Chapter 7
Chapter 7 MCQs
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
1
A car travels around a circular track at constant speed.
Which statement is correct?
A The velocity is constant because the speed is constant.
B The resultant force is zero because the speed is constant.
C The car is accelerating because its direction of motion is changing.
D The car is not accelerating because its kinetic energy is constant.
2
A stone is tied to a string and whirled in a horizontal circle at constant speed.
What is the direction of the resultant force on the stone?
A away from the centre of the circle
B towards the centre of the circle
C in the direction of motion
D opposite to the direction of motion
3
A satellite moves in a circular orbit around Earth at constant speed.
Which row is correct?
| speed | velocity | resultant force | |
|---|---|---|---|
| A | constant | constant | zero |
| B | constant | changing | towards Earth |
| C | changing | constant | towards Earth |
| D | changing | changing | zero |
4
A cyclist moves around a circular bend at constant speed. The frictional force between the tyres and the road provides the force needed for circular motion.
What happens if the cyclist enters the bend at a greater speed, with the radius of the bend unchanged?
A less frictional force is needed
B no frictional force is needed
C greater frictional force is needed
D the resultant force becomes zero
5
A car moves around a circular bend. The radius of the path is decreased while the speed and mass remain constant.
What happens to the force needed to keep the car moving in the circle?
A it decreases
B it increases
C it remains unchanged
D it becomes zero
6
Two objects move in circular paths with the same speed and radius. Object X has twice the mass of object Y.
Which statement is correct?
A X needs a smaller resultant force than Y.
B X needs the same resultant force as Y.
C X needs a larger resultant force than Y.
D X needs no resultant force because speed is constant.
7
A child moves on a roundabout at constant speed. She moves from a position near the centre to a position near the edge. The speed remains the same.
What happens to the force needed for circular motion?
A it decreases
B it increases
C it stays the same
D it becomes equal to weight
8
A ball moves in a vertical circle on a string. At the top of the circle, the ball is moving horizontally.
Which force or forces may contribute to the resultant force towards the centre at the top?
A tension only
B weight only
C tension and weight
D normal contact force only
9
A stone tied to a string is moving in a circle. The string suddenly breaks.
Ignoring air resistance, what is the initial path of the stone after the string breaks?
A directly towards the centre
B directly away from the centre
C tangent to the circle
D vertically downwards immediately
10
A car travels at constant speed around a circular track. Which quantity remains constant?
A velocity
B acceleration direction
C speed
D resultant force direction relative to Earth
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
11
A force of 18 N acts at a perpendicular distance of 0.25 m from a pivot.
What is the moment of the force about the pivot?
A 4.5 N m
B 18.25 N m
C 72 N m
D 7.2 N m
12
A force of 40 N produces a moment of 12 N m about a pivot.
What is the perpendicular distance from the pivot to the line of action of the force?
A 0.30 m
B 3.3 m
C 28 m
D 480 m
13
A force acts at a distance of 60 cm from a pivot and produces a moment of 9.0 N m.
What is the force?
A 0.15 N
B 5.4 N
C 15 N
D 540 N
14
A spanner is used to turn a nut. The force is applied at 30° to the spanner. The force is 80 N and the length of the spanner is 0.25 m.
The perpendicular distance from the pivot to the line of action of the force is 0.125 m.
What is the moment?
A 10 N m
B 20 N m
C 40 N m
D 160 N m
15
A student pushes a door with a force of 25 N. The force acts perpendicular to the door at a distance of 0.80 m from the hinge.
What moment is produced about the hinge?
A 20 N m
B 31 N m
C 45 N m
D 0.032 N m
16
A force of 50 N is applied to a door at a distance of 0.90 m from the hinge. The force is directed straight towards the hinge.
What is the moment about the hinge?
A 0 N m
B 45 N m
C 56 N m
D 450 N m
17
A uniform beam is pivoted at its centre. A 12 N weight is placed 0.40 m to the left of the pivot. A 16 N weight is placed 0.20 m to the right of the pivot.
Which statement is correct?
A clockwise moment is greater
B anticlockwise moment is greater
C moments are equal
D beam weight must be included
18
A uniform metre rule is pivoted at the 50 cm mark. A 3.0 N weight is placed at the 20 cm mark.
Where should a 2.0 N weight be placed to balance the rule?
A 5 cm mark
B 35 cm mark
C 80 cm mark
D 95 cm mark
19
A uniform metre rule is pivoted at the 50 cm mark. A 4.0 N weight is placed at the 10 cm mark. A 6.0 N weight is placed on the other side.
