Selina Concise Solutions for ICSE Class 10 Physics Chapter 3 Machines

Question 1. What do you understand by a simple machine?

Answer: A machine is a device by which we can either overcome a large resistive force at some point by applying a small force at a convenient point and in a desired direction or by which we can obtain a gain in the speed.

In simple words: A simple machine helps us do work more easily, either by letting us use less force to move something heavy, or by changing where or how we push, or by making things move faster.

📝 Teacher's Note: Emphasize the three main functions: force multiplication, change in direction/point of application, and gain in speed. Provide simple examples for each.

🎯 Exam Tip: Define a simple machine by its ability to change the magnitude, direction, or point of application of force, or to gain speed.

Question 2. State four ways in which machines are useful to us?

Answer: Machines are useful to us in the following ways:
(1) In lifting a heavy load by applying a less effort.
(2) In changing the point of application of effort to a convenient point.
(3) In changing the direction of effort to a convenient direction.
(4) For obtaining a gain in speed.

In simple words: Machines help us lift heavy things easily, push from a better spot, pull in a more comfortable direction, or make things move faster.

📝 Teacher's Note: Discuss real-world examples for each point, like a crowbar for lifting (force multiplication), bicycle pedals (change point of application), a pulley for a well (change direction), and scissors (gain in speed).

🎯 Exam Tip: List the four primary uses: force multiplication, change in point of application of effort, change in direction of effort, and gain in speed.

Question 3. Name a machine for each of the following use: (a) to multiply force (b) To change the point of application of force (c) To change the direction of force (d) To obtain gain in speed

Answer:
(a) To multiply force: A jack is used to lift a car.
(b) To change the point of application of force: The wheel of a cycle is rotated with the help of a chain by applying the force on the pedal.
(c) To change the direction of force: A single fixed pulley is used to lift a bucket full of water from the well by applying the effort in the downward direction instead of applying it upwards when the bucket is lifted up without the use of pulley.
(d) To obtain gain in speed: When a pair of scissors is used to cut the cloth, its blades move longer on cloth while its handles move a little.

In simple words:
(a) A car jack helps lift a car with less effort.
(b) A bicycle's pedals and chain let you push from your feet to turn the wheel.
(c) A pulley on a well lets you pull down to lift a bucket up.
(d) Scissors make the blades move a lot for a small squeeze of the handles.

📝 Teacher's Note: Encourage students to think of other examples for each category. For instance, a crowbar for force multiplication, a screwdriver for changing point of application (though less direct), a flagpole pulley for changing direction, and a fishing rod for gain in speed.

🎯 Exam Tip: Provide clear, distinct examples for each function. Ensure the example clearly illustrates the specific use of the machine.

Question 4. What is the purpose of a jack in lifting a car by it?

Answer: The purpose of a jack is to make the effort less than the load so that it works as a force multiplier.

In simple words: A car jack helps you lift a very heavy car using only a small push or turn, making it much easier.

📝 Teacher's Note: Explain the concept of "force multiplier" in simple terms: a machine that allows a small input force (effort) to overcome a large output force (load). Relate it to mechanical advantage.

🎯 Exam Tip: State that the jack's purpose is to act as a force multiplier, enabling a small effort to lift a large load.

Question 5. What do you understand by an ideal machine? How does it differ from a practical machine?

Answer: An ideal machine is a machine whose parts are weightless and frictionless so that there is no dissipation of energy in any manner. Its efficiency is 100%, i.e., the work output is equal to work input.

Differences between Ideal Machine and Practical Machine:

In simple words: An ideal machine is like a perfect dream machine that never loses any energy to things like friction or heavy parts, so it gives back exactly as much work as you put in. A practical machine is real, so it always loses a little energy and gives back slightly less work.

📝 Teacher's Note: Emphasize that ideal machines are theoretical concepts used for simplifying calculations and understanding fundamental principles. Practical machines always have inefficiencies due to friction, air resistance, and the weight of their parts. Discuss how engineers try to make practical machines as close to ideal as possible.

🎯 Exam Tip: Clearly define an ideal machine (100% efficiency, no energy loss, weightless/frictionless parts). For the comparison, use a clear tabular format and highlight key differences like efficiency, energy loss, and work input/output.

Question 6. Explain the term mechanical advantage. State its unit.

Answer: The ratio of the load to the effort is called mechanical advantage (MA) of the machine. Mathematically, it is expressed as: \( MA = \frac{\text{Load}}{\text{Effort}} \). It has no unit because it is a ratio of two forces.

In simple words: Mechanical advantage tells us how much easier a machine makes it to lift or move something. If it's more than 1, the machine helps you use less force.

📝 Teacher's Note: Explain that MA > 1 means force multiplication, MA < 1 means gain in speed, and MA = 1 means change in direction. Use simple examples like a crowbar (MA > 1) vs. tweezers (MA < 1).

🎯 Exam Tip: Define mechanical advantage as the ratio of load to effort. Crucially, state that it is a dimensionless quantity (has no unit).

Question 7. Define the term velocity ratio, state its unit.

Answer: The ratio of the velocity of effort to the velocity of the load is called the velocity ratio (VR) of the machine. Mathematically, it is expressed as: \( VR = \frac{\text{Velocity of Effort}}{\text{Velocity of Load}} \). It has no unit because it is a ratio of two velocities.

In simple words: Velocity ratio tells us how much faster the effort moves compared to the load.

📝 Teacher's Note: Explain that velocity ratio is a theoretical value determined by the machine's geometry and is constant for a given machine, unlike mechanical advantage which can vary slightly with friction.

🎯 Exam Tip: Define velocity ratio as the ratio of the velocity of effort to the velocity of the load. State that it is a dimensionless quantity (has no unit).

Question 8. How is mechanical advantage related to the velocity ratio for an ideal machine?

Answer: For an ideal machine, mechanical advantage (MA) is numerically equal to the velocity ratio (VR). This can be written as: \( MA = VR \).

In simple words: In a perfect machine, how much easier it makes work (mechanical advantage) is exactly the same as how much faster the effort moves compared to the load (velocity ratio).

📝 Teacher's Note: This relationship (MA = VR) is a key characteristic of an ideal machine and directly relates to its 100% efficiency. For practical machines, MA is always less than VR due to energy losses.

🎯 Exam Tip: State clearly that for an ideal machine, Mechanical Advantage (MA) is equal to Velocity Ratio (VR). This is a fundamental concept for understanding machine efficiency.

Question 9. Define the term efficiency of a machine. Why is a machine not 100% efficient?

Answer: It is the ratio of the useful work done by the machine to the work put into the machine by the effort. In an actual machine, there is always some loss of energy due to friction and the weight of moving parts, thus the output energy is always less than the input energy.

In simple words: Efficiency tells us how much useful work we get out of a machine compared to the energy we put in. Machines are never 100% efficient because some energy is always lost as heat due to rubbing parts (friction) or moving the machine's own parts.

📝 Teacher's Note: Explain that energy is conserved, but not all of it is converted into *useful* work. The "lost" energy often turns into heat or sound. Use examples like a bicycle chain getting warm after a long ride.

🎯 Exam Tip: Define efficiency as the ratio of useful work output to total work input. State that efficiency is always less than 1 (or 100%) for real machines due to energy losses, primarily friction and the weight of moving parts.

Question 10. When does a machine act as (a) a force multiplier (b) a speed multiplier Can a machine act as a force multiplier and a speed multiplier simultaneously?

Answer: (a) A machine acts as a force multiplier when the effort arm is longer than the load arm. The mechanical advantage of such machines is greater than 1. (b) A machine acts as a speed multiplier when the effort arm is shorter than the load arm. The mechanical advantage of such machines is less than 1. It is not possible for a machine to act as a force multiplier and speed multiplier simultaneously. This is because machines which are force multipliers cannot gain in speed and vice-versa.

In simple words: A machine helps you lift heavy things with less effort (force multiplier) if your pushing/pulling point is far from the pivot. It helps you move things faster (speed multiplier) if your pushing/pulling point is close to the pivot. You can't have both at the same time – if you gain force, you lose speed, and if you gain speed, you lose force.

📝 Teacher's Note: Use a seesaw as an analogy. If a small child wants to lift a heavy adult, the child needs to sit further from the pivot (force multiplier). If the adult wants to make the child move up and down quickly, the adult pushes closer to the pivot (speed multiplier). Emphasize the trade-off between force and distance/speed.

🎯 Exam Tip: For a force multiplier, M.A. > 1 and effort arm > load arm. For a speed multiplier, M.A. < 1 and effort arm < load arm. Clearly state that a machine cannot be both simultaneously, linking it to the principle of conservation of energy (work input = work output for an ideal machine).

Question 11. State the relationship between mechanical advantage, velocity ratio and efficiency. Name the term that will not change for a machine of a given design.

Answer: Mechanical advantage is equal to the product of velocity ratio and efficiency. M.A = \( \eta \) x V.R For a machine of a given design, the velocity ratio does not change.

In simple words: Mechanical Advantage (how much a machine multiplies your force) is found by multiplying its Velocity Ratio (how much faster the load moves compared to your effort) by its Efficiency (how good it is at using energy without waste). For a specific machine, its Velocity Ratio always stays the same, no matter how much load you put on it.

📝 Teacher's Note: Explain that Velocity Ratio depends only on the geometry (design) of the machine, like the number of pulleys or the lengths of arms, which doesn't change. Mechanical Advantage and Efficiency, however, can vary slightly with load due to changing friction.

🎯 Exam Tip: The formula is M.A = \( \eta \) x V.R. Remember that Velocity Ratio (V.R.) is a theoretical value based on the machine's design and is constant, while Mechanical Advantage (M.A.) and efficiency (\( \eta \)) are practical values that can be affected by real-world conditions like friction.

Question 12. Derive the relationship between mechanical advantage, velocity ratio and efficiency of a machine.

Answer: Let a machine overcome a load L by the application of an effort E. In time t, let the displacement of effort be d_E and the displacement of load be d_L. Work input = Effort x displacement of effort = E x d_E Work output = Load x displacement of load = L x d_L Efficiency \( \eta \) = \( \frac{\text{work output}}{\text{work input}} \) \( \eta \) = \( \frac{L \times d_L}{E \times d_E} \) = \( \frac{L}{E} \times \frac{1}{d_E / d_L} \) But \( \frac{L}{E} \) = M.A And \( \frac{d_E}{d_L} \) = V.R So, \( \eta \) = \( \frac{M.A}{V.R} \) Rearranging, M.A = \( \eta \) x V.R Thus, mechanical advantage of a machine is equal to the product of its efficiency and velocity ratio.

In simple words: We start by thinking about the energy we put into a machine (Effort x distance moved by effort) and the useful energy we get out (Load x distance moved by load). Efficiency is the useful energy out divided by the energy in. If we then replace the "Load/Effort" part with Mechanical Advantage and "distance of effort/distance of load" part with Velocity Ratio, we find that Mechanical Advantage is simply Efficiency multiplied by Velocity Ratio.

📝 Teacher's Note: Emphasize the definitions of work input, work output, M.A., V.R., and efficiency. Guide students through the algebraic manipulation step-by-step. Highlight that this derivation connects the practical (M.A., \( \eta \)) and theoretical (V.R.) aspects of a machine.

🎯 Exam Tip: Clearly define all terms (L, E, d_L, d_E, M.A., V.R., \( \eta \)) at the beginning. Show the steps for work input, work output, and then substitute the definitions of M.A. and V.R. into the efficiency formula. Ensure the final relationship M.A = \( \eta \) x V.R is stated clearly.

Question 13. How is the mechanical advantage related with the velocity ratio for an actual machine? State whether the efficiency of such a machine is equal to 1, less than 1 or more than 1.

Answer: The mechanical advantage for an actual machine is equal to the product of its efficiency and velocity ratio. M.A = V.R x \( \eta \) The efficiency of such a machine is always less than 1, i.e. \( \eta < 1 \). This is because there is always some loss in energy in the form of friction etc.

In simple words: For a real machine, the force it helps you with (Mechanical Advantage) is its design advantage (Velocity Ratio) multiplied by how good it is at not wasting energy (Efficiency). A real machine always wastes some energy, so its efficiency is always less than 1 (or 100%).

📝 Teacher's Note: Reiterate that "actual machine" implies friction and other energy losses. Contrast this with an "ideal machine" where efficiency would be 1. Discuss how M.A. is always less than V.R. for an actual machine because \( \eta \) is less than 1.

🎯 Exam Tip: State the relationship M.A = V.R x \( \eta \). Crucially, state that for an actual machine, efficiency \( \eta \) is always less than 1 (or 100%) due to unavoidable energy losses like friction, heat, and the weight of moving parts.

Question 14. State reason why is mechanical advantage less than the velocity ratio for an actual machine.

Answer: This is because the output work is always less than the input work, so the efficiency is always less than 1 because of energy loss due to friction. M.A = V.R x \( \eta \) Since \( \eta < 1 \), it follows that M.A < V.R.

In simple words: Mechanical Advantage is less than Velocity Ratio for a real machine because some of the energy we put in is always wasted (like turning into heat from rubbing parts), meaning the machine isn't perfectly efficient. Since efficiency is always less than 1, it makes the Mechanical Advantage smaller than the Velocity Ratio.

📝 Teacher's Note: Connect this directly to the definition of efficiency. If efficiency is work output / work input, and work output is always less than work input for a real machine, then efficiency must be less than 1. Since M.A. = V.R. x \( \eta \), and \( \eta < 1 \), then M.A. must be less than V.R.

🎯 Exam Tip: The key reason is that for an actual machine, efficiency (\( \eta \)) is always less than 1 due to energy losses (e.g., friction). Since M.A. = V.R. x \( \eta \), if \( \eta \) is less than 1, then M.A. will necessarily be less than V.R.

Question 15. What is a lever? State its principle.

Answer: A lever is a rigid, straight or bent bar which is capable of turning about a fixed axis. Principle: A lever works on the principle of moments. For an ideal lever, it is assumed that the lever is weightless and frictionless. In the equilibrium position of the lever, by the principle of moments, Moment of load about the fulcrum = Moment of the effort about the fulcrum.

In simple words: A lever is a strong stick or bar that can spin around a fixed point. Its main rule is that for it to be balanced, the "turning effect" caused by the heavy thing (load) on one side must be equal to the "turning effect" caused by your push (effort) on the other side.

📝 Teacher's Note: Use a crowbar or a seesaw as a simple example of a lever. Explain "fulcrum" as the pivot point. Define "moment" as force x perpendicular distance from the pivot. Emphasize that the principle of moments is fundamental to understanding how levers work.

