Torque — the rotational analogue of force
Torque (moment of force) is defined as the cross product of the position vector r and the applied force F, measured from the chosen pivot point (NCERT Class 11 Physics Chapter 6, page 95):
τ = r × F
The magnitude is τ = r F sin θ, where θ is the angle between r and F. The direction follows the right-hand rule: curl fingers from r toward F, and the thumb points along τ.
Three things that control torque magnitude:
The perpendicular-distance shortcut: τ = F × d, where d = r sin θ is the perpendicular distance from the pivot to the line of action of F. Equivalently, τ = r × F⊥, where F⊥ = F sin θ is the component of force perpendicular to r. Both give the same result — pick whichever the problem makes easier.
SI unit: N·m. Note: this is dimensionally identical to the joule (J = N·m), but torque and energy are distinct physical quantities. NEET will never accept "joule" as a unit of torque.
Torque as a vector: For a fixed rotation axis (the NEET scenario), only the component of torque along the axis matters. A torque that tends to produce counterclockwise rotation is taken positive by convention.
Watch out: When multiple forces act on a body, compute the torque of each force about the same pivot, then sum algebraically. Forgetting to use the same pivot point is a silent error that produces plausible-looking wrong answers.
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: B. Torque is measured in newton-metres (N·m). Although N·m is dimensionally identical to the joule, "joule" is reserved for energy. The SI unit of torque is written as N·m, not J. (NCERT Class 11 Physics Chapter 6, page 95.)
Why A is wrong: A is wrong because joule (J) is the unit of energy/work, not torque. Despite N·m and J having the same dimensions, they name different physical quantities.
Why C is wrong: C is wrong because kg·m²/s² is the dimensional equivalent expressed in base units — it is not the named SI unit of torque. NEET expects N·m.
Why D is wrong: D is wrong because N/m is the unit of spring constant (force per unit length), not torque.
Torque is defined as:
Answer: D. By definition, τ = r × F, where r is the position vector from the pivot to the point of application of force. The order matters: r × F gives the correct direction by the right-hand rule. (NCERT Class 11 Physics Chapter 6, page 95.)
Why A is wrong: A is wrong because the dot product r·F gives a scalar (work), not a vector quantity like torque.
Why B is wrong: B is wrong because F × r reverses the direction (F × r = −r × F). The correct definition is τ = r × F, not F × r.
Why C is wrong: C is wrong because torque is a vector quantity obtained via a cross product, not a scalar product.
A force of 10 N acts on a rigid body. If the line of action of the force passes through the pivot point, the torque about that pivot is:
Answer: D. When the force passes through the pivot, the angle θ between r and F is 0° (or 180°), so sin θ = 0, giving τ = r F sin θ = 0. The perpendicular distance from the pivot to the line of action is zero. (NCERT Class 11 Physics Chapter 6, page 95.)
Why A is wrong: A is wrong because this assumes the moment arm is 1 m and the angle is 90°. When the force passes through the pivot, sin θ = 0, so torque is zero regardless of force magnitude.
Why B is wrong: B is wrong because no matter how far the point of application is from the pivot, if the force's line of action passes through the pivot, the perpendicular distance d = 0, making torque zero.
Why C is wrong: C is wrong because torque depends on r, F, and sin θ — not on mass. Here sin θ = 0, so torque is zero.
A force of 5.0 N is applied at the end of a 0.40 m wrench, perpendicular to the wrench handle. The torque about the pivot is:
Answer: B. τ = r F sin θ = 0.40 × 5.0 × sin 90° = 0.40 × 5.0 × 1 = 2.0 N·m. The force is perpendicular, so sin 90° = 1. (Application of τ = r × F; NCERT Class 11 Physics Chapter 6, page 95.)
Why A is wrong: A is wrong. This results from computing 0.40 × 5.0 = 2.0 and then mistakenly dividing by 2.5 or from an arithmetic error. The correct product is 2.0 N·m.
Why C is wrong: C is wrong. This would be the torque if r were 0.20 m (half the wrench length) — a common error of using the midpoint instead of the full length where force is applied.
Why D is wrong: D is wrong. This results from dividing F by r (5.0/0.40 = 12.5) instead of multiplying. Torque is r × F, not F/r.
A 6.0 N force acts at a point 0.50 m from the pivot, making an angle of 30° with the position vector. The magnitude of the torque is:
Answer: A. τ = r F sin θ = 0.50 × 6.0 × sin 30° = 0.50 × 6.0 × 0.50 = 1.5 N·m. Note sin 30° = 0.50. (Application of τ = r F sin θ; NCERT Class 11 Physics Chapter 6, page 95.)
Why B is wrong: B is wrong. This results from using cos 30° ≈ 0.866 instead of sin 30° = 0.50, giving 0.50 × 6.0 × 0.866 ≈ 2.6. Torque uses sin θ, not cos θ.
Why C is wrong: C is wrong. This is r × F without the sin θ factor (0.50 × 6.0 = 3.0). Forgetting the angle is a common error — torque requires the perpendicular component.
Why D is wrong: D is wrong. This results from computing r F sin θ with an extra factor of 0.50, perhaps squaring sin 30° (giving 0.25 instead of 0.50).
