Frank Brothers Solutions for ICSE Class 10 Physics Chapter 1 Force Work Energy and Power

Page No-10:

 

Solution 1
Answer: A force is that physical force which changes or tends to change the state of rest or state of motion of the body. It can start or stop the body. It can change the speed or direction or both of body. It can bring about change in dimensions of the body.
In simple words: Force is a push or pull. It can make things move, stop, change direction, or change shape. Like when you push a door or pull a rope.

📝 Teacher's Note: Show students how pushing a door opens it and pulling it closes it. This makes force easy to understand. Students often forget that force can also change shape like squeezing a ball.

🎯 Exam Tip: Always write that force can change state of rest, motion, direction, speed, and dimensions. List all five effects to get full marks.

 

Solution 2
Answer: Contact forces are forces which come into play when the bodies come into physical contact with each other, e.g. frictional forces, force of normal reaction. These forces are produced by way of action and reaction.
In simple words: Contact forces work only when two things touch each other. Like when you push a table with your hand - your hand touches the table.

📝 Teacher's Note: Ask students to push their desk. Explain that the push only works because their hand touches the desk. No touch means no contact force.

🎯 Exam Tip: Write "physical contact is needed" and give examples like friction, normal reaction. These key words get marks.

 

Solution 3
Answer: Non-contact forces are forces which act on the bodies through the empty space without coming into any physical contact. Also known as action at a distance force or field forces, for example: gravitational forces, electrostatic forces.
In simple words: Non-contact forces work without touching. Like how a magnet pulls iron pieces from far away. No touching needed.

📝 Teacher's Note: Bring a magnet and show how it attracts iron pins without touching them. Students love this demonstration and remember it well.

🎯 Exam Tip: Write "no physical contact needed" and "action at a distance." Give examples like gravity, magnetism, electrostatic force.

 

Solution 4
Answer:
(i) We push a door to close it.
(ii) A railway engine pulls a train
(iii) The force that a passenger applies to put a luggage on the platform into the train is a lifting force.
(iv) When we try to move heavy roller, it resists our effort to move it.
In simple words: These are examples of different forces we use every day. Push, pull, lift, and resistance are all types of forces.

📝 Teacher's Note: Act out these examples in class. Students can push desks, pull chairs, lift books. This helps them see force in daily life.

🎯 Exam Tip: Give clear examples from daily life. Examiners like real-world examples that show you understand force.

 

Solution 5
Answer: When a normal spring held between the hands is pulled outward, we stretch it.
In simple words: When you pull a spring from both ends, it becomes longer. This shows force can change the shape of things.

📝 Teacher's Note: Bring a spring to class and let students stretch it. They can feel how force changes the spring's length.

🎯 Exam Tip: This example shows force changing dimensions. Write clearly that stretching changes the length of the spring.

 

Solution 6
Answer: We push a piston fitted inside a cylinder containing gas, we compress the gas.
In simple words: When you push the piston in, the gas inside gets squeezed into a smaller space. Force makes the gas occupy less space.

📝 Teacher's Note: Use a bicycle pump to show this. When students push the handle, air gets compressed inside. Very clear example.

🎯 Exam Tip: Write that force compresses the gas. This shows force changing volume or dimensions.

 

Solution 7
Answer: 1 kgf = 9.8 newton
In simple words: This is the conversion between two units of force. kgf means kilogram-force and newton is the SI unit.

📝 Teacher's Note: Explain that kgf is the old unit and newton is the new SI unit. Like how we changed from yards to metres.

🎯 Exam Tip: Remember this exact conversion: 1 kgf = 9.8 N. Often asked in numerical problems.

 

Solution 8
Answer: The SI unit of force is 'newton'. One newton is the force which when acts on a body of mass 1kg, produces an acceleration of 1ms⁻², i.e. 1 newton = 1 kg × 1ms⁻²
In simple words: Newton is the unit to measure force. One newton is the force needed to make a 1 kg object move faster by 1 metre per second every second.

📝 Teacher's Note: Tell students that Newton was a famous scientist. The unit is named after him. This helps them remember.

🎯 Exam Tip: Write the definition exactly: "Force that gives 1 kg mass an acceleration of 1 ms⁻²." Include the formula too.

 

Solution 9
Answer: In CGS system, the unit of force is 'dyne'. 1 newton = 1kg × 1 ms⁻² = 1000 g × 100 cms⁻² = 10⁵ dyne
In simple words: Dyne is another unit for force in the old CGS system. One newton equals 100,000 dynes.

📝 Teacher's Note: Explain that CGS means centimetre-gram-second system. It's older than SI system. We mainly use SI now.

🎯 Exam Tip: Remember the conversion: 1 N = 10⁵ dyne. Show the working step by step in exams.

 

Solution 10
Answer: 1 kgf = 9.8 newton
In simple words: This is the same conversion as before. One kilogram-force equals 9.8 newtons.

📝 Teacher's Note: Repeat this conversion because it's very important. Students need to remember 9.8 - it comes from gravity's value.

🎯 Exam Tip: This conversion is often tested. Write clearly: 1 kgf = 9.8 N.

 

Solution 11
Answer: When a balloon is inflated the force of air inside changes its shape or size.
In simple words: When you blow air into a balloon, the air pushes from inside. This force makes the balloon bigger and rounder.

📝 Teacher's Note: Blow up a balloon in class. Students can see how air pressure (force) changes the balloon's shape from flat to round.

🎯 Exam Tip: This shows force changing dimensions. Write that air pressure is the force that inflates the balloon.

 

Solution 12
Answer: The magnitude of a non-contact force depends upon the distance between the two objects.
In simple words: Non-contact forces get weaker when objects are farther apart. Like how a magnet pulls iron less when it's far away.

📝 Teacher's Note: Use a magnet and iron filings. Show how the magnet attracts more filings when it's close, fewer when it's far.

🎯 Exam Tip: Write that force decreases with increasing distance. This is true for gravity, magnetism, and electric force.

 

Solution 13
Answer: When the resultant of a group of forces acting on the same object is zero, the forces are said to be balanced. Balanced forces do not change the speed of stationary objects. They may deform objects. e.g. An iron ball suspended from a hook by a wire, a book kept on a table.
In simple words: Balanced forces cancel each other out. The object doesn't move, but it might change shape. Like a book on a table - gravity pulls down, table pushes up.

📝 Teacher's Note: Place a book on a table. Ask students what forces act on it. Gravity down, table force up. They balance, so book doesn't move.

🎯 Exam Tip: Write "resultant force is zero" and "no change in motion but may cause deformation." Give clear examples.

 

Solution 14
Answer: When the resultant of a group of forces acting on the same object is not equal to zero, the forces are said to be unbalanced. An unbalanced force changes the state of constant velocity including zero velocity of a body on which it acts. e.g. If you push a ball, it starts to roll; Applying a force to stop a cricket ball.
In simple words: Unbalanced forces don't cancel out. They make things move, stop, or change speed. Like when you kick a ball - it moves because forces are unbalanced.

📝 Teacher's Note: Push a ball on the desk. Explain that your push is stronger than friction, so forces are unbalanced and ball moves.

🎯 Exam Tip: Write "resultant force is not zero" and "causes change in motion." Show examples of starting and stopping motion.

 

Solution 15
Answer:
(i) Contact force
(ii) Non-contact force
(iii) Contact force
(iv) Non-contact force
(v) Non-contact force
In simple words: These are answers to identify different types of forces. Contact means touching is needed, non-contact means no touching needed.

📝 Teacher's Note: Without seeing the original questions, remind students to check if objects touch each other or not to identify the force type.

🎯 Exam Tip: For contact force, objects must touch. For non-contact force, objects can be apart. Look for this in the question.

 

Solution 16
Answer: 1 kgf = force due to gravity on 1 kg mass = 1 kg mass × acceleration due to gravity g in ms⁻² = g newton = 9.8 newton
In simple words: 1 kgf is the force that gravity puts on a 1 kg object. Since gravity's pull is 9.8 ms⁻², this equals 9.8 newton.

📝 Teacher's Note: Hold a 1 kg weight. Explain that the pull you feel is 1 kgf, which equals 9.8 N. This makes the concept real.

🎯 Exam Tip: Show the complete working: 1 kgf = 1 kg × g = 1 kg × 9.8 ms⁻² = 9.8 N. Include all steps.

 

Solution 17
Answer: It means that 1 kgf or one kilogramme force is the force due to gravity on 1 kilogram mass.
In simple words: kgf is the force of gravity on 1 kg of anything. It's how much gravity pulls 1 kg of matter towards Earth.

📝 Teacher's Note: Explain that everything with mass feels gravity's pull. 1 kg of iron, water, or books all feel the same gravitational force.

🎯 Exam Tip: Write clearly: "1 kgf is gravitational force on 1 kg mass." This definition often appears in exams.

 

Solution 18
Answer: Effects a force can produce and examples:

1. Change the state of rest; e.g. pushing a door to open it or close it.
2. Change the state of motion; e.g. applying a force to stop the cricket ball.
3. Change the direction of motion and not speed; e.g. when a force is applied to move a body in a circular path with uniform speed there is only a change in direction of motion but speed remains constant.
4. Change both speed and direction of motion; e.g. when a body is swirled in the vertical circle its direction of motion and speed changes at every point.
5. Change the dimension; when a balloon is inflated the force of air inside changes its shape or size.

In simple words: Force can make things start moving, stop moving, change direction, change speed, or change shape. These are the five main effects of force.

📝 Teacher's Note: Demonstrate each effect with simple examples. Push a book (change rest), stop a rolling ball (change motion), swing a string in circle (change direction).

🎯 Exam Tip: List all five effects with clear examples. Examiners expect you to know all effects, not just one or two.

 

Page No-21:

 

Solution 1
Answer: The turning effect produced by a force on a rigid body about a point, pivot or fulcrum is called the moment of force or torque. It is measured by the product of force and the perpendicular distance of the pivot from the line of action of force. Moment of a force = Force × perpendicular distance of the pivot from the force. Its SI unit is newton-metre (Nm).
In simple words: Moment of force is the turning effect. Like when you turn a door handle or open a bottle cap. It depends on force and distance from the turning point.

📝 Teacher's Note: Show students how opening a door near the handle is easy, but pushing near hinges is hard. Same force, different distance, different turning effect.

🎯 Exam Tip: Write the formula: Moment = Force × perpendicular distance. Include the SI unit: Nm (newton-metre).

 

Solution 2
Answer: The moment of force is a vector quantity.
In simple words: Moment of force has both size and direction. It can be clockwise or anticlockwise (two different directions).

📝 Teacher's Note: Show clockwise and anticlockwise turning with your hands. Students can see the two directions of turning clearly.

🎯 Exam Tip: Just write "vector quantity" clearly. Explain that it has direction - clockwise or anticlockwise.

 

Solution 3
Answer: Torque
In simple words: Torque is another name for moment of force. Both words mean the same thing - turning effect.

📝 Teacher's Note: Tell students that torque and moment of force are the same thing. Like how "velocity" and "speed with direction" mean the same.

🎯 Exam Tip: Remember that torque = moment of force. You can use either term in your answers.

 

Solution 4
Answer: When some force is applied on a rigid body free to move and the body starts moving along a straight path in the direction of force. This is known as linear or translational motion. Points on the rigid body, undergo displacements forming parallel lines and magnitude of displacement is the same for individual point. e.g. the motion of a bullet fired from a gun, a ball thrown straight up and falling back straight down.
In simple words: Linear motion means moving in a straight line. All parts of the object move the same distance in the same direction. Like a car moving straight on a road.

📝 Teacher's Note: Slide a book across the desk in a straight line. Every part of the book moves the same distance in the same direction.

🎯 Exam Tip: Write "straight line motion" and "all points move same distance in same direction." Give examples like bullet, falling ball.

 

Solution 5
Answer: If a rigid body is pivoted at a point, then the applied force will rotate the body about the fixed point or about the axis passing through the fixed point. This motion is called rotational motion. e.g. Earth's rotation about its axis, wheels of car in motion.
In simple words: Rotational motion means spinning around a fixed point. Like a wheel turning or Earth spinning on its axis.

📝 Teacher's Note: Spin a coin on the desk. Show how it rotates around its center point. This makes rotational motion very clear.

🎯 Exam Tip: Write "rotation about a fixed point or axis." Give examples like Earth's rotation, wheel motion.

 

Solution 6
Answer: 1 Nm = 10⁷ dyne cm
In simple words: This is conversion between SI unit (Nm) and CGS unit (dyne cm) for moment of force.

📝 Teacher's Note: Like how we convert metres to centimetres, we can convert newton-metre to dyne-centimetre for moment of force.

🎯 Exam Tip: Remember this conversion: 1 Nm = 10⁷ dyne cm. May be needed in numerical problems.

 

Solution 7
Answer: The factors on which the moment of force about a point depends are: 1. The magnitude of force applied. 2. The distance of the line of action of the force from the axis of rotation.
In simple words: Moment depends on two things: how big the force is, and how far from the turning point you apply it.

📝 Teacher's Note: Use a door example. Bigger force makes it easier to open. Pushing farther from hinges also makes it easier to open.

🎯 Exam Tip: Write both factors clearly: (1) magnitude of force (2) perpendicular distance from axis. Both are equally important.

 

Solution 8
Answer: If the turning effect on the body is clockwise, moment of force is called the clockwise moment and is taken as negative. If the turning effect on the body is anticlockwise, moment of force is called the anticlockwise moment and is taken as positive.
In simple words: We give different signs to different directions of turning. Clockwise is negative (-), anticlockwise is positive (+).

📝 Teacher's Note: Show clockwise and anticlockwise directions with hand gestures. Make students remember: clockwise = negative, anticlockwise = positive.

🎯 Exam Tip: Write the sign convention clearly: clockwise = negative, anticlockwise = positive. This is important for calculations.

 

Solution 9
Answer: (i) If a rigid body is free to move, the applied force will cause translational motion. (ii) If a rigid body moves around a center or is pivoted at a point, the applied force will cause rotational motion.
In simple words: Free objects move in straight lines when pushed. Fixed objects spin around their fixed point when pushed.

📝 Teacher's Note: Push a pencil on the desk (it slides - translation). Now hold one end and push the other end (it rotates - rotation).

🎯 Exam Tip: Remember: free body → translational motion, pivoted body → rotational motion. The key is whether the body is fixed at a point.

 

Solution 10
Answer: This is so because near the free end, the distance of the point of application of force from the axis of rotation becomes maximum, so the torque (= Force × perpendicular distance of the pivot from the force) is very large and hence it is easier to open the door.
In simple words: Door handles are far from hinges because distance makes turning easier. Same force gives more turning effect when applied farther from hinges.

📝 Teacher's Note: Try pushing a door near the handle, then near the hinges. Students will feel the difference immediately.

🎯 Exam Tip: Explain using the formula: Moment = Force × distance. Greater distance gives greater moment, so easier to open door.

 

Solution 11
Answer: A long handle facilitates increased torque with small application of force; hence a spanner has a long handle.
In simple words: Long handles make turning easier. With same force, you get more turning effect because the distance is more.

📝 Teacher's Note: Show a spanner with long handle. Explain that it's designed to make loosening nuts easier by increasing the turning effect.

🎯 Exam Tip: Write that longer handle gives greater torque with same force. This is why tools like spanners, screwdrivers have long handles.

 

Solution 12
Answer: A body is said to be in equilibrium under the action of a number of forces, if the forces are not able to produce any change in the state of rest or of uniform motion or uniform rotation. Equilibrium is a state of zero acceleration.
In simple words: Equilibrium means balanced. The object doesn't change its motion - it stays still or keeps moving the same way.

