NEET Physics — Rotational Motion
Torque – Complete Notes, Revision, Important Questions & Downloads
Torque (τ) is the rotational equivalent of force — the turning effect of a force about an axis. NEET tests two subtopics: Definition and Components (τ = r × F, τ = rF sinφ, maximum when φ = 90°, zero when φ = 0° or 180°) and Couple (two equal, opposite, non-collinear forces that produce pure rotation with no net translation). Common NEET questions include identifying the condition for maximum torque, comparing the torque of two forces acting at different angles, and applying the couple concept to distinguish net force from net torque.
5 SubtopicsMedium FrequencyCross-Product Based
Expected QuestionsQ
1
Torque typically generates 1 NEET question per year, often testing the angle dependence of torque or the couple concept.
Time Required⏱
1–2 hours
One session to derive the cross-product formula and understand angle dependence, one short session to distinguish couple from single-force torque.
Difficulty⚡
Medium
The cross-product direction (right-hand rule for axial vector) needs practise; the rest is algebraic substitution into τ = rF sinφ.
NRI USA Curriculum GapUS
Low
AP Physics covers torque (τ = rF sinθ) thoroughly. The couple concept is less emphasised in AP but is conceptually straightforward.
5Subtopics
8+Practice Questions
4Free Downloads
1–2 hrsPrep Time
NEET Weightage — Torque
Rotational Motion (Chapter 7)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 1 | 4 | |
| 2023 | 1 | 4 | |
| 2022 | 0 | 0 | |
| 2021 | 1 | 4 | |
| 2020 | 1 | 4 | |
| 2019 | 0 | 0 | |
| 6-Year Total (2019–2024) | 3–5 | 12–20 |
Angle dependence of torque (τ = rF sinφ) appears as 'find the angle for maximum torque' — answer is always 90°, giving τ_max = rF.
Couple questions test whether net force = 0 even though net torque ≠ 0 — do not confuse the two conditions for equilibrium.
Direction of torque (using right-hand rule or screw rule) is tested conceptually — if r and F are in the x-y plane, τ points along ±z.
Couple questions test whether net force = 0 even though net torque ≠ 0 — do not confuse the two conditions for equilibrium.
Direction of torque (using right-hand rule or screw rule) is tested conceptually — if r and F are in the x-y plane, τ points along ±z.
~0.8
Avg Questions / Year
12–20
Total Marks (6 yrs)
Direct
Pattern
Medium
Difficulty
Exam Strategy for Torque
1
Identify the pivot axis and perpendicular distance first τ = F × d where d is the perpendicular distance from the axis to the line of action of the force (the moment arm). Equivalently, τ = rF sinφ where φ is the angle between r and F vectors. Always draw a free-body diagram and mark the pivot, the force vector, and angle φ before computing.
2
Apply the right-hand rule to find torque direction Curl the fingers of the right hand from r toward F through the smaller angle. The extended thumb points in the direction of τ = r × F. If r = xî and F = yĵ, then τ = xy(î × ĵ) = xyk̂ (out of the page). NEET tests this as 'what is the direction of the torque' — give it as +z, −z, into the page, or out of the page.
3
For a couple: compute torque = F × d (separation), not F × 0 A couple consists of two forces of equal magnitude but opposite direction, acting along different lines separated by distance d. Net force = 0 (forces cancel). Net torque = F × d about any point — the torque of a couple is the same about any axis, which is a key property NEET has tested.
Download Study Notes — Torque
PDF · Cheat Sheet · MCQ Set · PYQTorque — Full Notes
Complete notes covering cross-product derivation of τ = r × F, angle dependence, units (N·m), dimensional analysis, direction rules, couple definition, torque of a couple about any axis, and 6 worked NEET-style problems.
Download PDFTorque — Formula Sheet
One-page reference: τ = r × F, |τ| = rF sinφ, τ_max = rF (φ = 90°), torque direction (right-hand rule), couple torque = F × d, relation τ = Iα.
Download PDFTorque — MCQ Practice
10 NEET-style MCQs on torque direction, angle dependence, couple problems (net force vs net torque), and torque in equilibrium conditions.