At which mark should the 6.0 N weight be placed for equilibrium?
A 23 cm
B 77 cm
C 83 cm
D 90 cm
20
A uniform beam is pivoted at its centre. A force of 30 N acts downwards 0.20 m to the left of the pivot. A force of 20 N acts downwards 0.40 m to the right of the pivot.
Which way does the beam turn?
A clockwise
B anticlockwise
C it remains balanced
D it moves vertically upwards
21
A beam is pivoted at one end. A load of 80 N is placed 0.60 m from the pivot. An effort is applied 1.5 m from the pivot to balance the load.
What effort is needed?
A 32 N
B 48 N
C 120 N
D 200 N
22
A non-uniform beam is balanced on a pivot. No extra weights are attached.
What can be concluded?
A the pivot is at the centre of gravity of the beam
B the pivot is at the centre of the beam
C the beam has no weight
D the beam must be uniform
23
A uniform ruler of weight 1.2 N is pivoted at the 40 cm mark. The ruler is 100 cm long.
What is the moment of the ruler’s weight about the pivot?
A 0.012 N m clockwise
B 0.12 N m clockwise
C 0.12 N m anticlockwise
D 1.2 N m clockwise
24
A uniform metre rule of weight 2.0 N is pivoted at the 30 cm mark. A weight of 3.0 N is placed at the 10 cm mark.
What weight must be placed at the 70 cm mark to balance the rule?
A 0.50 N
B 1.0 N
C 1.5 N
D 2.5 N
25
A uniform beam of weight 10 N is 2.0 m long and is pivoted at one end. A force F is applied upwards at the other end. The beam is horizontal and in equilibrium.
What is F?
A 2.5 N
B 5.0 N
C 10 N
D 20 N
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
26
A uniform plank of length 4.0 m and weight 120 N is supported at its centre. A child of weight 300 N stands 0.80 m to the left of the support.
Where must a second child of weight 200 N stand for equilibrium?
A 0.53 m to the right
B 1.2 m to the right
C 1.6 m to the right
D 2.4 m to the right
27
A uniform beam of weight 50 N is 3.0 m long. It is pivoted 1.0 m from the left end. A 30 N weight is hung from the left end.
What downward force must be applied at the right end to balance the beam?
A 5.0 N
B 10 N
C 15 N
D 20 N
28
A uniform metre rule of weight 1.0 N is pivoted at the 60 cm mark. A 2.0 N weight is attached at the 20 cm mark.
Where must a 5.0 N weight be attached to balance the rule?
A 34 cm mark
B 44 cm mark
C 76 cm mark
D 82 cm mark
29
A beam is in equilibrium under the action of several parallel forces.
Which condition must be true?
A clockwise moments are greater than anticlockwise moments
B anticlockwise moments are greater than clockwise moments
C total clockwise moment equals total anticlockwise moment
D total upward force must be zero
30
A metre rule is pivoted at the 50 cm mark. A 2.0 N weight is placed at 30 cm. A 3.0 N weight is placed at 70 cm.
Which extra force balances the rule?
A 1.0 N at 10 cm
B 1.0 N at 90 cm
C 2.0 N at 90 cm
D 3.0 N at 30 cm
31
A uniform rod of weight 6.0 N is pivoted at one end. A force of 4.0 N acts vertically upwards at the other end. The rod is horizontal.
Which statement is correct?
A The rod turns clockwise.
B The rod turns anticlockwise.
C The rod is in rotational equilibrium.
D The rod has zero weight moment.
32
A uniform beam of length 2.4 m and weight 18 N is pivoted at its centre. A 12 N load is placed 0.80 m from the pivot on the left side.
A 6.0 N load is placed on the right side.
Where must the 6.0 N load be placed for equilibrium?
A 0.40 m from pivot
B 0.80 m from pivot
C 1.2 m from pivot
D 1.6 m from pivot
33
A beam is balanced. A student calculates clockwise and anticlockwise moments using distances from the left end of the beam rather than distances from the pivot.
Why is this wrong?
A Moment depends on distance from the pivot.
B Moment depends on total length only.
C Moment depends on mass only.
D Moment must always use distance from the centre of gravity.
34
A force of 100 N acts on a lever. The line of action of the force passes through the pivot.
What is the moment of the force about the pivot?