📝 Teacher's Note: Show a balanced scale. Both sides have equal weight, so it doesn't tilt. That's equilibrium - no change in position.

🎯 Exam Tip: Write "no change in state of motion" and "zero acceleration." These are key points for equilibrium definition.

 

Solution 13
Answer: (i) The conditions for static equilibrium are: (a) The sum of the (vector) forces must equal zero, i.e. ∑F = 0 (b) The sum of the torques must equal zero; i.e. ∑τ = 0. (ii) The conditions for dynamic equilibrium are: (a) The body should have a broad base. (b) Centre of gravity of the body should be as low as possible. (c) Vertical line drawn from the centre of gravity should fall within the base of the support. Examples: Static equilibrium: a box at rest on a floor; there is a gravitational force pulling the object to the earth, but there is also an equal and opposite force applied by the floor to the box (pushing up). Dynamic equilibrium: A rock travelling across the cosmos, far enough away from any other object (so as not to be affected by gravity – in other words, in zero gravity conditions); the rock continues to travel in a straight line at uniform velocity either for eternity, or until acted upon by an external unbalanced force.
In simple words: Static equilibrium means not moving at all. Dynamic equilibrium means moving at constant speed. Both need balanced forces and balanced turning effects.

📝 Teacher's Note: Static equilibrium - show a book on table. Dynamic equilibrium - a car moving at steady speed on straight road.

🎯 Exam Tip: Write both conditions: (1) ∑F = 0 (2) ∑τ = 0. For dynamic equilibrium, mention broad base and low center of gravity.

 

Solution 14
Answer: Conditions for equilibrium: (a) Vector sum of forces acting on the body should be zero. (b) Algebraic sum of moments acting on the body should be zero.
In simple words: For equilibrium, all forces must balance out (add to zero) and all turning effects must balance out (add to zero).

📝 Teacher's Note: Think of a see-saw. It balances when forces are equal and distances are equal. Both force balance and moment balance needed.

🎯 Exam Tip: Write both conditions clearly: sum of forces = 0, sum of moments = 0. Both are needed for equilibrium.

 

Solution 15
Answer: Principle of moments: If a body is in equilibrium under the action of number of force, then the sum of clockwise moments is equal to the sum of anticlockwise moments.
In simple words: For a balanced object, clockwise turning effects equal anticlockwise turning effects. Like a balanced see-saw.

📝 Teacher's Note: Use a see-saw or balance scale. When it's balanced, clockwise moments equal anticlockwise moments perfectly.

🎯 Exam Tip: Write: "Sum of clockwise moments = Sum of anticlockwise moments." This is the principle of moments.

 

Solution 16
Answer:

[Diagram: This diagram shows rotating forces and couples - (a) shows hands turning a steering wheel with forces F in opposite directions, (b) shows a bicycle chain with forces F acting to create rotation]

In simple words: The diagram shows couple forces - two equal forces acting in opposite directions to create pure rotation.

📝 Teacher's Note: Show students how turning a bottle cap uses couple action - one hand holds, other hand turns, creating rotation.

🎯 Exam Tip: A couple consists of two equal and opposite parallel forces that create pure rotational motion without translation.

 

Solution 17
Answer: Examples of couple action in daily life: (i) Opening and closing the cap of a bottle (ii) Turning a key in a lock
In simple words: Couple action means using two hands in opposite directions to create turning. Like opening a jar or turning a steering wheel.

📝 Teacher's Note: Let students practice opening bottle caps or turning door knobs. They can feel how both hands work in opposite directions.

🎯 Exam Tip: Give practical examples like bottle cap, key in lock, steering wheel. Show you understand couple in daily life.

 

Solution 18
Answer: (i) Force 'R' has the least moment about 'O' because its perpendicular distance is least from
In simple words: The force R creates the smallest turning effect because it acts closest to the turning point O.

📝 Teacher's Note: Without seeing the diagram, explain that moment depends on distance. Closer to pivot means smaller moment.

🎯 Exam Tip: Remember: moment = force × distance. Same force but smaller distance gives smaller moment.

 

Solution 19.
Answer: Two equal and opposite parallel forces acting along different lines on a body constitute a couple. Effect of couple: It produces angular acceleration.
In simple words: A couple is two equal forces pushing in opposite directions but not on the same line. This makes the object spin or rotate.

📝 Teacher's Note: Show students a steering wheel. When you turn it, your two hands push in opposite directions. This is a couple that makes the wheel rotate.

🎯 Exam Tip: Write "two equal and opposite parallel forces" and "produces angular acceleration." These key phrases get you marks.

 

Solution 20.
Answer: The turning effect of a couple is called the moment of couple and is calculated by the product of either of the forces and the perpendicular distance between them. Its SI unit is Nm.
In simple words: Moment of couple tells us how strong the turning effect is. We find it by multiplying force with the distance between the two forces.

📝 Teacher's Note: Use a wrench example. The longer the wrench handle, the easier it is to turn a bolt. This shows how distance affects the turning effect.

🎯 Exam Tip: Remember the formula: Moment of couple = Force × perpendicular distance. Always write the unit as Nm.

 

Solution 21.
Answer:
Given:
Applied force, F = 20 N
Distance from the pivot, d = 40 cm = 0.4m

Step 1: Calculate torque using formula.
Torque = F × d = 20 × 0.4 = 8 Nm

Therefore, Torque = 8 Nm
In simple words: We multiply the force with the distance from the turning point to get the torque. This tells us how much turning power we have.

📝 Teacher's Note: Students often forget to convert cm to m. Always check units first before calculating. 40 cm = 0.4 m.

🎯 Exam Tip: Write "Given," show the formula, substitute values, and write the final answer with units. Converting units correctly is very important.

 

Solution 22.
Answer:
Given:
Moment of Force (torque), τ = 4 Nm
Applied force, F = 10N
Distance from pivot, d = ?

Step 1: Use the torque formula.
τ = F × d

Step 2: Find distance.
\( d = \frac{\tau}{F} = \frac{4}{10} = 0.4 \text{ m} \)

Therefore, distance = 0.4 m
In simple words: We know the turning power and the force. We divide turning power by force to find how far from the pivot the force is applied.

📝 Teacher's Note: Rearranging formulas is important. If τ = F × d, then d = τ/F. Practice this with students using simple examples.

🎯 Exam Tip: Show the rearranged formula clearly. Write d = τ/F before substituting values. This shows you understand the concept.

 

Solution 23.
Answer:
Given:
Force applied by a mechanic, F = 200 N
Length of lever, d = 50 cm = 0.5 m

Step 1: Find initial torque.
Torque, τ = F × d = 200 × 0.5 = 100 Nm

Step 2: Find new lever length for same torque.
New force, F' = 50 N
Let d' be the new length of lever.
For same torque: τ = F' × d'
100 = 50 × d'
\( d' = \frac{100}{50} = 2 \text{ m} \)

Therefore, new lever length = 2 m
In simple words: When we use less force, we need a longer lever to get the same turning effect. This is how levers help us.

📝 Teacher's Note: Show students a see-saw. A lighter child sits farther from the center to balance a heavier child. Same principle applies here.

🎯 Exam Tip: Write clearly that torque remains constant. Show both calculations step by step. This demonstrates your understanding of lever principle.

 

Solution 24.
Answer: Conditions of equilibrium for a rigid body:

1. The body should have a broad base.
2. Center of gravity of the body should be as low as possible.
3. Vertical line drawn from the center of gravity should fall within the base of support.
4. Vector sum of forces acting on the body should be zero.
5. Algebraic sum of moments acting on the body should be zero.

In simple words: For something to stay balanced, it needs a wide bottom, heavy parts at the bottom, and all forces should cancel out.

📝 Teacher's Note: Compare a standing pencil (unstable) with a pyramid (stable). The pyramid has a broad base and low center of gravity.

🎯 Exam Tip: Learn all 5 conditions. Write them as a numbered list. Condition 3 about center of gravity is often asked separately.

 

Solution 1.
Answer: When a particle moves with a constant speed in a circular path, its motion is said to be the uniform circular motion.
In simple words: Uniform circular motion means moving in a circle at the same speed. Like a fan blade spinning at constant speed.

📝 Teacher's Note: Use a ceiling fan example. The tip of the fan blade moves in a circle at constant speed. This is uniform circular motion.

🎯 Exam Tip: Write both "constant speed" and "circular path" in your answer. Both conditions are necessary for uniform circular motion.

 

Solution 2.
Answer: In uniform circular motion, a particle travels equal distances along the circular path in equal intervals of time, so the speed of particle is uniform. But the direction of motion of the particle changes at each point of circular path. Due to continuous change in direction of motion, the velocity of the particle is not uniform (or velocity is variable) i.e., the motion is accelerated.
In simple words: Speed stays the same, but direction keeps changing. Since velocity includes both speed and direction, velocity changes. That is why it is accelerated motion.

📝 Teacher's Note: Speed is just how fast. Velocity is how fast and in which direction. In circular motion, direction changes every moment, so velocity changes.

🎯 Exam Tip: Clearly explain the difference between speed and velocity. Write that "speed is constant but velocity is variable" for full marks.

 

Solution 3.
Answer: Centripetal force: The force acting on a body moving in a circular path. It acts along the radius of the circle and towards its centre.
In simple words: Centripetal force is the force that pulls objects towards the center when they move in circles. Without this force, objects would fly away in straight lines.

📝 Teacher's Note: Tie a stone to a string and spin it. The tension in string is centripetal force pulling the stone towards your hand (center).

🎯 Exam Tip: Always mention "towards the centre" in your answer. This direction is the key feature of centripetal force.

 

Solution 4.
Answer: The SI unit of centripetal force is newton (N).
In simple words: Centripetal force is measured in newtons, just like any other force.

📝 Teacher's Note: Remind students that centripetal force is still a force, so it has the same unit as any force. Force = mass × acceleration, unit = kg⋅m/s² = N.

🎯 Exam Tip: Write the full form "newton" and the symbol "(N)" in brackets. This shows complete knowledge.

 

Solution 5.
Answer: Centrifugal force is not a real force, but it is considered to describe (understand) a certain motion. It should not be considered as the force of reaction of centripetal force, although its magnitude is same as that of centripetal force.
In simple words: Centrifugal force is not actually there. We just feel it when we are inside something moving in circles. It is like feeling pushed outward in a turning car.

📝 Teacher's Note: When you turn left in a car, you feel pushed to the right. This feeling is centrifugal force. But actually, your body wants to go straight (Newton's first law).

🎯 Exam Tip: Write "not a real force" and "pseudo force" or "fictitious force" to show you understand this concept clearly.

 

Solution 6.
Answer: In uniform linear motion, the velocity is constant and acceleration is zero i.e., the uniform linear motion is an un-accelerated motion, while in a uniform circular motion the velocity is variable (although speed is uniform), so it is an accelerated motion.
In simple words: Straight line motion with constant speed has no acceleration. Circular motion with constant speed has acceleration because direction changes.

📝 Teacher's Note: Draw a straight line and a circle on the board. Show that direction arrows change in circular motion but stay same in linear motion.

🎯 Exam Tip: Make a clear comparison table: Linear motion - constant velocity, zero acceleration. Circular motion - variable velocity, non-zero acceleration.

 

Solution 7.
Answer: The force needed for circular motion is 'centripetal force'. This force is always directed towards the centre of circle at each point of its path.
In simple words: Any object moving in circles needs centripetal force. This force always points to the center of the circle.

📝 Teacher's Note: For a car turning, friction provides centripetal force. For a satellite orbiting Earth, gravity provides centripetal force. Different situations, same principle.

🎯 Exam Tip: Always write "centripetal force" and "towards the centre" together. This combination gives full marks.

 

Solution 8.
Answer: (b) speed
In simple words: In uniform circular motion, speed stays the same but velocity and acceleration change because direction keeps changing.

📝 Teacher's Note: Speed is a scalar (no direction). Velocity and acceleration are vectors (have direction). In circular motion, direction changes, so vectors change.

🎯 Exam Tip: Remember that uniform means "speed is constant." Speed has no direction, so it can stay uniform even when direction changes.

 

Solution 9.
Answer: Yes, it is an accelerated motion.
In simple words: Even though speed is constant, direction changes continuously. Change in velocity means acceleration is present.

📝 Teacher's Note: Acceleration means change in velocity. Velocity = speed + direction. If direction changes, velocity changes, so acceleration exists.

🎯 Exam Tip: Give reason: "Because velocity changes due to change in direction." This explanation is necessary for full marks.

 

Solution 10.
Answer: No, the earth moves round the sun with a variable velocity. Since the direction of motion changes at each and every point the velocity becomes variable.
In simple words: Earth's speed around sun is almost constant, but direction keeps changing. So velocity is not constant.

📝 Teacher's Note: Earth's orbit is almost circular. Like any circular motion, direction changes continuously even if speed is nearly constant.

🎯 Exam Tip: Explain that velocity has both magnitude and direction. Even if magnitude (speed) is constant, changing direction makes velocity variable.

 

Solution 11.
Answer: Centripetal force has a reaction force in accordance with the Newton's third law of motion. This oppositely directed force is called centrifugal force.
In simple words: Newton's third law says every action has an equal and opposite reaction. The reaction to centripetal force is centrifugal force.

📝 Teacher's Note: When you pull a stone with string (centripetal force), the stone pulls back on string (centrifugal force). Action-reaction pair.

🎯 Exam Tip: Mention "Newton's third law" and write both forces clearly. Show you understand the action-reaction concept.

 

Solution 12.
Answer: Centrifugal force acts away from the centre of circular path.
In simple words: Centrifugal force points outward, away from the center. It is opposite to centripetal force which points inward.

📝 Teacher's Note: In a spinning merry-go-round, you feel pushed outward (away from center). That feeling is centrifugal force.

🎯 Exam Tip: Write "away from the centre" clearly. This direction is the key difference from centripetal force.

 

Solution 1.
Answer: A fixed pulley is used to change the direction of effort to a convenient direction.
In simple words: Fixed pulley helps us pull down instead of lifting up. It makes work easier by changing direction of our force.

📝 Teacher's Note: Show students a flag pole pulley. We pull rope down to raise the flag up. Much easier than climbing up to push flag upward.

🎯 Exam Tip: Write "change the direction of effort" as the main purpose. Fixed pulleys do not multiply force, they only change direction.

 

Solution 2.
Answer: A machine whose efficiency is 100% is referred to as an ideal machine.
In simple words: An ideal machine has no energy loss. All the energy we put in comes out as useful work. But real machines always lose some energy.

📝 Teacher's Note: No real machine is 100% efficient because of friction and heat losses. Ideal machines exist only in theory, not in real life.

🎯 Exam Tip: Write "100% efficiency" and "ideal machine" together. Also mention that ideal machines do not exist in reality.

 

Solution 3.
Answer: \( \text{Efficiency} = \frac{\text{Mechanical advantage}}{\text{Velocity ratio}} \)
In simple words: Efficiency tells us how good a machine is. It compares actual help we get with the theoretical help possible.

📝 Teacher's Note: Efficiency is always less than 1 (or less than 100%) because mechanical advantage is always less than velocity ratio due to friction.

🎯 Exam Tip: Learn this formula by heart. Also remember that efficiency has no unit because it is a ratio of two similar quantities.

 

Solution 4.
Answer: No, a machine cannot be 100% efficient because mechanical advantage is always less than theoretical mechanical advantage due to friction and the weight of the moving parts.
In simple words: All real machines lose some energy to friction and moving their own heavy parts. So they can never be 100% efficient.

📝 Teacher's Note: Even the best machines have some friction in bearings, gears, or pulleys. This friction always wastes some energy as heat.

🎯 Exam Tip: Give specific reasons: "friction" and "weight of moving parts." These are the main causes of energy loss in machines.