🎯 Exam Tip: Define a lever as a rigid bar pivoting around a fixed point (fulcrum). State the principle of moments clearly: "For a lever in equilibrium, the sum of clockwise moments about the fulcrum equals the sum of anticlockwise moments about the fulcrum," or more simply, "Moment of load = Moment of effort" for an ideal lever.

Question 16. Write down a relation expressing the mechanical advantage of a lever.

Answer:

\( M.A = \frac{\text{Effort arm}}{\text{Load arm}} \)

This is the expression of the mechanical advantage of a lever.

In simple words: Mechanical Advantage (M.A.) tells us how much easier a lever makes work. It's found by dividing the length of the arm where you push (effort arm) by the length of the arm where the weight is (load arm).

📝 Teacher's Note: Emphasize that M.A. is a ratio of distances, not forces, for a lever. This formula is fundamental for understanding lever mechanics.

🎯 Exam Tip: Remember the formula \( M.A = \frac{\text{Effort arm}}{\text{Load arm}} \). Clearly state the definition of effort arm and load arm for full marks.

Question 17. Name the three classes of levers and distinguish between them. Give two examples of each class.

Answer:

The three classes of levers are:

(i) Class I levers: In these types of levers, the fulcrum (F) is in between the effort (E) and the load (L).
Example: a seesaw, a pair of scissors, crowbar.

(ii) Class II levers: In these types of levers, the load (L) is in between the effort (E) and the fulcrum (F). The effort arm is thus always longer than the load arm.
Example: a nut cracker, a bottle opener.

(iii) Class III levers: In these types of levers, the effort (E) is in between the fulcrum (F) and the load (L) and the effort arm is always smaller than the load arm.
Example: sugar tongs, forearm used for lifting a load.

In simple words: Levers are like simple machines that help us lift or move things. We sort them into three groups based on where the push (effort), the thing being moved (load), and the pivot point (fulcrum) are located. Class I has the pivot in the middle, Class II has the load in the middle, and Class III has the push in the middle.

📝 Teacher's Note: Use a simple mnemonic like "FLE" (Fulcrum-Load-Effort) to help students remember the order for each class (F-L-E for Class I, F-L-E for Class II, F-L-E for Class III, where the middle letter is the one "in between"). Demonstrate with physical examples in the classroom.

🎯 Exam Tip: Clearly define the position of the fulcrum, load, and effort for each class. Provide at least two distinct examples for each class to secure full marks.

Question 18. Give one example each of a class I lever where mechanical advantage is (a) more than one, and (b) less than one. What is the use of the lever if its mechanical advantage is less than I?

Answer:

(a) More than one: Shears used for cutting thin metal sheets.

(b) Less than one: A pair of scissors whose blades are longer than its handles.

When the mechanical advantage is less than 1, the levers are used to obtain gain in speed. This implies that the displacement of load is more as compared to the displacement of effort.

In simple words: Some levers make it easier to push heavy things (M.A. > 1), like big metal cutters. Others make things move faster or further, even if you have to push harder (M.A. < 1), like some scissors where the blades move a lot for a small hand movement. When M.A. is less than 1, you get speed or distance, not force.

📝 Teacher's Note: Explain the trade-off: gain in force means loss in distance/speed, and vice-versa. M.A. < 1 is not "bad"; it serves a different purpose (speed/range of motion).

🎯 Exam Tip: For M.A. > 1, the effort arm is longer than the load arm (gain in force). For M.A. < 1, the effort arm is shorter than the load arm (gain in speed/distance). State the specific use of M.A. < 1 levers (gain in speed or displacement) for full credit.

Question 19. A pair of scissors and a pair of pliers both belongs to the same class of levers. Name the class of lever. Which one has the mechanical advantage less than 1?

Answer:

A pair of scissors and a pair of pliers both belong to Class I lever.

A pair of scissors has mechanical advantage less than 1.

In simple words: Both scissors and pliers are Class I levers because their pivot point (fulcrum) is in the middle. But scissors often have blades longer than their handles, meaning you push harder to cut, but the blades move a greater distance, so their M.A. is less than 1. Pliers usually have longer handles for a stronger grip, giving them an M.A. greater than 1.

📝 Teacher's Note: Use a physical pair of scissors and pliers to demonstrate the fulcrum position and the relative lengths of effort and load arms. This helps visualize why M.A. differs even within the same class.

🎯 Exam Tip: Identify Class I lever correctly. For M.A. < 1, remember it's when the effort arm is shorter than the load arm, which is typical for scissors designed for speed/range of cut rather than brute force.

Question 20. Explain why scissors for cutting cloth may have blades longer than the handles, but shears for cutting metals have short blades and long handles.

Answer:

A pair of scissors used to cut a piece of cloth has blades longer than the handles so that the blades move longer on the cloth than the movement at the handles. This provides a gain in speed and range of motion, suitable for cutting fabric quickly.

While shears used for cutting metals have short blades and long handles because this configuration makes the effort arm longer than the load arm. This enables us to overcome large resistive force by a small effort, providing a gain in force necessary for cutting tough materials like metal.

In simple words: Cloth scissors have long blades so you can cut a lot of cloth with a small hand movement – you get speed. Metal shears have long handles and short blades so you can push with less effort to cut hard metal – you get strength. It's a trade-off between how far or fast something moves versus how much force you need.

📝 Teacher's Note: This question perfectly illustrates the practical application of mechanical advantage. Emphasize the concept of "gain in speed" vs. "gain in force" and how lever design is optimized for specific tasks.

🎯 Exam Tip: Clearly link the design (blade/handle length) to the purpose (cutting cloth vs. metal) and the resulting mechanical advantage (gain in speed vs. gain in force). Use terms like "effort arm" and "load arm" to explain the difference in M.A. for each case.

Question 21. Fig 3.12, shows a uniform metre scale of weight W supported on a fulcrum at the 60 cm mark by applying the effort E at the 90 cm mark.

[Diagram: A uniform metre scale with fulcrum at 60 cm mark, weight W acting at 50 cm mark, and effort E applied at 90 cm mark.]

(a) state with reason whether the weight W of the scale is greater than, less than or equal to the effort E.

(b) Find the mechanical advantage in an ideal case.

Reason: The weight W of a uniform metre scale acts at 50 cm mark. Since distance of weight of scale from fulcrum F is less then that of the effort E., so the weight W of scale is greater than the effort E.

Answer:

(a) The weight W of the scale is greater than E.

It is because the effort arm (distance from fulcrum to effort) is 30 cm (90 cm - 60 cm) and the load arm (distance from fulcrum to weight of scale) is 10 cm (60 cm - 50 cm). Since the load arm is shorter than the effort arm, to balance the scale, the weight W of the scale should be more than the effort E.

(b) The mechanical advantage (M.A.) is calculated as:

\( M.A = \frac{\text{Effort arm}}{\text{Load arm}} \)

Here, effort arm = \( (90 - 60) \text{ cm} = 30 \text{ cm} \)

Load arm = \( (60 - 50) \text{ cm} = 10 \text{ cm} \)

Therefore, \( M.A = \frac{30 \text{ cm}}{10 \text{ cm}} = 3 \)

In simple words: Imagine a seesaw. If you sit closer to the middle (fulcrum) and someone else sits further away, you'd need to be heavier to balance them. Here, the scale's weight (W) is closer to the pivot (fulcrum) than where you push (effort E). So, the scale's weight must be bigger than your push to balance it. The "mechanical advantage" tells you how much easier it is to lift or balance something. In this case, it's 3, meaning your push feels three times stronger.

📝 Teacher's Note: This problem is excellent for teaching moments and principle of moments. Emphasize that for a uniform metre scale, the weight acts at its center (50 cm mark). Guide students to correctly identify the fulcrum, load, and effort points, and then calculate the respective arms. Reinforce that M.A. > 1 means a gain in force.

🎯 Exam Tip: For problems involving uniform scales, always assume the weight acts at the center of gravity (50 cm mark for a metre scale). Clearly show the calculation of both effort arm and load arm. State the reason for W > E based on the principle of moments or the relationship between M.A. and arm lengths. Ensure units cancel out for M.A. as it's a ratio.

Question 22. Which type of lever has a mechanical advantage always more than one? Give one example. What change can be made in this lever to increase its mechanical advantage?

Answer: Class II lever always have a mechanical advantage more than one. Example: a nut cracker. To increase its mechanical advantage we can increase the length of effort arm.

In simple words: A Class II lever, like a nutcracker, always makes things easier to lift or move. To make it even easier, you can make the handle (where you push) longer.

📝 Teacher's Note: Emphasize the relationship between effort arm length and mechanical advantage. A longer effort arm means less effort is needed to overcome a given load. This is a fundamental concept in levers.

🎯 Exam Tip: For full marks, clearly state "Class II lever" and provide a correct example. Also, mention "increasing the length of the effort arm" as the method to increase mechanical advantage.

Question 23. Draw a diagram of a lever which is always used as a force multiplier. How is the effort arm related to the load arm in such a lever?

Answer: [Diagram: Lever used as a force multiplier] The effort arm is longer than load arm in such a lever.

In simple words: A force multiplier lever helps you lift heavy things with less effort. In these levers, the part where you push (effort arm) is always longer than the part that lifts the weight (load arm).

📝 Teacher's Note: A force multiplier lever is typically a Class II or Class I lever where the effort arm is longer than the load arm. Encourage students to draw a simple diagram showing F, L, E positions and the relative lengths of the arms. A wheelbarrow or a bottle opener are good examples.

🎯 Exam Tip: When asked to draw, ensure the diagram is clearly labelled with Fulcrum (F), Load (L), and Effort (E). Crucially, state that "the effort arm is longer than the load arm" for a force multiplier lever.

Question 24. Explain why the mechanical advantage of a Class II type of lever is always more than 1.

Answer: In these types of levers, the load L is in between the effort E and the fulcrum F. So, the effort arm is thus always longer than the load arm. Therefore M.A > 1.

In simple words: In a Class II lever, the weight you're trying to lift is always between where you push and the pivot point. Because of this setup, the distance from where you push to the pivot is always greater than the distance from the weight to the pivot. This makes it easier to lift, so its mechanical advantage is always more than 1.

📝 Teacher's Note: Use a simple diagram of a wheelbarrow or nutcracker to visually explain the positions of F, L, and E. Emphasize that Mechanical Advantage (M.A) = Effort Arm / Load Arm. If Effort Arm > Load Arm, then M.A > 1.

🎯 Exam Tip: To score full marks, clearly state the relative positions of F, L, and E (Load is between Fulcrum and Effort). Then, explain that this arrangement makes the effort arm longer than the load arm, leading to M.A > 1.

Question 25. Draw a labelled diagram of a class II lever. Give one example of such a lever.

Answer: [Diagram: Labelled diagram of a Class II lever] Example: a bottle opener.

In simple words: A Class II lever is like a bottle opener or a wheelbarrow. The weight you're moving is in the middle, between the pivot point and where you apply force. A bottle opener is a good example.

📝 Teacher's Note: When drawing, ensure the fulcrum (F) is at one end, the effort (E) at the other, and the load (L) is in between. Common examples include a wheelbarrow, nutcracker, and bottle opener. Discuss how these examples fit the F-L-E arrangement.

🎯 Exam Tip: For the diagram, clearly label F, L, and E in their correct Class II positions. Provide a specific, correct example of a Class II lever.

Question 26. Fig 3.13 shows a nut cracker. (a) In the diagram, mark the position of the fulcrum F and the line of action of load L and effort E. (b) Name the class of lever .

Answer: (a) [Diagram: Positions of fulcrum F, load L, and effort E marked on the nut cracker (Fig 3.13)] (b) The nut cracker is class II lever.

In simple words: Imagine a nutcracker. The hinge is the pivot (fulcrum). The nut you're cracking is the load, and where you squeeze the handles is the effort. It's a Class II lever.

📝 Teacher's Note: Use a physical nutcracker or a clear diagram to demonstrate the F-L-E positions. The fulcrum is the hinge, the load is the nut, and the effort is applied at the handles. This reinforces the definition of a Class II lever.

🎯 Exam Tip: For part (a), ensure F, L, and E are correctly identified and marked on the diagram (if provided in the exam). For part (b), correctly identify the nutcracker as a "Class II lever" for full marks.

Question 27. The diagram below shows a wheel barrow. Mark position of fulcrum F and draw arrows to show the directions of load L and effort E. What class of lever is the whell barrow? Give one more example of the same class of lever.

Answer: [Diagram: Wheelbarrow with fulcrum F, load L, and effort E marked with arrows]

[Solution content for Question 27 is missing in the provided text. Assuming it would identify the class of lever and provide an example.]

In simple words: Think of a wheelbarrow. The wheel is the pivot point (fulcrum). The stuff you put in it is the load, and where you lift the handles is the effort. It's a Class II lever, just like a nutcracker.

📝 Teacher's Note: A wheelbarrow is an excellent real-world example of a Class II lever. Guide students to identify the wheel as the fulcrum, the contents of the barrow as the load (acting downwards), and the lifting force on the handles as the effort (acting upwards). Discuss other Class II examples like a bottle opener.

🎯 Exam Tip: When marking the diagram, ensure F is at the wheel, L is in the middle (the load in the tray), and E is at the handles. Arrows should indicate the direction of L (downwards) and E (upwards). Correctly identify it as a "Class II lever" and provide another valid example for full marks.

Answer: The wheel barrow is a class II lever. One more example of this class is a nut cracker.

In simple words: A wheelbarrow helps carry heavy things easily. It's like a seesaw where the load is in the middle, making it easier to lift. A nutcracker works the same way to crack nuts.

📝 Teacher's Note: Emphasize the position of load, effort, and fulcrum for Class II levers. Real-world examples like bottle openers or paper cutters (some types) can be discussed.

🎯 Exam Tip: For Class II levers, the load is always between the fulcrum and the effort. Mechanical advantage is always greater than 1.

Question 28. State the kind of lever which always has the mechanical advantage less than 1. Draw a labelled diagram of such lever.

Answer: Classes III levers always have mechanical advantage less than one. [Diagram: Labelled diagram of a Class III lever showing fulcrum, effort, and load positions]

In simple words: Levers where you push or pull in the middle to move something at the end, like using tongs or a fishing rod, always make it harder to lift but let you move things a longer distance.

📝 Teacher's Note: Use a pair of tongs or a broom to demonstrate a Class III lever. Highlight that the effort is applied between the fulcrum and the load. Discuss the trade-off: less force gain, but greater displacement/speed gain.