A 0 N m
B 1 N m
C 100 N m
D cannot be determined
35
A force is applied to a lever. The distance from the pivot to the point where force is applied is 0.50 m. The perpendicular distance from the pivot to the force’s line of action is 0.30 m.
The force is 60 N.
What is the moment?
A 18 N m
B 30 N m
C 36 N m
D 120 N m
36
A non-uniform plank of weight 80 N is supported by a pivot. Its centre of gravity is 0.30 m to the right of the pivot. A weight of 20 N is placed 1.2 m to the left of the pivot.
Which additional force is needed to balance the plank?
A 40 N downwards 0.60 m to the right
B 40 N upwards 0.60 m to the right
C 80 N downwards 0.30 m to the left
D no additional force is needed
37
A uniform beam is balanced by two forces. A 5.0 N force acts 0.40 m to the left of the pivot. A force F acts 0.25 m to the right.
What is F?
A 2.0 N
B 3.1 N
C 8.0 N
D 12.5 N
38
A beam is balanced. A 10 N force acts 0.30 m left of a pivot. A 6.0 N force acts 0.40 m right of the pivot.
Where should a 3.0 N force act to balance the beam?
A 0.20 m to the left
B 0.20 m to the right
C 2.0 m to the left
D 2.0 m to the right
39
A uniform ladder leans against a wall. The ladder is about to topple.
Which change makes it more stable?
A raise its centre of gravity
B narrow its base
C lower its centre of gravity
D move its centre of gravity outside the base
40
An object is stable when tilted slightly and released.
What happens to its centre of gravity?
A it rises, so the object returns to its original position
B it falls, so the object returns to its original position
C it stays outside the base
D it becomes zero
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
41
A tall narrow object is less stable than a short wide object because it has:
A lower centre of gravity and wider base
B higher centre of gravity and narrower base
C lower centre of gravity and narrower base
D higher centre of gravity and wider base
42
A bus is more likely to topple when passengers stand on the upper deck because:
A the total weight becomes zero
B the centre of gravity is raised
C the base area becomes wider
D the normal reaction disappears
43
A block rests on a horizontal surface. It is slowly tilted. It topples when:
A its weight becomes zero
B its centre of gravity rises
C the line of action of its weight passes outside its base
D the normal contact force becomes larger than weight
44
A cone rests on its circular base.
Which change makes it least stable?
A placing it on its side
B placing it on its circular base
C placing it on its pointed tip
D increasing the radius of its base
45
A plane lamina is suspended freely from a pin and a plumb line is hung from the same point.
Where is the centre of gravity?
A at the suspension point
B somewhere on the vertical line below the suspension point
C always at the geometrical centre of the lamina
D at the lowest point of the lamina
46
A student suspends an irregular lamina from point P and marks the vertical line shown by a plumb line. The student then suspends it from point Q and marks a second vertical line.
Where is the centre of gravity?
A at point P
B at point Q
C where the two vertical lines intersect
D halfway between P and Q
47
A uniform triangular lamina is suspended from one corner.
Which statement is correct when it comes to rest?
A Its centre of gravity lies vertically below the suspension point.
B Its centre of gravity lies horizontally from the suspension point.
C Its centre of gravity must be at the lowest corner.
D Its centre of gravity moves to the suspension point.
48
An irregular lamina is suspended from a point. It comes to rest.
Why does the centre of gravity lie vertically below the suspension point?
A otherwise the weight would produce a turning effect
B otherwise the weight would become zero
C otherwise the mass would change
D otherwise the lamina would have no area
49
A uniform square sheet has a small circular hole cut near one corner.
What happens to the centre of gravity of the remaining sheet?
A it moves towards the hole
B it moves away from the hole
C it remains at the centre of the original square
D it moves to the nearest edge automatically
50
A rectangular block is placed on a slope. The block is just about to topple but has not yet slipped.
Which statement is correct?
A The line of action of weight passes through the centre of the base.
B The line of action of weight passes through the lower edge of the base.
C The frictional force is zero.
D The centre of gravity is outside the object.