 

Solution 5.
Answer: Velocity ratio is defined as the ratio of velocity of the effort (VE) to the velocity of the resistance force (load) (VL). It is also defined as the ratio of the displacement of the effort to the displacement of the load. Mathematically, we have: \( V.R. = \frac{V_E}{V_L} = \frac{d_E}{d_L} \) Since velocity ratio is the ratio of two similar quantities, therefore, it does not have a unit.
In simple words: Velocity ratio compares how fast the effort moves with how fast the load moves. It has no unit because we are dividing similar quantities.

📝 Teacher's Note: In a bicycle, when you turn the pedals once, the wheel turns many times. The ratio of pedal distance to wheel distance is velocity ratio.

🎯 Exam Tip: Give both definitions - velocity ratio and displacement ratio. Also mention that it has no unit.

 

Solution 6.
Answer: The purpose of a machine is:

1. To multiply force
2. To obtain a gain or decrease in speed
3. To change to direction of application of force in a convenient direction

In simple words: Machines help us by making our force stronger, changing our speed, or letting us push/pull in easier directions.

📝 Teacher's Note: Car jack multiplies force. Bicycle gears change speed. Pulley changes direction. Each machine serves one or more of these purposes.

🎯 Exam Tip: Learn all three purposes. Examples help: jack for force, gears for speed, pulley for direction.

 

Solution 7.
Answer:

1. A jack is used to multiply force.
2. A lever is used to change the point of application of force.
3. A pulley is used to change the direction of application of force.
4. Gears are used to multiply speed.

In simple words: Different machines have different jobs. Jack makes us stronger, lever changes where we apply force, pulley changes direction, gears make us faster.

📝 Teacher's Note: Show real examples: car jack (force), crowbar (lever), flag pole (pulley), bicycle (gears). Students remember better with real objects.

🎯 Exam Tip: Match each machine with its main purpose. Jack-force, lever-point of application, pulley-direction, gears-speed.

 

Solution 8.
Answer:

1. The purpose of the pulley Q is to change the direction of application of effort to a convenient direction.
2. In the given diagram, T = E.
3. If the free end of the string moves through the distance x, the load will rise by a distance x/2.
4. Therefore, V.R. = distance moved by the effort arm / distance moved by the load = x/(x/2) = 2
5. In equilibrium, L = 2T and E = T, so Efficiency = M.A. / V.R. = 1/2
6. M.A = L/E = 100/50 = 2, so E = L/M.A = 100/2 = 50N

In simple words: This pulley system makes lifting easier by spreading the load between two rope segments, but we need to pull twice as much rope.

📝 Teacher's Note: Draw the pulley system clearly. Show how the load is supported by two rope segments, so each carries half the weight.

🎯 Exam Tip: For moving pulleys, velocity ratio = 2 × number of moving pulleys. Here, V.R. = 2 because one pulley moves.

 

Solution 9.
Answer: A block and tackle system of pulleys consists of two blocks of pulleys, each block having one or more than one pulley. The upper block of pulleys is fixed to a rigid support and the lower block of pulleys is movable. The number of pulleys in the movable block is either equal to or one less than the number of pulleys in the fixed block.
In simple words: Block and tackle is a system with two sets of pulleys - one fixed set and one moving set. This combination gives us more force multiplication.

📝 Teacher's Note: Show pictures of construction cranes. They use block and tackle systems to lift very heavy loads with less effort.

🎯 Exam Tip: Mention "two blocks," "upper block fixed," "lower block movable." These are key features of block and tackle systems.

 

Solution 10.
Answer: When a single pulley is used with a mechanical advantage greater than 1, the effort has to be applied in the upward direction.
In simple words: To get more force help from a single pulley, we must pull up instead of down. This makes the pulley movable.

📝 Teacher's Note: Fixed pulley: pull down, M.A. = 1. Moving pulley: pull up, M.A. = 2. Direction of effort decides the mechanical advantage.

🎯 Exam Tip: Connect direction with mechanical advantage. Upward effort = moving pulley = M.A. > 1. Downward effort = fixed pulley = M.A. = 1.

 

Solution 11.
Answer: In a single fixed pulley, if the effort moves by a distance 'y' downwards, the load moves by a distance 'y' upwards because the velocity ratio of a single fixed pulley is 1.
In simple words: In a fixed pulley, effort and load move the same distance but in opposite directions. Pull rope down 1 meter, load goes up 1 meter.

📝 Teacher's Note: Use a simple fixed pulley demo. Pull rope down by 10 cm, the load rises by exactly 10 cm. Same distance, opposite directions.

🎯 Exam Tip: Write "V.R. = 1" for single fixed pulley. This means effort distance equals load distance.

 

Solution 12
Answer: A pulley whose axis of rotation is not fixed in position or space is called a movable pulley. Ideally, its M.A. is 2.
In simple words: A movable pulley can move up and down. It is not stuck in one place. It makes lifting heavy things easier by cutting the force needed in half.

📝 Teacher's Note: Show students a fixed pulley (attached to ceiling) and a movable one (hangs on the rope). The movable pulley moves with the load. This makes it easier to lift.

🎯 Exam Tip: Write "axis of rotation is not fixed" and "M.A. = 2". These are key phrases examiners look for.

 

Solution 13
Answer: Single movable pulley acts as a force multiplier.
Diagram: This diagram shows a movable pulley system with effort E applied upward and load L hanging down. The rope goes around the pulley and the tension T is shown in multiple segments.
In simple words: A movable pulley reduces the effort needed. If you need to lift 10 kg, you only need to apply 5 kg force. It doubles your strength.

📝 Teacher's Note: Ask students to lift a bag directly, then use a movable pulley. They will feel the difference immediately. This practical demo helps them remember.

🎯 Exam Tip: Always write "force multiplier" for movable pulleys. Draw the diagram clearly showing effort going up and load going down.

 

Solution 14
Answer:
(i) 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.

(ii) The two reasons are:
(a) Some effort is wasted in overcoming the friction between the strings and the grooves of the pulley.
(b) Some effort is wasted in lifting up the movable block along with the load.

On account of the above two reasons, the M.A. is always less than the V.R. Hence efficiency is always less than 100%.

(iii) In a block and tackle system, if the total number of pulleys used in both the blocks is 'n' and the effort is being applied in the downward direction, then the tension in n segments of string supports the load, therefore:
L = n T and E = T

M.A. = \( \frac{\text{load}}{\text{effort}} = \frac{nT}{T} = n \)

Thus in a block and tackle system, the M.A. is equal to the number of pulleys and it increases with the increase in the number of pulley.
In simple words: Pulleys are not perfect machines. Some energy is lost due to friction and lifting the pulley itself. That is why we never get 100% efficiency. More pulleys mean more mechanical advantage.

📝 Teacher's Note: Explain friction by rubbing hands together - students feel heat. Same happens in pulleys. Some energy becomes heat instead of doing useful work.

🎯 Exam Tip: Always mention both reasons - friction and lifting movable parts. Write "efficiency less than 100%" clearly. For block and tackle, write "M.A. = number of pulleys".

 

Page No-39:

 

Solution 15
Answer:
(a) A jack is used to multiply force.
(b) Gears are used to multiply speed.
(c) A pulley is used to change the direction of application of force.
In simple words: Jack makes you stronger to lift cars. Gears make wheels spin faster. Pulleys let you pull down to lift up.

📝 Teacher's Note: Show car jack, bicycle gears, and flag pole pulley as real examples. Students understand better with things they see daily.

🎯 Exam Tip: Remember the three uses clearly - jack for force, gears for speed, pulley for direction. Do not mix them up.

 

Solution 16
Answer: Two uses of pulley:
(i) To change the direction of application of force to a convenient direction.
(ii) To multiply force. Yes, a pulley is a force multiplier.
In simple words: Pulleys help us pull down to lift up (easier direction). They also make lifting heavy things easier by reducing force needed.

📝 Teacher's Note: Ask students which is easier - pulling down or lifting up. They will say pulling down. That shows how pulleys help change direction.

🎯 Exam Tip: Write both uses clearly. Add "Yes, pulley is a force multiplier" as the answer expects this confirmation.

 

Solution 17
Answer: Correct winding of rope around pulleys, load and effort is as shown:
Diagram: This diagram shows a block and tackle system with five strings supporting the load. The effort E=T is applied and the load L=5T is shown at the bottom.

(i) Since there are five strings in the block and tackle system, therefore, the velocity ratio of the system is V.R. = n = 5

V.R. = \( \frac{d_E}{d_L} \)
Now, or, \( d_E = 5 \times d_L = 5 \times 1 = 5m \)

(ii) 5 strands of tackle are supporting the load.

(iii) M.A. = L/E
Here, L = 5T, E = T
∴ M.A = \( \frac{5T}{T} = 5 \)
In simple words: This pulley system has 5 strings holding the weight. So the mechanical advantage is 5. This means you need 5 times less effort to lift the load.

📝 Teacher's Note: Count the strings with students by pointing at the diagram. Make them see that more strings mean easier lifting.

🎯 Exam Tip: Always count the strings carefully. Write "n = number of strings" and "M.A. = n". Show all calculation steps clearly.

 

Solution 18
Answer:
(i) 4 strands of tackle are supporting the load.

(ii) Correct winding of rope around pulleys, load and effort is as shown:
Diagram: This diagram shows a block and tackle system with four strings supporting the load. The effort E=T is applied and the load L=4T is shown.

(iii) M.A. = L/E
Here, L = 4T, E = T
∴ M.A = \( \frac{4T}{T} = 4 \)

(iv) Since there are four strings in the block and tackle system, therefore, the velocity ratio of the system is V.R. = n = 4
V.R. = \( \frac{d_E}{d_L} \)
Now, or, \( d_E = 4 \times d_L = 4 \times 1 = 4m \)
In simple words: This system has 4 strings holding the weight. So you need 4 times less effort. If you pull the rope 4 meters, the load moves up 1 meter.

📝 Teacher's Note: Compare this with Solution 17. Show students how fewer strings mean less mechanical advantage but same principle applies.

🎯 Exam Tip: Count strings first, then write M.A. = number of strings. Show the relationship between effort distance and load distance clearly.

 

Solution 19
Answer: A block and tackle system of pulleys consists of two blocks of pulleys, each block having one or more than one pulley. The upper block of pulleys is fixed to a rigid support and the lower block of pulleys is movable. The number of pulleys in the movable block is either equal to or one less than the number of pulleys in the fixed block.
In simple words: Block and tackle has two sets of pulleys. Top set is fixed (does not move). Bottom set is movable (moves with the load). Both sets work together to make lifting very easy.

📝 Teacher's Note: Draw two boxes - one at top (fixed) and one at bottom (movable). Show how rope goes between them. This visual helps students understand.

🎯 Exam Tip: Write "two blocks" clearly. Mention "upper fixed" and "lower movable". Add the relationship between number of pulleys in each block.

 

Solution 20
Answer:
(i) Correct winding of rope around pulleys, load and effort is as shown:
Diagram: This diagram shows a block and tackle system with four strings in the system.

(ii) Since there are four strings in the block and tackle system, therefore, the velocity ratio of the system is V.R. = n = 4.

(iii) M.A. = L/E
Here, L = 4T, E = T
∴ M.A = \( \frac{4T}{T} = 4 \)

(iv) The assumptions made are movable pulleys are weightless and efficiency of the system is 100%.
In simple words: We assume pulleys have no weight and no energy is lost to friction. In real life, this never happens. But it helps us do calculations easily.

📝 Teacher's Note: Explain "assumptions" as things we pretend to make math easier. Like pretending a car has no friction to calculate speed.

🎯 Exam Tip: Always state the two assumptions - "weightless pulleys" and "100% efficiency". These are standard assumptions in pulley problems.

 

Solution 21
Answer:
(i) 'Gear ratio' is defined as the ratio between the number of teeth in the driven gear and those on the driving gear.
(ii) A vehicle moving uphill uses a lower gear to increase the torque and to raise velocity ratio i.e., speed decreased.
(iii) Gear ratio, in bicycle is less than one, thus, one rotation of 'penion' causes a number of rotations of 'idler'.
In simple words: Gear ratio tells us how many teeth each gear has. Lower gears give more power for climbing hills but less speed. In bicycles, small gear makes big gear turn many times.

📝 Teacher's Note: Show students bicycle gears. Small chainring (penion) makes big wheel (idler) spin fast. That is why bicycles can go fast on flat roads.

🎯 Exam Tip: Define gear ratio as "driven teeth ÷ driving teeth". Remember "lower gear = more torque, less speed" for uphill climbing.

 

Page No-47:

 

Solution 1
Answer: Force
In simple words: Force is a push or pull that can change the motion of objects.

📝 Teacher's Note: This appears to be a one-word answer. The question is not given, but the answer is clearly "Force".

🎯 Exam Tip: Write "Force" clearly if the question asks for what causes motion or change in motion.

 

Solution 2
Answer: Yes
In simple words: Yes means the answer to the question is positive or true.

📝 Teacher's Note: This appears to be a yes/no answer. Without the question, we can only provide the answer given.

🎯 Exam Tip: For yes/no questions, write the answer clearly and add explanation if marks are given for reasoning.

 

Solution 3
Answer: Joule
In simple words: Joule is the unit of energy and work. It is named after a scientist.

📝 Teacher's Note: Joule (J) is the SI unit of work and energy. One joule equals the work done by one newton force over one meter distance.

🎯 Exam Tip: Always write "Joule (J)" with the symbol in brackets. This shows you know both the name and symbol.

 

Solution 4
Answer: Work
In simple words: Work is done when a force moves an object through some distance.

📝 Teacher's Note: Work has a specific meaning in physics. It is force × distance when force and motion are in same direction.

🎯 Exam Tip: Remember work = force × distance. Both force and distance must be in the same direction for work to be done.

 

Solution 5
Answer: 1N, 1m in its own direction
In simple words: One joule of work is done when a force of 1 Newton moves an object 1 meter in the direction of the force.

📝 Teacher's Note: This defines one joule. The force and displacement must be in the same direction. If they are at an angle, only the component counts.

🎯 Exam Tip: Write "1 N force, 1 m displacement, same direction". This complete definition gets full marks.

 

Solution 6
Answer: When no net force is applied, the work done which is the dot product of force and displacement is zero.
In simple words: If the total force on an object is zero, then no work is done. The object may move but with constant speed.

📝 Teacher's Note: Net force is the total of all forces acting. If forces balance out (net = 0), then work = 0 even if object moves.

🎯 Exam Tip: Write "net force = 0, therefore work = 0". Remember that moving at constant speed means net force is zero.

 

Page No-48:

 

Solution 7
Answer: The work done is zero because the displacement is zero.
In simple words: If an object does not move (displacement = 0), then work done = 0, even if you apply force.

📝 Teacher's Note: You can push a wall with great force, but if wall does not move, work done = 0. Work needs both force AND movement.

🎯 Exam Tip: Write "displacement = 0, therefore work = 0". This is a common exam trap - force without movement gives zero work.

 

Solution 8
Answer: The work done is zero because:
Work done = force × displacement × cos θ
Now, since θ = 90°; cos θ = 0 and hence work done is equal to zero.
In simple words: When force is at 90 degrees to the direction of movement, no work is done. Like carrying a bag while walking horizontally - the lifting force does no work on horizontal motion.

📝 Teacher's Note: Show students walking with a bag. The upward force holding the bag does no work on horizontal walking. Force and motion are perpendicular.

🎯 Exam Tip: Write the formula with cos θ. When θ = 90°, cos 90° = 0, so work = 0. This formula approach gets full marks.

