Question

Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side/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 ABC, in gram-cm² units will be(a) $\frac{3}{2}$ml²(b) $\frac{3}{4}$ml²(c) 2ml²(d) $\frac{5}{4}$ml²

Answer 1

Answer:

5ml²/4

Explanation:

Mass of three articles = m (Given)

Following the moment of inertia -

System = mara²+mbrb²+mcrc²

= ma(0)²+m(l)²+m(lsin30°)²

Sin 30° = d/l

or 1/2 = d/l

or d = l/2

Thus, ml²+ml²/4

= 5ml²/4

Therefore, the moment of inertia of the system about a line AX perpendicular to AB and in the plane of ABC, in gram-cm² units will be 5ml²/4

Answer 2

Answer:5ml^2/4

Explanation: