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Torque As A Vector Cross Product

Hardphysics

A force is applied to a particle whose position vector relative to the origin is given, and the force is directed parallel to that position vector. What is the magnitude of the torque about the origin?

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About This Question

Subject
physics
Chapter
rotational motion
Topic
torque as a vector cross product
Difficulty
Hard
Year
2025
Tags
torque vectorcross productradial forceangle dependenceperpendicular component

Solution

Correct Answer:

Zero

As presented in NCERT Class 11, Chapter 7 (System of Particles and Rotational Motion), torque is defined as the vector cross product τ = r × F, whose magnitude is given by rF sinθ, where θ is the angle between the position vector r and the applied force F. Being a cross product, the torque depends only on the component of the force that is perpendicular to the position vector, since the cross product extracts exactly that perpendicular contribution. The direction of the torque is given by the right-hand rule, perpendicular to the plane containing r and F. When the force is parallel to the position vector, the angle between them is θ = 0°, so sinθ = 0 and the torque magnitude is exactly zero, because a purely radial force exerts no turning effect about the origin. The option 'Maximum' applies only when the force is perpendicular to the position vector, that is θ = 90°, not parallel. The option 'Equal to rF' would require sinθ = 1, which again is the perpendicular case and not the situation described. The option 'Equal to r + F' is dimensionally meaningless, since one cannot add a length to a force. A consistency check confirms that a purely radial force cannot rotate a particle about the very point it points toward, so a torque of exactly zero is the physically correct answer.

This hard difficulty physics question is from the chapter rotational motion, covering the topic of torque as a vector cross product. It appeared in the 2025 exam.

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