Torque — the moment of a force — is the rotational analogue of force. NCERT Class 11 Physics Chapter 6 (System of Particles and Rotational Motion), page 95 defines it as the cross product of the position vector and the applied force: τ = r × F. The magnitude is τ = rF sin θ, where θ is the angle between r and F.
Three quantities control torque magnitude: (1) the force magnitude F, (2) the distance r from the pivot, and (3) the angle θ between them. Maximum torque occurs at θ = 90° (force perpendicular to the lever arm). Zero torque occurs when the force passes through the pivot (r = 0) or acts along the line of r (θ = 0° or 180°).
The perpendicular-distance shortcut. You can write τ = F × d⊥, where d⊥ = r sin θ is the perpendicular distance from the pivot to the line of action of F (the "moment arm"). This form is often faster for NEET problems because you read d⊥ directly from the geometry instead of computing sin θ.
Sign convention matters. For fixed-axis rotation, torques producing anticlockwise rotation are conventionally positive. Mixing sign conventions mid-problem is a common source of wrong answers — fix the convention at the start and stick with it.
SI unit: N·m (newton-metre). This is dimensionally the same as a joule, but torque is NOT energy; the unit is written N·m to avoid confusion.
Direction (vector form). τ = r × F follows the right-hand rule: curl fingers from r toward F, and the thumb gives the torque direction. For planar problems (the majority on NEET), direction reduces to clockwise/anticlockwise.
Watch out: NEET stems sometimes give angles measured from the lever arm rather than from the force direction. Sketch the geometry, identify θ between r and F, and verify which angle the problem actually specifies before substituting.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
What is the SI unit of torque?
Answer: A. Torque has the SI unit newton-metre (N·m). Although dimensionally identical to the joule, the unit N·m is used for torque to distinguish it from energy (NCERT Class 11 Physics Chapter 6, page 95).
Why B is wrong: B is wrong because the joule is the unit of energy and work. Torque is not energy, even though the dimensions match — the unit N·m is used to maintain the distinction.
Why C is wrong: C is wrong because kg·m/s² is the newton (unit of force), not torque. Torque requires an additional length factor: N × m = kg·m²/s².
Why D is wrong: D is wrong because the watt is the unit of power (energy per unit time), unrelated to torque.
Which of the following correctly expresses the vector form of torque about a point O?
Answer: B. Torque is defined as τ = r × F, the cross product of the position vector from the pivot to the point of application of force, with the force vector (NCERT Class 11 Physics Chapter 6, page 95).
Why A is wrong: A is wrong because F × r = −(r × F). The order of the cross product matters — reversing it gives the opposite direction.
Why C is wrong: C is wrong because the dot product r · F gives a scalar (related to work), not a vector. Torque is a vector quantity requiring the cross product.
Why D is wrong: D is wrong because F · r is the same scalar dot product as option C, just written in reversed order. It yields a scalar, not a torque vector.
A force acts on a rigid body. The torque about a given axis is zero when the line of action of the force:
Answer: C. When the line of action of the force passes through the axis (pivot), the position vector r and force F are collinear, so r × F = 0. Equivalently, the perpendicular distance (moment arm) from the axis to the line of action is zero (NCERT Class 11 Physics Chapter 6, page 95).
Why A is wrong: A is wrong because a force perpendicular to the axis of rotation generally produces non-zero torque — it has a non-zero moment arm unless it also passes through the axis.
Why B is wrong: B is wrong because a tangential force has the maximum moment arm (d⊥ = r), producing maximum torque, not zero torque.
Why D is wrong: D is wrong because a 45° angle gives τ = rF sin 45° = rF/√2, which is non-zero.
A force of 10 N acts at the end of a 0.50 m long wrench. The angle between the force and the wrench handle is 30°. What is the magnitude of the torque about the pivot?
Answer: B. τ = rF sin θ = 0.50 × 10 × sin 30° = 0.50 × 10 × 0.5 = 2.5 N·m (NCERT Class 11 Physics Chapter 6, page 95).
Why A is wrong: A is wrong because it uses sin 90° = 1 instead of sin 30° = 0.5 — this is the result when you ignore the angle entirely and compute rF.
Why C is wrong: C is wrong because it uses sin 60° ≈ 0.866 instead of sin 30° = 0.5. This error arises from confusing the complement of the given angle.
Why D is wrong: D is wrong because it appears to halve the correct answer again (perhaps using sin 30° twice or halving r). Only one factor of sin θ enters τ = rF sin θ.
A 20 N force is applied perpendicular to a door at a distance of 0.80 m from the hinge. A second force, also 20 N, is applied at 0.40 m from the hinge but at an angle of 90° to the door. What is the ratio of the torque due to the first force to the torque due to the second force?
Answer: B. Both forces act at 90° to the door, so sin θ = 1 for both. τ₁ = 20 × 0.80 = 16 N·m; τ₂ = 20 × 0.40 = 8 N·m. Ratio = 16/8 = 2 : 1. Torque scales linearly with the distance from the pivot when force and angle are identical.
Why A is wrong: A is wrong because it ignores the different distances from the hinge. Equal forces at different lever arms produce different torques.
Why C is wrong: C is wrong because it inverts the ratio — the force farther from the hinge produces larger torque, not smaller.
