JEE Main 2019PhysicsRotational MotionTorquemediumMCQ

JEE Main 2019Rotational Motion Question with Solution

From: JEE Main 2019 (Online) 11th January Morning Slot

Question

A slab is subjected to two forces and of same magnitude F as shown in the figure. Force is in XY-plane while force acts along z = axis at the point . The moment of these forces about point O will be :

JEE Main 2019 (Online) 11th January Morning Slot Physics - Rotational Motion Question 184 English

This question includes a diagram. The text above accompanies the figure.

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Show full solutionCorrect option: D
Correct answer
D

Step-by-step explanation

To determine the moment of the forces about point O, we need to calculate the moments of each force and then combine them vectorially.

First, consider force . This force acts along the z-axis at the point .

The position vector of the point of application of relative to the origin O is:

Since acts along the z-axis, we can express it as:

The moment of force about point O is given by the cross product:

Substitute the values:

Compute the cross product:

So, the moment due to force is:



Torque for F2 force

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About this question

This is a previous-year question from JEE Main 2019, covering the Rotational Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.