Chapter 4: Turning Effect of ForcesMCQs with Answers
All 24 MCQs of 9th Class Physics Chapter 4, Turning Effect of Forces, with answers and the reason each option is right or wrong: the 10 textbook exercise MCQs plus 14 more board-style MCQs. Punjab Board new syllabus.
Try each one first: tap an option to check it, then read the reasoning and why the others are wrong.
A note about the questions
The wording follows published solved exercises and every answer was re-checked by hand, so the wording or the order of options may differ slightly from your print.
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Set A: Textbook exercise 10 MCQs
The multiple-choice questions at the end of Chapter 4 in your book, in the book's option order.
Q4.1Parallel Forces and Adding Forces
A particle is simultaneously acted upon by two forces of 4 N and 3 N. The net force on the particle is:
Answer(b) between 1 N and 7 N
Step by step
Forces in the same direction add: 4 + 3 = 7 N (the largest possible resultant).
Forces in opposite directions subtract: 4 − 3 = 1 N (the smallest possible resultant).
At any other angle the resultant lies between these, for example 5 N at right angles.
The angle is not given, so the answer is "between 1 N and 7 N".
Why the others are wrong
(a) 1 N is only the special case when the forces are exactly opposite.
(c) 5 N is only the case when the forces are at right angles (3-4-5 triangle).
(d) 7 N is only the case when both forces act in the same direction.
Remember
Resultant of two forces: from (difference) to (sum).
Q4.2Resolution of Forces
A force F is making an angle of 60° with the x-axis. Its y-component is equal to:
Answer(b) F sin 60°
Step by step
Draw F at θ = 60° to the x-axis and drop a perpendicular to make a right-angled triangle.
The side along the x-axis is F cos θ; the side parallel to the y-axis (opposite θ) is F sin θ.
So Fy = F sin 60° (about 0.866 F).
Why the others are wrong
(a) F is the whole force; a component is always smaller (unless the force lies along that axis).
(c) F cos 60° is the x-component.
(d) tan is not used to find a component; tan θ = Fy ÷ Fx gives the direction.
Remember
Angle from the x-axis: x gets cos, y gets sin.
Q4.3Torque (Moment of a Force)
Moment of force is called:
Answer(d) torque
Step by step
Moment of a force = force × moment arm (τ = F × l).
This turning effect is named torque, written τ (tau), SI unit N m.
Why the others are wrong
(a) the moment arm is the perpendicular distance l, only one part of the moment.
(b) a couple is a pair of equal, opposite, parallel forces; it produces torque but is not its name.
(c) the couple arm is the distance between the two forces of a couple.
Remember
Moment of force = torque.
Q4.4Equilibrium
If F₁ and F₂ are the forces acting on a body and τ is the torque produced in it, the body will be completely in equilibrium when:
Answer(a) ΣF = 0 and Στ = 0
Step by step
First condition: ΣF = 0, so the body has no linear acceleration.
Second condition: Στ = 0, so the body has no rotational (angular) acceleration.
Only when both hold is the body completely in equilibrium.
Why the others are wrong
(b) with a resultant torque the body would start to rotate (for example a couple).
(c) with a resultant force the body would accelerate along a line.
(d) neither condition is satisfied.
Remember
Complete equilibrium: both sums zero.
Q4.5Principle of Moments
A shopkeeper sells his articles by a balance having unequal arms of the pans. If he puts the weights in the pan having the shorter arm, then the customer:
Answer(a) loses
Step by step
Let the weights W hang on the short arm a, and the goods G on the long arm b (b > a).
Principle of moments at balance: W × a = G × b.
So G = W × (a ÷ b). Since a ÷ b is less than 1, G is less than W.
The customer pays for W but receives the smaller G: he loses.
Why the others are wrong
(b) the customer would gain only if the weights were on the longer arm.
(c) that is true only for equal arms.
(d) the principle of moments tells us exactly who loses.
Remember
The longer arm needs less weight to balance; goods on the longer arm means fewer goods.
Q4.6Principle of Moments
A man walks on a tight rope. He balances himself by holding a bamboo stick horizontally. It is an application of:
Answer(c) principle of moments
Step by step
A tipping body has an unbalanced turning moment about the rope.
The long stick lets a small shift of weight act at a large distance, making a large opposite moment.
He keeps clockwise and anticlockwise moments equal: the principle of moments.
Why the others are wrong
(a) no collision or explosion is involved.
(b) the second law links force and acceleration; balancing is about turning effects.
(d) action and reaction do not explain how he stays upright.
Remember
Balancing = equal clockwise and anticlockwise moments.
Q4.7States of Equilibrium and Stability
In stable equilibrium, the centre of gravity of the body lies:
Answer(b) at the lowest position
Step by step
Tilt a body in stable equilibrium slightly: its centre of gravity rises.