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
Answer Key
| Q | Ans | Q | Ans | Q | Ans | Q | Ans | Q | Ans |
|---|---|---|---|---|---|---|---|---|---|
| 1 | C | 11 | A | 21 | A | 31 | B | 41 | B |
| 2 | B | 12 | A | 22 | A | 32 | D* | 42 | B |
| 3 | B | 13 | C | 23 | B | 33 | A | 43 | C |
| 4 | C | 14 | A | 24 | A | 34 | A | 44 | C |
| 5 | B | 15 | A | 25 | B | 35 | A | 45 | B |
| 6 | C | 16 | A | 26 | B | 36 | D | 46 | C |
| 7 | A | 17 | B | 27 | No correct option: 2.5 N | 37 | C | 47 | A |
| 8 | C | 18 | D | 28 | No correct option: 78 cm | 38 | B | 48 | A |
| 9 | C | 19 | B | 29 | C | 39 | C | 49 | B |
| 10 | C | 20 | A | 30 | No correct option | 40 | A | 50 | B |
*Q32 gives 1.6 m from the pivot, but this is outside a 2.4 m beam because maximum distance from centre pivot is 1.2 m. Mathematically D, physically impossible on that beam.
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
Detailed Explanations
1. C
- The car is travelling at constant speed, but its direction keeps changing.
- Velocity depends on speed and direction.
- Since velocity changes, the car is accelerating.
- A and B are wrong because constant speed does not mean constant velocity in circular motion.
2. B
- In circular motion, the resultant force acts towards the centre of the circle.
- It is not outwards.
- It is not along the direction of motion.
- The force changes the direction of motion, not necessarily the speed.
3. B
- Satellite speed may be constant.
- Velocity changes because the direction changes continuously.
- The resultant force is gravitational force, directed towards Earth.
- So: speed constant, velocity changing, resultant force towards Earth.
4. C
- Greater speed requires greater force for circular motion.
- Same mass and same radius, but speed increases.
- Therefore greater frictional force is needed.
- If friction is insufficient, the cyclist skids outwards.
5. B
- Smaller radius means the object turns more sharply.
- With same mass and speed, more resultant force is needed.
- So the required force increases.
6. C
- Same speed and radius.
- Greater mass requires greater force for circular motion.
- Object X has twice the mass, so it needs a larger resultant force.
7. A
- Moving closer to the edge means radius increases.
- Same speed and mass.
- Larger radius means less force is needed for circular motion.
- So the force decreases.
8. C
- At the top of a vertical circle, the centre of the circle is below the ball.
- Weight acts downwards.
- Tension also acts downwards towards the centre.
- Both tension and weight may contribute to the resultant force towards the centre.
9. C
- If the string breaks, the force towards the centre disappears.
- The stone continues in the direction of its velocity at that instant.
- That direction is tangent to the circle.
- It does not fly directly outwards. That’s the cartoon-physics trap.
10. C
- Speed remains constant.
- Velocity changes because direction changes.
- The resultant force direction changes as the car moves around the circle.
- So only speed remains constant.
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
11. A
- Moment = force × perpendicular distance from pivot
- Moment = 18 × 0.25
- Moment = 4.5 N m
12. A
- Moment = force × perpendicular distance
- Distance = moment / force
- Distance = 12 / 40
- Distance = 0.30 m
13. C
- Distance = 60 cm = 0.60 m
- Moment = force × distance
- Force = moment / distance
- Force = 9.0 / 0.60
- Force = 15 N
14. A
- Use the perpendicular distance, not the full spanner length.
- Moment = 80 × 0.125
- Moment = 10 N m
- The 0.25 m is a trap because the force is not perpendicular to the spanner.
15. A
- Moment = force × perpendicular distance
- Moment = 25 × 0.80
- Moment = 20 N m
16. A
- The force is directed straight towards the hinge.
- Its line of action passes through the pivot.
- Perpendicular distance from pivot = 0
- Moment = 0 N m
17. B
- Left side:
- moment = 12 × 0.40 = 4.8 N m anticlockwise
- Right side:
- moment = 16 × 0.20 = 3.2 N m clockwise
- Anticlockwise moment is greater.
- Beam turns anticlockwise.
18. D
- 3.0 N weight at 20 cm.
- Pivot at 50 cm.
- Distance = 30 cm = 0.30 m
- Moment = 3.0 × 0.30 = 0.90 N m
- For 2.0 N weight:
- distance = 0.90 / 2.0 = 0.45 m = 45 cm
- It must be placed 45 cm to the right of 50 cm.
- Position = 95 cm mark
19. B
- 4.0 N weight at 10 cm.
- Pivot at 50 cm.
- Distance = 40 cm.
- Moment = 4.0 × 40 = 160 N cm
- For 6.0 N:
- distance = 160 / 6.0 = 26.7 cm
- Position = 50 + 26.7 = 76.7 cm
- Closest = 77 cm
20. A
- Left moment:
- 30 × 0.20 = 6.0 N m anticlockwise
- Right moment:
- 20 × 0.40 = 8.0 N m clockwise
- Clockwise moment is greater.