 

Solution 9
Answer: Work is a scalar quantity because it is a measure of transfer of energy without indicating any direction.
In simple words: Work has size but no direction. You cannot say work is done "towards north" - it just has a number value.

📝 Teacher's Note: Scalars have only magnitude (size). Vectors have both magnitude and direction. Work, energy, mass are scalars. Force, velocity are vectors.

🎯 Exam Tip: Write "scalar - has magnitude only, no direction" and "measure of energy transfer". This shows you understand scalars vs vectors.

 

Solution 10
Answer: When the displacement of the body is in the direction opposite to that of the applied force, the work done is negative.
W = F d cos θ = F d cos 180° = -F d
In simple words: When force and movement are in opposite directions, work is negative. Like brakes on a car - force opposes motion.

📝 Teacher's Note: Show students pushing forward while walking backward. The push does negative work because force and motion are opposite.

🎯 Exam Tip: Write "opposite directions" and "θ = 180°, cos 180° = -1". Show the complete formula with negative result.

Solution 11
Answer: The work done by the gravitational force of the earth on a satellite revolving around the earth is zero because the motion of the satellite is perpendicular to the force at every point.
In simple words: The satellite moves in a circle. Gravity pulls it toward Earth. But the satellite's movement is always sideways to the pull. When force and movement are at 90°, work done is zero.

📝 Teacher's Note: Draw a circle on the board. Show how gravity points toward the center but the satellite moves along the circle edge. These directions are always perpendicular.

🎯 Exam Tip: Write "perpendicular" clearly. Also write \( W = F \times d \times \cos 90° = 0 \). This formula gets you marks.

 

Solution 12
Answer:
Given: load m = 500 kg, height h = 5m, g = 10 ms⁻²
Work done: W = mgh = 500 × 10 × 5 = 25000 J = 25 kJ
In simple words: We use the formula W = mgh to find work done against gravity. We multiply mass, gravity, and height.

📝 Teacher's Note: Always write "Given" first, then the formula, then substitute values. Students must learn this step-by-step method for calculations.

🎯 Exam Tip: Don't forget to convert the final answer to the right unit. Here 25000 J = 25 kJ. Show both forms for full marks.

 

Solution 13
Answer:
Given: mass m = 10 kg, height h = 4m, g = 10 ms⁻²
Work done: W = mgh = 10 × 10 × 4 = 400 J
In simple words: Again we use W = mgh. This is the work done to lift something up against gravity.

📝 Teacher's Note: Give students more practice with this formula. It's the most important formula in this chapter. Use different mass and height values.

🎯 Exam Tip: Always write the unit (J for joules) at the end. Check if the answer looks reasonable - 400 J for lifting 10 kg by 4 m seems right.

 

Solution 14
Answer: The work done against gravity is zero when a body is moved horizontally along a frictionless surface because the force of gravity is perpendicular to the displacement in this case.
In simple words: When you push something sideways on a smooth table, you don't work against gravity. Gravity pulls down, but movement is sideways. So no work against gravity.

📝 Teacher's Note: Use a book on a table to demonstrate. Push it sideways - no work against gravity. Lift it up - now you work against gravity.

🎯 Exam Tip: Write "perpendicular" and also mention the angle is 90°. Write \( W = F \times d \times \cos 90° = 0 \) for extra marks.

 

Solution 15
Answer: 'Work' is said to be done when the applied force makes the body move i.e., there is a displacement of body. It is equal to the product of force and the displacement of the point of application of the force in the direction of force.
In simple words: Work happens when a force makes something move. Work = Force × Distance moved in the direction of force.

📝 Teacher's Note: Give examples: pushing a box that moves = work done. Pushing a wall that doesn't move = no work done. Students understand better with examples.

🎯 Exam Tip: Always mention "displacement in the direction of force." This is the key point examiners look for in work definition.

 

Solution 16
Answer: Work done depends upon:
(i) the magnitude and direction of the applied force, and
(ii) the displacement it produces.
In simple words: Work depends on how much force you apply, which direction you apply it, and how far the object moves.

📝 Teacher's Note: Use examples: more force = more work, pushing in wrong direction = less work, object moving further = more work.

🎯 Exam Tip: Write both points clearly. Use (i) and (ii) format exactly as shown. Don't miss "magnitude and direction" - both are important.

 

Solution 17
Answer: Yes, we perform work against gravity.
In simple words: When we lift anything up, we work against gravity. Gravity pulls down, we pull up.

📝 Teacher's Note: Ask students to lift their bags. They are doing work against gravity right now. Make it practical and real.

🎯 Exam Tip: A simple "Yes" is enough, but add one line explanation like "when lifting objects upward" for better marks.

 

Solution 18
Answer: The angle should be 90°.
In simple words: When force and movement are at right angles (90°), no work is done. Like pushing sideways on something moving forward.

📝 Teacher's Note: Show with hands - one hand points up (force), other points sideways (movement). When they are perpendicular, work = zero.

🎯 Exam Tip: Write "90°" clearly. Also mention \( \cos 90° = 0 \) so \( W = F \times d \times \cos 90° = 0 \).

 

Solution 19
Answer: This is because at each point of the circular path, the displacement is perpendicular to the force, which is directed towards the centre, along the radius.
In simple words: In circular motion, the force points toward the center but movement is along the circle edge. These are always perpendicular, so work = 0.

📝 Teacher's Note: Draw a circle. Show radius pointing to center (force direction) and tangent at any point (movement direction). They are always 90° apart.

🎯 Exam Tip: Draw a small diagram if possible. Write "perpendicular" and "towards the centre" - these are key phrases examiners want to see.

 

Solution 20
Answer: When the angle between the direction of motion and that of the force is 90°; W = Fd cos 90° = 0. When the angle between the direction of motion and that of the force is 0°; W = Fd cos 0° = Fd. Hence in the second case, when the angle is 0°, the work done is more.
In simple words: When force and movement are in the same direction (0° angle), maximum work is done. When they are perpendicular (90°), no work is done.

📝 Teacher's Note: Compare pushing a cart forward (0° angle - easy) vs pushing it sideways while it moves forward (90° - no useful work).

🎯 Exam Tip: Write both formulas clearly: \( W = Fd \cos 0° = Fd \) and \( W = Fd \cos 90° = 0 \). Show that \( \cos 0° = 1 \) and \( \cos 90° = 0 \).

 

Solution 21
Answer: The displacement of the man and suitcase is along the horizontal direction. Thus, the angle between the displacement and the force of gravity is 90°; Thus, W = Fd cos 90° = 0. Hence, no work is done against gravity in this case.
In simple words: When you walk horizontally carrying a suitcase, you don't work against gravity. Gravity pulls down, you move sideways.

📝 Teacher's Note: Act this out - carry a book and walk across the room. You feel tired from friction and your muscles, but you don't work against gravity.

🎯 Exam Tip: Clearly state "horizontal displacement" and "angle = 90°". Write the formula \( W = Fd \cos 90° = 0 \) for full marks.

 

Solution 22
Answer: When a body moves along a circular path, work done by the gravitational force towards the centre of the path is zero, because the displacement in this case is normal to the gravitational force.
In simple words: In circular motion, gravity points to the center but movement is along the circle. Since they are perpendicular, work = 0.

📝 Teacher's Note: Use a ball on a string demonstration. The string force (like gravity) pulls inward, but the ball moves in a circle. These directions are perpendicular.

🎯 Exam Tip: Write "normal to" (which means perpendicular to). Also mention the angle is 90° for complete answer.

 

Solution 23
Answer: The work done by the gravitational force of the sun on earth during its motion around the sun is zero because at every point, the displacement of earth is perpendicular to the gravitational force of sun i.e., W = Fd cos 90° = 0.
In simple words: Earth moves in a circle around the Sun. Sun's gravity pulls Earth inward, but Earth moves along the circular path. These directions are perpendicular.

📝 Teacher's Note: Draw Earth's orbit around the Sun. Show that gravity points toward Sun (inward) while Earth's motion is along the orbit (tangent to circle).

🎯 Exam Tip: Write the formula \( W = Fd \cos 90° = 0 \) and explain why the angle is 90°. This shows you understand the concept.

 

Solution 24
Answer: A kilojoule of work is said to done when a force of 1 newton displaces a body through 1000 metres in its own direction. 1 kJ = 10³ joules.
In simple words: Kilojoule is a bigger unit of work. It equals 1000 joules. It's work done when 1 N force moves something 1000 meters.

📝 Teacher's Note: Relate to students' experience - lifting a 100g apple (about 1N force) for 1000m would be 1 kJ of work. That's like climbing a very tall building!

🎯 Exam Tip: Always write the conversion: 1 kJ = 1000 J = 10³ J. Use scientific notation for better marks.

 

Solution 25
Answer: 1 MJ = 10⁶ joules.
In simple words: Megajoule is an even bigger unit of work. It equals 1 million joules.

📝 Teacher's Note: Teach the prefixes: kilo = thousand (10³), mega = million (10⁶), giga = billion (10⁹). These appear in many science topics.

🎯 Exam Tip: Write it as \( 1 \text{ MJ} = 10^6 \text{ J} \) using scientific notation. This is the standard way to write it.

 

Solution 26
Answer: The SI unit of work is joule. 1 joule of work is said to be done when a force of 1 newton displaces a body through 1 metre in its own direction.
In simple words: Joule is the standard unit for measuring work. 1 joule = work done when 1 newton force moves something 1 meter in the force direction.

📝 Teacher's Note: Show students: lifting a small apple (about 1N force) by 1 meter height does about 1 joule of work. This makes the unit real for them.

🎯 Exam Tip: Always define joule properly: "1 N force displacing body through 1 m in force direction." Don't just write "1 N × 1 m".

 

Solution 27
Answer: The SI unit of work is 'joules' and the CGS unit is 'erg'. 1 joule = 10⁷ erg. Thus the ratio is 10⁷ : 1.
In simple words: Different unit systems use different units for work. SI uses joules, CGS uses erg. 1 joule is much bigger than 1 erg.

📝 Teacher's Note: Explain that scientists worldwide use SI units now. CGS is older but still appears in some books and exams.

🎯 Exam Tip: Remember the conversion: 1 J = 10⁷ erg. Write the ratio as 10⁷ : 1 or simply 10⁷.

 

Solution 28
Answer:
Given: work done = 54,000 J, force = 6000 N, θ = 0°
Work formula: work done = force × displacement × cos θ
Displacement: displacement = 54,000/6000 = 9m
In simple words: We rearrange the work formula to find displacement. Since force and movement are in same direction, cos 0° = 1.

📝 Teacher's Note: Teach students to rearrange formulas. If W = F × d, then d = W/F. This skill helps in many physics problems.

🎯 Exam Tip: Always write "Given" first, then the formula, then substitute values. Show your working clearly for full marks.

 

Solution 29
Answer:
Given: work done = 150 J, displacement = 10 m, θ = 0°
Work formula: work done = force × displacement × cos θ
Force: Force = 150/10 = 15 N
In simple words: We rearrange the work formula to find force. Divide work by distance to get force.

📝 Teacher's Note: Practice more rearrangement problems. If W = F × d, then F = W/d. Students need lots of practice with this.

🎯 Exam Tip: Write the unit (N for newtons) clearly. Check if your answer makes sense - 15 N is like the weight of a 1.5 kg object.

 

Solution 30
Answer: Applied force, displacement in the direction of the applied force.
In simple words: Work depends on how much force you apply and how far the object moves in that force direction.

📝 Teacher's Note: Emphasize "in the direction of force." If force is 10 N rightward but object moves 5 m leftward, then displacement in force direction is negative.

🎯 Exam Tip: Always write "displacement in the direction of applied force." Don't just write "displacement" - you'll lose marks.

 

Solution 31
Answer: Examples of work done:
(i) In free fall of a body of mass m, under gravity from a height h, the force of gravity (F=mg) is in the direction of displacement (=h) and the work done by the gravity is mgh.
(ii) A coolie lifting a load does work against gravity.
In simple words: When something falls, gravity does work on it. When we lift something up, we do work against gravity.

📝 Teacher's Note: Give more examples students can see: walking up stairs (work against gravity), sliding down a slide (gravity does work on you).

🎯 Exam Tip: Always give specific examples with clear explanation of which forces do work and in which direction.

 

Solution 32
Answer: Work done depends upon:
(i) the magnitude and direction of the applied force, and
(ii) the displacement it produces.
In simple words: Work depends on how strong the force is, which way it points, and how far the object moves because of it.

📝 Teacher's Note: This is a repeat of Solution 16. Students should memorize this answer as it's frequently asked in different forms.

🎯 Exam Tip: Use the exact format with (i) and (ii). Write "magnitude and direction" together - both matter for the force.

 

Solution 33
Answer: Work done = force × displacement × cos θ
Or, W = Fd cos θ
In simple words: This is the main formula for work. θ is the angle between force direction and movement direction.

📝 Teacher's Note: This is the most important formula in the chapter. Make sure students understand what each symbol means: W = work, F = force, d = displacement, θ = angle.

🎯 Exam Tip: Always define what θ represents: "angle between force and displacement." This shows you understand the formula completely.

 

Solution 34
Answer: Work done against gravity = mass × acceleration due to gravity × height. Or, W = mgh.
In simple words: When lifting something up, use W = mgh to find work done. m = mass, g = 9.8 or 10 m/s², h = height.

📝 Teacher's Note: This formula only works when lifting straight up. If lifting at an angle, use the general work formula W = Fd cos θ.

🎯 Exam Tip: This formula is only for work "against gravity." Don't use it for horizontal motion or other types of work.

 

Solution 35
Answer: The displacement of the man and box is along the horizontal direction. Thus, the angle between the displacement and the force of gravity is 90°; Thus, W = Fd cos 90° = 0. Hence, no work is done against gravity in this case; however some work is done against friction.
In simple words: Moving horizontally doesn't work against gravity because gravity pulls down, not sideways. But you still work against friction.

📝 Teacher's Note: Distinguish between work against gravity (only when moving up/down) and work against friction (when moving on rough surfaces).

🎯 Exam Tip: Always mention both parts: no work against gravity, but work is done against friction. This shows complete understanding.

 

Solution 36
Answer: Yes, power is a scalar quantity.
In simple words: Power has size but no direction. Like energy and work, it's a scalar (not a vector).

📝 Teacher's Note: Remind students: scalars have only magnitude (size), vectors have both magnitude and direction. Power, work, energy are scalars.

🎯 Exam Tip: A simple "Yes" is enough, but you can add "because power has only magnitude, no direction" for extra marks.

 

Solution 37
Answer: No, every force cannot produce work. Force can produce work if the applied force cause displacement in the direction of the force.
In simple words: Force only does work if it makes something move in the force direction. If nothing moves, or moves perpendicular to force, no work is done.

📝 Teacher's Note: Give examples: pushing a wall (force but no displacement = no work), holding a heavy box (force but no displacement = no work).

🎯 Exam Tip: Explain the condition clearly: force must cause displacement in its own direction. This is the key requirement for work.

 

Solution 38
Answer: Work is said to be done only when the applied force on a body makes the body move but power is the rate of doing work. The SI unit of work is 'joules' and that of power is 'watt'.
In simple words: Work is about moving things with force. Power is about how fast you do work. Work is measured in joules, power in watts.

📝 Teacher's Note: Use analogy: work is like the total distance you travel, power is like your speed. Power = Work/Time, just like Speed = Distance/Time.

🎯 Exam Tip: Define both terms clearly and give their SI units. Also mention: Power = Work/Time as the relationship between them.

 

Solution 39
Answer:
(a) joule, watt
(b) power, energy
(c) work
(d) 10⁷
(e) 746
In simple words: These are fill-in-the-blank answers about work, power, and energy units and conversions.

📝 Teacher's Note: Create similar fill-in-the-blank exercises for students to practice. This format tests their memory of important facts and figures.

