with “alcoholic fermentation.”

6. In which part of the chloroplast does the light-dependent reaction occur?

  • Options:
    a) Stroma
    b) Thylakoid membrane
    c) Outer membrane
    d) Matrix
  • Answer: b) Thylakoid membrane
  • Explanation: The light-dependent reactions take place in the thylakoid membrane, where light is absorbed to produce ATP and NADPH.
  • Tip: Light = “Thylakoid”; Calvin cycle = “Stroma.”

7. Which molecule donates electrons in the light-dependent reactions of photosynthesis?

  • Options:
    a) NADPH
    b) Water
    c) Oxygen
    d) Carbon dioxide
  • Answer: b) Water
  • Explanation: Water (H₂O) is split during photolysis in the light-dependent reactions, releasing electrons, protons, and oxygen as a by-product.
  • Tip: Remember that “water splitting” produces the electrons needed for photosynthesis.

8. Which process in aerobic respiration produces the most ATP?

  • Options:
    a) Glycolysis
    b) Electron transport chain
    c) Fermentation
    d) Krebs cycle
  • Answer: b) Electron transport chain
  • Explanation: The electron transport chain (ETC) produces the most ATP (approximately 34 molecules per glucose molecule) during aerobic respiration, using the energy from NADH and FADH₂.
  • Tip: The ETC is the “energy powerhouse” of respiration.

9. In yeast cells, anaerobic respiration leads to the production of:

  • Options:
    a) Lactic acid
    b) Ethanol
    c) Acetic acid
    d) Glucose
  • Answer: b) Ethanol
  • Explanation: Under anaerobic conditions, yeast converts glucose into ethanol and CO₂ via fermentation.
  • Tip: Recall the industrial use of yeast in brewing and alcohol production.

Answers to MCQs:

  1. How many ATP molecules are produced from one glucose molecule during anaerobic respiration?
    Answer: a) 2
  2. What is a common byproduct of anaerobic respiration in animal cells?
    Answer: c) Lactic acid

Short Answers:

  1. Importance of oxidation-reduction reactions:
    Oxidation-reduction reactions are essential in cellular respiration and photosynthesis. They help transfer energy by moving electrons between molecules.
  2. Meaning and roles of ATP and ADP:
  • ATP (Adenosine Triphosphate): The main energy carrier in cells. It stores and provides energy for cellular activities.
  • ADP (Adenosine Diphosphate): A lower-energy molecule that is converted back to ATP during cellular respiration.
  1. Word equation for photosynthesis:
    Carbon dioxide + Water → Glucose + Oxygen
    (In the presence of sunlight and chlorophyll)
  2. Why is chlorophyll important for photosynthesis?
    Chlorophyll absorbs light energy from the sun, which is needed to drive the process of photosynthesis.
  3. How is oxygen produced during photosynthesis?
    Oxygen is produced as a byproduct when water molecules are split during the light-dependent reactions of photosynthesis.
  4. Organisms that carry out photosynthesis and responsible organelle:
    Plants, algae, and some bacteria carry out photosynthesis. The chloroplast is the organelle responsible for absorbing light.
  5. Main purpose of cellular respiration:
    To produce energy in the form of ATP, which is used for various cellular activities.
  6. Equation for aerobic respiration:
    Glucose + Oxygen → Carbon dioxide + Water + Energy (ATP)
  7. Role of oxygen in aerobic respiration:
    Oxygen is essential for breaking down glucose completely to release energy efficiently.
  8. Definition of anaerobic and aerobic respiration:
  • Anaerobic respiration: Respiration that occurs without oxygen, producing less ATP.
  • Aerobic respiration: Respiration that requires oxygen, producing more ATP.
  1. End products of anaerobic respiration in animals and yeast:
  • Animals: Lactic acid and ATP.
  • Yeast: Ethanol, carbon dioxide, and ATP.
  1. How muscles respond to oxygen deficiency during exercise:
    Muscles switch to anaerobic respiration, producing lactic acid and causing fatigue.
  2. Ways respiratory energy is used in the body:
  • Muscle contraction
  • Cell division
  • Active transport of molecules
  • Maintaining body temperature

Detailed Answers:

  1. Explain ATP as the chief energy currency of all cells:
    ATP is the primary molecule that stores and transfers energy in cells. It powers cellular processes like muscle contraction, nerve impulses, and biosynthesis. ATP releases energy when its phosphate bonds are broken, turning into ADP.
  2. Outline the processes involved in photosynthesis:
    Photosynthesis occurs in two stages:
  • Light-dependent reactions: Light energy splits water into oxygen, hydrogen ions, and electrons.
  • Light-independent reactions (Calvin cycle): Carbon dioxide combines with hydrogen ions to form glucose using energy from ATP and NADPH.
  1. Write a note on the intake of carbon dioxide and water by plants:
    Plants absorb carbon dioxide through tiny pores called stomata in their leaves. Water is absorbed by roots from the soil and transported to leaves via xylem vessels.
  2. Explain the types and importance of anaerobic respiration:
  • Types:
    • Lactic acid fermentation in animals.
    • Alcoholic fermentation in yeast.
  • Importance:
    • Provides energy in low-oxygen conditions.
    • Used in industries like brewing and baking.
  1. Outline the mechanism of aerobic respiration:
    Aerobic respiration occurs in three stages:
  • Glycolysis: Glucose is broken down into pyruvate, producing 2 ATP.
  • Krebs cycle: Pyruvate is broken down further, releasing carbon dioxide and energy-rich molecules.
  • Electron transport chain: Oxygen accepts electrons, forming water and generating a large amount of ATP (about 36 molecules).
  1. Compare the processes of respiration and photosynthesis:
  • Photosynthesis:
    • Occurs in chloroplasts.
    • Converts light energy into chemical energy.
    • Reactants: Carbon dioxide and water.
    • Products: Glucose and oxygen.
  • Respiration:
    • Occurs in mitochondria.
    • Converts chemical energy (glucose) into usable energy (ATP).
    • Reactants: Glucose and oxygen.
    • Products: Carbon dioxide and water.

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