Torque — the moment of a force — is the rotational analogue of force. NCERT Class 11 Physics Chapter 7 (System of Particles and Rotational Motion), page 4 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?
Which of the following correctly expresses the vector form of torque about a point O?
A force acts on a rigid body. The torque about a given axis is zero when the line of action of the force:
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?
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?
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?
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?
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 at the free end. Taking g = 10 m/s², find the net torque about the hinge.
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 7, page 4).
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.) ---
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. 4For a particle: L = r x p. For a rigid body about its rotation axis: L = I omega. Vector quantity.
| Symbol | Quantity | SI Unit |
|---|---|---|
| L | angular momentum | kg*m^2/s |
| I | moment of inertia | kg*m^2 |
| omega | angular velocity | rad/s |
The position of the centre of mass equals the mass-weighted average of particle positions. For continuous bodies use integral form.
| Symbol | Quantity | SI Unit |
|---|---|---|
| R_cm | CoM position | m |
| m_i | mass of i-th particle | kg |
| r_i | position of i-th particle | m |
Standard moments of inertia about the symmetry axis. For other axes use parallel/perpendicular axes theorems.
| Symbol | Quantity | SI Unit |
|---|---|---|
| M | mass | kg |
| R | radius | m |
| L | length | m |
| I | moment of inertia | kg*m^2 |
Moment of inertia about any axis = moment about parallel axis through CM + Md^2.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | MOI about given axis | kg*m^2 |
| I_cm | MOI about parallel CM axis | kg*m^2 |
| M | total mass | kg |
| d | perpendicular distance | m |
For planar lamina: MOI about axis perpendicular to plane = sum of MOI about two perpendicular in-plane axes through same point.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I_z | MOI perp to plane | kg*m^2 |
| I_x, I_y | MOI in plane | kg*m^2 |
Rotational analogues of linear kinematic equations under constant angular acceleration.
| Symbol | Quantity | SI Unit |
|---|---|---|
| omega | angular velocity | rad/s |
| alpha | angular acceleration | rad/s^2 |
| theta | angular displacement | rad |
| t | time | s |
Energy of rotation about an axis. Adds to translational KE for rolling bodies.
| Symbol | Quantity | SI Unit |
|---|---|---|
| I | moment of inertia | kg*m^2 |
| omega | angular velocity | rad/s |
Cross 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 |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Student answers L/2 for two-particle CoM regardless of mass ratio.
Question gives two masses on rigid rod and asks for CoM distance.
R_cm from m1 = m2*L/(m1+m2). Heavier mass pulls CoM closer to it.
Category: Similar Terms
Student conserves rotational KE when angular momentum is conserved (or vice versa). When I changes, L = Iω is conserved but KE = ½Iω² is NOT (it depends on I and ω together).
Question describes a body whose moment of inertia changes (skater pulling arms in, star collapsing).
L conservation requires zero external torque. KE conservation requires no work done — different criteria. When I changes via internal forces, L conserved, ω increases, KE increases.
Category: Similar Terms
Student confuses 2/5 (solid sphere) with 2/3 (hollow sphere) or 1/2 (disc) with 1 (ring).
Question gives a specific geometry and asks for I or radius of gyration.
Memorise: solid sphere 2/5, hollow sphere 2/3, disc/cylinder 1/2, ring/hoop 1, rod-centre 1/12, rod-end 1/3.
Category: Unit Conversion
Student plugs rpm directly into formulas requiring rad/s. 1 rpm = 2π/60 rad/s.
Question gives ω in rpm and asks for kinematic quantities in SI units.
Convert: ω(rad/s) = (2π/60) × rpm. Always check units before substituting.
Root cause: concept gap
L = Iω is conserved when external torque is zero. KE = ½Iω² is NOT conserved when I changes (since ω changes too). When skater pulls arms in, L conserved, ω increases, KE increases (work done by muscles).
Root cause: concept gap
Memorise standard moments of inertia. Solid sphere has more mass near axis (smaller MOI = 2MR²/5); hollow sphere has all mass at radius R (larger MOI = 2MR²/3).
Root cause: unit error
Convert: ω(rad/s) = (2π/60) × ω(rpm). For example, 1200 rpm = 1200 × 2π/60 = 125.66 rad/s.
The angular acceleration of a body, moving along the circumference of a circle, is
forgets conversion rpm to rad s
Treats rpm as rad/s without 2*pi/60 conversion
uses equal distribution
Default to L/2 regardless of mass ratio
uses energy conservation instead
Confusing L conservation with KE conservation
swap solid hollow coefficients
Confusing 2/5 (solid sphere) with 2/3 (hollow sphere)
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