Download PDFTorque — NEET-Style PYQ Practice
NEET-style practice questions on torque — including cross-product, couple definition, distinguishing torque from force, and application to rotating bodies.
Download PDFSubtopics in Torque
2-Column TableRapid Revision — Torque
Concept → Trap → Example1) Definition and Components
Cross-Product + Angle Dependenceτ = r × F (vector). |τ| = rF sinφ, where φ = angle between r and F. Unit: N·m. Dimension: [ML²T⁻²]. τ is an axial vector perpendicular to the plane of r and F.
- τ_max = rF when φ = 90° (force perpendicular to position vector). τ = 0 when φ = 0° or 180° (force along or opposite to r — passes through the pivot).
- The perpendicular distance from pivot to line of action of force is called the moment arm (d = r sinφ). Thus τ = Fd — multiply force by its moment arm.
- Common NEET trap: treating N·m (torque) as the same unit as Joule (energy). Both have the same dimensional formula [ML²T⁻²], but torque and work are different physical quantities — torque is a vector, work is a scalar.
Example (NEET-style)A force of 10 N acts at a position vector r = 2 m at 60° to the force. τ = 2 × 10 × sin60° = 20 × 0.866 = 17.32 N·m. If the force were perpendicular to r, τ_max = 2 × 10 = 20 N·m. So the 60° case gives 86.6% of maximum torque.
2) Couple
Pure Rotation, Zero Net ForceA couple = two equal and opposite forces (magnitude F each) separated by perpendicular distance d. Net force = 0. Net torque = F × d (same about any point in any frame).
- A couple produces pure rotation without any translational acceleration — the body's CM does not accelerate, only rotates.
- The torque of a couple is the same about any axis (frame-independence): τ_couple = F × d for any choice of pivot. This distinguishes it from a single force whose torque depends on the chosen axis.
- Common NEET trap: stating that a couple has net torque zero because two forces are equal and opposite. Net force = 0, but net torque ≠ 0 unless d = 0 (i.e., the forces are collinear). A couple's torque is non-zero precisely because the forces act along different parallel lines.
Example (NEET-style)Steering wheel: two hands apply forces of 20 N each at opposite ends of a 0.4 m diameter wheel (d = 0.4 m). Torque = 20 × 0.4 = 8 N·m. Net force on the steering column = 0 (hands push in opposite directions). The wheel rotates without the column being pushed sideways.
US Curriculum Gaps — Torque
Students from the US system studying for NEET should note these specific coverage gaps:AP Physics 1 covers basic torque but not the couple concept
AP Physics 1 defines torque as τ = rF sinθ and introduces static equilibrium using Στ = 0. However, the couple — two equal-opposite forces separated by a distance producing pure rotation — is generally not covered in AP Physics 1 and is rarely tested in AP Physics C Mechanics either.
- Study the couple definition separately: Couple = forces of equal magnitude, opposite directions, along parallel but non-collinear lines of action.
- Practise the key property: torque of a couple is the same about any point — this is distinctly different from a single force whose torque depends on the pivot.
- NEET has tested 'what is the net force on a body subject to a couple?' — answer is zero; this is the defining property of a couple.
AP Physics C does not emphasise dimensional analysis of torque vs work
In US courses, the distinction between N·m (torque, a vector) and J (work/energy, a scalar) is not explicitly drilled as an exam topic. NEET has historically used this as a distractor in multiple-choice options.
- Torque: N·m = kg·m²·s⁻², an axial vector, perpendicular to the plane of rotation.
- Work: J = N·m = kg·m²·s⁻², a scalar, equal to F·d cosθ (dot product).
- Both have the same SI unit symbol (N·m = J) and dimensional formula [ML²T⁻²], but they are physically distinct — torque is not energy.
NEET-Style Practice Questions — Torque
5 NEET-style questions1A force F⃗ = 3î + 4ĵ N acts at a position r⃗ = 2î + 3ĵ m. What is the torque about the origin?Cross-Product Calculation
9k̂ − 8k̂ = k̂ N·m
(9 − 8)k̂ = 1k̂ N·m
−(9 − 8)k̂ = −1k̂ N·m
(3×4 − 2×3 ... = (12 − 6 − 8 + 6) — actually (2×4 − 3×3)k̂ = (8−9)k̂ = −1k̂ N·m
τ = r × F: τz = rx·Fy − ry·Fx = 2×4 − 3×3 = 8 − 9 = −1 N·m. So τ⃗ = −1k̂ N·m (into the page). The trap is computing r × F as rx·Fx + ry·Fy (which is the dot product, r·F, not the cross product). For 2D problems in the x-y plane, only the z-component of torque is non-zero, and τz = rx·Fy − ry·Fx.
