Question

Three particles, each of mass $m$ grams situated at the vertices of an equilateral $\Delta ABC$ of side $a$ cm (as shown in the figure). The moment of inertia of the system about a line $AX$ perpendicular to $AB$ and in the plane of $\Delta ABC$ in $g$-cm² units will be:

• A. 2 ma2
• B. 3/2 ma2
• C. 3/4 ma2
• D. 5/4 ma2

Answer 1

The Correct Option is D

1. Given: Three particles of mass m grams each are placed at the vertices of an equilateral triangle ABC with side length a cm.
2. Required: Find the moment of inertia of the system about the line AX, which is:
   • Perpendicular to side AB
   • Lies in the plane of triangle
3. Coordinate Geometry Setup:
   • Let point \( A = (0, 0) \)
   • Then \( B = (a, 0) \)
   • Point \( C \) lies above the midpoint of \( AB \), so:
     Midpoint of \( AB = \left(\frac{a}{2}, 0\right) \)
     Height of equilateral triangle = \( \frac{\sqrt{3}}{2}a \)
     So \( C = \left(\frac{a}{2}, \frac{\sqrt{3}}{2}a\right) \)
4. Moment of Inertia Calculation:
   • About point A: Distance = 0 → \( I_A = 0 \)
   • Point B: Perpendicular distance from AX = \( a \) → \( I_B = m a^2 \)
   • Point C: Perpendicular distance from AX = \( \frac{a}{2} \) → \( I_C = m \left(\frac{a}{2}\right)^2 = \frac{1}{4}ma^2 \)
5. Total Moment of Inertia:
   \[ I = I_A + I_B + I_C = 0 + ma^2 + \frac{1}{4}ma^2 = \frac{5}{4}ma^2 \]
6. Final Answer: \( \boxed{\frac{5}{4}ma^2} \)