🎯 Exam Tip: Clearly state "Class III levers" and mention that the effort is between the fulcrum and the load, leading to the effort arm being shorter than the load arm, hence M.A. < 1. A well-labelled diagram showing F, E, L positions is crucial for full marks.

Question 29. Explain why the mechanical advantage of the class III type of lever is always less than 1.

Answer: In these types of levers, the effort is in between the fulcrum F and the load L and so the effort arm is always smaller than the load arm. Therefore M.A. < 1.

In simple words: For these levers, you push closer to the pivot point than where the heavy thing is. Because your push is closer, you need to push harder than the weight of the object, so it doesn't make lifting easier.

📝 Teacher's Note: Reinforce the definition of mechanical advantage (M.A. = Load / Effort = Effort Arm / Load Arm). Visually demonstrate with a ruler and weights that if the effort arm is shorter than the load arm, M.A. must be less than 1. Connect this to the concept of torque/moments.

🎯 Exam Tip: The key explanation is that in Class III levers, the effort (E) is located between the fulcrum (F) and the load (L). This configuration inherently makes the effort arm shorter than the load arm, directly resulting in a mechanical advantage less than 1. Stating the relationship M.A. = Effort Arm / Load Arm is essential.

Question 30. Class III levers have mechanical advantage less than one. Why are they then used?

Answer: With levers of class III, we do not get gain in force, but we get gain in speed, that is a longer displacement of load is obtained by a smaller displacement of effort.

In simple words: Even though these levers don't make things feel lighter, they are useful because a small movement from you can make the other end move a much bigger distance or faster. Think of sweeping with a broom – a small hand movement makes the bristles move a lot.

📝 Teacher's Note: This is a crucial concept: the trade-off between force and distance/speed. Use examples like fishing rods (small wrist movement, large lure movement), baseball bats, or human limbs (e.g., forearm lifting) to illustrate gain in speed or displacement. Emphasize that machines don't create energy, they transform it.

🎯 Exam Tip: The primary reason for using Class III levers is to achieve a "gain in speed" or "gain in displacement." A smaller displacement of effort results in a larger displacement of the load. This allows for quick movements or covering larger distances with minimal effort movement, despite the force disadvantage.

Question 31. What type of lever is formed by the human body while (a) raising a load on the palm, (b) raising the weight of body on toes?

Answer:

(a) Class III. Here, the fulcrum is the elbow of the human arm. Biceps exert the effort in the middle and load on the palm is at the other end.

(b) Class II. Here, the fulcrum is at toes at one end, the load (i.e. weight of the body) is in the middle and effort by muscles is at the other end.

In simple words: (a) When you lift something with your hand, your elbow is the pivot, your bicep muscle pulls in the middle, and the weight is at your hand – that's like a fishing rod. (b) When you stand on your tiptoes, your toes are the pivot, your body's weight is in the middle, and your calf muscles push up at the back – that's like a wheelbarrow.

📝 Teacher's Note: This question is excellent for demonstrating levers using the human body. Have students physically perform the actions. For (a), emphasize the biceps (effort) between elbow (fulcrum) and hand (load). For (b), highlight toes (fulcrum), body weight (load) through the ankle, and calf muscles (effort) at the heel. This helps solidify the F-L-E positions for each class.

🎯 Exam Tip: For human body examples, clearly identify the fulcrum, load, and effort for each scenario. For (a) raising a load on the palm, it's F (elbow), E (biceps), L (load on palm) = Class III. For (b) raising body weight on toes, it's F (toes), L (body weight), E (calf muscles) = Class II. Correctly identifying the positions is key.

Question 32. Indicate the positions of load, effort and fulcrum in the forearm shown below in Fig 3.15 Name class of lever.

Answer: It is Class III lever. [Diagram: Forearm showing load, effort, and fulcrum positions as per Fig 3.15]

In simple words: Your forearm, when lifting something, works like a Class III lever. Your elbow acts as the pivot point (fulcrum), the muscle in your arm (biceps) pulls in the middle (effort), and the object you're lifting is at your hand (load).

📝 Teacher's Note: Emphasize that in a Class III lever, the effort is applied between the fulcrum and the load. This arrangement allows for a greater range of motion and speed, though it typically requires more effort force than the load force.

🎯 Exam Tip: For full marks, clearly state the class of lever and correctly identify the relative positions of the fulcrum, effort, and load. If a diagram is provided, ensure your identification aligns with it.

Question 33. Draw a labelled sketch of a class III lever. Give one example of this kind of lever.

Answer: [Diagram: Labelled sketch of a Class III lever showing fulcrum at one end, load at the other, and effort in the middle] Examples: foot treadle.

In simple words: Imagine a fishing rod. Your hand holding the rod near your body is the pivot (fulcrum), your other hand pulling the line is the effort in the middle, and the fish at the end of the line is the load. A foot treadle works similarly, with the pivot at the back, your foot applying effort in the middle, and the action happening at the front.

📝 Teacher's Note: When drawing a Class III lever, ensure the fulcrum (F) is at one end, the load (L) is at the other, and the effort (E) is applied somewhere in between them. Common examples include fishing rods, sugar tongs, and the human forearm.

🎯 Exam Tip: A labelled diagram must clearly show F (Fulcrum), E (Effort), and L (Load) in their correct relative positions for a Class III lever. Providing a relevant, distinct example is crucial for completing the answer.

Question 34. Give example of each class of lever in a human body?

Answer:

(i) Class I lever in the action of nodding of the head: In this action, the spine acts as the fulcrum, load is at its front part, while effort is at its rear part.

(ii) Class II lever in raising the weight of the body on toes: The fulcrum is at toes at one end, the load is in the middle and effort by muscles is at the other end.

(iii) Class III lever in raising a load by forearm: The elbow joint acts as fulcrum at one end, biceps exerts the effort in the middle and a load on the palm is at the other end.

In simple words: Your body uses all three types of levers! Nodding your head is a Class I lever (spine is pivot, head is load, neck muscles are effort). Standing on your tiptoes is a Class II lever (toes are pivot, body weight is load, calf muscles are effort). Lifting something with your arm is a Class III lever (elbow is pivot, bicep muscle is effort, object in hand is load).

📝 Teacher's Note: Use simple demonstrations in class for each lever type in the human body. For Class I, have students nod their heads. For Class II, have them stand on tiptoes. For Class III, have them lift a light object with their forearm. This makes the concepts very tangible and relatable.

🎯 Exam Tip: For each class of lever, clearly identify the specific body action and precisely state the location of the fulcrum, load, and effort. Using distinct and accurate examples for each class is key to scoring full marks.

Question 35. State the class of levers and the relative positions of load (L) effort (E) and fulcrum (F) in (a) a bottle opener, (b) sugar tongs.

Answer:

(a) A bottle opener is a lever of the second order, as the load is in the middle, fulcrum at one end and effort at the other. Bottle opener

(b) Sugar tongs is a lever of the third order as the effort is in the middle, load at one end and fulcrum at the other end. Sugar tongs

In simple words: A bottle opener is like a wheelbarrow – the pivot (fulcrum) is at one end, the bottle cap (load) is in the middle, and you push down (effort) at the other end. Sugar tongs are like tweezers – the pivot (fulcrum) is at the hinge, you squeeze in the middle (effort), and the sugar cube (load) is at the end.

📝 Teacher's Note: Bring actual bottle openers and sugar tongs to class to demonstrate. This hands-on approach helps students visualize the positions of F, E, and L much more effectively than just descriptions. Emphasize the order of F-L-E for Class II and F-E-L for Class III.

🎯 Exam Tip: Clearly state the class of lever (first, second, or third order) for each item. Then, explicitly describe the relative positions of the load (L), effort (E), and fulcrum (F) to earn full credit. Use the F-L-E or F-E-L notation if helpful.

Question 36. Draw Diagrams to illustrate the position of fulcrum load and effort, in each of the following: (a) A seesaw (b) A common balance (c) A nut cracker (d) Forceps

Answer:

(a) A seesaw [Diagram: A seesaw illustrating fulcrum in the middle, load at one end, and effort at the other end]

In simple words: A seesaw is the classic example of a Class I lever. The pivot point (fulcrum) is right in the middle, the person sitting on one end is the load, and the person pushing down on the other end is the effort.

📝 Teacher's Note: For Class I levers like a seesaw, emphasize that the fulcrum is always between the effort and the load. The mechanical advantage can be greater than, less than, or equal to 1, depending on the distances of the effort and load from the fulcrum.

🎯 Exam Tip: When asked to draw diagrams, ensure they are clearly labelled with F (Fulcrum), E (Effort), and L (Load) in their correct positions. For a seesaw, the fulcrum must be central, with load and effort on opposite sides.

Question 37. Classify the following into levers as class I, class II or class III.

(a) a door

(b) a catapult

(c) a wheel barrow

(d) a fishing rod.

Answer:

a. Class II

b. Class I

c. Class II

d. Class III

In simple words: Levers are tools that help us lift or move things. They are grouped into three types based on where the effort (where you push), the load (what you're moving), and the pivot point (fulcrum) are located. For example, a door is a Class II lever because the load (the door's weight) is between the fulcrum (hinges) and the effort (where you push).

📝 Teacher's Note: To help students remember, use simple analogies. Class I: Seesaw (fulcrum in middle). Class II: Wheelbarrow (load in middle). Class III: Fishing rod (effort in middle). Have students identify everyday objects for each class.

🎯 Exam Tip: For full marks, clearly state the class of lever and be prepared to justify your answer by identifying the relative positions of the fulcrum, load, and effort for each example.

Question 1. Mechanical advantage (M.A), load (L) and effort (E) are related as:

(a) M.A. = x E

(b) M.A x E = L

(c) E = M.A. x L

(d) None of these

Answer: M.A. x E = L

In simple words: Mechanical advantage tells us how much a machine multiplies the force we put in. If you push with a small force (effort) and the machine lifts a heavy object (load), it has a high mechanical advantage. The formula shows that if you multiply the mechanical advantage by the effort you put in, you get the load that can be moved.

📝 Teacher's Note: Emphasize that M.A. is a ratio (Load/Effort) and has no units. Use examples like a crowbar or a car jack to illustrate how a small effort can move a large load due to high M.A.

🎯 Exam Tip: Remember the fundamental relationship: Mechanical Advantage (M.A.) = Load (L) / Effort (E). This can be rearranged as L = M.A. x E or E = L / M.A. Knowing this formula is crucial for solving problems related to machines.

Question 2. The correct relationship between the mechanical advantage (M.A), the velocity ratio (V.R) and the efficiency (n) is:

(a) M.A. = \( \eta \) x V.R.

(b) V.R. = \( \eta \) x M.A.

(c) \( \eta \) = M.A. x V.R.

(d) None of these

Answer: M.A. = \( \eta \) x V.R.

In simple words: Efficiency tells us how good a machine is at turning the energy we put in into useful work. It's like how much of your effort actually helps move the load, compared to how much is lost to things like friction. This formula shows that the mechanical advantage (how much force is multiplied) is equal to the efficiency multiplied by the velocity ratio (how much faster the effort moves compared to the load).

📝 Teacher's Note: Explain that efficiency is always less than 1 (or 100%) for real machines due to energy losses (friction, air resistance). Use a bicycle as an example: some energy is lost in the chain and gears, so not all your pedaling effort goes into moving the bike forward.

🎯 Exam Tip: The formula for efficiency is \( \eta = \frac{M.A.}{V.R.} \). Rearranging this gives M.A. = \( \eta \) x V.R. or V.R. = M.A. / \( \eta \). Ensure you can recall and apply this relationship, and remember that \( \eta \) is usually expressed as a decimal or percentage.

Question 3. Which of the following statements is not true for a machine:

(a) It always has efficiency less than 100%

(b) its mechanical advantage can be less than 1.

(c) It can also be used as a speed multiplier

(d) It can have a mechanical advantage greater than the velocity ratio.

Answer: (d) It can have a mechanical advantage greater than the velocity ratio.

Reason: If the mechanical advantage of a machine is greater than its velocity ratio, then it would mean that the efficiency of a machine is more than 100%, which is practically not possible.

In simple words: Machines help us do work, but they aren't perfect. They always lose a little bit of energy, usually as heat from friction. So, a machine can never give you more work out than you put in. If a machine's mechanical advantage (how much force it multiplies) was more than its velocity ratio (how much faster the effort moves), it would mean it's giving you more energy than you put in, which is like magic and doesn't happen in the real world.

📝 Teacher's Note: This question tests the understanding of the practical limits of machines and the concept of efficiency. Emphasize that efficiency \( \eta = \frac{M.A.}{V.R.} \) and for any real machine, \( \eta < 1 \) (or < 100%). Therefore, M.A. must always be less than or equal to V.R. (M.A. \( \le \) V.R.).

🎯 Exam Tip: A key concept in machines is that efficiency cannot exceed 100%. This implies that the Mechanical Advantage (M.A.) can never be greater than the Velocity Ratio (V.R.) for a real machine. If M.A. > V.R., then \( \eta \) > 100%, which violates the law of conservation of energy. This is a common trick question to test fundamental understanding.

Question 4. The lever for which the mechanical advantage is less than 1 has:
(a) fulcrum at mid - point between load and effort.
(b) Load between effort and fulcrum
(c) effort between fulcrum and load
(d) Load and effort acting at the same point

Answer: (c) effort is between fulcrum and load. Hint: Levers, for which the mechanical advantage is less than 1, always have the effort arm shorter than the load arm.

In simple words: If a lever makes it harder to lift something (you need to push or pull with more force than the weight you're lifting), it means the part where you push is shorter than the part that lifts the weight. This happens when your push point is in the middle, between the pivot (fulcrum) and the weight.

📝 Teacher's Note: Emphasize the relationship between effort arm, load arm, and mechanical advantage (MA). MA < 1 implies a gain in speed/distance, but a loss in force. Examples include tweezers, fishing rods, and human forearm when lifting.

🎯 Exam Tip: For MA < 1, the effort arm is always shorter than the load arm. This configuration is characteristic of a Class III lever, where the effort is applied between the fulcrum and the load.

Question 5. Class II levers are designed to have:
(a) M.A. = V.R.
(b) M.A. > V.R.
(c) M.A > 1
(d) M.A < 1

Answer: (c) M.A > 1. Hint: In class II levers, the load is in between the effort and fulcrum. Thus, the effort arm is always longer than the load arm and less effort is needed to overcome a large load. Hence, M.A > 1.

In simple words: Class II levers are like a wheelbarrow or a bottle opener. They are designed so that you don't have to push or pull as hard as the weight you're lifting. This means they make your job easier by multiplying your force, so their mechanical advantage is always greater than 1.

