Problem 11 - Master Resource Book in JEE Main Physics

Question

If two masses \(M\) and \(m\) are tied to two ends of a meter scale. If a balanced point is obtained at point \(P\) and if \(M>m\), then (a) \(P A=P B\) (b) \(P A>P B\) (c) \(P B>P A\) (d) \(P A=3 P B\)

Step-by-Step Solution

Answer

The correct answer is (c) \(PB>PA\).

Step 1: Understand the Problem

We have two masses, \(M\) and \(m\), attached to the ends of a meter scale. The scale balances at point \(P\). We need to determine the relationship between the distances \(PA\) and \(PB\) given that \(M > m\).

Step 2: Apply the Principle of Moments

The principle of moments states that for the system to be in equilibrium (balanced at \(P\)), the clockwise moment about \(P\) must equal the counterclockwise moment about \(P\). This gives us the equation: \[ M \times PA = m \times PB \] where \(PA\) is the distance from \(P\) to the mass \(M\), and \(PB\) is the distance from \(P\) to the mass \(m\).

Step 3: Compare the Moments and Relationships

Since \(M > m\), for the equation \(M \times PA = m \times PB\) to hold, \(PA\) must be less than \(PB\), i.e., \(M\) must be closer to \(P\) to balance the larger weight. Thus, \(PB > PA\).

Step 4: Conclusion

From the analysis, we see that the correct relationship is \(PB > PA\). Therefore, the correct answer is option (c) \(P B > P A\).

Key Concepts

Moment of Force

The moment of force, often simply referred to as the "moment," is a key concept when analyzing objects in equilibrium. It represents the force's ability to cause an object to rotate around a pivot point. Think of it as the force's "twisting strength," which depends on two main factors:

• Magnitude of the Force: The larger the force applied, the greater the moment. It's like when you push harder on a door; it swings open faster.
• Distance from the Pivot: The farther away the force is applied from the pivot point, the greater its moment. For instance, pushing a door open is easier when you push at the edge, rather than near the hinges.

This relationship can be mathematically expressed as: \[\text{Moment} = \text{Force} \times \text{Distance from pivot}\]In the exercise, the masses generate moments about point \(P\). To maintain balance, the moments created by each mass around this point must be equal. That's why we have the equation \(M \times PA = m \times PB\), ensuring the system stays in equilibrium.

Equilibrium

Equilibrium in physics refers to the state where all forces and moments acting on a body are balanced, resulting in no movement. In our exercise, the system is in rotational equilibrium, which means all turning forces (moments) are perfectly balanced around the pivot point \(P\).
To achieve equilibrium, the principle of moments must be satisfied, meaning the sum of the clockwise moments around \(P\) must equal the sum of the counterclockwise moments. If this condition is met, the object remains stable and does not rotate.
In the scenario with the meter scale, the mass \(M\) on one side creates a larger force but must be placed closer to the point \(P\) than the smaller mass \(m\) to balance these moments. This ensures that both sides generate equal moments about \(P\), even though \(M > m\), leading to stable equilibrium.

Balance of Forces

The balance of forces concept is essential in understanding how objects maintain their position without moving when pinned or balanced at a point. In the context of the exercise, this concept extends to how forces acting on either side of the meter scale achieve harmony.

• Static Balance: When forces are perfectly balanced, the object remains stationary.
• Dynamic Consideration: When external conditions or forces change, the system may attempt to reach a new equilibrium point.

In our problem, the mass \(M\) and mass \(m\) exert gravitational forces that pull down on either side of the pivot \(P\). For the forces to balance, the meter scale must adjust so that the reactant forces attribute to \(PA\) and \(PB\) in a way where their turning moments match.
This balance means even though mass \(M\) is heavier, it is positioned closer to point \(P\) compared to mass \(m\), resulting in equal moments of the opposing forces, ensuring the system remains balanced without tipping over.