AIPMT PRELIMS 2004 Physics Moment of Inertia MCQ Question

To solve the problem of finding the moment of inertia of three particles situated at the vertices of an equilateral triangle about a line AX that is perpendicular to the base AB, let's break down the solution step by step.

Step 1: Understanding the Configuration

We have three particles, each of mass $m$ grams, located at the vertices of an equilateral triangle ABC with side length $\ell$ cm. The vertices can be represented in a coordinate system as follows:

• Vertex A: $(0, 0)$
• Vertex B: $(\ell, 0)$
• Vertex C: $\left( \frac{\ell}{2}, \frac{\sqrt{3}}{2} \ell \right)$

Step 2: Moment of Inertia Definition

The moment of inertia $I$ of a system of point masses about an axis is given by the formula:

$$I = \sum m_i r_i^2$$

where $m_i$ is the mass of each particle and $r_i$ is the perpendicular distance of each particle from the axis of rotation.

Step 3: Calculate Distances from AX

The line AX is perpendicular to AB and can be considered as a vertical line passing through point A. The distances of each particle from this line (AX) are as follows:

1. For particle A at $(0, 0)$:

   • Distance from AX = 0 cm

2. For particle B at $(\ell, 0)$:

   • Distance from AX = $\ell$ cm

3. For particle C at $\left( \frac{\ell}{2}, \frac{\sqrt{3}}{2} \ell \right)$:

   • The distance from AX (the x-coordinate distance) = $\frac{\ell}{2}$ cm

Step 4: Calculate Moment of Inertia for Each Particle

Now we can calculate the moment of inertia for each particle about the axis AX:

1. For particle A:

   $$I_A = m \cdot (0)^2 = 0$$

2. For particle B:

   $$I_B = m \cdot (\ell)^2 = m\ell^2$$

3. For particle C:

   $$I_C = m \cdot \left( \frac{\ell}{2} \right)^2 = m \cdot \frac{\ell^2}{4} = \frac{m\ell^2}{4}$$

Step 5: Total Moment of Inertia

Now, summing up the contributions from all three particles, we have:

$$I_{total} = I_A + I_B + I_C = 0 + m\ell^2 + \frac{m\ell^2}{4}$$

Combining the terms gives:

$$I_{total} = m\ell^2 + \frac{m\ell^2}{4} = m\ell^2 \left( 1 + \frac{1}{4} \right) = m\ell^2 \left( \frac{4}{4} + \frac{1}{4} \right) = m\ell^2 \cdot \frac{5}{4}$$

Therefore, the moment of inertia $I$ is:

$$I = \frac{5}{4} m \ell^2$$

Conclusion

The correct answer to the moment of inertia of the system about the line AX is indeed:

Correct Answer: B) $\frac{5}{4} m \ell^2$

Why Other Options Are Incorrect

• Option A (2 mℓ²): This overestimates the contributions by not accounting for the distances correctly.
• Option C (3/2 mℓ²): This also does not correctly represent the summed distances from the axis.
• Option D (3/4 mℓ²): This underestimates the contributions from the distances of particles B and C.

In summary, the calculations show that the total moment of inertia correctly corresponds to option B, and the reasoning for the spatial arrangement of the particles is crucial for understanding the calculation of distances.