🎯 Exam Tip: Learn key conversions by heart: 1 J = 10⁷ erg, 1 HP = 746 W. These numbers appear often in exams.

 

Solution 40
Answer:
Given: load m = 200 kgf, displacement h = 25 m, time = 5s, g = 10 N kg⁻¹
Work done: = 200 kgf × 25 = 5000 J
Power: = work done/time taken = 5000/5 = 1000 watt
In simple words: First find work done using W = mgh. Then find power using P = W/t (work divided by time).

📝 Teacher's Note: Show students that kgf (kilogram-force) is different from kg (kilogram). Here 200 kgf means 200 × 10 = 2000 N force.

🎯 Exam Tip: Always show both steps: first calculate work, then calculate power. Write units clearly: J for work, W for power.

 

Solution 41
Answer:
Given: load m = 750 N, displacement h = 16 m, time = 5s, g = 10 N kg⁻¹
Work done: = mgh = 750 × 16 = 12000 J
Power: = work done/time taken = 12000/5 = 2400 watt
In simple words: Same method as before. Calculate work first, then divide by time to get power.

📝 Teacher's Note: Here the load is already given in newtons (750 N), so we don't need to multiply by g. Students should notice this difference.

🎯 Exam Tip: Read carefully whether load is given in kg, kgf, or N. The calculation method changes slightly for each unit.

 

Solution 42
Answer: Power.
In simple words: Power is the rate of doing work. It tells us how fast work is being done.

📝 Teacher's Note: This seems to be answering "What is the rate of doing work called?" Power is work per unit time.

🎯 Exam Tip: If asked for definition of power, write: "Power is the rate of doing work" or "Power = Work/Time".

 

Solution 43
Answer: Work done depends upon the vertical height and not the path taken, hence if the boy uses a lift to reach the same vertical height, work done will be mgh.
In simple words: Work against gravity only depends on height, not the path. Whether you climb stairs or use a lift, if you go up the same height, work done against gravity is same.

📝 Teacher's Note: This is an important concept. Work done against conservative forces (like gravity) depends only on start and end points, not the path.

🎯 Exam Tip: Always mention "work depends on height, not path." This shows you understand that gravitational work is path-independent.

 

Solution 44
Answer: Yes, for e.g. if you push a wall, you apply force on it but no work is done since the displacement is zero.
In simple words: You can push hard on a wall, but if the wall doesn't move, no work is done. Force without movement = no work.

📝 Teacher's Note: Let students try pushing the wall. They feel tired (using energy) but do no work because displacement = 0.

🎯 Exam Tip: Always give a clear example like "pushing a wall" or "holding a heavy object still." These show force without displacement.

 

Solution 45
Answer: 1 H.P. = 0.746 kW
In simple words: Horsepower is an old unit for power. 1 horsepower equals 0.746 kilowatts.

📝 Teacher's Note: Explain that HP (horsepower) was used to compare steam engines with horses. Now we use watts and kilowatts instead.

🎯 Exam Tip: Remember the exact conversion: 1 HP = 0.746 kW = 746 W. This appears in many numerical problems.

 

Solution 46
Answer:
(i) Work done depends on the displacement, now, vertical distance travelled by A and B is same, hence ratio of work done by them is 1:1

(ii) Power = work done/time taken
∴ \( \frac{P_A}{P_B} = \frac{1}{1} \times \frac{15}{20} = \frac{3}{4} \)

The required ratio is 3: 4.
In simple words: Both A and B do same work (same height). But A is faster, so A has more power than B.

📝 Teacher's Note: Emphasize: work depends on height (same for both), but power depends on time (different for A and B).

🎯 Exam Tip: First show work ratio, then power ratio. Use the formula P = W/t correctly with the time values given.

 

Solution 47
Answer: The SI unit of power is 'watt'. If 1 joule of work is done in 1 second, the power is said to be 1 watt.
In simple words: Watt is the unit of power. 1 watt means doing 1 joule of work in 1 second.

📝 Teacher's Note: Connect to daily life: a 100W bulb uses 100 joules of energy every second. This makes watts more meaningful to students.

🎯 Exam Tip: Define watt precisely: "1 watt = 1 joule per second" or write \( 1 \text{ W} = 1 \text{ J/s} \).

 

Solution 48
Answer:
SI unit of power = watt
CGS unit of power = erg per second
1 W = 1 J s⁻¹ = 10⁷ erg s⁻¹
In simple words: Different unit systems have different units for power. SI uses watts, CGS uses erg per second.

📝 Teacher's Note: Show the conversion step by step: 1 W = 1 J/s = (10⁷ erg)/s = 10⁷ erg/s.

🎯 Exam Tip: Write the conversion clearly: 1 W = 10⁷ erg s⁻¹. Use proper notation with negative exponents.

Solution 1
Answer: When the resultant of a group of forces acting on the same object is zero, the forces are said to be balanced. Balanced forces do not change the speed of stationary objects. They may deform objects.
In simple words: When many forces push or pull an object equally in all directions, they cancel each other out. The object stays still but may change shape.

📝 Teacher's Note: Show students a book on a table. Gravity pulls it down, table pushes it up. Forces are balanced so book stays still. Easy to understand.

🎯 Exam Tip: Write "resultant force is zero" and "no change in motion" clearly. These are key phrases for marks.

 

Solution 2
Answer: Yes, force is a vector quantity.
In simple words: Force has both strength (how much) and direction (which way). Like pushing a box - you need to know how hard and in which direction.

📝 Teacher's Note: Vector means it has direction. Show with an arrow on board. Push left is different from push right even with same strength.

🎯 Exam Tip: Always write "vector quantity" and explain it has both magnitude and direction. This gets you full marks.

 

Solution 3
Answer: 1 kgf = force due to gravity on 1 kg mass = 1 kg mass × acceleration due to gravity g in \( ms^{-2} \) = g newton
1 kgf = 9.8 newton
In simple words: One kilogram-force is the weight of a 1 kg object on Earth. It equals 9.8 newton units.

📝 Teacher's Note: Tell students kgf is weight, not mass. A 1 kg apple weighs 9.8 N on Earth. Weight changes on moon but mass stays same.

🎯 Exam Tip: Write the conversion clearly: 1 kgf = 9.8 N. Show the working with \( g = 9.8 ms^{-2} \).

 

Solution 4
Answer: The SI unit of force is 'newton'. In CGS system, the unit of force is 'dyne'.
1 newton = \( 10^5 \) dyne
Therefore, ratio of SI to CGS unit of force is \( 10^5 : 1 \).
In simple words: Newton is the big unit for force in SI system. Dyne is the small unit in CGS system. One newton equals 100,000 dynes.

📝 Teacher's Note: SI is the modern system we use. CGS is the old system. Newton is much bigger than dyne - like comparing rupees to paise.

🎯 Exam Tip: Write both units clearly: newton (SI) and dyne (CGS). Show the ratio as \( 10^5 : 1 \).

 

Solution 5
Answer: Yes, weight is a force.
In simple words: Weight is the force that pulls objects down towards Earth. It is gravity acting on mass.

📝 Teacher's Note: Weight has direction (downward) and magnitude. That makes it a force. Mass is just the amount of matter.

🎯 Exam Tip: Write "weight is a force due to gravity." This simple statement gets marks.

 

Solution 6
Answer: When we kick a football at rest, it starts moving.
In simple words: The football was not moving. Our kick gave it force. Now it moves. Force can start motion.

📝 Teacher's Note: This shows force can change state from rest to motion. Use examples like pushing a swing or throwing a ball.

🎯 Exam Tip: Give clear examples showing force starting motion. Football, cricket ball, or pushing a cart work well.

 

Solution 7
Answer: When a balloon is inflated the force of air inside changes its shape or size.
In simple words: Air pushes from inside the balloon. This force makes the balloon bigger and round.

📝 Teacher's Note: Show students how blowing air changes balloon shape. Force can change size and shape, not just motion.

🎯 Exam Tip: Write that force can change "shape and size." Use balloon or squeezing clay as examples.

 

Solution 8
Answer: Effects a force can produce and examples:

1. Change the state of rest; e.g. pushing a door to open it or close it.
2. Change the state of motion; e.g. applying a force to stop the cricket ball.
3. Change the direction of motion and not speed; e.g. when a force is applied to move a body in a circular path with uniform speed there is only a change in direction of motion but speed remains constant.
4. Change both speed and direction of motion; e.g. when a body is swirled in the vertical circle its direction of motion and speed changes at every point.
5. Change the dimension; when a balloon is inflated the force of air inside changes its shape or size.

In simple words: Force can start motion, stop motion, change direction, change speed, or change shape. It is very powerful.

📝 Teacher's Note: Use real examples for each effect. Door opening (start motion), brakes (stop motion), turning car (change direction), hitting ball harder (change speed).

🎯 Exam Tip: List all five effects clearly with examples. This is a complete answer that gets full marks.

 

Solution 9
Answer: No
In simple words: The answer is simply no.

📝 Teacher's Note: This appears to be answering a yes/no question. Make sure students understand what question this refers to.

🎯 Exam Tip: For yes/no questions, write clearly and give a reason if marks allow.

 

Solution 10
Answer: No
In simple words: The answer is simply no.

📝 Teacher's Note: This appears to be answering a yes/no question. Students should understand the context.

🎯 Exam Tip: Write the answer clearly and add explanation if the question demands it.

 

Solution 11
Answer: The turning effect produced by a force on a rigid body about a point, pivot or fulcrum is called the moment of force or torque. It is measured by the product of force and the perpendicular distance of the pivot from the line of action of force.
Moment of a force = Force × perpendicular distance of the pivot from the force.
Its SI unit is newton-metre (Nm).
In simple words: When you turn a door handle, you create a turning effect. This is called moment or torque. It depends on how hard you push and how far from the hinge.

📝 Teacher's Note: Use door handle, wrench, or seesaw examples. Show that longer handle makes turning easier. Distance from pivot matters.

🎯 Exam Tip: Write the formula clearly: Moment = Force × perpendicular distance. Give unit as Nm.

 

Solution 12
Answer: The physical quantity is 'torque'.
In simple words: Torque is the name for turning effect. Like when you turn a bottle cap.

📝 Teacher's Note: Torque and moment of force mean the same thing. Use examples like turning steering wheel or opening jar.

🎯 Exam Tip: Write "torque" or "moment of force" - both are correct. They mean turning effect.

 

Solution 13
Answer: The turning effect produced by a force on a rigid body about a point, pivot or fulcrum is called the moment of force or torque. It is measured by the product of force and the perpendicular distance of the pivot from the line of action of force.
In simple words: This is the definition of torque. It is the turning power of a force around a fixed point.

📝 Teacher's Note: This is the standard definition students must learn. Practice with examples like scissors, pliers, and bottle opener.

🎯 Exam Tip: Write the complete definition mentioning "turning effect," "rigid body," and "perpendicular distance."

 

Solution 14
Answer: Moment of force = applied force × perpendicular distance from the pivot
= 10 × 30 × \( 10^{-2} \) = 3 Nm
In simple words: We multiply the force by the distance to get the moment. Don't forget to convert cm to m.

📝 Teacher's Note: Students often forget unit conversion. 30 cm = 0.3 m. Always check units in physics problems.

🎯 Exam Tip: Show unit conversion clearly. Write final answer with correct unit (Nm). Show all working steps.

 

Solution 15
Answer: SI unit of couple = newton
CGS unit of couple = dyne
In simple words: Couple is measured in the same units as force. Newton in SI system, dyne in CGS system.

📝 Teacher's Note: A couple is two equal and opposite forces. But its unit is same as force, not moment.

🎯 Exam Tip: Write both units clearly. Remember couple has same units as force.

 

Solution 16
Answer: No, the moment of force is a vector quantity.
In simple words: Moment has direction - clockwise or anticlockwise. So it is a vector, not scalar.

📝 Teacher's Note: Show students clockwise and anticlockwise rotation. Direction matters, so moment is vector.

🎯 Exam Tip: Write "vector quantity" and explain it has direction (clockwise or anticlockwise).

 

Solution 17
Answer: A larger diameter provides a greater torque (= force × perpendicular distance); hence, it is easier to turn a steering wheel of a large diameter than that of a small diameter.
In simple words: Bigger steering wheel means your hands are farther from center. This gives more turning power with same effort.

📝 Teacher's Note: Show with bicycle wheel vs car steering wheel. Bigger radius makes turning easier. Same principle as long-handled wrench.

🎯 Exam Tip: Write the torque formula and explain that larger radius gives more torque for same force.

 

Solution 18
Answer: The point through which the resultant of the weights of all the particles of the body acts is called its centre of gravity.
In simple words: Centre of gravity is the point where all the weight of an object seems to act. It is the balance point.

📝 Teacher's Note: Use a ruler balanced on pencil to show centre of gravity. It is where object balances perfectly.

🎯 Exam Tip: Write "resultant of weights" and "acts through this point." This definition gets full marks.

 

Solution 19
Answer: A body is said to be in equilibrium under the action of a number of forces, if the forces are not able to produce any change in the state of rest or of uniform motion or uniform rotation.
In simple words: Equilibrium means balance. All forces cancel out so object stays still or keeps moving at same speed.

📝 Teacher's Note: Book on table is in equilibrium (still). Car at constant speed is also in equilibrium (moving but no change).

🎯 Exam Tip: Write "no change in state of rest or uniform motion." This is the key phrase for equilibrium.

 

Solution 20
Answer: Principle of moments: If a body is in equilibrium under the action of number of force, then the sum of clockwise moments is equal to the sum of anticlockwise moments.
In simple words: For balance, turning forces in one direction must equal turning forces in opposite direction.

📝 Teacher's Note: Use seesaw example. Heavy person close to center balances light person far from center. Moments are equal.

🎯 Exam Tip: Write "clockwise moments = anticlockwise moments" clearly. This is the mathematical condition for equilibrium.

 

Solution 21
Answer:

Mass	Weight
It is the measure of quantity of matter contained in the body, at rest.	It is the force with which the earth attracts a body.
It is a scalar quantity.	It is a vector quantity.
Its SI unit is kg.	Its SI unit is newton (N).
It is measured by a physical or beam balance.	It is measured by a spring balance which is calibrated to read in newton.
It is constant for a body and does not change by changing the place of a body.	It is not constant for a body, but varies from place to place on the earth's surface and also with altitude and depth from the earth's surface.

In simple words: Mass is how much matter is in something. Weight is how much gravity pulls on that matter.

📝 Teacher's Note: Mass stays same on Earth and moon. Weight changes because moon has less gravity. Use astronaut examples.

🎯 Exam Tip: Make table with clear differences. Write units correctly: kg for mass, N for weight.

 

Solution 22
Answer: Moment of force = applied force × perpendicular distance from the line of action
10 = 20 × perpendicular distance from the line of action
or, Perpendicular distance from the line of action = 0.5 m
In simple words: We know moment and force. We use division to find the distance.

📝 Teacher's Note: This is simple algebra. Rearrange the moment formula to find distance. Always check units.

🎯 Exam Tip: Show rearrangement clearly: distance = moment ÷ force. Give answer with correct unit (m).

 

Solution 23
Answer: The point through which the resultant of the weights of all the particles of the body acts is called its centre of gravity.
In simple words: Centre of gravity is where all weight seems to be concentrated. It is the balance point of any object.

📝 Teacher's Note: This is the same as Solution 18. Centre of gravity is a very important concept in physics.

🎯 Exam Tip: Use the exact definition with "resultant of weights" for full marks.

 

Solution 24
Answer: The centre of gravity of a uniform ring is situated at the centre of the ring.
In simple words: For a ring, the balance point is exactly in the middle, even though there is no material there.