📝 Teacher's Note: Use real-world examples like a nutcracker or a wheelbarrow to illustrate Class II levers. Highlight that the load is always between the fulcrum and the effort, making the effort arm longer than the load arm, thus providing a force multiplier.

🎯 Exam Tip: In Class II levers, the load is always between the fulcrum and the effort. This arrangement ensures the effort arm is longer than the load arm, resulting in a mechanical advantage (MA) greater than 1, meaning they are force multipliers.

Question 1. A crowbar of length 120 cm has its fulcrum situated at a distance of 20 cm from the load. Calculate the mechanical advantage of the crowbar.

Answer:
Total length of crowbar = 120 cm
Load arm = 20 cm
Effort arm = 120 - 20 = 100 cm
Mechanical advantage M.A = \( \frac{\text{Effort arm}}{\text{Load arm}} \)
M.A = \( \frac{100}{20} \) = 5

In simple words: Imagine a long stick (crowbar) used to lift something heavy. If the part of the stick that pushes the heavy thing is short (20 cm) and the part where you push down is long (100 cm), then the crowbar makes it 5 times easier to lift the heavy thing. That "5 times easier" is its mechanical advantage.

📝 Teacher's Note: Reinforce the definition of mechanical advantage as the ratio of effort arm to load arm for levers. Stress the importance of consistent units (cm in this case). Discuss how a higher MA means less effort is required.

🎯 Exam Tip: Clearly identify the effort arm and load arm from the problem description. The formula for mechanical advantage of a lever is MA = (Effort Arm) / (Load Arm). Ensure units are consistent before calculation.

Question 2. A 4 m long rod of negligible weight is to be balanced about a point 125 cm from one end. A load of 18 kgf is suspended at a point 60 cm from the support on the shorter arm. (a) a weight W is placed 250 cm from the support on the longer arm Find W. (b) If W = 5 kgf, where must it be kept to balance the rod? (c) To which class of lever does it belong?

Answer:
Total length of rod = 4 m = 400 cm

(a) 18 kgf load is placed at 60 cm from the support. W kgf weight is placed at 250 cm from the support.
By the principle of moments:
18 x 60 = W x 250
W = \( \frac{18 \times 60}{250} \)
W = 4.32 kgf

(b) Given W = 5 kgf
18 kgf load is placed at 60 cm from the support.
Let 5 kgf of weight is placed at d cm from the support.
By the principle of moments:
18 x 60 = 5 x d
d = \( \frac{18 \times 60}{5} \)
d = 216 cm from the support on the longer arm

(c) It belongs to class I lever.

In simple words: Imagine a seesaw. If a heavy person (18 kgf) sits close to the middle (60 cm), a lighter person (W) needs to sit further away (250 cm) to balance it. We calculate how heavy that lighter person needs to be. Then, if we know the lighter person's weight (5 kgf), we figure out exactly where they need to sit to balance the seesaw. This setup, where the pivot is in the middle, is like a Class I lever.

📝 Teacher's Note: This problem applies the principle of moments (or torque balance): Clockwise Moment = Anti-clockwise Moment. Emphasize that moment = force x perpendicular distance from the fulcrum. For part (c), identify the fulcrum between the load and effort for a Class I lever.

🎯 Exam Tip: For balancing problems, apply the principle of moments: \( F_1 d_1 = F_2 d_2 \). Clearly label forces and distances from the fulcrum. For lever classification, identify the relative positions of fulcrum, load, and effort.

Question 3. A pair of scissors has its blades 15 cm long, while its handles are 7.5 cm long. What is its mechanical advantage?

Answer:
Effort arm = 7.5 cm
Load arm = 15 cm
Mechanical advantage M.A = \( \frac{\text{Effort arm}}{\text{Load arm}} \) = \( \frac{7.5}{15} \) = 0.5

In simple words: Scissors are a type of lever. The handles are where you put your effort, and the blades are where the cutting force acts (the load). If the handles are shorter (7.5 cm) than the blades (15 cm), it means you have to squeeze harder than the force the blades apply. So, the mechanical advantage is less than 1 (0.5), meaning it's designed for speed or range of motion, not to make cutting easier in terms of force.

📝 Teacher's Note: Scissors are a pair of Class I levers. The pivot is the screw. The handles are the effort arm, and the blades are the load arm. When the effort arm is shorter than the load arm (as in this case), MA < 1, indicating a gain in speed/distance but a loss in force.

🎯 Exam Tip: For scissors, the fulcrum is the pivot screw. The distance from the pivot to where you apply force on the handles is the effort arm. The distance from the pivot to the cutting point on the blades is the load arm. Calculate MA using the ratio of these arms.

Question 4. A force of 5 kgf is required to cut a metal sheet. A shears used for cutting the metal sheet has its blades 5 cm long, while its handles is 10 cm long. What effort is needed to cut the sheet?

Answer:
Effort arm = 10 cm
Load arm = 5 cm
Mechanical advantage M.A = \( \frac{\text{Effort arm}}{\text{Load arm}} \) = \( \frac{10}{5} \) = 2
Load = 5 kgf
Effort = \( \frac{\text{Load}}{\text{M.A}} \) = \( \frac{5}{2} \) = 2.5 kgf

In simple words: A metal shear is like a super-strong pair of scissors. If its handles are long (10 cm) and its blades are short (5 cm), it means it makes cutting much easier. It multiplies your effort. If you need 5 kgf of force to cut the metal, and the shear makes your effort twice as strong (MA=2), then you only need to push with half that force, which is 2.5 kgf.

📝 Teacher's Note: This problem demonstrates how a lever can be a force multiplier (MA > 1). Shears are typically Class I levers. Emphasize the inverse relationship between effort and mechanical advantage: Effort = Load / MA.

🎯 Exam Tip: Identify the effort arm (handles) and load arm (blades). Calculate MA first. Then, use the relationship \( \text{Effort} = \frac{\text{Load}}{\text{Mechanical Advantage}} \) to find the required effort. Ensure consistent units.

Question 5. [Diagram: Lever in use, Fig 3.16] below shows a lever in use. (a) To which class of lever does it belong? (b) If AB = 1m, AF = 0.4 m, find its mechanical advantage, (c) calculate the value of E.

Answer:

(a) This is a class I lever.

(b) Given AB=1m, AF=0.4m and BF=0.6 m
Mechanical advantage M.A = \( \frac{BF}{AF} = \frac{0.6}{0.4} = 1.5 \)

(c) Load = 15 kgf
Effort = \( \frac{\text{Load}}{\text{M.A}} = \frac{15}{1.5} = 10 \text{ kgf} \)

In simple words: A Class I lever has the fulcrum in the middle, between the effort and the load. Its mechanical advantage tells us how much easier it makes lifting something. Here, it makes it 1.5 times easier, so a 15 kgf load needs only 10 kgf of effort.

📝 Teacher's Note: Emphasize the position of the fulcrum, load, and effort for each class of lever. Use real-world examples like a seesaw (Class I), wheelbarrow (Class II), and fishing rod (Class III) to make it relatable. Mechanical advantage is a ratio, so it's unitless, but the effort and load are in kgf.

🎯 Exam Tip: Clearly state the class of lever based on the relative positions of F (Fulcrum), L (Load), and E (Effort). For calculations, remember the formulas for mechanical advantage (MA = Load/Effort or MA = Effort arm/Load arm) and ensure units are consistent. Show all steps for full marks.

Question 6. A man uses a crowbar of length 1.5 m to raise a load of 75 kgf by putting a sharp edge below the bar at a distance 1 m from his hand. Draw a diagram of the arrangement showing the fulcrum (F), load (L) and effort (E) with their directions. State the kind of lever. Calculate: (i) load arm, (ii) effort arm, (iii) mechanical advantage and (iv) the effort needed.

Answer:

[Diagram: Crowbar arrangement showing fulcrum (F), load (L), and effort (E) with directions]
Crowbar is a class I lever.

(i) Total length of crowbar = 1.5 m
Effort arm = 1 m
Load arm = 1.5 - 1 = 0.5 m

(ii) Effort arm = 1 m

(iii) Mechanical advantage M.A = \( \frac{\text{Effort arm}}{\text{Load arm}} = \frac{1}{0.5} = 2 \)

(iv) The effort needed
Effort = \( \frac{\text{Load}}{\text{M.A}} = \frac{75}{2} = 37.5 \text{ kgf} \)

In simple words: A crowbar works like a seesaw (Class I lever). You push down on one end (effort), the fulcrum (the sharp edge) is in the middle, and the other end lifts the heavy thing (load). Because your hand is further from the fulcrum than the load, it makes lifting the heavy load much easier.

📝 Teacher's Note: When drawing the diagram, ensure the fulcrum is between the effort and the load for a Class I lever. Clearly label the effort arm (distance from fulcrum to effort) and load arm (distance from fulcrum to load). Discuss how a longer effort arm increases mechanical advantage.

🎯 Exam Tip: For diagram questions, accurately label F, L, E, and their directions. Correctly identify the class of lever. Show calculations for load arm, effort arm, MA, and effort clearly, including units. Remember that for a Class I lever, MA can be greater than, less than, or equal to 1.

Question 7. A pair of scissors is used to cut a piece of a cloth by keeping it at a distance 8.0 cm from its rivet and applying an effort of 10 kgf by fingers at a distance 2.0 cm from the rivet. (a) Find : (i) the mechanical advantage of scissors and (ii) the load of fered by the cloth (b) How does the pair of scissors act: as a force multiplier or as speed multiplier?

Answer:

Effort arm = 2 cm
Load arm = 8.0 cm
Given effort = 10 kgf

(i) Mechanical advantage M.A = \( \frac{\text{Effort arm}}{\text{Load arm}} = \frac{2}{8} = 0.25 \)

(ii) Load = M.A × effort = 0.25 × 10 = 2.5 kgf

The pair of scissors acts as a speed multiplier because MA < 1.

In simple words: Scissors are like two levers joined together. When you cut cloth, your fingers (effort) move a short distance, but the blades (load) move a longer distance to cut. This means you need more force from your fingers, but the cutting action is faster over a larger area. Since the mechanical advantage is less than 1, it's a speed multiplier, not a force multiplier.

📝 Teacher's Note: Explain that scissors are a Class I lever, with the rivet as the fulcrum. Discuss the trade-off between force multiplication and speed multiplication. When MA < 1, it's a speed multiplier (or distance multiplier), meaning you apply more force but get a larger movement at the load end. When MA > 1, it's a force multiplier.

🎯 Exam Tip: Correctly identify the effort arm (distance from fulcrum to effort) and load arm (distance from fulcrum to load). Calculate MA accurately. Remember that if MA < 1, the machine is a speed multiplier; if MA > 1, it's a force multiplier. Clearly state the reason for your conclusion.

Question 8. [Diagram: Lever in use, Fig 3.17] below shows a lever in use. (a) To which class of lever does it belong? (b) If FA = 80 cm, AB = 20 cm, find its mechanical advantage. (c) Calculate the value of E.

Answer:

(a) This is a class II lever.

(b) Given: FA = 80 cm, AB = 20 cm, BF = FA + AB = 100 cm
Mechanical advantage M.A = \( \frac{BF}{AF} = \frac{100}{80} = 1.25 \)

(c) Effort (E) = \( \frac{\text{Load (L)}}{\text{M.A}} = \frac{5}{1.25} = 4 \text{ Kgf} \)

In simple words: A Class II lever has the load in the middle, between the fulcrum and the effort. Think of a wheelbarrow. Here, the mechanical advantage is greater than 1, meaning it helps you lift a heavy load with less effort.

📝 Teacher's Note: For Class II levers, the load is always between the fulcrum and the effort. This configuration always results in a mechanical advantage greater than 1, making it a force multiplier. Use examples like a nutcracker or bottle opener to illustrate.

🎯 Exam Tip: Accurately identify the class of lever based on the F-L-E arrangement. Correctly identify the load arm and effort arm from the given distances. For Class II levers, the effort arm is always greater than the load arm, ensuring MA > 1. Show all calculation steps and units clearly.

Question 9. [Diagram: Lever setup] shows the use of a lever. (a) State the principle of moments as applied to the above lever. (b) Give an example of this class of lever. (c) If FA = 10 cm, AB = 500 cm calculate : (i) the mechanical advantage and (ii) the minimum effort required to lift the load.

Answer:

(a) The principle of moments: Moment of the load about the fulcrum = moment of the effort about the fulcrum

\( FB \times Load = FA \times Effort \)

(b) Sugar tongs is an example of this class of lever.

(c) Given: FA = 10 cm, AB = 500 cm. Assuming F is the fulcrum and B is the load point, then BF = AB - FA = 500 - 10 = 490 cm. (Note: The solution uses BF = 500 + 10 = 510 cm, implying A is the fulcrum, B is the effort, and F is the load, or a different lever configuration. Following the solution's calculation for BF=510cm and assuming a load of 50N as used in the solution):

Given: FA = 10 cm, AB = 500 cm. From the solution's calculation, BF = 500 + 10 = 510 cm. (This implies F is the fulcrum, A is the effort point, and B is the load point, or vice-versa, with AB being the total length and FA being one arm, and FB the other. Let's follow the solution's values directly.)

(i) The mechanical advantage:

\( M.A = \frac{AF}{BF} = \frac{10}{510} = \frac{1}{51} \)

(ii) The minimum effort required to lift the load (assuming Load = 50 N, as used in the solution):

\( Effort = \frac{Load}{M.A} = \frac{50}{\frac{1}{51}} = 2550 \ N \)

In simple words: Levers help us lift heavy things or apply force easily by balancing the turning effect of the load with our effort. Like a seesaw, if you push down far from the middle, you can lift someone heavy on the other side.

📝 Teacher's Note: Emphasize the concept of fulcrum, load arm, and effort arm. Discuss the three classes of levers and their practical applications. Highlight that the principle of moments is fundamental to understanding rotational equilibrium. Note that the load value was not explicitly given in the question but was used as 50 N in the solution.

🎯 Exam Tip: Define the principle of moments clearly: "For a body to be in rotational equilibrium, the sum of clockwise moments about the fulcrum must equal the sum of anticlockwise moments about the fulcrum." State the formula: Load × Load Arm = Effort × Effort Arm. Ensure correct identification of load arm and effort arm based on the fulcrum's position.

Question 10. [Diagram: Wheelbarrow setup] shows a wheel barrow of mass 15 kg carrying a load of 30 kgf with its centre of gravity at A. The points B and C are the centre of wheel and tip of the handle such that the horizontal distance AB = 20 cm and AC = 40 cm. Calculate: (i) the load arm, (ii) the effort arm, (iii) the mechanical advantage and (iv) the minimum effort required to keep the leg just off the ground.