2A pivoted rod of length 2 m experiences a force of 6 N at its free end. At what angle to the rod should the force be applied to produce the maximum torque?Maximum Torque Condition
30°
45°
60°
90°
τ = rF sinφ is maximum when sinφ = 1, i.e., φ = 90°. At 90°, all of the force acts perpendicular to the lever arm with no wasted component along the rod. τ_max = 2 × 6 × sin90° = 12 N·m. At 30°: τ = 12 × 0.5 = 6 N·m (half the maximum). At 60°: τ = 12 × 0.866 = 10.4 N·m.
3Two equal and opposite forces of magnitude 15 N act on a rod at points 0.4 m apart. What is the torque of the couple about the midpoint of the rod?Couple Torque
0 N·m
6 N·m
3 N·m
12 N·m
Torque of a couple = F × d = 15 × 0.4 = 6 N·m. This is independent of the choice of pivot — the torque about any point equals 6 N·m. The trap is computing each force's torque about the midpoint separately and then cancelling them: from the midpoint, each force has moment arm 0.2 m and both torques act in the same rotational direction (both clockwise or both counterclockwise). Total = 15×0.2 + 15×0.2 = 6 N·m. They add, not cancel.
4A spanner applies a torque of 12 N·m to a bolt. If the wrench is 0.3 m long and the force applied is perpendicular to the wrench, what is the magnitude of force?Torque Formula Application
36 N
3.6 N
40 N
4 N
τ = rF sinφ. With φ = 90°: F = τ/r = 12/0.3 = 40 N. The force is perpendicular to the wrench (sinφ = 1), so this is the minimum force needed to produce 12 N·m. Any other angle would require a larger force. The trap is computing 12 × 0.3 = 3.6 N (multiplying instead of dividing).
5A body is acted upon by a couple. Which of the following statements is correct?Couple — Conceptual
Net force = 0, net torque = 0
Net force ≠ 0, net torque = 0
Net force = 0, net torque ≠ 0
Net force ≠ 0, net torque ≠ 0
A couple consists of two equal and opposite forces, so net force = F + (−F) = 0. The forces act along different parallel lines separated by distance d, so net torque = F × d ≠ 0 (assuming d ≠ 0). A body under a couple has no translational acceleration (ΣF = 0) but does have rotational acceleration (Στ = Iα ≠ 0). This makes a couple useful for starting rotation without shifting the centre of mass.
Practice Questions — Torque
Click "Reveal Answer" after attempting1A force of 8 N acts at a point 1.5 m from a pivot at an angle of 30° to the line joining the pivot to the point. Find the torque.
6 N·m
12 N·m
8 N·m
4 N·m
👁 Reveal Answer
Answer: 6 N·m. τ = rF sinφ = 1.5 × 8 × sin30° = 12 × 0.5 = 6 N·m.
2A couple consists of two forces of 25 N each, separated by 0.6 m. What is the torque of the couple? Does it change if the pivot is moved to one end of the separation distance?
15 N·m, no change
15 N·m, changes to 15 N·m still
7.5 N·m, no
15 N·m, doubles to 30 N·m
👁 Reveal Answer
Answer: 15 N·m, and it does not change regardless of pivot location. Torque of couple = F × d = 25 × 0.6 = 15 N·m. At one end: torque₁ = 25×0 = 0 (force passes through pivot), torque₂ = 25×0.6 = 15 N·m. Total = 15 N·m — same value. This axis-independence is the key property of a couple.
3Two forces act on a rod: 10 N upward at x = 0 and 10 N downward at x = 0.5 m. What is the net force and net torque about x = 0?