- Beam turns clockwise.
21. A
- Load moment = effort moment
- 80 × 0.60 = F × 1.5
- 48 = 1.5F
- F = 48 / 1.5
- F = 32 N
22. A
- If a beam balances on a pivot with no extra weights, the pivot is under its centre of gravity.
- For a non-uniform beam, the centre of gravity is not necessarily at the geometrical centre.
23. B
- Uniform ruler weight acts at 50 cm.
- Pivot is at 40 cm.
- Distance = 10 cm = 0.10 m
- Moment = 1.2 × 0.10
- Moment = 0.12 N m
- Weight is to the right of pivot, so it causes a clockwise moment.
24. A
- Pivot = 30 cm.
- 3.0 N at 10 cm:
- distance = 20 cm
- anticlockwise moment = 3.0 × 20 = 60 N cm
- Ruler weight 2.0 N acts at 50 cm:
- distance = 20 cm
- clockwise moment = 2.0 × 20 = 40 N cm
- Unknown weight W at 70 cm:
- distance = 40 cm
- clockwise moment = 40W
- Balance:
- 40 + 40W = 60
- 40W = 20
- W = 0.50 N
25. B
- Beam weight = 10 N acting at centre.
- Centre is 1.0 m from pivot.
- Upward force F acts 2.0 m from pivot.
- Balance moments:
- F × 2.0 = 10 × 1.0
- F = 5.0 N
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
26. B
- Plank is supported at its centre, so plank weight causes no moment.
- Left child moment:
- 300 × 0.80 = 240 N m
- Right child:
- 200 × d = 240
- d = 240 / 200
- d = 1.2 m to the right
27. No correct option
- Beam length = 3.0 m.
- Pivot is 1.0 m from left end.
- Beam centre is 1.5 m from left end.
- Beam weight acts 0.5 m to the right of pivot.
- Clockwise moment from beam:
- 50 × 0.5 = 25 N m
- 30 N weight at left end is 1.0 m from pivot.
- Anticlockwise moment:
- 30 × 1.0 = 30 N m
- Extra downward force at right end is 2.0 m from pivot.
- Required extra clockwise moment:
- 30 − 25 = 5 N m
- F × 2.0 = 5
- F = 2.5 N
- 2.5 N is not listed, so the MCQ options are faulty.
28. No correct option
- Pivot = 60 cm.
- 2.0 N at 20 cm:
- distance = 40 cm
- anticlockwise moment = 2.0 × 40 = 80 N cm
- Metre rule weight = 1.0 N at 50 cm:
- distance = 10 cm
- anticlockwise moment = 1.0 × 10 = 10 N cm
- Total anticlockwise moment = 90 N cm.
- 5.0 N must be on the right:
- 5.0 × d = 90
- d = 18 cm
- Position = 60 + 18 = 78 cm
- 78 cm is not listed, so the options are faulty.
29. C
- For rotational equilibrium:
- total clockwise moment = total anticlockwise moment
- Upward force does not have to be zero.
- The correct condition for moments is equality of clockwise and anticlockwise moments.
30. No correct option
- Pivot = 50 cm.
- 2.0 N at 30 cm:
- distance = 20 cm
- anticlockwise moment = 2 × 20 = 40 N cm
- 3.0 N at 70 cm:
- distance = 20 cm
- clockwise moment = 3 × 20 = 60 N cm
- Clockwise moment is greater by 20 N cm.
- Need an extra anticlockwise moment of 20 N cm.
- None of the listed options provides exactly 20 N cm in the correct direction.
- Example correction:
- 1.0 N at 30 cm would give 20 N cm anticlockwise.
- or 0.5 N at 10 cm would give 20 N cm anticlockwise.
31. B
- Beam weight acts at the centre:
- clockwise moment = 6.0 × L/2 = 3L
- Upward force at the end:
- anticlockwise moment = 4.0 × L = 4L
- Anticlockwise moment is greater.
- Rod turns anticlockwise.
32. D
- Left moment:
- 12 × 0.80 = 9.6 N m
- Right load:
- 6.0 × d = 9.6
- d = 1.6 m
- So mathematically answer = D
- But the beam length is 2.4 m and pivot is at centre, so the maximum distance to an end is 1.2 m.