Its weight then makes a moment that brings it back down.
So the resting position is the one with the centre of gravity at its lowest.
Why the others are wrong
(a) the highest position belongs to unstable equilibrium (a pencil on its tip).
(c) any position at the same height is neutral equilibrium.
(d) G may be outside some bodies (a ring), but that is not what makes them stable.
Remember
Stable = lowest G; unstable = highest G; neutral = G stays at the same height.
Q4.8Centre of Mass and Centre of Gravity
The centre of mass of a body:
Answer(d) may lie within, outside or on the surface
Step by step
Solid sphere or block: the centre of mass is inside the material.
Ring or hollow pipe: the centre of mass is at the centre, where there is no material.
So it may lie within, outside, or on the surface.
Why the others are wrong
(a) a ring's centre of mass is not in the material.
(b) a solid ball's centre of mass is inside it.
(c) it is on the surface only for some special shapes.
Remember
Think of a ring: its centre of mass is in the hole.
Q4.9States of Equilibrium and Stability
A cylinder resting on its circular base is in:
Answer(a) stable equilibrium
Step by step
Standing on its flat base, the cylinder's centre of gravity is above the middle of the base.
Tilt it slightly onto an edge: the centre of gravity rises and stays above the base.
Its weight brings it back to the upright position, so it is in stable equilibrium.
Why the others are wrong
(b) it does not fall over when tilted slightly.
(c) neutral is a cylinder lying on its curved side, which rolls and stays wherever it is put. The question says "on its circular base", which is the flat end.
(d) it is stable.
Remember
Upright on its base: stable. Lying on its side: neutral.
Q4.10Centripetal Force
Centripetal force is given by:
Answer(c) mv²/r
Step by step
Centripetal force grows with mass and with the square of the speed.
It is smaller for a larger circle.
So Fc = mv² ÷ r.
Why the others are wrong
(a) rF (with r and F perpendicular) is a torque.
(b) torque uses sin θ (rF sin θ), and it is still a torque, not a centripetal force.
(d) mv/r has the wrong units for a force; the speed must be squared.
Remember
Centripetal: m v squared over r.
Set B: More MCQs from this chapter 14 MCQs
Extra questions on the same ideas, of the kind set in board papers and class tests.
QB1Torque (Moment of a Force)
The SI unit of torque is:
Answer(c) N m
Step by step
τ = F × l.
Unit: N × m = N m (newton metre).
Why the others are wrong
(a) N m⁻¹ divides by the distance instead of multiplying.
(b) N s is the unit of impulse.
(d) kg m s⁻¹ is the unit of momentum.
Remember
Force × distance: newton metre.
QB2Torque (Moment of a Force)
A force of 20 N is applied at right angles to the end of a spanner 0.5 m long. The torque on the nut is:
Answer(b) 10 N m
Step by step
The force is perpendicular, so the moment arm is the whole length, 0.5 m.
τ = F × l = 20 × 0.5 = 10 N m.
Why the others are wrong
(a) 40 divides 20 by 0.5 instead of multiplying.
(c) 20.5 adds the numbers.
(d) 0.025 divides 0.5 by 20.
Remember
Torque = force × moment arm (in metres).
QB3Torque (Moment of a Force)
A force produces no torque about an axis when:
Answer(a) its line of action passes through the axis
Step by step
τ = F × l, where l is the perpendicular distance from the axis to the line of action.
If the line of action passes through the axis, l = 0.
So τ = 0: pushing a door straight at its hinges does not turn it.
Why the others are wrong
(b) a larger force gives a larger torque, unless the moment arm is zero.
(c) a long moment arm gives a large torque.
(d) a perpendicular force gives the largest torque.
Remember
Push at the hinge: nothing turns.
QB4Resolution of Forces
A force of 100 N acts at 30° to the horizontal. Its horizontal component is:
Answer(d) 86.6 N
Step by step
Fx = F cos θ.
cos 30° = 0.866.
Fx = 100 × 0.866 = 86.6 N.
Why the others are wrong
(a) 50 N is F sin 30°, the vertical component.
(b) 57.7 N is F tan 30°, which is not a component.
(c) 100 N is the whole force.
Remember
cos 30° ≈ 0.866, sin 30° = 0.5.
QB5Resolution of Forces
The rectangular components of a force are Fx = 6 N and Fy = 8 N. The magnitude of the force is:
Answer(d) 10 N
Step by step
The components are at right angles, so use Pythagoras.
F = √(6² + 8²) = √(36 + 64) = √100.
F = 10 N.
Why the others are wrong
(a) 14 N adds the components as if they were in the same direction.
(b) 2 N subtracts them.
(c) 48 multiplies them.
Remember
6-8-10 is a doubled 3-4-5 triangle.