Answer:

Assuming B is the fulcrum (centre of the wheel):

(i) Load arm AF = AB = 20 cm

(ii) Effort arm CF = CB = AB + AC = 20 cm + 40 cm = 60 cm

(iii) Mechanical advantage:

\( M.A = \frac{CF}{AF} = \frac{60}{20} = 3 \)

(iv) Total load = Load carried + Mass of wheelbarrow = 30 kgf + 15 kgf = 45 kgf

Minimum effort required:

\( Effort = \frac{Load}{M.A} = \frac{30 + 15}{3} = \frac{45}{3} = 15 \ kgf \)

In simple words: A wheelbarrow makes it easier to carry heavy stuff. The wheel acts as a pivot, and because the load is closer to the wheel than where you lift, you don't have to push as hard.

📝 Teacher's Note: Explain why a wheelbarrow is a Class 2 lever (Load between Fulcrum and Effort). Discuss how the mechanical advantage is always greater than 1 for Class 2 levers, making them force multipliers. Remind students to include the mass of the wheelbarrow itself in the total load if specified.

🎯 Exam Tip: Correctly identify the fulcrum, load, and effort points. Calculate load arm and effort arm accurately from the given distances. Remember to sum all loads (object + machine's own weight) when calculating total load.

Question 11. A fire tongs has arms 20 cm long. Its is used to lift a coal of weight 1.5 kgf by applying an effort at a distance 15 cm from the fulcrum. Find: (i) the mechanical advantage of fire tongs and (ii) the effort needed.

Answer:

For fire tongs (a Class 3 lever):

Load arm = Length of arms = 20 cm

Effort arm = 15 cm (distance from fulcrum where effort is applied)

(i) Mechanical advantage:

\( M.A = \frac{Effort \ arm}{Load \ arm} = \frac{15}{20} = 0.75 \)

(ii) Effort needed:

\( Effort = \frac{Load}{M.A} = \frac{1.5}{0.75} = 2.0 \ kgf \)

In simple words: Fire tongs are good for picking up small, hot things, but you have to squeeze harder than the weight of the object. This is because your hand is closer to the pivot than the coal you're picking up.

📝 Teacher's Note: Explain why fire tongs are a Class 3 lever (Effort between Fulcrum and Load). Discuss that for Class 3 levers, the mechanical advantage is always less than 1, meaning they are not force multipliers but are used for speed or convenience.

🎯 Exam Tip: Clearly identify the fulcrum, load, and effort positions. Understand that for Class 3 levers, the effort arm is always shorter than the load arm, resulting in MA < 1. Be careful with units (kgf vs N).

Question 1. What is an inclined plane? Give two examples where it us used to raise a heavy load with less effort?

Answer:

Inclined plane: An inclined plane is a sloping surface that behaves like a simple machine whose mechanical advantage is always greater than 1.

Example: The inclined plane is used to load a truck or to take the scooter from road into the house on a higher level. Inclined planes are used to reach the bridge over the railway tracks at a railway station.

In simple words: An inclined plane is just a ramp! It helps us move heavy things up without lifting them straight up, making it feel lighter because we push over a longer distance.

📝 Teacher's Note: Introduce the inclined plane as one of the six simple machines. Explain how it reduces the effort required by increasing the distance over which the force is applied. Discuss the trade-off between force and distance.

🎯 Exam Tip: Define an inclined plane as a sloping surface that allows a heavy load to be moved to a higher level with less effort. Provide clear, real-world examples like ramps, slides, or screw threads.

Question 2. ‘The force needed to push a load up an inclined plane is less than the force needed to lift it directly’ Give a reason.

Answer:

Since less effort is needed in lifting a load to a higher level by moving over an inclined plane as compared to that in lifting the load directly, an inclined plane acts as a force multiplier. This is because the mechanical advantage of an inclined plane is always greater than 1.

In simple words: Pushing something up a ramp is easier than lifting it straight up because the ramp spreads out the work over a longer path, so you don't have to push as hard at any one moment.

📝 Teacher's Note: Reinforce the concept of work done (Work = Force × Distance). Explain that the work done against gravity is the same whether lifting directly or using an inclined plane, but the inclined plane increases the distance, thus decreasing the required force (effort). This is the principle of a force multiplier.

🎯 Exam Tip: The key reason is that an inclined plane acts as a force multiplier because its mechanical advantage is always greater than 1. This means the effort required is less than the load, as the load is moved over a greater distance.

Question 3. Write an expression for the mechanical advantage of an inclined plane in terms of its length I and vertical height h.

Answer: Solution not provided in the source text.

In simple words: If you have a ramp, its "helpfulness" (mechanical advantage) is how long the ramp is divided by how high it goes. A longer, less steep ramp is more helpful.

📝 Teacher's Note: Derive the formula for MA of an ideal inclined plane: MA = Length (l) / Height (h). Explain that this assumes an ideal inclined plane (no friction). Discuss how friction would reduce the actual mechanical advantage.

🎯 Exam Tip: The ideal mechanical advantage (IMA) of an inclined plane is given by the ratio of its length (l) to its vertical height (h), i.e., MA = l/h. Ensure 'l' is the length along the slope and 'h' is the vertical rise.

Question 4. State whether the mechanical advantage of an inclined plane is equal to 1, less than 2 or greater than 1?

Answer: Mechanical advantage of an inclined plane is always greater than 1.

In simple words: An inclined plane (like a ramp) makes it easier to lift things. This "easier" part means you need less force, which is why its mechanical advantage is always more than 1.

📝 Teacher's Note: Discuss how an inclined plane trades distance for force. The longer the ramp for the same height, the less force is needed, illustrating MA > 1. Use examples like pushing a box up a ramp vs. lifting it directly.

🎯 Exam Tip: For full marks, state clearly that MA > 1 for an inclined plane and briefly explain why (force required is less than the weight lifted, due to the angle). Mentioning the formula \( MA = \frac{L}{h} \) where L > h is also good.

Question 5. What is a gear system? Explain its working.

Answer:

Gear system: A gear system is a device to transfer precisely the rotational motion from one point to the other. A gear is a wheel with teeth around its rim. The teeth act as the components of a machine and they transmit rotational motion to the wheel by successively engaging the teeth of the other rotating gear.

Working: Each tooth of a gear acts like a small lever of class I. A gear when in operation, can be considered as a lever with an additional property that it can be continuously rotated instead of moving back and forth as is the case with an ordinary lever. Each gear wheel is mounted on an axle which rotates at a speed depending upon the motion transmitted to it. The gear wheel closer to the source of power is called the driver, while the gear wheel which receives motion from the driver is called the driven gear. The driven gear rotates in a direction opposite to the driving gear when the two gears make an external contact. On the other hand, if the gears make an internal contact, both gears rotate in the same direction.

In simple words: A gear system is like a set of toothed wheels that fit together. When one wheel turns, its teeth push the teeth of the next wheel, making it turn too. This helps move power or change how fast or strong something spins.

📝 Teacher's Note: Use physical gear models or animations to demonstrate external vs. internal contact and the resulting direction of rotation. Emphasize the concept of force transmission and how gears are essentially continuous levers.

🎯 Exam Tip: Define a gear system clearly, mentioning "toothed wheels" and "transfer rotational motion." For working, explain the engagement of teeth and the roles of "driver" and "driven" gears, including the direction of rotation for external and internal contacts.

Question 6. Explain how a gear system can be used to obtain: (a) gain in speed (b) gain in torque and (c) Change in direction of rotation. Given one example for each.

Answer:

(a) Gain in speed: A gear system can be used to obtain gain in speed when the bigger wheel drives the smaller wheel, i.e., when the driving gear has more number of teeth than the driven gear. To obtain gain in speed, the gear ratio should be more than one. Mathematically, \( \text{Gain in speed} = \frac{\text{Number of teeth in driving wheel}}{\text{Number of teeth in driven wheel}} \)

Example: A toy motor car uses the gear principle to obtain gain in speed. It has a key and spring on the axle fitted with a driving gear having more teeth which engages the driven gear having fewer teeth. The wheels of the car are fitted on the axle of the driven gear. When the key is turned clockwise (or the toy car is pulled back by hand) the spring is wound up. On releasing the key (or the toy car), the spring turns the driving gear anti-clockwise, which in turn rotates the wheels of the toy car clockwise and the car moves forward at a greater speed.

(b) Gain in torque: A gear system can be used to obtain gain in torque when the smaller wheel drives the bigger wheel, i.e., when the driving gear has less number of teeth than the driven gear. To obtain gain in torque, the gear ratio should be less than one. Mathematically, \( \text{Gain in torque} = \frac{\text{Number of teeth in driven wheel}}{\text{Number of teeth in driving wheel}} \)

Example: While ascending a hill, an automobile driver changes gears and puts the driving gear of less number of teeth with a driven gear of more number of teeth. By doing so, he obtains a gain in torque, as more torque is required to go up the hill than to move along a level road.

(c) Change in direction of rotation: A gear system can be used to obtain change in direction when both the wheels of the gear system have the same number of teeth. Two gears mesh together in such a way that the driven gear rotates in direction opposite to the driving gear without any gain in speed or torque. So, if the driving gear turns clockwise, the driven gear turns counterclockwise. To obtain change in direction, the gear ratio should be equal to 1.

Example: In a car, the differential (a gearbox in the middle of the rear axle of a rear-wheel drive car) uses a cone-shaped bevel gear to turn the driveshaft's power through 90 degrees and turn the back wheels.

In simple words: Gears can be used to make things spin faster (gain speed, like in a toy car), spin with more power (gain torque, like when a car climbs a hill), or simply change the direction of spin (like in a car's rear axle).

📝 Teacher's Note: Use diagrams or animations to show different gear configurations for speed gain, torque gain, and direction change. Emphasize the inverse relationship between speed and torque. A smaller driving gear turning a larger driven gear gives torque gain (and speed loss), and vice-versa. For direction change, two meshing gears always rotate in opposite directions.

🎯 Exam Tip: Clearly state the condition for each (driving gear teeth vs. driven gear teeth). Provide the mathematical expression for gain in speed and torque. Give a relevant, concise example for each. For direction change, mention that gears rotate in opposite directions when externally meshed.

Question 7. Define the following terms in reference to a gear system: (a) Driving gear (b) Driven gear (c) Gear ratio (d) Gain in speed (e) Gain in torque

Answer:

(a) Driving gear: The gear wheel closer to the source of power is called the driving gear.

In simple words: In a gear system, the 'driving gear' is the one that starts the movement.

📝 Teacher's Note: Emphasize that the driving gear is directly connected to the power source (e.g., motor, engine, or hand crank) and initiates the motion transfer.

🎯 Exam Tip: Define the driving gear as the one connected to the power source that transmits motion to other gears.

(b) Driven gear: The gear wheel which receives motion from the driver is called the driven gear.

(c) Gear ratio: The ratio of number of teeth in the driving wheel to the number of teeth in the driven wheel is called the gear ratio.

(d) Gain in speed: The gain in speed is equal to the ratio of speed of rotation of driven wheel to the speed of rotation of the driving wheel.

(e) Gain in torque: The gain in torque is equal to the ratio of number of teeth in driven gear to the number of teeth in driving gear gives the gain in torque.

Question 8. What should be the gear ration of a car: equal to 1, less than 1 or greater than 1, while (a) gaining speed on the road, (b) ascending a hill

Answer:

(a) While gaining speed on the road, the gear ratio should be more than 1. That is, the driving gear should have more number of teeth than the driven gear.

(b) While ascending a hill more torque is required; thus, the gear ratio should be less than 1. That is, the driving gear should have less number of teeth than the driven gear.

In simple words: When a car needs to go fast on a flat road, it uses a gear setup where the engine's gear spins faster than the wheel's gear (gear ratio > 1). When it needs more power to climb a hill, it uses a gear setup where the engine's gear spins slower but with more force (torque) than the wheel's gear (gear ratio < 1).

📝 Teacher's Note: Explain the concept of mechanical advantage and disadvantage in terms of speed and torque. Use bicycle gears as an analogy: high gear for speed (flat road), low gear for power (uphill).

🎯 Exam Tip: For gaining speed, gear ratio > 1 (driving gear has more teeth). For ascending a hill (gaining torque), gear ratio < 1 (driving gear has fewer teeth). State the relationship between gear ratio, speed, and torque clearly.

Question 9. In a gear system, the gear ration of the driving wheel A and driven wheel B is 10:1. To rotate the driven wheel B in the direction of driving wheel A, the driving wheel A is engaged with other wheel C. what should be the gear ratio of the wheels A and C?

Answer:

Given, gear ratio = \( \frac{N_A}{N_B} = \frac{10}{1} \)

It is possible to obtain a change in direction, Using wheel C, if the number of teeth in Wheel C is equal to the number of teeth in wheel A.

∴ \( N_C = N_A = 10 \)

Hence, the gear ratio of wheels A and C: Gear ratio = \( \frac{N_A}{N_C} = \frac{10}{10} \)

∴ The required gear ration is 1:1

In simple words: If you want two gears to spin in the same direction, you need an extra gear in between them. If the first gear (A) has 10 teeth and the middle gear (C) also has 10 teeth, then the ratio between A and C is 10:10, which simplifies to 1:1. This middle gear C helps change the direction of the final driven wheel B to match A.

📝 Teacher's Note: Emphasize the role of an idler gear (wheel C) in changing the direction of rotation without altering the speed ratio between the driver and the final driven gear if its size is equal to the driver. The gear ratio between the driving gear and the idler gear is calculated as requested.

🎯 Exam Tip: Understand that an intermediate gear (idler) is used to achieve the same direction of rotation between the driving and driven gears. If the idler gear has the same number of teeth as the driving gear, their ratio is 1:1. Clearly state the given ratios and the final calculated ratio.

Question 1. The mechanical advantage of an inclined place is always: (a) less than 1 (b) equal to 1 (c) greater than 1 (d) nothing can be said

Answer:

Greater than 1

Hint: M.A = \( \frac{1}{\sin\theta} \)

In simple words: An inclined plane (like a ramp) always makes it easier to lift something, meaning you need less force. This "making it easier" is called mechanical advantage, and it's always more than 1 for a ramp.

📝 Teacher's Note: Discuss how an inclined plane reduces the effort required by increasing the distance over which the force is applied. Explain that for any angle \( \theta \) between 0 and 90 degrees, \( \sin\theta \) is between 0 and 1, making \( \frac{1}{\sin\theta} \) always greater than 1.

