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JEE Advanced 1998
LEVELJEE Main

Animated Solution for Physics - Rotational Motion: The torque on a body about a given point is found to be equal to , where is a constant vector and is the angular momentum of the body about that point. From this it follows that

Select Answer:

* Multiple Correct

Visualized Solution

The Sigma Insight: Torque and Angular Momentum

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The Elegance of Vector Dynamics

Imagine a spinning top defying gravity, or an electron dancing in a magnetic field. These mesmerizing phenomena are governed by a single, elegant mathematical relationship.
The problem presents us with a fascinating scenario where the torque on a body is the cross product of a constant vector and its angular momentum.
This might look like a simple abstract equation, but it holds the secret to one of the most beautiful motions in physics: precession. Let's decode this vector puzzle piece by piece.

The Perpendicular Push

We are given the master equation for the torque: .
From Newton's second law applied to rotational motion, we also know that torque is the rate of change of angular momentum. Mathematically, this is written as .
By equating these two expressions, we arrive at the core differential equation governing the system: .
Now, we must invoke a fundamental property of the vector cross product. The result of a cross product is always strictly perpendicular to both of the original vectors being multiplied.
Therefore, the rate of change of angular momentum, , must be perpendicular to the angular momentum vector itself at every single instant.
This brilliant geometric fact immediately confirms that our first option is absolutely correct.

The Circular Motion Analogy

But what does it physically mean for the rate of change of a vector to be perpendicular to the vector itself?
To build intuition, let's take a detour to a familiar concept: uniform circular motion. In circular motion, the acceleration vector is always directed towards the center, making it perfectly perpendicular to the velocity vector .
What is the consequence of this perpendicular push? The acceleration can only change the direction of the velocity, never its magnitude. The speed remains perfectly constant.
We can apply this exact same logic to our angular momentum vector. Because is always perpendicular to , the "push" on the angular momentum vector only twists its direction.
It can never stretch or shrink the vector. Consequently, the magnitude of the angular momentum, , remains absolutely constant over time.
This confirms our third option and simultaneously disproves the fourth option, as the vector is indeed changing its direction.

The Invariant Projection

Finally, we need to investigate the component of the angular momentum along the direction of the constant vector .
The component of one vector along another is mathematically captured by their dot product. To see if this component changes, we must take the time derivative of the dot product .
Using the product rule for differentiation, we get: .
Here is where the magic happens. We are explicitly told that is a constant vector. Therefore, its rate of change, , is exactly zero.
This simplifies our derivative to just .
But wait! We already established from the cross product that is perpendicular to .
The dot product of any two perpendicular vectors is always zero. Thus, .
Because the time derivative of the dot product is zero, the dot product itself must be a constant.
This proves that the component of the angular momentum along the constant vector never changes, confirming our second option.
This beautiful conservation law is the mathematical heartbeat of Larmor precession, a concept that echoes through classical mechanics and quantum physics alike.

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