Physics, asked by ranjithmulki766, 10 months ago

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) 3/2 m????²
(b) 3/4 m????²
(c) 2 m????²
(d) 5/4 m????²

Answers

Answered by dhareaveer
4

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

Answered by Anonymous
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

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