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Question: Answered & Verified by Expert
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

PhysicsRotational MotionBITSATBITSAT 2021
Options:
  • A $\frac{3}{2} m l^{2}$
  • B $\frac{3}{4} m l^{2}$
  • C $2 m l^{2}$
  • D $\frac{5}{4} m l^{2}$
Solution:
1101 Upvotes Verified Answer
The correct answer is: $\frac{5}{4} m l^{2}$



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

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