SATHEE: Rotational Motion Question 192

Rotational Motion Question 192

Question: Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side f. cm . 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

Options:

A) $\frac{3}{2} m ℓ^{2}$

B) $\frac{3}{4} m ℓ^{2}$

C) $2 m ℓ^{2}$

D) $\frac{5}{4} m ℓ^{2}$

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Answer:

Correct Answer: D

Solution:

[d]

$I_{A X} = m {(A B)}^{2} + m {(O C)}^{2}$

$= m ℓ^{2} + m {(ℓ \cos 60^{o})}^{2}$

$= m ℓ^{2} + m ℓ^{2} / 4 = 5 / 4 m ℓ^{2}$

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Rotational Motion Question 192

Question: Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side f. cm . 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

Options:

Answer:

Solution:

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