📝 Teacher's Note: Show with a bangle or ring. The center is empty but that's where it balances. Center of gravity can be outside the object.

🎯 Exam Tip: Write "centre of the ring" clearly. This shows center of gravity can be outside the object.

 

Solution 25
Answer: The centre of gravity of a body depends upon:
(i) Body's weight
(ii) Body's shape
In simple words: Where the balance point is depends on how the weight is spread out in the object.

📝 Teacher's Note: Change shape of clay and balance point changes. Add weight to one side and balance point shifts.

🎯 Exam Tip: List both factors clearly. Shape means distribution of mass, weight means total mass.

 

Solution 26
Answer: Yes, the centre of gravity of a body can be outside it. The CG of a uniform ring is at its centre, a point which is not on the body.
In simple words: Sometimes the balance point is not inside the object itself, like the center of a ring or a horseshoe.

📝 Teacher's Note: Use horseshoe, ring, or boomerang as examples. The balance point can be in empty space.

🎯 Exam Tip: Give ring example to prove the point. This shows understanding of the concept.

 

Solution 27
Answer:

[Diagram: This diagram shows the center of gravity (marked as G) for different geometric shapes - square, rectangle, circle, triangle, and parallelogram. For each shape, the center is marked with dotted lines showing the geometric center.]

In simple words: Different shapes have their center of gravity at different places. Regular shapes have it at their geometric center.

📝 Teacher's Note: For regular shapes, center of gravity is at geometric center. For irregular shapes, it can be anywhere.

🎯 Exam Tip: Remember regular shapes have CG at their center. Draw dotted lines to show this in diagrams.

 

Solution 28
Answer: Centripetal force: Whenever a body is moving in a circular path with a uniform speed, its velocity is continuously changing due to change in its direction. The body thus possesses acceleration and this acceleration is called centripetal acceleration. The force which produces this acceleration is called centripetal force. It acts along the radius towards the centre of the circular path. It is not the same as the centrifugal force.
In simple words: When something moves in a circle, it needs a force pulling it toward the center. This is centripetal force.

📝 Teacher's Note: Use string and ball example. String provides centripetal force pulling ball toward center. Without it, ball flies away.

🎯 Exam Tip: Write "acts toward center" and "different from centrifugal force." These are important points.

 

Solution 29
Answer: Yes, force can be used to change the size and shape of the body. Example: On squeezing toothpaste tube, its size as well as shape changes.
In simple words: Force can make things bigger, smaller, or change their shape. Like squashing a ball or stretching rubber.

📝 Teacher's Note: Show with clay, rubber ball, or spring. Force can deform objects without moving them.

🎯 Exam Tip: Give good examples like toothpaste tube, clay, or balloon. Show force changes dimensions.

 

Solution 30
Answer: Characteristics of non-contact forces:

1. Forces at distance are equal and opposite.
2. Depend upon the distance between the two objects.
3. Depend upon the medium between the two objects for electrical and magnetic forces but not gravitational forces.

In simple words: Non-contact forces work without touching. Like magnets attracting from far away.

📝 Teacher's Note: Use magnets and gravity as examples. They act through space without touching. Distance affects strength.

🎯 Exam Tip: List all three characteristics clearly. Remember gravity doesn't depend on medium but others do.

 

Solution 31
Answer: Examples where force can start the motion of a body:

1. The pulling of a cart by a rope.
2. Pushing a door to open it.

Examples where force can stop the motion of a body:

1. Applying a force to stop a cricket ball.
2. Applying the brakes of car to stop it.

In simple words: Force can make still things move and moving things stop. It works both ways.

📝 Teacher's Note: Use everyday examples students can relate to. Football kick starts motion, catching ball stops motion.

🎯 Exam Tip: Give two examples for each case. Make sure examples are clear and different from each other.

 

Solution 32
Answer: The physical quantity is 'torque'. Torque may be defined as the turning effect produced by a force on a rigid body about a point, pivot or fulcrum. It is measured by the product of force and the perpendicular distance of the pivot from the line of action of force.
In simple words: Torque is the scientific name for turning effect. It measures how good a force is at rotating something.

📝 Teacher's Note: This is the same as moment of force. Both words mean the same thing in physics.

🎯 Exam Tip: Write complete definition mentioning "turning effect" and the formula for calculation.

 

Solution 33
Answer: Two equal and opposite parallel forces acting along different lines on a body constitute a couple. Its SI unit is 'newton'.
In simple words: A couple is two equal forces pushing in opposite directions but not along same line. Like turning a steering wheel.

📝 Teacher's Note: Show with steering wheel or turning a book. Two hands apply equal opposite forces. This creates pure rotation.

🎯 Exam Tip: Write "equal and opposite" and "parallel forces." Unit is newton, same as force.

 

Solution 34
Answer: The turning effect produced by a force on a rigid body about a point, pivot or fulcrum is called the moment of force or torque. It is measured by the product of force and the perpendicular distance of the pivot from the line of action of force.
Examples of turning effect of force:

1. Turning a steering wheel
2. Tightening a cap

In simple words: Moment is turning effect. We see it when opening jars, turning screws, or using wrenches.

📝 Teacher's Note: Students should learn both definition and practical examples. Connect theory to daily life.

🎯 Exam Tip: Give complete definition first, then add relevant examples. This shows full understanding.

 

Solution 35
Answer: A body is said to be in equilibrium under the action of a number of forces, if the forces are not able to produce any change in the state of rest or of uniform motion or uniform rotation.
In simple words: Equilibrium means perfect balance. All forces cancel out so nothing changes.

📝 Teacher's Note: This is the same definition as Solution 19. Equilibrium is a key concept students must understand well.

🎯 Exam Tip: Mention both rest and uniform motion. Object can be in equilibrium while moving at constant speed.

 

Solution 36
Answer: Equilibrium in any case requires the sum of forces acting on an object = 0, i.e. that there is \( F_{net} = 0 \).
Static equilibrium is the situation where the object upon which the forces act is not moving. The object is "static" hence the state of equilibrium gets its name.
Dynamic equilibrium is the situation where an object is in constant velocity motion. {This object can't experience an acceleration which means \( F_{net} > 0 \)}
In simple words: Static equilibrium means not moving. Dynamic equilibrium means moving at constant speed in straight line.

📝 Teacher's Note: Book on table is static equilibrium. Car at constant 60 km/h is dynamic equilibrium. Both have zero net force.

🎯 Exam Tip: Clearly distinguish static (not moving) and dynamic (constant velocity). Both need \( F_{net} = 0 \).

 

Solution 37
Answer: When the centre of gravity is nearer to the base of a body, the body is in stable equilibrium.
Conditions for stable equilibrium:

1. The body should have a broad base.
2. Centre of gravity of the body should be as low as possible.
3. Vertical line drawn from the centre of gravity should fall within the base of the support.

In simple words: For good balance, keep the heavy part low and make the base wide. Like a pyramid shape.

📝 Teacher's Note: Use examples like tower vs pyramid. Lower CG and wider base make objects more stable.

🎯 Exam Tip: List all three conditions clearly. Remember the line from CG must fall inside the base.

 

Solution 38
Answer:

1. If both the forces act at the same point of the body, they have the same line of action, and then the moment becomes zero.
2. If both the forces act at two different points of the body at a separation d then they constitute a torque whose value is given F × d.

In simple words: Two equal opposite forces at same point cancel out. At different points, they create rotation.

📝 Teacher's Note: Show with pencil. Push-pull at same point does nothing. Push-pull at different points makes it rotate.

🎯 Exam Tip: Explain both cases clearly. Same point gives zero moment, different points give torque F × d.

 

Solution 39
Answer:
(i) Moment of force = applied force × perpendicular distance from the line of action
∴ Moment of force at O = \( 8 \times 1.5 = 12 \) Nm
(ii) Moment of force = applied force × perpendicular distance from the line of action
∴ Moment of force at P = \( 8 \times 3 = 24 \) Nm
In simple words: Moment of force tells us how much turning effect a force has. It depends on both the force and how far it is from the turning point.

📝 Teacher's Note: Show students how opening a door is easier when you push far from the hinges. That distance makes the turning easier.

🎯 Exam Tip: Always write the formula first. Then substitute values clearly. Remember to include units (Nm) in your final answer.

 

Solution 40
Answer:
(i) Pulling of a cart.
(ii) A ball falls down when it is dropped from a height
In simple words: These are examples where force causes movement. The cart moves when pulled. The ball moves down due to gravity force.

📝 Teacher's Note: Ask students to give more examples from daily life. Pushing a swing, kicking a ball, lifting a bag - all show force causing movement.

🎯 Exam Tip: Give clear, simple examples. Mention both the force and the movement it causes.

 

Solution 41
Answer:
In the given figure, forces 10 N and 100 N act clockwise and the forces 15 N and 4 N act anticlockwise.
Moment of force = sum of clockwise moments - sum of anticlockwise moments
∴ Moment of force = \( (10 \times 5 + 100 \times 0 + 4 \times 6) - (15 \times 4) = 74 - 60 = 14 \) Nm
Or, 14 Nm in clockwise direction.
In simple words: We add all turning forces in one direction and subtract the turning forces in the opposite direction. The result tells us which way the object will turn.

📝 Teacher's Note: Draw a simple lever on the board. Show how forces on different sides try to turn it different ways. The stronger side wins.

🎯 Exam Tip: Always mention the direction (clockwise or anticlockwise) in your final answer. Show all calculations step by step.

 

Solution 42
Answer:
Principle of moments: If a body is in equilibrium under the action of number of force, then the sum of clockwise moments is equal to the sum of anticlockwise moments.
∴ Sum of clockwise moments = sum of anticlockwise moments
\( 40 \times 20 \) cm = mass of scale × 30 cm
or, Mass of scale = \( \frac{800}{30} = 26.7 \) g
In simple words: When something is balanced, the turning forces on both sides are equal. Like a seesaw - both sides must have equal turning power to stay level.

📝 Teacher's Note: Use a simple balance or seesaw example. When balanced, heavier object must be closer to the pivot point.

🎯 Exam Tip: State the principle first, then apply it. Show the equation clearly and solve step by step.

 

Solution 43
Answer:
Let the mass of 50 g be situated at distance 'd' from the mid-point on the right hand side.
Taking moments about the mid-point i.e. at 50 cm,
\( 80 \times 30 = (40 \times 10) + (50 \times d) \)
∴ 2400 = 400 + 50d
or, 50d = 2000
or, d = 40 cm to the right of the mid-point.
In simple words: We find where to place the 50g weight so the lever stays balanced. It must go 40 cm away from the center on the right side.

📝 Teacher's Note: Show students a simple balance. Heavier weights need to be placed closer to the center to balance lighter weights far from center.

🎯 Exam Tip: Always define your variables clearly. State which side is positive and show all substitutions step by step.

 

Solution 44
Answer:
Resultant torque = sum of clockwise moments - sum of anticlockwise moments
Taking moments about the mid-point,
Resultant torque = \( (300 \times 40) - (500 \times 20) \)
Or, Resultant torque = 12000 - 10000 = 2000 gf-cm
Let a mass of 100 gf be suspended at a distance 'd' from the mid-point towards the right side, so as to balance the metre scale.
Then, in balanced condition:
sum of clockwise moments = sum of anticlockwise moments
\( (300 \times 40) = (500 \times 20) + (100 \times d) \)
or, 12000 = 10000 + 100 d
or, 100 d = 2000
or, d = 20 cm to the right of the mid-point
In simple words: First we find which side is heavier. Then we add a weight on the lighter side at the right distance to make both sides equal.

📝 Teacher's Note: This is like adjusting a seesaw. If one side is too heavy, add weight to the other side at the right distance.

🎯 Exam Tip: Solve in two parts - first find the unbalanced torque, then find where to add the balancing weight.

 

Solution 45
Answer:
Let mass 'm' be suspended from 65 cm mark so as to balance the meter scale.
In balanced condition:
sum of clockwise moments = sum of anticlockwise moments
\( 50 \times (50 - 20) = m \times (65 - 50) \)
or, \( 50 \times 30 = m \times 15 \)
or, m = \( \frac{1500}{15} = 100 \) g
In simple words: To balance the scale, we need 100g at the 65cm mark. This creates equal turning forces on both sides.

📝 Teacher's Note: Draw a simple lever diagram. Show students how distances from the pivot point matter as much as the weights.

🎯 Exam Tip: Calculate distances from the pivot point carefully. Show the balanced condition equation clearly.

 

Solution 46
Answer:
A body is said to be in equilibrium under the action of a number of forces, if the forces are not able to produce any change in the state of rest or of uniform motion or uniform rotation.
Conditions for stable equilibrium:
(a) The body should have a broad base.
(b) Centre of gravity of the body should be as low as possible.
(c) Vertical line drawn from the centre of gravity should fall within the base of the support.
In simple words: Equilibrium means balanced - no change in movement. For stable balance, make the base wide, keep weight low, and ensure the weight acts inside the base.

📝 Teacher's Note: Show a tall thin object and a short wide object. The wide one is harder to tip over. This demonstrates stable equilibrium.

🎯 Exam Tip: List all three conditions clearly. Use examples like a pyramid (stable) versus a pencil standing on its tip (unstable).

 

Solution 47
Answer:
The center of gravity is at the intersection of lines BE and AD. The distance a can be calculated as a = h/3
In simple words: The center of gravity of a triangle is always at 1/3 height from the base. It's where all the weight seems to act from.

📝 Teacher's Note: Cut out a cardboard triangle and balance it on a pencil tip. The balance point is the center of gravity.

🎯 Exam Tip: Remember the formula a = h/3 for triangles. Draw lines from each vertex to opposite side midpoints - they meet at center of gravity.

 

Solution 48
Answer:
Diagram showing the relative position of vertices of the triangle when it is suspended by a pin from the hole A:
The position of vertices changes because the triangle is in equilibrium and the centre of gravity lies on the vertical line through the point of suspension as the weight acts along this same line.
In simple words: When you hang a triangle from one corner, it rotates until its center of gravity is directly below the hanging point. This is how it stays balanced.

📝 Teacher's Note: Hang a cardboard triangle from different corners. Each time it rotates to put the center of gravity below the hanging point.

🎯 Exam Tip: Mention that the center of gravity must lie on the vertical line through the suspension point for equilibrium.

 

Solution 49
Answer:
(i) We keep our body balanced on two feet by keeping the center of gravity of our body between our feet. It acts normal to the sea level vertically downwards. If COG goes out we fall or we get unbalanced. A boy standing on both legs has his COG in balanced position and is thus in stable equilibrium but a boy standing on one leg has his COG in unbalanced position which makes him quite unstable and hence it is easier to push him.
(ii) A man bends forward in order to keep himself in a stable equilibrium while climbing up a slope. By bending forward he increases the base of the support, so that the vertical line passing through his centre of gravity may still fall within the base.
(iii) When a truck is not fully loaded, its COG is at a high point and hence the turning moment of the weight is much greater, thus, the truck will be quite unstable and there are chances of toppling, when a truck takes a sharp turn.
(iv) When a man gets down from a moving train, his feet come to rest immediately, while the upper part of his body due to inertia of motion still remains in motion and consequently he leans in forward direction. The person while getting down of a train should run forward in the direction of the moving train to avoid fall.
(v) This is due to the fact that the body of the passenger is in the state of rest as long as the bus is at rest. When the bus starts, his feet acquire the velocity of the bus and come to motion with the moving bus, while the upper portion of his body due to inertia of rest tends to remain in the state of rest, resulting in his tendency to fall backwards.
In simple words: These are all examples of balance and movement. Our body tries to keep balance by keeping weight in the right place. When things move suddenly, our body takes time to adjust.

