**(1) Three identical thin rods each having mass 2 kg and length 1 m are joined to form an equilateral triangle. The moment of inertia of this system about an axis perpendicular to the plane of the triangle and passing through one corner of the triangle is (in kg m**

(a) 1 (b) 2 (c) 3 (d) 4 (e) 5

^{2})(a) 1 (b) 2 (c) 3 (d) 4 (e) 5

The moment of inertia of the rods AB and CB about an axis perpendicular to the plane of the triangle and passing through the corner B is ML

^{2}/3 + ML

^{2}/3 = 2 ML

^{2}/3 where M is the mass of each rod and L is the length. [Note that the moment of inertia of a thin rod about a normal axis through its centre of mass is ML

^{2}/12 and hence its moment of inertia about a normal axis through one end, according to the theorem of parallel axes, is (ML

^{2}/12) + M(L/2)

^{2}= ML

^{2}/3].

The moment of inertia of the rod AC about the axis through B is

ML

^{2}/12 + M[(√3/2)L]

^{2}= ML

^{2}/12 + 3 ML

^{2}/4 = 5ML

^{2}/6, in accordance with the theorem of parallel axes. [The distance of AC from B is L sin60° = (√3/2)L].

The moment of inertia of the entire triangle about the axis through B is the sum of the moments of inertia of the three rods and is equal to 2 ML

^{2}/3 +5ML

^{2}/6 = 9ML

^{2}/6 = 3ML

^{2}/2. Substituting for M (=2kg) and L (=1m), we obtain the answer as 3 kgm

^{2}.

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