Three particles,each of mass m,are placed at | vedclass

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(D) Let the particles be at positions $A, B, C$. Axis '$1$' passes through one particle and is perpendicular to the side connecting the other two. The

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All of these.

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(D) Let the particles be at positions $A, B, C$. Axis '$1$' passes through one particle and is perpendicular to the side connecting the other two. The distances of the three particles from axis '$1$' are $0, l/2, l/2$. Thus,$I_1 = m(0)^2 + m(l/2)^2 + m(l/2)^2 = m(l^2/4 + l^2/4) = \frac{1}{2} ml^2$. Wait,re-evaluating: Axis '$1$' passes through one particle and is parallel to the opposite side. The distances are $0, l\sin(60^\circ), l\sin(60^\circ)$ is incorrect. Looking at the diagram,axis '$1$' passes through one vertex and is parallel to the opposite side. The distances are $0, l\frac{\sqrt{3}}{2}, l\frac{\sqrt{3}}{2}$. So $I_1 = m(0)^2 + m(l\frac{\sqrt{3}}{2})^2 + m(l\frac{\sqrt{3}}{2})^2 = m(\frac{3l^2}{4} + \frac{3l^2}{4}) = \frac{3}{2} ml^2$. Actually,based on standard problems of this type: For axis '$1$' (passing through one vertex and parallel to the opposite side): $I_1 = m(0)^2 + m(l\frac{\sqrt{3}}{2})^2 + m(l\frac{\sqrt{3}}{2})^2 = \frac{3}{2} ml^2$. For axis '$2$' (passing through one side): $I_2 = m(0)^2 + m(0)^2 + m(l\frac{\sqrt{3}}{2})^2 = \frac{3}{4} ml^2$. For axis perpendicular to the plane through a vertex: $I_3 = m(0)^2 + m(l)^2 + m(l)^2 = 2 ml^2$. Given the options provided,option $(d)$ is the intended answer.

Three particles,each of mass $m$,are placed at the corners of an equilateral triangle of side $l$. Which of the following is/are correct?

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