🎯 Exam Tip: Remember that the mechanical advantage (MA) of an ideal inclined plane is \( \frac{1}{\sin\theta} \) or \( \frac{\text{length}}{\text{height}} \). Since the length is always greater than the height (for a non-vertical ramp), MA is always greater than 1. State the formula and its implication.

Question 1. A boy has to lift a load of mass 50 kg to a height of 1 m. (a) what effort is required if he lifts it directly? Take g = 10 N kg^-1 (b) If the boy can exert a maximum effort of 250 N, so he uses an inclined plane to lift the load up. What should be the minimum length of the plank used by him?

Answer:

Mass of load m = 50 kg

Force required to lift a load directly (Load L) = m x g = 50 kg x 10 N kg^-1 = 500 N

The maximum effort exerted by boy E = 250 N

Load L = 500 N

Mechanical advantage M.A = \( \frac{\text{Load}}{\text{Effort}} = \frac{500 \text{ N}}{250 \text{ N}} = 2 \)

Also, for an inclined plane, M.A = \( \frac{l}{h} \)

Height (h) = 1 m

Minimum length of plank l = M.A x h = 2 x 1 m = 2 m

In simple words: (a) To lift a 50 kg object straight up, you need a force of 500 N. (b) If the boy can only push with 250 N, he needs a ramp (inclined plane). Since he needs to reduce his effort by half (500 N / 250 N = 2), the ramp needs to be twice as long as the height he wants to lift the object. So, for a 1 m height, the ramp must be at least 2 m long.

📝 Teacher's Note: Emphasize the difference between lifting directly (effort = load) and using a simple machine like an inclined plane (effort < load). Highlight the principle of conservation of work: Work input = Work output (ignoring friction). Work = Force x Distance. So, if force is reduced, distance must increase proportionally.

🎯 Exam Tip: Clearly state the given values and the formulas used. For part (a), calculate the load (weight) as effort. For part (b), calculate MA first, then use MA = \( \frac{l}{h} \) to find the length. Ensure units are consistent and correctly stated.

Question 2. A coolie uses a sloping wooden plank of length 2.0 m to push up a drum of mass 100 kg into the truck at a height 1.0 m. (a) What is the mechanical advantage of the sloping plank? (b) How much effort is needed to push the drum up into the truck? What assumption have you made in arriving at the answer in part (b) above?

Answer:

Length of sloping wooden plank l = 2.0 m

Mass of drum = 100 kg

Height of inclined plane h = 1.0 m

(a) The mechanical advantage of the sloping plank M.A = \( \frac{l}{h} = \frac{2.0 \text{ m}}{1.0 \text{ m}} = 2 \)

(b) To find the effort needed, we first calculate the load (weight of the drum). Assuming g = 10 N kg^-1 (standard for such problems unless specified):

Load L = mass x g = 100 kg x 10 N kg^-1 = 1000 N

We know M.A = \( \frac{\text{Load}}{\text{Effort}} \)

So, Effort = \( \frac{\text{Load}}{\text{M.A}} = \frac{1000 \text{ N}}{2} = 500 \text{ N} \)

The assumption made in part (b) is that there is no friction between the drum and the plank, meaning the inclined plane is ideal (100% efficient).

In simple words: A coolie uses a 2-meter long ramp to push a 100 kg drum up to a 1-meter high truck. (a) The ramp makes it twice as easy (Mechanical Advantage = 2). (b) To lift the 100 kg drum directly would need 1000 N of force. Since the ramp makes it twice as easy, the coolie only needs to push with 500 N. This calculation assumes the ramp is perfectly smooth and there's no friction making it harder.

📝 Teacher's Note: This question is excellent for discussing the concept of an ideal machine versus a real machine. Explain that in reality, friction would always be present, requiring more effort than calculated for an ideal machine. This leads to the concept of efficiency.

🎯 Exam Tip: For part (a), directly use the formula M.A = \( \frac{l}{h} \). For part (b), first calculate the load (weight) if not given, then use M.A = \( \frac{\text{Load}}{\text{Effort}} \) to find the effort. Crucially, remember to state the assumption of an ideal machine (no friction) for full marks when asked.

Question 3. A gear system has one wheel with 10 teeth and the other wheel with 50 teeth. Calculate the gain in speed and the gain in torque that you can obtain using them. What will be the gear ration in each case?

Answer:

Number of teeth in first wheel = 10

Number of teeth in second wheel = 50

For gain in speed, the second wheel of 50 teeth (N_A = 50) is used as driving wheel and the first wheel of 10 teeth (N_B = 10) is used as driven wheel.

Gear ratio = \( \frac{N_A}{N_B} \) = \( \frac{10}{50} \) = \( \frac{1}{5} \) = 1 : 5

Gain in speed = \( \frac{N_A}{N_B} \) = \( \frac{50}{10} \) = \( \frac{5}{1} \) = 5

For gain in torque, the second wheel of 50 teeth (N_B = 50) is used as driven wheel and the first wheel of 10 teeth (N_A = 10) is used as driving

Question 3. Name the pulley which has no gain in mechanical advantage. Explain, why is such a pulley then used?

Answer: There is no gain in mechanical advantage in the case of a single fixed pulley. A single fixed pulley is used only to change the direction of the force applied, that is, with its use, the effort can be applied in a more convenient direction. To raise a load directly upwards is difficult.

In simple words: A single fixed pulley doesn't make lifting easier in terms of force, but it lets you pull downwards instead of upwards, which is much more comfortable and practical, like pulling a bucket from a well.

📝 Teacher's Note: Emphasize that while MA=1, the change in direction is a significant practical advantage. Discuss examples like flagpoles or drawing water from a well.

🎯 Exam Tip: For full marks, state that a single fixed pulley has a Mechanical Advantage (MA) of 1 and its primary function is to change the direction of effort, making the task more convenient. Mention that it does not act as a force multiplier.

Question 4. What is the velocity ratio of a single fixed pulley?

Answer: The velocity ratio of a single fixed pulley is 1.

In simple words: For a single fixed pulley, if you pull the rope down by one meter, the load goes up by one meter. So, the "speed ratio" is 1.

📝 Teacher's Note: Explain that Velocity Ratio (VR) is the ratio of the distance moved by effort to the distance moved by load. For a single fixed pulley, these distances are equal, hence VR = 1. This is an ideal value, not affected by friction.

🎯 Exam Tip: State clearly that the velocity ratio (VR) of a single fixed pulley is 1. No further explanation is usually required unless asked.

Question 5. In a single fixed pulley, if the effort moves by a distance x downwards, by what height is the load raised upwards?

Answer: The load rises upwards with the same distance x.

In simple words: If you pull the rope down by a certain amount (x), the thing you're lifting goes up by the exact same amount (x).

📝 Teacher's Note: This question directly relates to the concept of Velocity Ratio. Since VR = 1 for a single fixed pulley, the distance moved by effort equals the distance moved by load. Use a simple diagram to illustrate this.

🎯 Exam Tip: The key is to state that the load is raised by the *same distance x*. This demonstrates understanding of the 1:1 displacement ratio in a single fixed pulley.

Question 6. What is a single movable pulley? What is its mechanical advantage in the ideal case?

Answer: Single movable pulley: A pulley, whose axis of rotation is not fixed in position, is called a single movable pulley. Mechanical advantage in the ideal case is 2.

In simple words: A single movable pulley is one that moves up and down with the load you're lifting. It helps you lift things by making them feel half as heavy, so its "lifting power" is 2 times better.

📝 Teacher's Note: Contrast this with a fixed pulley. Explain that the load is supported by two rope segments, effectively halving the effort required. Demonstrate with a simple setup if possible. Ideal case assumes no friction and massless pulley.

🎯 Exam Tip: Define a single movable pulley by stating its axis of rotation moves with the load. For ideal mechanical advantage, state MA = 2. Mentioning that it acts as a force multiplier is also good.

Question 7. Name the type of single pulley that can act as a force multiplier. Draw a labelled diagram of the pulley mentioned by you.

Answer: The single pulley that can act as a force multiplier is called a single movable pulley. It is supported by two ropes and has a mechanical advantage of two. [Diagram: A single movable pulley system showing the load, effort, and two supporting rope segments]

In simple words: The single movable pulley is like a helper that makes heavy things feel lighter. It uses two parts of the rope to hold the weight, so you only have to pull with half the force.

📝 Teacher's Note: Explain how the load is shared between two segments of the rope, leading to MA=2. Emphasize that while force is multiplied, the distance moved by effort is also doubled (VR=2), so work done remains the same (ignoring friction).

🎯 Exam Tip: Identify the single movable pulley as the force multiplier. State its ideal mechanical advantage is 2. For the diagram, ensure clear labels for load, effort, and the fixed support point.

Question 8. Give two reasons why the efficiency of a single movable pulley system is not 100%

Answer: The efficiency of a single movable pulley system is not 100% because: (i) The friction of the pulley bearing is not zero. (ii) The weight of the pulley and string is not zero.

In simple words: A real pulley system isn't perfect because the parts rub against each other (friction), and the pulley itself and the rope have some weight, which you also have to lift. These things waste a little bit of your effort.

📝 Teacher's Note: Discuss the concept of 'ideal' vs. 'actual' machines. Explain that efficiency is always less than 100% in real-world scenarios due to energy losses. Relate friction to heat generation and the weight of components to additional work done.

🎯 Exam Tip: The two primary reasons for efficiency being less than 100% are: 1. Friction in the pulley bearings/axle. 2. The weight of the movable pulley itself and the string/rope. Both require extra effort, reducing the useful work output.

Question 9. In which direction the force need to applied, when a single pulley is used with a mechanical advantage greater than one? How can you change the direction of force applied without altering its mechanical advantage? Draw a labelled diagram of the system.

Answer: The force should be applied in an upward direction when a single movable pulley (which has MA > 1) is used alone. The direction of force applied can be changed without altering its mechanical advantage by using a single movable pulley along with a single fixed pulley to change the direction of applied force. [Diagram: A block and tackle system with a single movable pulley and a single fixed pulley, showing the load, effort, and direction of forces]

In simple words: If you use a pulley that makes lifting easier (like a movable pulley), you usually have to pull the rope upwards. But if you want to pull downwards (which is easier for you) and still get the same lifting power, you can add another fixed pulley to just change the direction of your pull.

📝 Teacher's Note: This question introduces the concept of combining pulleys. Explain that a single movable pulley provides MA but requires upward effort. Adding a fixed pulley (MA=1, VR=1) changes the direction of effort without changing the overall MA or VR of the movable pulley, making the system more convenient. This is a basic block and tackle system.

🎯 Exam Tip: For a single movable pulley (MA > 1), the effort is applied upwards. To change the direction of effort without changing MA, combine a single movable pulley with a single fixed pulley. The diagram should clearly show the movable pulley supporting the load and the fixed pulley redirecting the effort downwards. Label load, effort, and directions.

Question 10. What is the velocity ratio of a single movable pulley?

Answer: The velocity ratio of a single movable pulley is always 2.

In simple words: For a single movable pulley, if you pull the rope a certain distance, the load moves up half that distance. So, the velocity ratio is 2.

📝 Teacher's Note: Emphasize that the velocity ratio is a theoretical value and does not account for friction. It's determined by the number of rope segments supporting the movable pulley.

🎯 Exam Tip: State "Velocity ratio = 2" clearly. Mention that it's constant for a single movable pulley and is independent of friction or load.

Question 11. In a single movable pulley, if the effort moves by a distance x upwards, by what height is the load raised?

Answer: The load is raised to a height of x/2.

In simple words: If you pull the rope up by a distance 'x', the thing you're lifting (the load) will only go up by half that distance, which is 'x/2'.

📝 Teacher's Note: This demonstrates the principle of velocity ratio. For every unit distance the load moves, the effort end of the rope moves twice that distance. Use a physical demonstration with a string and a weight to illustrate this.

🎯 Exam Tip: The relationship is direct: distance moved by load = (distance moved by effort) / velocity ratio. For a single movable pulley, VR = 2, so load distance = effort distance / 2.

Question 12. Draw a labelled diagram of an arrangement of two pulleys, one fixed and other movable. In the diagram, mark the directions of all forces acting on it. What is the ideal mechanical advantage of the system? How can it be achieved?

Answer: [Diagram: Arrangement of two pulleys, one fixed and one movable, with forces marked] Ideal mechanical advantage of this system is 2. This can be achieved by assuming that string and the pulley are massless and there is no friction in the pulley bearings or at the axle or between the string and surface of the rim of the pulley.

In simple words: Imagine one pulley fixed to the ceiling and another pulley moving up and down with the weight. This setup helps lift things. Ideally, it makes lifting twice as easy. To get this perfect "twice as easy" effect, we pretend the rope and pulleys have no weight and there's no rubbing (friction) anywhere.

📝 Teacher's Note: When drawing the diagram, ensure clear labels for fixed pulley, movable pulley, load, effort, and tension in each segment of the string. Explain that 'ideal' conditions are theoretical and help understand the maximum potential of the machine.

🎯 Exam Tip: For full marks, clearly state the Ideal Mechanical Advantage (IMA) as 2. List the ideal conditions: massless string, massless pulley, and no friction at any point (bearings, axle, rim).

Question 13. The diagram below shows a pulley arrangement. (a) In the diagram, mark the direction of tension on each strand of string. (b) What is the purpose of the pulley B? (c) If the tension is T, Deduce the relation between T and E. (d) What is the velocity ratio of the arrangement? (e) Assuming that the efficiency of the system is 100%, What is the mechanical advantage?

Answer: (a) [Diagram: Pulley arrangement with tension directions marked] (b) The fixed pulley B is used to change the direction of effort to be applied from upward to downward. (c) The effort E balances the tension T at the free end, so E=T. (d) The velocity ratio of this arrangement is 2. (e) The mechanical advantage is 2 for this system (if efficiency is 100%).

In simple words: This pulley system helps lift things. Part (a) asks to show how the rope pulls on itself. Part (b) explains that one pulley (B) is just there to make it easier to pull down instead of up. Part (c) says the force you pull with (Effort, E) is equal to the pull in the rope (Tension, T). Part (d) means if you pull the rope 2 meters, the load goes up 1 meter. Part (e) says if the system is perfect (100% efficient), it makes lifting twice as easy.

📝 Teacher's Note: For part (a), ensure students understand that tension is uniform throughout a continuous string and acts away from the point of contact. For part (b), highlight the practical advantage of changing effort direction. For (c), explain that E=T is true for the effort end of the string. For (d) and (e), connect VR and MA to the number of supporting strands and the efficiency concept.