Why D is wrong: D is wrong because torque is proportional to r (first power), not r². Doubling the distance doubles the torque, not quadruples it.
A force F acts on a rigid body. The perpendicular distance from the axis of rotation to the line of action of the force is d. Which expression gives the torque magnitude?
Answer: A. When d is the perpendicular distance (moment arm) from the axis to the line of action, the torque magnitude is simply τ = F × d. The sin θ factor is already absorbed into d = r sin θ (NCERT Class 11 Physics Chapter 6, page 95).
Why B is wrong: B is wrong because dividing force by distance gives units of N/m (not N·m) and has no physical meaning in this context.
Why C is wrong: C is wrong because multiplying by cos θ double-counts the angular factor incorrectly. d is already the perpendicular distance; no additional trigonometric factor is needed.
Why D is wrong: D is wrong because multiplying d (which already equals r sin θ) by another sin θ would give rF sin²θ, overcounting the angular projection.
Two forces of equal magnitude act on a rigid body. Force P passes through the centre of mass, and force Q acts at the rim, perpendicular to the radius. Which statement is correct?
Answer: D. A force whose line of action passes through the pivot (here, the centre of mass) has zero moment arm and therefore zero torque. Force Q acts at a non-zero perpendicular distance from the centre of mass, producing non-zero torque (NCERT Class 11 Physics Chapter 6, page 95).
Why A is wrong: A is wrong because equal forces do not guarantee equal torques — the moment arm (perpendicular distance from the pivot to the line of action) also matters. Force P has zero moment arm.
Why B is wrong: B is wrong because force Q clearly acts at a distance from the centre of mass with a 90° angle, producing τ = FR (where R is the rim radius).
Why C is wrong: C is wrong because it reverses the situation. Force Q has a non-zero moment arm (the rim radius), so it produces torque. Force P passes through the pivot, giving zero torque.
A uniform rod of length 1.0 m and mass 2.0 kg is hinged at one end and held horizontal. A vertical force of 5.0 N is applied UPWARD at the free end. Taking g = 10 m/s², find the net torque about the hinge.
Answer: B. B is correct. The weight acts at the centre of the rod, 0.50 m from the hinge: τ_weight = 2.0 × 10 × 0.50 = 10.0 N·m clockwise. The applied force acts at the free end, 1.0 m from the hinge, and points upward, so it turns the rod the other way: τ_applied = 5.0 × 1.0 = 5.0 N·m anticlockwise. Net = 10.0 − 5.0 = 5.0 N·m, and it keeps the direction of the LARGER torque, which is the weight's: clockwise. Getting the magnitude right and the direction wrong is the whole trap here — subtracting the two numbers is the easy half (NCERT Class 11 Physics, Chapter 7).
Why A is wrong: A is wrong on direction only. 5.0 N·m is the right magnitude, but anticlockwise is the direction of the SMALLER torque. The 10.0 N·m clockwise weight dominates the 5.0 N·m anticlockwise applied torque, so what is left over turns clockwise.
Why C is wrong: C is wrong because 15.0 N·m adds the two torques (10 + 5) instead of subtracting. That would be the answer if the 5.0 N force pointed downward, adding to the weight rather than opposing it.
Why D is wrong: D is wrong because the two torques are 10.0 N·m and 5.0 N·m. They oppose each other but are not equal, so they cannot cancel. Cancellation would need the applied force to be 20 N.
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Given
• F = 25 N• r = 0.60 m• θ = 60° (angle between r and F)
Required
Magnitude of torque τ about the pivot.
Concept
Torque is the cross product of the position vector from the pivot to the point of force application, and the force vector. Its magnitude is τ = rF sin θ (NCERT Class 11 Physics Chapter 6, page 95).
Formula
τ = rF sin θ
Substitution
τ = 0.60 × 25 × sin 60°
Calculation
sin 60° = √3/2 ≈ 0.8660
τ = 0.60 × 25 × 0.8660 = 15 × 0.8660 = 12.99 N·m
Note on exact values: 25 N and 0.60 m are given data (2 significant figures each). sin 60° = √3/2 is an exact trigonometric value and does not limit the sig-fig count.
Final answer
τ ≈ 13 N·m (2 significant figures, matching the precision of the given data).
Common trap
A frequent error is using cos 60° = 0.5 instead of sin 60°. This happens when students confuse the angle between the force and the lever with the angle between the force and the perpendicular to the lever. Using cos 60° would give τ = 7.5 N·m — roughly half the correct value.
Similar NEET-style question
A spanner of length 0.25 m is used to tighten a bolt. A force of 40 N is applied at the end, making an angle of 45° with the spanner. Find the torque about the bolt. (Answer: τ = 0.25 × 40 × sin 45° ≈ 7.1 N·m.)
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The moment of a force (torque) about a point is τ = r × F, where r is the position vector from the pivot to the point of force application. Magnitude |τ| = r F sin θ. SI unit: N·m.
-- NCERT Class 11 Physics, Ch. 6, p. 106Cross product of position vector and force vector. Magnitude r F sin(theta).
| Symbol | Quantity | SI Unit |
|---|---|---|
| tau | torque | N*m |
| r | position from pivot | m |
| F | force | N |
| theta | angle between r and F | rad |
More in System of Particles and Rotational Motion: 7 exam traps and mistakes · 7 formulas · 4 question patterns from its other lessons.
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