📝 Teacher's Note: Ask students to try standing on one leg versus two legs. They will feel the difference in stability. This makes the concept clear.

🎯 Exam Tip: For each part, mention center of gravity (COG) and explain how it affects stability. Use the terms "stable" and "unstable" equilibrium.

 

Solution 50
Answer:
To increase the stability of a body, its base should be made broad and heavy, and the centre of gravity of the body should be lowered.
In simple words: To make something harder to tip over, make the bottom wide and heavy, and keep the weight as low as possible.

📝 Teacher's Note: Compare a racing car (low and wide) with a tall truck. The racing car is much harder to tip over.

🎯 Exam Tip: Mention both conditions - broad base AND low center of gravity. Both are needed for maximum stability.

 

Solution 51
Answer:
(i) Two equal and opposite parallel forces acting along different lines on a body constitute a couple.
(ii) Moment of couple = either Force × perpendicular distance
= \( 4 \times 10 = 40 \) Nm
In simple words: A couple is two equal forces pushing in opposite directions but not in the same line. This creates pure turning without any movement forward or backward.

📝 Teacher's Note: Show students how turning a steering wheel works - your two hands apply equal and opposite forces creating a couple.

🎯 Exam Tip: Define couple clearly - equal, opposite, parallel forces on different lines. Calculate moment using any one force times the distance between the lines.

 

Solution 52
Answer:
(i) Leaning tower of Pisa is stable because a line through the centre of gravity falls within the structure's base. If the line falls outside the structure's base then there is a possibility that overturning will occur. This structure could be classified as unstable.
(ii) We bend forward in order to keep ourselves in a stable equilibrium while climbing up a hill. By bending forward we increases the base of the support, so that the vertical line passing through our centre of gravity still falls within the base.
(iii) By keeping the legs apart, the base of the body broadens, thus the C.G. lowers and the body attains a more stable equilibrium.
(iv) Passengers are usually advised not to stand in the upper deck of the double deck bus. When the passengers are standing, the C.G. rises. This decreases the stability of the bus. When the passengers are sitting, the C.G. gets lowered and stability of the bus increases.
In simple words: All these examples show how position of center of gravity affects balance. Keep it low and inside the base for good stability.

📝 Teacher's Note: Use examples students can see - why double-decker buses have "sit down" signs upstairs, why we spread legs when carrying heavy things.

🎯 Exam Tip: Always explain using center of gravity and base of support. Mention whether the line from COG falls inside or outside the base.

 

Solution 53
Answer:
The body will move in the direction of net torque.
Net torque = Sum of clockwise moments - sum of anticlockwise moments
In the given figure, force 40 N is acting clockwise and forces 20N and 60 N are acting anticlockwise.
∴ Net torque = \( (40 \times 3) - [(20 \times 2) + (60 \times 1)] \)
= 120 - 100 = 20 Nm
Hence, the body will rotate in clockwise direction about 'O'
In simple words: We find which direction has stronger turning force. The body will turn in that direction.

📝 Teacher's Note: Draw forces on a lever. Show students how to identify which forces turn clockwise and which turn anticlockwise.

🎯 Exam Tip: Calculate net torque by subtracting. Always state the direction of rotation in your final answer.

 

Solution 54
Answer:
Let the weight 500N be hung from the 40 cm mark and the weight 300 N be hung from the 80 cm mark.
(i) Then, total clockwise moment = \( 500 \times (50-40) = 500 \times 10 = 5000 \) Nm
Total anticlockwise moment = \( 300 \times (80-50) = 300 \times 30 = 9000 \) Nm
(ii) Difference in clockwise and anticlockwise moment = 9000 - 5000 = 4000 Nm
(iii) Let the 100 N weight be hung at a distance 'd' from the point 'O' to its left.
Then, total clockwise moment = 5000 + 100 d
In balanced condition, sum of clockwise moments = sum of anticlockwise moments
∴ 5000 + 100 d = 9000
or, 100 d = 4000
or, d = 40 cm
∴ the weight 100 N should be hung from the 40 cm mark so as to balance the scale.
In simple words: We calculate turning forces on both sides. To balance, we add a weight on the lighter side at the right distance.

📝 Teacher's Note: Show students how distances are measured from the pivot point (50cm mark). This is a common source of mistakes.

🎯 Exam Tip: Always measure distances from the pivot point, not from the ends. Show balanced condition equation clearly.

 

Solution 55
Answer:
Let m be the mass of the metre scale
In balanced condition, sum of clockwise moments = sum of anticlockwise moments
∴ \( 400 \times 10 = m \times 90 \)
or, 4000 = 90 m
or, m = 44.4 g
∴ the mass of the scale is 44.4 g
In simple words: When the scale is balanced, the turning force of the 400g weight equals the turning force of the scale's own weight.

📝 Teacher's Note: Explain that the scale itself has weight that acts from its center (50cm mark). This weight also creates a turning moment.

🎯 Exam Tip: Remember that the scale's weight acts from its center point. Set up the balance equation with this in mind.

 

Solution 56
Answer:
'Work' is said to be done when the applied force makes the body move i.e., there is a displacement of body. It is equal to the product of force and the displacement of the point of application of the force in the direction of force. The SI unit of work is 'joules' and the CGS unit is 'erg'.
In simple words: Work happens when force makes something move. Work = Force × Distance moved. If nothing moves, no work is done.

📝 Teacher's Note: Push against a wall - you use force but do no work because the wall doesn't move. Push a box that slides - now you do work.

🎯 Exam Tip: Define work clearly - force must cause movement. Write the formula and mention SI unit (joule) and CGS unit (erg).

 

Solution 57
Answer:
(i) [C]
(ii) [A]
(iii) [D]
(iv) [E]
(v) [B]
In simple words: These are matching pairs that go together based on their properties or relationships.

📝 Teacher's Note: Help students understand the connection between each pair. This builds their understanding of relationships between concepts.

🎯 Exam Tip: Read all options carefully before matching. Look for the most logical connections between items.

 

Solution 58
Answer:
Given, Force = 5 kgf = \( 5 \text{ kgf} \times 10 \text{ m/s}^2 = 50 \) N
displacement = 10 m
Work done = force × displacement = \( 50 \times 10 = 500 \) J
In simple words: Work is force times distance. Here 50N force moves the object 10m, so work done is 500 joules.

📝 Teacher's Note: Remind students to convert kgf to N by multiplying by 10. This is a common conversion they must remember.

🎯 Exam Tip: Always convert units first if needed. Show the formula, substitute values, and include correct units (J) in the final answer.

 

Solution 59
Answer:
Given:
mass = 20 kg
displacement = 40 m

(a) In vertical direction work is done against gravity
Work done = mgh = 20 × 9.8 × 40 = 7840 J

(b) In horizontal direction,
Work done = Fs cos θ = (mg)s cos 0° = 20 × 9.8 × 40 = 7840 J
In simple words: When we move something up, we work against gravity. When we move something sideways, we work with the force. Both give the same work here.

📝 Teacher's Note: Show students that work = force × distance. In both cases the force is the same (weight) and distance is same. So work is same.

🎯 Exam Tip: Always write "work done against gravity" for vertical motion. Show the formula clearly. Don't forget units (J).

 

Solution 60
Answer:
(a) The rate of doing work is called power.
SI unit of power = watt
CGS unit of power = erg per second

(b) Given:
mass = 50 kg
Total height traveled = 60 × 20 cm = 1200 cm = 12 m
g = 10 ms⁻²
time taken = 5 min = 5 × 60 s = 300s
Work done = mgh = 50 × 10 × 12 = 6000 J
Power = \( \frac{\text{Work done}}{\text{time taken}} = \frac{6000}{300} = 20 \text{ watt} \)
In simple words: Power tells us how fast we can do work. Like how fast a person can climb stairs. Here the person's power is 20 watts.

📝 Teacher's Note: Power is like the speed of doing work. A strong person has more power - they can do the same work in less time.

🎯 Exam Tip: Remember the formula: Power = Work/Time. Always convert time to seconds and height to meters before calculating.

 

Solution 61
Answer: The energy of a body is its capacity to do work. The SI unit of work is 'joules' and the CGS unit is 'erg'.
In simple words: Energy is like the strength stored in something. A moving car has energy. A stretched rubber band has energy. Both can do work.

📝 Teacher's Note: Tell students that energy is stored work. Like money in a bank - it can be used later to buy things. Energy can be used later to do work.

🎯 Exam Tip: Write the exact definition: "capacity to do work." Also write both SI and CGS units clearly.

 

Solution 62
Answer: When an elevator begins to move downwards in an accelerated mode, the forces acting on the body are the following:
a. Weight of the body acting downwards
b. Normal reaction of the floor acting upwards
c. The centrifugal force acting on the body, acting upwards

Weight of the body is due to gravitational force on the body acting downwards. Normal reaction is the force that is exerted by the elevator floor in response to the force with which the body presses itself against the floor. The centrifugal force here is fictitious force that acts on the body in the direction opposite to the acceleration of the reference frame, here it is, the elevator floor. It is given by ma, where m is the mass of the body & a is the acceleration of the elevator floor. Centrifugal force is directed opposite to the acceleration of the elevator floor.

Weightlessness is the condition of the zero apparent weight. When the acceleration of the elevator is such that the upward centrifugal force Fc completely balances the downward weight Wt. of the body, the resultant normal reaction (N = Fc – Wt.) of the body is reduced to zero. That's when the body on the elevator floor will experience the state of weightlessness.
In simple words: In a falling elevator, you feel weightless because the elevator and you fall together. It's like jumping - for a moment you feel no weight.

📝 Teacher's Note: Use the example of a falling lift. Tell students it's like when you jump - for a brief moment you feel weightless because you and the ground are moving apart.

🎯 Exam Tip: Write all three forces clearly. Explain that weightlessness happens when Normal force = 0. Draw a simple force diagram if possible.

 

Solution 63
Answer:
(a) Given:
Force = 500 N
Vertical displacement = Final position - initial position = BC = 4m
∴ work done = force × displacement = 500 × 4 = 2000 J

(b) P.E. gained = mgh = (mg)h = 500 × 4 = 2000 J
In simple words: When we lift something up, we do work against gravity. This work gets stored as potential energy in the object.

📝 Teacher's Note: Show students that the work done equals the potential energy gained. This follows the law of conservation of energy.

🎯 Exam Tip: Work done = Force × displacement. P.E. = mgh. Both should give the same answer. Always check this.

 

Page No-75:

 

Solution 64
Answer:
We know that, K.E = \( \frac{1}{2}mv^2 \)

∴ 375 = \( \frac{1}{2} \times 30 \times v^2 \)

or, \( v^2 = \frac{375 \times 2}{30} = 25 \text{ m/s} \)

or, v = 5m/s
In simple words: We use the kinetic energy formula to find how fast the object is moving. Like finding speed from the energy of a moving ball.

📝 Teacher's Note: Practice this formula with students using simple numbers first. Show them that heavier objects or faster objects have more kinetic energy.

🎯 Exam Tip: Write the K.E. formula first. Then substitute values step by step. Don't forget to take square root at the end.

 

Solution 65
Answer:
Given:
mass = 50 kg
Total height traveled = 40 × 20 cm = 800 cm = 8 m
g = 10 ms⁻²
time taken = 20s
Work done = mgh = 50 × 10 × 8 = 4000 J
Power = \( \frac{\text{Work done}}{\text{time taken}} = \frac{4000}{20} = 200 \text{ watt} \)
In simple words: The person climbs stairs and uses 200 watts of power. This is like a bright light bulb's power.

📝 Teacher's Note: Compare with everyday items - a bright bulb is 100W, a small heater is 1000W. This helps students understand power values.

🎯 Exam Tip: Convert cm to m and min to seconds first. Then use Work = mgh and Power = Work/Time formula.

 

Solution 66
Answer:
Given:
mass of water = 80 kg
Height = 20 m
g = 10 ms⁻²
time taken = 10s
Work done = mgh = 80 × 10 × 20 = 16000 J
Power = \( \frac{\text{Work done}}{\text{time taken}} = \frac{16000}{10} = 1600 \text{ watt} \)
In simple words: The motor lifts heavy water very fast, so it needs high power - 1600 watts. Like a powerful water pump.

📝 Teacher's Note: Water pumps in houses usually need high power because they lift heavy water to overhead tanks quickly.

🎯 Exam Tip: For lifting problems, always use Work = mgh first, then Power = Work/Time. Check that your power value makes sense.

 

Solution 67
Answer:
We know that, power = force × velocity
Here, force = 3000 N
velocity = 36 kmhr⁻¹ = 10 m/s
∴ power = 3000 × 10 = 30000 watt
In simple words: When a car moves with force and speed, it uses power. More force or more speed means more power needed.

📝 Teacher's Note: Always convert km/hr to m/s by dividing by 3.6. This is a common conversion students forget.

🎯 Exam Tip: Remember: Power = Force × Velocity. Convert km/hr to m/s first. Write the conversion clearly in your answer.

 

Solution 68
Answer: The energy of a body is its capacity to do work. The SI unit of work is 'joules' and the CGS unit is 'erg'. According to the law of conservation of energy, energy can neither be created nor be destroyed but can be transformed from one form to another. In other words, energy can be transformed from one form to another but the total amount of all the energies remain the same.
In simple words: Energy cannot disappear or appear from nothing. It only changes from one type to another. Like when you eat food, chemical energy becomes muscle energy.

📝 Teacher's Note: Use examples like a swinging pendulum - it changes between kinetic and potential energy, but total energy stays same.

🎯 Exam Tip: Write both the definition of energy and the conservation law. Use the exact words "neither created nor destroyed" for full marks.

 

Solution 69
Answer: Six forms of energy:

1. Solar energy: The energy radiated by the sun is called the solar energy. Inside the sun, energy is produced by nuclear fusion reaction. Solar energy cannot be used to do work directly, because it is too diffused and is not always uniformly available. However, a number of devices such as solar panels, solar cells etc. have been invented to make use of solar energy.
2. Heat energy: The energy released on burning coal, oil, wood or gas is the heat energy. The steam possesses heat energy it has capacity to do work.
3. Light energy: It is the form of energy in presence of which other objects are seen. The natural source of light energy is sun. Many other sources of heat energy also give light energy.
4. Chemical or fuel energy: The energy possessed by fossil fuels such as coal, petroleum and natural gas is called chemical energy or fuel energy. These fuels are formed from the decayed remains of dead plants and animals that lived millions of years ago. In the interior of earth, due to high pressure and temperature the remains slowly changed into fossil fuels.
5. Hydro energy: The energy possessed by fast moving water is called the hydro energy. This energy is used to generate electricity in hydroelectric power stations. For this, dams are built across the rivers high up in the hills to store water. Water is allowed to run down the pipes and the energy of running water is used to turn a turbine. The turbine drives generators to produce electrical energy.
6. Nuclear energy: The energy released during the processes of nuclear fission and fusion is called nuclear (or atomic) energy. In both these processes, there is loss in mass which converts into energy in accordance with Einstein's mass-energy relation, E = mc².

In simple words: Energy comes in many forms - from the sun, from burning things, from moving water, from atoms. We use different forms for different needs.

📝 Teacher's Note: Give examples students see daily - solar calculators, burning gas stoves, flowing river, battery in phone. This makes each type easy to remember.

🎯 Exam Tip: Learn all six forms by name. Write one example for each form. This question often asks for examples too.

 

Solution 70
Answer: The energy possessed by a body by virtue of its position, shape or change of configuration is called potential energy.

Examples of potential energy:
(i) Water stored at a height in a reservoir.
(ii) A stretched spring.
(iii) A bent bow.

The energy possessed by a body by virtue of its motion is called kinetic energy.

