Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle A B C of side cm (as shown in the figure). The moment of inertia of the system about a line A X perpendicular to A B and in

Correct answer: D. (5)/(4) m l^ 2 I_ AX =m(AB)^ 2 +m(OC)^ 2 =m l^ 2 +m(l 60^ ^ )^ 2 =m l^ 2 + m l^ 2 4 =(5)/(4) m l^ 2

BITSAT 2021

BITSAT 2021 · Physics · Rotational Motion

Three particles, each of mass m gram, are situated at the vertices...

Three particles, each of mass

m

gram, are situated at the vertices of an equilateral triangle

A B C

of side

\ell \mathrm{ 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 B C

, in gram

-\mathrm{cm}^2

units will be Three particles, each of mass m gram, are situated at the vertices of an equilateral triangle A B C of side cm (as

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

\mathrm{I}_{\mathrm{AX}}=\mathrm{m}(\mathrm{AB})^{2}+\mathrm{m}(\mathrm{OC})^{2}=\mathrm{m} l^{2}+\mathrm{m}\left(l \cos 60^{{}^{\circ}}\right)^{2}=m l^{2}+\frac{\mathrm{m} l^{2}}{4}=\frac{5}{4} \mathrm{ m} l^{2}