QB6Principle of Moments
A 20 N weight hangs 30 cm from the pivot of a balanced metre rule. What weight hung 20 cm from the pivot on the other side keeps it balanced?
Answer(c) 30 N
Step by step
Clockwise moment = anticlockwise moment.
20 × 30 = w × 20.
w = 600 ÷ 20 = 30 N. (The units of distance cancel, so cm can be used on both sides.)
Why the others are wrong
(a) 13.3 N puts the distances the wrong way round (20 × 20 ÷ 30).
(b) 600 is the moment in N cm, not the weight.
(d) 40 N does not balance: 40 × 20 = 800, not 600.
Remember
Closer to the pivot needs a bigger weight.
QB7Couple
A couple acting on a body produces:
Answer(b) rotational motion only
Step by step
ΣF = F − F = 0: the body does not move as a whole.
The forces are on different lines, so Στ = F × d ≠ 0.
So the body only rotates.
Why the others are wrong
(a) the resultant force of a couple is zero, so there is no translation.
(c) the zero resultant force rules out translation.
(d) the torque of the couple turns the body.
Remember
A couple turns but does not shift.
QB8Centre of Mass and Centre of Gravity
The centre of gravity of a uniform ring lies:
Answer(d) at its centre, where there is no material
Step by step
For a uniform symmetrical body, the centre of gravity is at its geometric centre.
The centre of a ring is in the hole.
So the centre of gravity lies outside the material.
Why the others are wrong
(a) no single point on the ring is the centre of the ring.
(b) the ring is symmetrical, so G is not at the top.
(c) the ring is symmetrical, so G is not at the bottom.
Remember
Ring, bangle, tyre: G is in the hole.
QB9Equilibrium
A car moving with uniform velocity on a straight road is in:
Answer(a) dynamic equilibrium
Step by step
Equilibrium means no acceleration.
Uniform velocity in a straight line: acceleration = 0, so ΣF = 0.
It is moving, so it is dynamic (not static) equilibrium.
Why the others are wrong
(b) static equilibrium is for a body at rest.
(c) unstable equilibrium is about what happens after a small tilt, not about moving at constant velocity.
(d) zero acceleration means it is in equilibrium.
Remember
At rest: static. Uniform velocity: dynamic.
QB10States of Equilibrium and Stability
A ball lying on a flat, horizontal floor is in:
Answer(c) neutral equilibrium
Step by step
Push the ball slightly so it rolls.
Its centre of gravity neither rises nor falls.
It stays wherever it stops: neutral equilibrium.
Why the others are wrong
(a) it does not come back to where it was.
(b) it does not move further away on its own.
(d) the forces on it balance, so it is in equilibrium.
Remember
Neutral: G at the same height, stays in its new place.
QB11States of Equilibrium and Stability
Racing cars are built low and wide mainly to:
Answer(d) lower their centre of gravity and widen their base
Step by step
A body topples when the vertical line from its centre of gravity falls outside its base.
A low G and a wide base keep that line inside the base even at a large tilt.
So the car stays stable on fast turns.
Why the others are wrong
(a) the design is for stability, not looks.
(b) racing cars are kept as light as possible.
(c) grip comes from tyres; being low and wide is about not toppling.
Remember
Stability: low G, wide base.
QB12Centripetal Force
A body moves in a circle of fixed radius. If its speed is doubled, the centripetal force needed becomes:
Answer(c) four times
Step by step
Fc = mv² ÷ r with m and r fixed, so Fc ∝ v².
New force = m(2v)² ÷ r = 4 × mv² ÷ r.
So the force becomes four times as large.
Why the others are wrong
(a) the force increases, it does not decrease.
(b) doubling would be true if Fc ∝ v; the speed is squared.
(d) Fc depends on speed.
Remember
Speed squared: double the speed, four times the force.
QB13States of Equilibrium and Stability
A body is tilted slightly. Its centre of gravity rises and it comes back to its original position. The body was in:
Answer(a) stable equilibrium
Step by step
Tilting raises G.
The weight then makes a moment that turns the body back.
It returns to its original position: stable equilibrium.
Why the others are wrong
(b) an unstable body moves further away when tilted, and its G falls.
(c) a neutral body stays in its new position.
(d) dynamic equilibrium is about moving with uniform velocity.
Remember
Comes back = stable.
QB14Torque (Moment of a Force)
Door handles are fixed far from the hinges so that:
Answer(b) a small force gives a large torque
Step by step
τ = F × l.
The farther the handle from the hinge, the larger l.
For the same torque, a larger l needs a smaller F.
Why the others are wrong
(a) the position of the handle does not change the door's weight.
(c) the hinges carry the same weight wherever the handle is.
(d) the handle is there to turn the door, not to keep it still.