🎯 Exam Tip: (a) Clearly mark tension arrows on all active strands. (b) State the purpose of the fixed pulley B as "changing the direction of effort". (c) The relation is E=T, as the effort directly balances the tension in the free end of the string. (d) Velocity ratio is 2, as there are two strands supporting the movable pulley. (e) If efficiency is 100%, then Mechanical Advantage (MA) = Velocity Ratio (VR), so MA = 2.

Question 14. Differentiate between a single fixed pulley and a single movable pulley.

Answer:

Single Fixed Pulley:

1. It is fixed to a rigid support.

2. Its mechanical advantage is one.

Single Movable Pulley:

1. It is not fixed to a rigid support; it moves with the load.

2. Its mechanical advantage is two.

In simple words: A "fixed pulley" stays in one place, like a flag pole pulley. It just changes the direction you pull. A "movable pulley" moves up and down with the weight you're lifting, making the lifting easier.

📝 Teacher's Note: Emphasize that a single fixed pulley acts as a force multiplier of 1 (MA=1), meaning it only changes direction, not magnitude of force. A single movable pulley acts as a force multiplier of 2 (MA=2), reducing the effort required but increasing the distance the effort must move. Use real-world examples for both.

🎯 Exam Tip: For differentiation questions, present points clearly, preferably in a comparative manner. Key points for a single fixed pulley are "fixed support" and "MA=1 (changes direction)". Key points for a single movable pulley are "moves with load" and "MA=2 (reduces effort)".

Question 15. The Diagram alongside shows an arrangement of three pulleys A, B, and C. The load is marked as L and the effort as E. (a) Name the Pulleys A, B, and C. (b) Mark in the diagram the directions of load (L), effort (E) and tension T₁ and T₂ in the two strings. (c) How are the magnitudes of L and E related to the tension T₁? (d) Calculate the mechanical advantage and velocity ratio of the arrangement. (e) What assumptions have you made in parts (c) and (d)?

Answer:

(a) Pulleys A and B are movable pulleys. Pulley C is a fixed pulley.

(b) [Diagram: Directions of load (L), effort (E), and tensions T₁ and T₂ in the strings are marked on the diagram.]

(c) The magnitude of effort E = T₁
And the magnitude of L = 2 * 2 * T₁ = 4 T₁

(d) The mechanical advantage = 2 * 2 = 4
The velocity ratio = 2 * 2 = 4

(e) Assumption: The pulleys A and B are weightless.

In simple words: This pulley system uses two movable pulleys (A and B) and one fixed pulley (C). The fixed pulley just changes the direction of the effort, while the movable pulleys help lift a heavy load with less effort. The system multiplies your force by 4, meaning you only need to pull with one-fourth of the load's weight. We assume the pulleys themselves have no weight and there's no friction to make calculations simple.

📝 Teacher's Note: Emphasize that each movable pulley ideally doubles the mechanical advantage. The fixed pulley (C) is crucial for convenience, allowing the effort to be applied downwards, using gravity to assist or making it easier to pull. Discuss the concept of tension in different parts of the string.

🎯 Exam Tip: For full marks, clearly state the type of each pulley (fixed/movable). When calculating MA and VR, show the steps (e.g., 2*2 or 2ⁿ). Always mention the assumptions made (weightless pulleys, no friction) when dealing with ideal situations.

Question 16. Draw a diagram of combination of three movable pulleys with a fixed pulley showing the directions of load, effort and tension in each strand. Find: (i) mechanical advantage, (ii) Velocity ratio and (iii) efficiency of the combination in ideal situation.

Answer:

[Diagram: Combination of three movable pulleys with a fixed pulley showing directions of load, effort, and tension in each strand.]

Tension T₁ in the string passing over the pulley A is given as:
2T₁ = L or T₁ = L/2

Tension T₂ in the string passing over the pulley B is given as:
2T₂ = T₁ or T₂ = T₁/2 = L/2²

Tension T₃ in the string passing over the pulley C is given as:
2T₃ = T₂ or T₃ = T₂/2 = L/2³

In equilibrium, T₃ = E

Therefore, E = L/2³

(i) Mechanical advantage (MA) = L/E = L / (L/2³) = 2³ = 8

As one end of each string passing over a movable pulley is fixed, the free end of the string moves twice the distance moved by the movable pulley.

If load L moves up by a distance x, then d_L = x.
The effort moves by a distance d_E = 2³x.

(ii) Velocity Ratio (VR) = \( \frac{\text{Distance moved by the effort } d_E}{\text{Distance moved by the load } d_L} = \frac{2^3 x}{x} = 2^3 = 8 \)

(iii) Efficiency (η) in ideal situation = \( \frac{\text{Mechanical Advantage}}{\text{Velocity Ratio}} = \frac{8}{8} = 1 \) or 100%.

In simple words: This system uses three movable pulleys and one fixed pulley. Each movable pulley helps to reduce the effort needed to lift the load. With three movable pulleys, the system multiplies your force by 8, meaning you only need to pull with one-eighth of the load's weight. In an ideal world, where there's no friction and pulleys have no weight, the system is 100% efficient, meaning all the effort you put in goes directly into lifting the load.

📝 Teacher's Note: Explain how the number of movable pulleys directly relates to the power of 2 in the MA and VR calculation (2ⁿ, where n is the number of movable pulleys). Highlight that the fixed pulley only changes the direction of effort, not the MA or VR. Discuss the difference between ideal and practical efficiency.

🎯 Exam Tip: When drawing the diagram, ensure all tensions (T₁, T₂, T₃), load (L), and effort (E) directions are clearly marked. For calculations, show the derivation of E in terms of L, and then use it to find MA. For VR, clearly state the relationship between d_L and d_E. Remember that in an ideal situation, efficiency is always 100% or 1.

Question 17. What is a block and tackle system of pulleys?

Answer: A block and tackle is a system of two or more pulleys with a rope or cable threaded between them, usually used to lift or pull heavy loads.

In simple words: Imagine a set of wheels (pulleys) with a rope going around them. This setup helps you lift very heavy things with less effort, like pulling a bucket from a deep well.

📝 Teacher's Note: Emphasize that the main purpose of a block and tackle system is to multiply force, making it easier to lift heavy objects, at the expense of increasing the distance the rope must be pulled.

🎯 Exam Tip: Define a block and tackle system as an arrangement of multiple pulleys (fixed and movable) used to achieve a high mechanical advantage, typically for lifting heavy loads. Mention the rope threaded between them.

Question 18. Draw a diagram of a block and tackle system of pulleys having a velocity ratio of 5. In your diagram indicate clearly t

Answer: [Diagram: A block and tackle system of pulleys with 5 strands supporting the load, indicating effort, load, and tension distribution for a velocity ratio of 5.]

In simple words: This question asks you to draw a picture of a special pulley system. It should have 5 ropes helping to lift something, showing how you pull and what gets lifted.

📝 Teacher's Note: When drawing, ensure students correctly identify the fixed and movable blocks, the number of strands supporting the load (which equals the velocity ratio), and the direction of effort and load. For VR=5, there should be 5 rope segments supporting the movable block/load.

🎯 Exam Tip: For full marks, the diagram must clearly show: 1) The correct number of pulleys for the specified velocity ratio (e.g., 3 in the upper block, 2 in the lower for VR=5, or 2 in upper, 3 in lower, depending on the threading, but 5 strands supporting the load is key). 2) The direction of effort (E) and load (L). 3) The threading of the rope. 4) Labels for fixed and movable pulleys/blocks.

Question 19. Give reasons for the following: (a) In a single fixed pulley, the velocity ratio is always more than the mechanical advantage. (b) The efficiency of a movable pulley is always less than 100% (c) In case of a block and tackle arrangement, the mechanical advantage increases with the increase in the number of pulleys. (d) The lower block of a block and tackle pulley system must be of negligible weight.

Answer: (a) In a single fixed pulley, some effort is wasted in overcoming friction between the strings and the grooves of the pulley; so the effort needed is greater than the load and hence the mechanical advantage is less than the velocity ratio. (b) This is because some effort is wasted in overcoming the friction between the strings and the grooves of the pulley. (c) This is because mechanical advantage is equal to the total number of pulleys in both the blocks. (d) The efficiency depends upon the mass of lower block; therefore efficiency is reduced due to the weight of the lower block of pulleys.

In simple words: (a) Even with a simple pulley, some energy is lost to rubbing, so you need to pull a bit harder than ideal. (b) No pulley is perfect; some energy is always lost to friction. (c) More pulleys mean you can lift heavier things with the same pull, so it's easier. (d) If the bottom part of the pulley system is heavy, you're wasting effort lifting that part instead of just your load, making the system less effective.

📝 Teacher's Note: This question covers fundamental concepts of ideal vs. actual machines. Emphasize that friction is the primary reason for efficiency being less than 100% and MA < VR. For part (c), clarify that MA is ideally equal to the number of supporting strands, which increases with the number of pulleys. For part (d), explain that the weight of the movable block adds to the load that the effort must overcome, reducing the net mechanical advantage for the actual load.

🎯 Exam Tip: When explaining reasons for efficiency less than 100% or MA < VR, always mention "friction" (between string and pulley groove, or axle friction) and "weight of movable parts" (especially the movable block). For block and tackle, state that MA (ideally) equals the number of supporting strands, which increases with the number of pulleys. For negligible weight, explain that the effort is then solely used to lift the load, maximizing efficiency.

Question 20. Name a machine which is used to: (a) multiply force, (b) multiply speed and (c) change the direction for force applied.

Answer: (a) Multiply force: a movable pulley. (b) Multiply speed: gear system or class III lever. (c) Change the direction of force applied: single fixed pulley.

In simple words: (a) A movable pulley helps you lift heavy things with less strength. (b) Gears or a specific type of lever (like tweezers) can make things move faster. (c) A simple pulley fixed to a ceiling lets you pull down to lift something up.

📝 Teacher's Note: This question tests the basic functions of simple machines. Clarify that force multiplication comes at the cost of distance (you pull more rope), and speed multiplication comes at the cost of force (you need more force). A single fixed pulley is ideal for changing direction without changing force or speed magnitude.

🎯 Exam Tip: Remember these classic examples: Force multiplier - movable pulley, block and tackle, hydraulic press, crowbar (Class I lever). Speed multiplier - gear systems, Class III lever (e.g., fishing rod, forearm). Change direction - single fixed pulley. Be precise with the machine names.

Question 21. State whether the following statements are true or false: (a) The velocity ratio of a single fixed pulley is always more than 1. (b) The velocity ratio of a single movable pulley is always 2. (c) The velocity ratio of a combination of n movable pulleys with a fixed pulley is always 2 n . (d) The velocity ratio of a block and tackle system is always equal to the number of strands of the tackle supporting the load.

Answer: (a) The velocity ratio of a single fixed pulley is always more than 1 .(false) (b) The velocity ratio of a single movable pulley is always 2 .(true) (c) The velocity ratio of a combination of n movable pulleys with a fixed pulley is always \(2^n\).(true) (d) The velocity ratio of a block and tackle system is always equal to the number of strands of the tackle supporting the load. (true)

In simple words: (a) For a simple fixed pulley, the rope you pull moves the same distance as the load, so its speed ratio is exactly 1, not more. (b) With one movable pulley, you pull twice the distance the load moves, so its speed ratio is 2. (c) If you have 'n' movable pulleys with a fixed one, the speed ratio grows very fast, like 2 multiplied by itself 'n' times. (d) For a block and tackle, if 'X' ropes are holding up the weight, then the speed ratio is 'X'.

📝 Teacher's Note: This question tests the understanding of velocity ratios for different pulley systems. Emphasize that VR is an ideal value, independent of friction. For a single fixed pulley, VR=1. For a single movable pulley, VR=2. For a system with 'n' movable pulleys (and one fixed), the VR is \(2^n\). For a block and tackle, VR equals the number of supporting strands.

🎯 Exam Tip: Memorize the standard velocity ratios: Single fixed pulley (VR=1), Single movable pulley (VR=2). For block and tackle, VR = number of supporting strands (or total number of pulleys if the effort is applied downwards and the number of pulleys is odd, or total pulleys + 1 if even, but "number of supporting strands" is the most robust definition). Be careful with the \(2^n\) formula; it applies to specific pulley arrangements, not all combinations.

Question 1. A Single fixed pulley is used because it: (a) has a mechanical advantage greater than 1. (b) has a velocity ratio less that 1 (c) gives 100% efficiency (d) helps to apply the effort in a convenient direction.

Answer: It helps in applying effort in a convenient direction. A single fixed pulley though does not reduce the effort but helps in changing the direction of effort applied. As it is far easier to apply effort in downward direction, the single fixed pulley is widely used.

In simple words: A fixed pulley makes it easier to pull things by letting you pull downwards instead of upwards, which feels more natural and uses your body weight. It doesn't make the load lighter, just the pulling direction more comfortable.

📝 Teacher's Note: Emphasize that a single fixed pulley has a Mechanical Advantage (MA) of 1 and a Velocity Ratio (VR) of 1. Its primary function is convenience, not force multiplication. Demonstrate with a simple setup in class.

🎯 Exam Tip: For a single fixed pulley, remember its MA = 1 and VR = 1. The key benefit to state is "change in direction of effort for convenience." This is a common conceptual question.

Question 2. The mechanical advantage of an ideal single movable pulley is: (a) 1 (b) 2 (c) less than 2 (d) less than 1

Answer: The mechanical advantage of an ideal single movable pulley is 2.

[Diagram: Single movable pulley with load L supported by two string segments, and effort E applied to the free end. Tension T is uniform throughout the string.]

Derivation: Here the load L is balanced by the tension in two segments of the string and the effort E balances the tension T at the free end, so L = T + T = 2T and E = T. Assumption: Weight of the pulley is negligible. We know that, \( M.A = \frac{\text{Load (L)}}{\text{Effort (E)}} = \frac{2T}{T} = 2 \). Thus, a single movable pulley has a M.A. equal to 2.

In simple words: A movable pulley helps you lift heavy things by making them feel half as heavy. You pull with half the force, but you have to pull twice the distance. So, it multiplies your force by two.

📝 Teacher's Note: Explain the concept of tension distribution in the string. Each segment supporting the movable pulley shares the load. An ideal pulley assumes no friction and negligible weight of the pulley itself. This is crucial for MA=2.

🎯 Exam Tip: Always state "ideal" conditions when discussing MA=2 for a single movable pulley. Clearly show the derivation L=2T and E=T, leading to MA=L/E = 2. This derivation is often asked.

Question 3. A movable pulley is used as: (a) force multiplier (b) speed multiplier (c) device to change the direction of effort (d) all the above

Answer: Force multiplier. The mechanical advantage of movable pulley is greater than 1. Thus, using a single movable pulley, the load can be lifted by applying an effort equal to half the load (in ideal situation), i.e. the single movable pulley acts as a force multiplier.

