Problem 125 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A bead of weight \(w\) can slide on smooth circular wire in a vertical plane. The bead is attached by a light thread to the highest point of the wire and in equilibrium, the thread is taut and make an angle \(\theta\) with the vertical then tension of the thread and reaction of the wire on the bead are (a) \(T=w \cos \theta \quad R=w \tan \theta\) (b) \(T=2 w \cos \theta \quad R=w\) (c) \(T=w R=w \sin \theta\) (d) \(T=w \sin \theta \quad R=n \cot \theta\)

Step-by-Step Solution

Answer

The correct answer is (a): \( T = w \cos \theta, R = w \tan \theta \).

Step 1: Understand the problem setup

We need to find the tension in the thread and the reaction of the wire on the bead when the bead is in equilibrium on a smooth circular wire. The light thread makes an angle \( \theta \) with the vertical, and the weight of the bead is \( w \).

Step 2: Analyze forces on the bead

The forces acting on the bead are its weight \( w \) acting vertically downward, the tension \( T \) in the thread, and the normal reaction \( R \) from the wire. In equilibrium, the net force in both the horizontal and vertical directions should be zero.

Step 3: Vertical component analysis

Resolve the tension \( T \) vertically, then we have \( T \cos \theta \) acting upward. The weight \( w \) acts downward, balancing this gives:\[ T \cos \theta = w \] Solve for \( T \): \[ T = \frac{w}{\cos \theta} \]

Step 4: Horizontal component analysis

Resolve the tension \( T \) horizontally, giving \( T \sin \theta \), and the reaction \( R \) from the wire balances this: \[ T \sin \theta = R \]

Step 5: Substitute tension in reaction equation

Substitute the expression for \( T \) obtained from the vertical component equation into the horizontal component equation: \[ \frac{w}{\cos \theta} \sin \theta = R \] Simplify to find the reaction \( R \): \[ R = w \tan \theta \]

Step 6: Match the findings with given options

From steps 3 and 5, we have \( T = \frac{w}{\cos \theta} \) and \( R = w \tan \theta \). Compare these with the given options: - (a) \( T = w \cos \theta, R = w \tan \theta \) - (b) \( T = 2w \cos \theta, R = w \) - (c) \( T = w, R = w \sin \theta \) - (d) \( T = w \sin \theta, R = n \cot \theta \)Only option (a) provides the correct values when \( T \) is expressed as \( w \) and \( R \) as \( w \tan \theta \), so the correct answer is option (a).

Key Concepts

Tension in Thread

When you hang something like a bead on a thread, tension is the force within the thread that holds up the bead. Imagine pulling on both ends of a rubber band. That stretching feel is tension. In this exercise, the tension in the thread is influenced by the weight of the bead and the angle the thread makes with the vertical.
The tension force can pull in different directions based on the angle \(\theta\). For the vertical part, tension is like an invisible helper pulling the bead upwards to balance its weight. The mathematical way to express this is through:

• Vertical Component: \( T \cos \theta = w \)
• Solving gives: \( T = \frac{w}{\cos \theta} \)

This tells us the tension adjusts based on how slanted the thread is, making the angle \(\theta\) crucial in finding the exact amount of tension needed to keep everything balanced.

Reaction Force

In physics, a reaction force is a counterforce that prevents an object from falling or sliding away. It's like how the floor stops you from sinking when you stand. For the bead, the wire provides a reaction force. This force acts perpendicular to the surface of the wire and is vital for maintaining equilibrium in both directions.
In this context, the reaction force is affected by the horizontal pull of the tension in the thread. The balance here is described by:

• Horizontal Component: \( T \sin \theta = R \)
• Resulting in reaction: \( R = w \tan \theta \)

This reaction force ensures the bead stays put and doesn't roll off or climb up the wire, highlighting the wire's role in balancing forces.

Trigonometry in Physics

Trigonometry helps us break down forces into different components. By understanding angles and triangles, we can easily dissect forces acting in various directions. In our bead scenario, trigonometry is especially important because it tells us how tension is divided between vertical and horizontal directions.
The angle \(\theta\) plays a major role here:

• Using \( \cos \theta \) and \( \sin \theta \) helps describe the components of tension.
• Combining these functions provides formulas \(T = \frac{w}{\cos \theta}\) and \(R = w \tan \theta\) from vertical and horizontal perspectives.

These simple trigonometric tools enable us to translate complex force problems into understandable measures, making it easier to solve.

Vertical and Horizontal Force Components

Every force can be broken down into vertical and horizontal components, making complex force systems easier to analyze. Think of separating tasks between two friends: one handles vertical, the other horizontal. For our bead, the forces involved are tension from the thread and the wire's reaction.

• Vertical forces include the weight \(w\) acting downwards and tension \(T\cos \theta\) acting upwards.
• Horizontal forces come from the tension's horizontal component \(T\sin \theta\) and the reaction \(R\).

Once we resolve these forces into components, they become independent sets of forces that we can set equations for, making it straightforward to see if the system stays in equilibrium. By analyzing these components, students can grasp how equilibrium is maintained and how each force contributes effectively to that balance.