- Therefore the 6.0 N load cannot actually be placed on the beam at 1.6 m. This is physically impossible as written.
33. A
- Moment depends on perpendicular distance from the pivot.
- Measuring from the left end gives wrong turning effects unless the pivot happens to be at the left end.
- The pivot is the turning point. No pivot, no moment calculation.
34. A
- If the line of action passes through the pivot, perpendicular distance = 0.
- Moment = force × perpendicular distance
- Moment = 100 × 0
- Moment = 0 N m
35. A
- Use perpendicular distance, not distance to where the force is applied.
- Moment = 60 × 0.30
- Moment = 18 N m
36. D
- Plank weight:
- 80 N acts 0.30 m right of pivot
- moment = 80 × 0.30 = 24 N m clockwise
- Extra 20 N weight:
- 20 N acts 1.2 m left of pivot
- moment = 20 × 1.2 = 24 N m anticlockwise
- Moments are already equal.
- No additional force is needed.
37. C
- Left moment = right moment
- 5.0 × 0.40 = F × 0.25
- 2.0 = 0.25F
- F = 8.0 N
38. B
- Left moment:
- 10 × 0.30 = 3.0 N m anticlockwise
- Right moment:
- 6.0 × 0.40 = 2.4 N m clockwise
- Need extra clockwise moment:
- 3.0 − 2.4 = 0.6 N m
- 3.0 N force needs distance:
- d = 0.6 / 3.0 = 0.20 m
- Clockwise means force should act to the right of pivot.
- Answer = 0.20 m to the right
39. C
- Lowering the centre of gravity makes an object more stable.
- Raising the centre of gravity makes toppling easier.
- A wider base also improves stability.
40. A
- A stable object returns to its original position when tilted slightly.
- This happens because tilting raises its centre of gravity.
- When released, the centre of gravity falls back to a lower position.
- So the object returns.
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
41. B
- Tall narrow objects are less stable because:
- centre of gravity is higher
- base is narrower
- The line of action of weight can pass outside the base more easily.
42. B
- Passengers on the upper deck raise the bus’s centre of gravity.
- A higher centre of gravity makes the bus less stable.
- It becomes more likely to topple when turning or tilting.
43. C
- A block topples when the line of action of its weight passes outside its base.
- Until then, it may return or remain supported.
- The weight does not become zero.
44. C
- A cone balanced on its pointed tip has a tiny base.
- Its centre of gravity is high above the support.
- This makes it least stable.
- Increasing base radius makes it more stable, not less.
45. B
- When suspended freely, a lamina comes to rest with its centre of gravity vertically below the suspension point.
- The centre of gravity lies somewhere on the plumb line.
- One suspension alone gives a line, not the exact point.
46. C
- Suspend the lamina from two different points.
- Draw the vertical line each time using a plumb line.
- The centre of gravity is where the two lines intersect.
47. A
- When any lamina is suspended and comes to rest, its centre of gravity lies vertically below the suspension point.
- It does not move to the suspension point.
- It is not necessarily the lowest corner.
48. A
- If the centre of gravity were not vertically below the suspension point, the weight would produce a turning effect.
- The lamina would rotate until the centre of gravity is vertically below the point of suspension.
- Then there is no turning effect.
49. B
- Removing material near one corner removes mass from that side.
- The centre of gravity shifts away from the removed mass.
- So it moves away from the hole.
50. B
- Just before toppling, the line of action of weight passes through the lower edge of the base.
- Once it passes outside the base, the block topples.
- This is the “edge of doom” moment. Proper physics villain arc.
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
Common Traps From This Chapter
| Trap | Correct Rule |
|---|---|
| Constant speed in circle | velocity changes because direction changes |
| Circular motion force | towards centre |
| String breaks | object moves tangentially |
| Larger speed in circle | greater force needed |
| Larger radius at same speed | smaller force needed |
| Larger mass in circle | greater force needed |
| Moment formula | force × perpendicular distance |
| Force through pivot | zero moment |
| Use distance from pivot | never from beam end unless pivot is there |
| Uniform beam weight | acts at centre |
| Principle of moments | clockwise moment = anticlockwise moment |
| Non-uniform beam balancing alone | pivot under centre of gravity |
| Stable object tilted | centre of gravity rises |
| Toppling condition | line of action of weight passes outside base |
| More stable object | lower centre of gravity and wider base |
| Plumb line method | centre of gravity at intersection of vertical lines |
| Hole cut in object | centre of gravity moves away from removed material |