JEE MAIN - Physics (2021 - 16th March Morning Shift - No. 7)
Four equal masses, m each are placed at the comers of a square of length (l) as shown in the figure. The moment of inertia of the system about an axis passing through A and parallel to DB would be :
_16th_March_Morning_Shift_en_7_1.png)
$$\sqrt 3 $$ ml2
2 ml2
ml2
3 ml2
Explanation
_16th_March_Morning_Shift_en_7_2.png)
$$AC = \sqrt {{l^2} + {l^2}} $$
$$AC = l\sqrt 2 $$
$$d = {{l\sqrt 2 } \over 2}$$
$$ \Rightarrow d = {1 \over {\sqrt 2 }}$$
Moment of inertia about the axis passing through A :
$$I = m{(O)^2} + m{(d)^2} + m{(d)^2} + M{(AC)^2}$$
$$ \Rightarrow I = O + m{\left( {{l \over {\sqrt 2 }}} \right)^2} + m{\left( {{l \over {\sqrt 2 }}} \right)^2} + m{\left( {l\sqrt 2 } \right)^2}$$
$$ \Rightarrow I = {{m{l^2}} \over 2} + {{m{l^2}} \over 2} + 2m{l^2}$$
$$ \Rightarrow I = 3m{l^2}$$
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