Net F = 0, τ = 5 N·m
Net F = 20 N, τ = 0
Net F = 0, τ = 0
Net F = 10 N, τ = 5 N·m
👁 Reveal Answer
Answer: Net F = 0, τ = 5 N·m. Net force = 10 N up + 10 N down = 0. Torque about x = 0: force at x = 0 contributes 0 (zero moment arm). Force at x = 0.5 m: 10 N × 0.5 m = 5 N·m. Total torque = 5 N·m. This is a couple — net force zero, non-zero torque.
4A 60 cm arm is used to open a valve. If the required torque is 30 N·m and the force is applied at 45° to the arm, what minimum force is needed?
50 N
70.7 N
35.4 N
42.4 N
👁 Reveal Answer
Answer: ~70.7 N. τ = rF sinφ → F = τ/(r sinφ) = 30/(0.6 × sin45°) = 30/(0.6 × 0.707) = 30/0.424 ≈ 70.7 N. At 90° (perpendicular), only 50 N would be needed — applying at 45° wastes half the force in the along-arm direction, requiring ×√2 more force.
Physics — Rotational Motion Revision Checklist
Check off chapters as you revise
Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
FAQ — Torque
Notes · Downloads · Revision · Important QuestionsWhy is torque an axial vector rather than a polar vector?
Torque τ = r × F is defined as the cross product of two polar vectors (r and F). Cross products of two polar vectors always produce axial (pseudovector) vectors: they transform like vectors under proper rotations but change sign under reflections/inversions. This means the direction of torque is determined by the right-hand rule and reverses under a mirror reflection of the coordinate system, unlike true (polar) vectors such as displacement or force.
If torque and work both have units N·m, why are they different quantities?
Torque is τ = r × F (cross product) — an axial vector of dimension [ML²T⁻²]. Work is W = F · d (dot product) — a scalar with the same dimensional formula and SI unit (J = N·m). They arise from different products of the same vectors: the cross product gives a directed quantity (torque) while the dot product gives a scalar (energy). NEET explicitly tests this distinction — do not call torque a unit of energy.
Can torque be zero even when both force and position vector are non-zero?
Yes — τ = rF sinφ = 0 whenever sinφ = 0, i.e., φ = 0° or 180°. This means the force acts along the line joining the point of application to the pivot. For example, a central force (like gravity from a planet) always passes through the chosen pivot (the planet's centre), giving zero torque and thus conserving angular momentum.
Why does a couple produce the same torque about any point?
τ_couple = r₁ × F + r₂ × (−F) = (r₁ − r₂) × F. Since r₁ − r₂ = the vector joining the two points of application (the separation vector d), and this vector is independent of the choice of origin, τ_couple = d × F is the same for any pivot. Numerically, τ = F × |d sinφ| = F × (perpendicular separation), which is a fixed geometric property of the couple.
What is the difference between torque and the moment of a force?
'Moment of a force' and 'torque' refer to the same physical quantity: τ = r × F. In engineering and statics, 'moment' is the more common term. In rotational dynamics, 'torque' is preferred. NEET uses both terms interchangeably. The moment arm d = r sinφ is the perpendicular distance from the pivot to the line of action, giving τ = F × d.
How does the Newton's second law for rotation relate to torque?
For a rigid body rotating about a fixed axis: Στ = Iα, where I = moment of inertia and α = angular acceleration. This is the rotational analogue of ΣF = ma. The total external torque produces angular acceleration inversely proportional to the moment of inertia — a body with greater I requires greater torque to achieve the same α.
Can a single force produce the same effect as a couple?
No. A couple's net force is zero, so it cannot produce translational acceleration. A single force always has a non-zero net force vector (unless F = 0), and thus always produces some translational effect in addition to any rotational effect. If you want pure rotation without translation, you must use a couple — this is why doorknobs and steering wheels are designed to apply a couple (force pair) rather than a single offset force.
What happens to the torque when the same force is applied at different points along the same line of action?
The torque changes because the moment arm changes. τ = rF sinφ — if the angle φ between r and F is fixed but r increases (force applied farther from pivot), τ increases proportionally. However, if the force moves along its own line of action without changing direction, the perpendicular distance from the pivot to the line of action (moment arm d) does NOT change, so the torque remains the same. This is the principle of transmissibility of force: a force can be moved along its line of action without changing the torque about any given pivot.
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