Two forces of equal magnitude F act on a rigid body at the same point, one at 90° to the position vector and the other at 45° to the position vector. The ratio of the torques τ₉₀ : τ₄₅ is:
Answer: C. τ₉₀ = r F sin 90° = r F. τ₄₅ = r F sin 45° = r F / √2. Ratio = r F : r F/√2 = √2 : 1. (Application of τ = r F sin θ; NCERT Class 11 Physics Chapter 6, page 95.)
Why A is wrong: A is wrong. This assumes torque doesn't depend on the angle, which contradicts τ = r F sin θ. Different angles give different torques.
Why B is wrong: B is wrong. This would be the ratio if sin 45° = 0.50, but sin 45° = 1/√2 ≈ 0.707. The ratio √2 : 1 ≈ 1.414 : 1, not 2 : 1.
Why D is wrong: D is wrong. This inverts the ratio. τ₉₀ > τ₄₅ (since sin 90° > sin 45°), so the ratio is √2 : 1, not 1 : √2.
A door of width w is pushed with force F at its outer edge, perpendicular to the door. If the same force is applied at w/2 from the hinge (still perpendicular), by what factor does the torque change?
Answer: C. At the edge: τ₁ = w × F × sin 90° = wF. At w/2: τ₂ = (w/2) × F × sin 90° = wF/2. The torque is halved. This is why door handles are placed at the outer edge — maximum lever arm. (NCERT Class 11 Physics Chapter 6, page 95.)
Why A is wrong: A is wrong. Doubling would occur if r increased, not decreased. Moving the point of application closer to the hinge reduces r, which reduces torque.
Why B is wrong: B is wrong. Torque depends on the distance from the pivot. Halving the distance halves the torque, since τ = r F sin θ and r changed.
Why D is wrong: D is wrong. Torque is zero only when the force passes through the pivot (θ = 0°) or has zero magnitude. Here the force is still perpendicular and non-zero, just applied closer to the hinge.
Two forces act on a rod pivoted at one end: a 4.0 N force acts perpendicularly at 0.30 m from the pivot (clockwise), and a 3.0 N force acts perpendicularly at 0.80 m from the pivot (counterclockwise). The net torque about the pivot is:
Answer: A. τ₁ = 0.30 × 4.0 × 1 = 1.2 N·m (clockwise, take as negative). τ₂ = 0.80 × 3.0 × 1 = 2.4 N·m (counterclockwise, take as positive). Net τ = 2.4 − 1.2 = 1.2 N·m counterclockwise. (Application of τ = r F sin θ with algebraic summation; NCERT Class 11 Physics Chapter 6, page 95.)
Why B is wrong: B is wrong. This is the magnitude of just the first torque (0.30 × 4.0 = 1.2 N·m) with only that force's direction. It ignores the second force's contribution entirely.
Why C is wrong: C is wrong. This is the magnitude of 3 × 1.2, possibly from adding the torques instead of subtracting opposing ones, or from an arithmetic error. The net torque requires subtracting the clockwise from the counterclockwise contribution.
Why D is wrong: D is wrong. This may result from subtracting forces first (4.0 − 3.0 = 1.0) then treating the result as torque. You cannot subtract forces directly — each torque must be computed separately using its own lever arm.
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Pattern: Torque calculation with multiple forces (no topic-specific PYQ pattern survived Rule 7 filtering; this worked example is constructed from the torque formula and NCERT definition to serve the lesson's scope)
Given
A uniform rod of length 1.00 m is pivoted at its left end. Two forces act on it:• Force F₁ = 8.0 N acts perpendicularly downward at 0.25 m from the pivot• Force F₂ = 5.0 N acts perpendicularly upward at 0.80 m from the pivot
Find the net torque about the pivot and state the direction of rotation.
Required
Net torque (τ_net) and its rotational direction (clockwise or counterclockwise).
Concept
Each force produces a torque about the pivot: τ = r F sin θ. Since both forces are perpendicular to the rod, sin θ = 1 for both. Assign signs by convention: counterclockwise = positive, clockwise = negative. Then sum algebraically.
Formula
τ = r F sin θ
Net torque: τ_net = τ₁ + τ₂ (with sign convention)
Substitution
F₁ acts downward at 0.25 m → produces clockwise rotation → negative:
τ₁ = −(0.25)(8.0)(1) = −2.0 N·m
F₂ acts upward at 0.80 m → produces counterclockwise rotation → positive:
τ₂ = +(0.80)(5.0)(1) = +4.0 N·m
Calculation
τ_net = −2.0 + 4.0 = +2.0 N·m
The integers 1 (from sin 90°) are exact values and do not limit significant figures.
Final answer
τ_net = 2.0 N·m, counterclockwise.
The answer has 2 significant figures, consistent with the least precise given data (2 sig figs in 8.0 N and 5.0 N). Sin 90° = 1 is an exact value and does not constrain the sig-fig count.
Common trap
A frequent error is subtracting forces first (8.0 − 5.0 = 3.0 N) and then computing torque with some average distance. This is wrong — each force has its own lever arm, and torques must be computed individually before summing.
Similar NEET-style question
A light rod of length 2.0 m is pivoted at the centre. A 6.0 N force acts perpendicularly at one end (clockwise) and a 4.0 N force acts perpendicularly at the other end (also clockwise). Find the net torque about the pivot.
*(Answer: Both torques are clockwise with lever arm 1.0 m each. τ_net = 1.0 × 6.0 + 1.0 × 4.0 = 10.0 N·m clockwise.)*
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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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