Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side l 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

Question

Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side l 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) (3/4)ml2 (B) 2ml2 (C) (5/4)ml2 (D) (3/2)ml2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/dfa839b9d102ce18ea6f6545544e3626-.png" /> I AX = m ℓ2+ m ((ℓ/2))2 = m ℓ2+( m ℓ2/4)=(5/4) m ℓ2

Q. Three particles, each of mass $m$ gram, are situated at the vertices of an equilateral triangle $A BC$ of side $l$ cm (as shown in the figure). The moment of inertia of the system about a line $A X$ perpendicular to $A B$ and in the plane of $A BC$ , in $g r am - c m^{2}$ units will be

$\frac{3}{4} m l^{2}$

$2 m l^{2}$

$\frac{5}{4} m l^{2}$

$\frac{3}{2} m l^{2}$

$I_{A X} = m ℓ^{2} + m (\frac{ℓ}{2})^{2}$
$= m ℓ^{2} + \frac{m ℓ ^{2}}{4} = \frac{5}{4} m ℓ^{2}$

Q. Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle ABC of side l 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

43​ml2

2ml2

45​ml2

23​ml2