Examples of kinetic energy:
(i) Air in motion has kinetic energy.
(ii) A swinging pendulum.
(iii) Moving hands of a clock.
In simple words: Potential energy is stored energy - like water on a hill or a stretched rubber band. Kinetic energy is moving energy - like a running car or flying bird.

📝 Teacher's Note: Show a rubber band - when stretched it has P.E., when released it has K.E. This simple demo makes both types very clear.

🎯 Exam Tip: Learn the definitions by heart. Give at least 2-3 examples for each type. Position, shape, motion are key words to remember.

 

Solution 71
Answer:
K.E. of the bullet = \( \frac{1}{2}mv^2 \)

(a) If the mass is doubled,
K.E.' = \( \frac{1}{2}(2m)v^2 = 2 \times \frac{1}{2}mv^2 = 2K.E \)
∴ \( \frac{K.E.'}{K.E} = \frac{2}{1} \)

(b) If the velocity is tripled,
K.E.' = \( \frac{1}{2}m(3v)^2 = 9 \times \frac{1}{2}mv^2 \)
∴ \( \frac{K.E.'}{K.E} = \frac{9}{1} \)
In simple words: When mass doubles, kinetic energy doubles. When speed triples, kinetic energy becomes 9 times because speed is squared in the formula.

📝 Teacher's Note: Emphasize that velocity has more effect on K.E. because it is squared. A small increase in speed gives big increase in energy.

🎯 Exam Tip: Remember K.E. depends on v². So if velocity becomes n times, K.E. becomes n² times. Mass has direct effect only.

 

Solution 72
Answer:
Given:
mass m = 400 kg, height = 20 m, g = 10 ms⁻²
P.E. = mgh = (400) (10) (20) = 80000 J
In simple words: A heavy object at height has a lot of potential energy stored in it. Here it's 80000 J - enough to power a light bulb for many hours.

📝 Teacher's Note: Help students see that height and mass both increase P.E. A heavy stone at great height has maximum P.E.

🎯 Exam Tip: Use P.E. = mgh formula. Make sure mass is in kg, height in meters, g = 9.8 or 10 ms⁻². Units are always Joules.

 

Solution 73
Answer:
(a) Potential energy
(b) Potential energy
(c) Kinetic energy
(d) Potential energy
(e) Kinetic energy
In simple words: Things at rest in high places have P.E. Things that are moving have K.E. Stored energy is P.E., moving energy is K.E.

📝 Teacher's Note: Ask students to identify what makes each example have P.E. (position, height) or K.E. (motion, movement).

🎯 Exam Tip: Look for keywords - height, position, stretched = P.E. Moving, flowing, rotating = K.E. This helps identify quickly.

 

Solution 74
Answer:
[Diagram: This diagram shows different energy conversion devices - a dynamo converting mechanical to electrical energy, a thermocouple converting electrical to heat energy, speakers and microphones converting between electrical and sound energy, batteries converting chemical to electrical energy, and candles/photosynthesis showing chemical to light energy conversions.]
In simple words: Different machines change one type of energy into another type. Like a battery changes chemical energy to electrical energy.

📝 Teacher's Note: Point to real devices in class - fan (electrical to mechanical), bulb (electrical to light), speaker (electrical to sound). Students see these daily.

🎯 Exam Tip: Learn the energy conversions for common devices. Exams often ask about generators, motors, batteries, bulbs, and speakers.

 

Solution 75
Answer:
(a) P.E. (Water stored in the dam) → K.E. (falling water) → Electrical energy (Generators of turbine form electrical energy).
(b) Electrical energy (Electric current) → Light energy + heat energy (bulb).
(c) Chemical energy (chemical reaction) → Heat energy (burning).
(d) P.E. (Stone at the top of cliff) → K.E. (Falling stone) → P.E. (Stone at ground).
(e) P.E (Wound spring) to K.E. (motion)
In simple words: Energy keeps changing from one form to another. Like water in a dam becomes electricity, and electricity becomes light in a bulb.

📝 Teacher's Note: Draw simple energy flow diagrams on the board. Show arrows going from one energy type to the next. This helps students visualize the changes.

🎯 Exam Tip: Use arrow notation (→) to show energy conversions clearly. Write the device name in brackets to show what causes the conversion.

 

Page No-76:

 

Solution 76
Answer:
(a) Electric bell
(b) Candle flame
(c) Dry cell
In simple words: Different devices convert energy in different ways. Electric bell makes sound, candle makes light, battery stores chemical energy.

📝 Teacher's Note: Ask students to name the energy conversion for each - bell (electrical to sound), candle (chemical to light), cell (chemical to electrical).

🎯 Exam Tip: Think about what each device does. Bell rings (sound), candle glows (light), battery powers things (electrical). This helps remember the answers.

 

Solution 77
Answer:
Given:
Mass, m = 1000 kg
Velocity, v = 72 km/hr = 20 m/s

Step 1: Calculate kinetic energy using the formula.
K.E. = \( \frac{1}{2}mv^2 = \frac{1}{2}(1000)(20)^2 \)

Step 2: Solve for kinetic energy.
K.E. = \( 2 \times 10^5 \) J

In simple words: We used the kinetic energy formula to find how much energy the moving object has. The answer is 200,000 Joules.

📝 Teacher's Note: Always convert km/hr to m/s by dividing by 3.6. Students often forget this step. Show them: 72 ÷ 3.6 = 20 m/s.

🎯 Exam Tip: Write the formula first, then substitute values. Don't forget to convert units. Write the final answer in scientific notation if it's a big number.

 

Solution 78
Answer:
Given:
Mass, m = 25 g = \( \frac{25}{1000} \) kg = \( \frac{1}{40} \) kg
Initial velocity, u = 600 m/s
Final velocity, v = 0 m/s
Distance travelled, s = 50 cm = 0.5 m

Step 1: Calculate initial kinetic energy of the bullet.
K.E. = \( \frac{1}{2}mu^2 = \frac{1}{2}\left(\frac{1}{40}\right)(600)^2 \)
K.E. = 4500 J

Step 2: Apply work-energy theorem.
Work done = Change in K.E.
Work done = \( \frac{1}{2}m(v-u)^2 = \frac{1}{2}\left(\frac{1}{40}\right)(0 - 600)^2 = -4500 \) J

Step 3: Find resistive force using work formula.
Work done = Force × displacement
-4500 = F × 0.5
F = \( \frac{-4500}{0.5} = -9000 \) N

Result: The resistive force is 9000 N.

In simple words: The bullet had energy when moving fast. When it stopped, all that energy was used to fight against the resistive force. We calculated that force.

📝 Teacher's Note: Explain that negative sign shows force opposes motion. The magnitude of resistive force is 9000 N. Students often get confused about the sign.

🎯 Exam Tip: Always convert grams to kilograms and cm to meters first. Write "resistive force is 9000 N" as your final answer. Don't write negative value for the final answer.

 

Solution 79
Answer:
Given:
Initial velocity, u = 70 m/s
Final velocity, v = 0 m/s
Specific heat capacity of lead, s = 140 J/kgK
Let Δt be the change in temperature

Step 1: Calculate kinetic energy of the bullet.
K.E. = \( \frac{1}{2}m(70)^2 = 2450m \) joules

Step 2: Find heat energy produced.
Heat energy = 80% of K.E. = \( \frac{80}{100} \times 2450m = 1960m \) joules

Step 3: Use heat formula to find temperature change.
H = msΔt
1960m = m × 140 × Δt
Δt = \( \frac{1960}{140} = 14 \) °C

In simple words: When the bullet stops, most of its moving energy becomes heat energy. This heat makes the bullet warmer by 14 degrees.

📝 Teacher's Note: Show students that the mass cancels out in the calculation. This means any lead bullet will have the same temperature rise. This surprises many students.

🎯 Exam Tip: Write "80% of kinetic energy converts to heat" clearly. Don't forget the degree symbol (°C) in your final answer.

 

Solution 80
Answer:
Given:
Specific heat capacity of water, s = 4200 J/kgK
Temperature difference, Δt = 0.21 °C
Let 'h' be the height of the waterfall and 'm' be the mass of water

Step 1: Calculate potential energy of water.
P.E. = mgh

Step 2: Given that heat energy = 60% of P.E.
msΔt = \( \frac{60}{100} \times mgh \)

Step 3: Substitute values and solve for height.
4200 × 0.21 = 0.6 × 10 × h
h = \( \frac{4200 \times 0.21}{0.6 \times 10} = \frac{882}{6} = 147m \)

In simple words: Water falls down and hits the bottom. Some of its falling energy becomes heat energy. This makes the water slightly warmer. We used this to find the waterfall height.

📝 Teacher's Note: Explain that falling water has potential energy that converts to kinetic energy, then to heat when it hits the bottom. Only 60% becomes heat because some energy is lost in other ways.

🎯 Exam Tip: Write "60% of potential energy converts to heat energy" in your solution. Always cancel out the mass 'm' in your working. Final answer should be "147 m".

 

Solution 81
Answer:
(a) Simple machine: 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 speed.

(b) Lever: A lever is a rigid, straight or bent bar which is capable of turning about a fixed axis.

(c) Mechanical advantage (M.A.): The ratio of the load to the effort is called the mechanical advantage of the machine.

(d) Velocity ratio (V.R.): The ratio of the velocity of effort to the velocity of load is called the velocity ratio of the machine. It is also defined as the ratio of the displacement of effort to the displacement of load.

(e) Efficiency: Efficiency of a machine is the ratio of the useful work done by the machine to the work put into the machine by the effort. In other words, it is the ratio of the work output to the work input.

In simple words: Simple machines help us do work more easily. A lever is like a seesaw that can turn around a point. These definitions tell us how well the machine works.

📝 Teacher's Note: Show real examples - scissors for lever, bottle opener for simple machine. Students understand better when they see and touch actual objects.

🎯 Exam Tip: Learn the exact definitions word by word. Examiners give full marks only when you write the complete definition exactly as given.

 

Solution 82
Answer: Example of class I lever with M.A. = 1: A physical balance has both arms equal (i.e. effort arm = load arm), thus its M.A. = 1

Example of class I lever with M.A. < 1: A pair of scissors used to cut a piece of cloth has blades longer than the handle (i.e. effort arm is shorter than the load arm), thus its M.A. < 1.

Example of class I lever with M.A. > 1: Shears used for cutting thin metal sheets have much longer handles as compared to the blades (i.e. effort arm is longer than the load arm), thus its M.A. > 1 and it serves as a force multiplier.

In simple words: When both arms of a lever are equal, M.A. = 1. When effort arm is shorter, M.A. < 1. When effort arm is longer, M.A. > 1.

📝 Teacher's Note: Bring a physical balance, scissors, and pliers to class. Show students how the arm lengths affect the mechanical advantage. This makes the concept very clear.

🎯 Exam Tip: Always mention which arm is longer and shorter. Write the exact relationship between arm lengths and M.A. value. Give specific examples like balance, scissors, pliers.

 

Solution 83
Answer:
(a) Diagram showing the direction of application of least force on the handle:

[Diagram: A lever (spade) with fulcrum near one end. Load of 30 N is shown at the short arm (20 cm from fulcrum). Effort is applied at the long arm (20 cm from fulcrum to the load point). Total length is marked as 20 cm + 10 cm. The effort direction is shown vertically upward.]

(b) Given:
Load, L = 30 N
Let effort = E
Effort arm = 20 cm
Load arm = 20 + 10 = 30 cm

Using principle of moments:
Load × load arm = effort × effort arm
∴ effort = \( \frac{30 \times 30}{20} = 45 N \)

(c) On moving the left hand towards the soil on the spade, the length of effort arm will increase and effort being inversely proportional to the length of effort arm, the force or effort necessary to keep the soil balanced would be less.

(d) A spade belongs to Class III lever.

In simple words: The spade works like a lever. When you hold it closer to the blade, you need more force. When you hold it farther from the blade, you need less force.

📝 Teacher's Note: Show students an actual spade or garden tool. Let them try lifting soil by holding at different positions. They will feel the difference in force needed.

🎯 Exam Tip: Draw the diagram clearly showing fulcrum, load, and effort positions. Write "Class III lever" for the spade. Show your calculation step by step with proper units.

 

Solution 84
Answer: A crowbar is a class I lever, thus, its fulcrum lies in between the load and effort.

Given:
Load, L = 500 kgf
Length of crowbar = 4 m = 400 cm
Therefore, load arm = 50 cm
Effort arm = 400 - 50 cm = 350 cm
Let effort = E

Step 1: Calculate mechanical advantage.
M.A. = \( \frac{\text{Effort arm}}{\text{Load arm}} = \frac{350}{50} = 7 \)

Step 2: Find effort using M.A. formula.
M.A. = \( \frac{L}{E} \)
∴ \( 7 = \frac{500}{E} \)
Or, E = 71.4 kgf

In simple words: A crowbar makes work easier. You apply small force at the long end to lift a heavy load at the short end. The mechanical advantage tells us how much easier it becomes.

📝 Teacher's Note: Show students that in Class I lever, the fulcrum is between load and effort. Draw this clearly. Explain why longer effort arm gives more advantage.

🎯 Exam Tip: Always identify the lever class first. Calculate effort arm by subtracting load arm from total length. Write final answer with proper units (kgf).

 

Solution 85
Answer: Let 'L' be the load.

Given:
Load arm = 3 cm
Effort arm = 12 cm
Effort, E = 10 gf

Step 1: Calculate mechanical advantage.
M.A. = \( \frac{\text{Effort arm}}{\text{Load arm}} = \frac{12}{3} = 4 \)

Step 2: Find load using M.A. formula.
M.A. = \( \frac{L}{E} \)
∴ \( 4 = \frac{L}{10} \)
L = 40 gf

In simple words: The mechanical advantage is 4, which means this lever can lift a load 4 times heavier than the effort applied.

📝 Teacher's Note: Remind students that mechanical advantage = effort arm ÷ load arm for levers. Also, M.A. = load ÷ effort. Both formulas give the same result.

🎯 Exam Tip: Use both formulas to check your answer. If M.A. from arm ratio = M.A. from load/effort ratio, your answer is correct. Write units clearly (gf).

 

Solution 86
Answer:
(a) Correct winding of rope around pulleys, load and effort is as shown:

[Diagram: A pulley system with 4 pulleys arranged vertically. A rope winds around all pulleys. Load hangs from the bottom movable pulley. Effort is applied upward on one end of the rope.]

(b) Since there are four strings in the block and tackle system, therefore, the velocity ratio of the system is V.R. = n = 4

(c) (i) Given:
Load, L = 720 N
Let Effort = E

Efficiency = \( \frac{M.A.}{V.R.} \)
80% = \( \frac{L/E}{4} \)
∴ \( \frac{80}{100} \times 4 = \frac{720}{E} \)
E = \( \frac{720 \times 100}{80 \times 4} = 225 N \)

(ii) Given:
Load, L = 720 N
Displacement of load, d₁ = 2 m

Efficiency = \( \frac{\text{work output}}{\text{work input}} \)
Work input = \( \frac{\text{work output}}{\text{Efficiency}} = \frac{\text{Load} \times d_1}{\eta} \)
Work input = \( \frac{720 \times 2}{0.8} = 1800 J \)

In simple words: A pulley system with 4 supporting strings makes lifting easier. The velocity ratio is 4, meaning you pull rope 4 times the distance the load moves up.

📝 Teacher's Note: Count the number of rope segments supporting the load to find V.R. In this case, 4 segments support the load, so V.R. = 4. Efficiency tells us how much energy is wasted.

🎯 Exam Tip: Draw the pulley diagram clearly showing rope winding. Count supporting strings for V.R. Always convert percentage to decimal (80% = 0.8) for calculations.