In simple words: A movable pulley makes you stronger! It lets you lift heavy objects with less effort, acting like a "force booster."

📝 Teacher's Note: Contrast this with a fixed pulley. A movable pulley's MA > 1 is its defining characteristic for force multiplication. Discuss the trade-off: less force, but greater distance of effort.

🎯 Exam Tip: When asked about the function of a movable pulley, "force multiplier" is the primary answer. Mentioning that MA > 1 (specifically 2 for a single movable pulley) reinforces the concept for full marks.

Question 1. A Woman draws water from a well using a fixed pulley. The mass of bucket and water together is 6 kg. The force applied by the woman is 70 N. Calculate the mechanical advantage. ( Take g = 10 m s^-2 )

Answer:

The force applied by the woman (Effort) = 70 N

The mass of bucket and water together = 6 kg

Total load (L) = mass × g = 6 kg × 10 m s^-2 = 60 N

Mechanical advantage (M.A) \( = \frac{\text{Load}}{\text{Effort}} = \frac{60 \text{ N}}{70 \text{ N}} = 0.857 \)

In simple words: The woman is pulling with 70 N of force, but the bucket and water weigh 60 N. Since she's pulling harder than the weight, the pulley isn't making it easier in terms of force, so its "advantage" is less than 1.

📝 Teacher's Note: This is a practical example where MA < 1 for a fixed pulley due to friction and the pulley's own weight (though not explicitly stated, it's implied by the effort being greater than the load). Remind students that ideal fixed pulleys have MA=1. Real-world scenarios often show MA < 1.

🎯 Exam Tip: Clearly identify Load and Effort. Remember to convert mass to weight (Load) using g. The formula for Mechanical Advantage is \( MA = \frac{\text{Load}}{\text{Effort}} \). Pay attention to units and significant figures in the final answer.

Question 2. A fixed pulley is driven by a 100 kg mass falling at a rate of 8.0 m in 4.0s. It lifts a load of 500 kgf. (a) Calculate the power input to the pulley taking the force of gravity on 1 kg as 10 N. (b) If the efficiency of the pulley is 75%, find the height to which the load is raised in 4.0s.

Answer:

Given:

Load = 500 kgf

Mass of falling object (Effort mass) = 100 kg

Displacement of effort = 8.0 m

Time taken = 4.0s

(a) Effort (force applied by falling mass) = 100 kg × 10 N/kg = 1000 N (or 1000 kgf, assuming 1 kgf = 10 N for consistency with the problem's g value)

Power Input \( = \frac{\text{displacement} \times \text{effort}}{\text{time}} = \frac{8.0 \text{ m} \times 1000 \text{ N}}{4.0 \text{ s}} = 2000 \text{ W} \)

(b) The efficiency of the pulley is = 75% = 0.75

In simple words: Part (a) asks how much "work per second" is put into the pulley by the falling weight. Part (b) tells us the pulley isn't perfect (75% efficient), so some energy is lost. We need to figure out how high the load actually goes with the remaining useful energy.

📝 Teacher's Note: Clarify the difference between kg and kgf. Here, 1 kgf is treated as 10 N. Power input is the rate at which effort does work. For part (b), students need to use the efficiency formula: Efficiency = (Power Output / Power Input) or (Work Output / Work Input). Power Output = Load × height / time.

🎯 Exam Tip: For power calculations, ensure units are consistent (Joules for work, Watts for power). Remember that 1 kgf is approximately 9.8 N, but the problem specifies 10 N for 1 kg, so use that. For efficiency, always express it as a decimal or fraction in calculations, not percentage. Clearly show steps for calculating power input and then use efficiency to find power output or work output to determine the height.

Question 3. In a block and tackle system consisting of 3 pulleys, a load of 75 kgf is raised with an effort of 25 kgf. Find: (i) the mechanical advantage, (ii) velocity ratio and (iii) efficiency.

Answer: Given:
Load (L) = 75 kgf
Effort (E) = 25 kgf
Number of pulleys (n) = 3

(i) Mechanical Advantage (M.A) = \( \frac{\text{Load}}{\text{Effort}} \) = \( \frac{75 \text{ kgf}}{25 \text{ kgf}} \) = 3
(For an ideal block and tackle system, M.A = n = 3)

(ii) Velocity Ratio (V.R) = n = 3 (for a block and tackle system)

(iii) Efficiency (\(\eta\)) = \( \frac{\text{M.A}}{\text{V.R}} \) = \( \frac{3}{3} \) = 1 or 100%

In simple words: Imagine lifting a heavy box. With this pulley system, you only need to pull with 1/3rd of the box's weight! It's super efficient, meaning almost all your effort goes into lifting the box.

📝 Teacher's Note: Emphasize that for an ideal block and tackle system, MA = VR = n (number of pulleys). Real-world efficiency is always less than 100% due to friction, but for theoretical problems, 100% is often assumed unless stated otherwise. Discuss the concept of 'kgf' as a unit of force.

🎯 Exam Tip: Remember the formulas: MA = Load/Effort, VR = n (for block and tackle), and Efficiency = MA/VR. Clearly state units where applicable. For 100% efficiency, MA = VR.

Question 4. In fig. 3.38, draw a tackle to lift the load by applying the force in the downward direction. Mark the position of load and effort. (a) If the load is raised by 1 m, through what distance will the effort move? (b) State how many strands of tackle are supporting the load? (c) What is the mechanical advantage of the system?

Answer: [Diagram: Block and tackle system, fig. 3.38, showing a system with 5 pulleys arranged to apply effort in the downward direction, with load and effort marked.]

(a) If the load is raised by 1 m, the effort will move a distance of 1 m × 5 = 5 m. (This implies a Velocity Ratio of 5 for the system).

(b) Five strands of tackle are supporting the load.

(c) The mechanical advantage of the system (M.A) = \( \frac{\text{load}}{\text{effort}} \) = \( \frac{5T}{T} \) = 5

In simple words: Imagine a pulley system that makes lifting things 5 times easier! If you want to lift something by 1 meter, you'll have to pull your rope 5 meters, but with much less force. This system uses 5 ropes to share the weight.

📝 Teacher's Note: Explain that the number of supporting strands directly relates to the Velocity Ratio (VR) and, ideally, the Mechanical Advantage (MA) of a block and tackle system. For downward effort, the number of supporting strands is usually equal to the total number of pulleys. Emphasize drawing the diagram correctly, showing the fixed and movable blocks and the rope reeving.

🎯 Exam Tip: For block and tackle systems, VR = number of supporting strands (or number of pulleys if the effort is downward). MA is ideally equal to VR. Clearly label load, effort, and direction of forces in diagrams. The distance moved by effort = VR × distance moved by load.

Question 5. A block and tackle system has 5 pulleys. If an effort of 1000 N is needed in the downward direction to raise a load of 4500 N, calculate: (a) the mechanical advantage (b) the velocity ratio, and (c) the efficiency of the system

Answer: Given:
Number of pulleys (n) = 5
Effort (E) = 1000 N
Load (L) = 4500 N

(a) The mechanical advantage (M.A) = \( \frac{\text{Load}}{\text{Effort}} \) = \( \frac{4500 \text{ N}}{1000 \text{ N}} \) = 4.5

(b) The velocity ratio (V.R) = n = 5 (for a block and tackle system with downward effort)

(c) The efficiency (\(\eta\)) of the system = \( \frac{\text{M.A}}{\text{V.R}} \) = \( \frac{4.5}{5} \) = 0.9 or 90%

In simple words: This pulley system makes lifting things 4.5 times easier! Even though it has 5 pulleys, some energy is lost (maybe due to friction), so it's not 100% perfect, but still very helpful. It's 90% efficient, meaning 90% of your effort goes into lifting the load.

📝 Teacher's Note: Highlight that the actual Mechanical Advantage (MA) can be less than the Velocity Ratio (VR) due to friction and the weight of the movable pulleys. This difference leads to an efficiency less than 100%. Discuss the practical implications of efficiency in real-world machines.

🎯 Exam Tip: Always calculate MA using the given Load and Effort. VR is determined by the number of pulleys (n) for a block and tackle system. Efficiency is a ratio, so it's unitless and often expressed as a percentage. Remember \( \eta = \frac{MA}{VR} \).

Question 6. A pulley system has a velocity ratio 3 and an efficiency of 80%. Draw a labelled diagram of this pulley system Calculate: (a) the mechanical advantage of the system and

Answer: Given:
Velocity Ratio (V.R) = 3
Efficiency (\(\eta\)) = 80% = 0.8

[Diagram: A labelled diagram of a pulley system with a Velocity Ratio of 3. This typically involves a block and tackle system with 3 pulleys (e.g., one fixed and two movable, or two fixed and one movable, depending on the configuration for downward effort). The diagram should show the load, effort, and rope reeving.]

(a) To calculate the mechanical advantage (M.A) of the system, we use the efficiency formula:
\( \eta = \frac{\text{M.A}}{\text{V.R}} \)
Therefore, M.A = \( \eta \) × V.R
M.A = 0.8 × 3
M.A = 2.4

In simple words: This pulley system helps you lift things, but it's not perfect. It makes things 2.4 times easier to lift. Even though it has a 'speed advantage' of 3, some energy is lost, so the actual 'lifting advantage' is a bit less.

📝 Teacher's Note: This question demonstrates the relationship between efficiency, MA, and VR. Remind students that efficiency is always less than or equal to 1 (or 100%). A VR of 3 implies a system with 3 supporting strands, typically a block and tackle with 3 pulleys. Students should be able to draw such a system. Note that the question seems incomplete, as it ends with "and" suggesting a part (b) which is not provided.

🎯 Exam Tip: Always remember the formula: Efficiency (\(\eta\)) = MA / VR. If any two values are given, the third can be calculated. Pay attention to units and percentage conversions (e.g., 80% = 0.8). For drawing diagrams, ensure correct reeving of the rope and clear labeling of load, effort, and fixed/movable pulleys.

Answer:

For part (b), the effort required to raise a load of 300 N:

A pulley system has a velocity ratio = 3
Efficiency of system = 80 % = 0.8
Mechanical advantage of the system
M.A = V.A × η = 3 × 0.8 = 2.4
Effort required to raise the load = Effort = \( \frac{Load}{M.A} = \frac{300}{2.4} = 125N \)

In simple words: To lift a 300 N weight with this pulley system, you only need to pull with a force of 125 N because the pulleys help you multiply your force, making the work easier.

📝 Teacher's Note: Emphasize that efficiency is always less than 100% in real systems due to friction and the weight of movable parts. Mechanical advantage (MA) is a practical measure of force multiplication, while velocity ratio (VR) is theoretical. The relationship MA = VR × η is crucial for understanding real-world pulley systems.

🎯 Exam Tip: Always state the formula for Mechanical Advantage (MA = Load/Effort) and the relationship between MA, VR, and efficiency (MA = VR × η). Ensure units are correctly stated (Newtons for force). Full marks are awarded for correct formula application and calculation, showing all steps clearly.

Question 7. Fig 3.39 shows a system of four pulleys, The upper two pulleys are fixed and the lower two are movable.

(a) Draw a string around the pulleys. Also show the place and direction in which the effort if applied. [Diagram: Pulley system with string and effort direction]

(b) What is the velocity ratio of the system?

(c) How are load and effort of the pulley system related?

(d) What assumption do you make in arriving at your answer in part(c)?

Answer:

(a) [Diagram: Pulley system with string drawn around the four pulleys, showing the effort applied downwards from the free end of the string.]

(b) Velocity ratio of the system = n = 4

(c) The relation between load and effort: MA = \( \frac{Load}{effort} = n = 4 \)

(d) The assumptions made are:

(i) There is no friction in the pulley bearings.

(ii) The weight of the lower (movable) pulleys is negligible.

(iii) The effort is applied downwards.

In simple words: This pulley system has 4 ropes supporting the load, so it makes lifting things 4 times easier (ideally). We assume the pulleys are perfect and light, and you pull straight down.

📝 Teacher's Note: When drawing the string, ensure the number of strands supporting the movable block (and thus the load) matches the velocity ratio. The direction of effort is crucial. Discuss how these ideal assumptions differ from real-world scenarios.

🎯 Exam Tip: For pulley diagrams, clearly show the string path and effort direction. Remember that for an ideal pulley system, the Mechanical Advantage (MA) equals the Velocity Ratio (VR), which is equal to the number of supporting strands (n). List all assumptions clearly, as they are often specifically asked for.

Question 8. Fig 3.40 shows a block and tackle system of pulleys used to lift a load.

(a) How many strands of tackle are supporting the load?

(b) Draw arrows to represent tension in each strand. [Diagram: Block and tackle system with tension arrows]

(c) What is the mechanical advantage of the system?

(d) When load is pulled up be a distance 1 m, how far does the effort end move?

Answer:

(a) There are 4 strands of tackle supporting the load.

(b) [Diagram: Block and tackle system with four strands supporting the movable block, each with an upward arrow representing tension (T). The effort end also has a downward arrow representing tension (T).]

In simple words: This pulley system uses 4 ropes to hold up the weight, so each rope shares the load. When you pull, each rope has the same pulling force (tension).

📝 Teacher's Note: Emphasize that in an ideal block and tackle system, the tension in each strand of the string is equal to the effort applied. The number of supporting strands directly determines the velocity ratio and, ideally, the mechanical advantage.

🎯 Exam Tip: To find the number of supporting strands, count the segments of the string that directly support the movable block. When drawing tension arrows, ensure they are consistent in direction (upwards for supporting strands, downwards for effort) and magnitude (all equal to effort in an ideal system).

Question 9. A block and tackle system has the velocity ratio 3. Draw a labelled diagram of the system indicating the points of application and the directions of load and effort. A man can exert a pull of 200 kgf. What is the maximum load he can raise with this pulley system is its efficiency is 60%?

Answer: [Diagram: Labelled block and tackle system with 3 pulleys (one fixed, two movable) showing load, effort, and points of application. The effort is applied downwards, and the load is lifted upwards. There are 3 strands supporting the movable block.]

A block and tackle system has the velocity ratio = 3

i.e., VR = n = 3

Efficiency of system \( \eta \) = 60% = 0.6

The mechanical advantage of the system MA = VR \( \times \eta \) = 3 \( \times \) 0.6 = 1.8

Man can exert a maximum effort = 200 kgf

Load = MA \( \times \) effort = 1.8 \( \times \) 200 = 360 kgf

In simple words: To lift something heavy with this pulley system