JEE MAIN - Physics (2020 - 4th September Evening Slot - No. 10)
For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes
perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is :
_4th_September_Evening_Slot_en_10_1.png)
$${1 \over 2}$$
$${1 \over 4}$$
$${1 \over 8}$$
$${2 \over 3}$$
Explanation
$${I_O} = {M \over {12}}\left( {{a^2} + {b^2}} \right)$$ = $${M \over {12}}\left[ {{{80}^2} + {{60}^2}} \right]$$
IO' = IO + Md2 {parallel axis theorem}
= $${M \over {12}}\left[ {{{80}^2} + {{60}^2}} \right]$$ + M[50]2
$$ \therefore $$ $${{{I_O}} \over {{I_{O'}}}}$$ = $${{{M \over {12}}\left[ {{{80}^2} + {{60}^2}} \right]} \over {{M \over {12}}\left[ {{{80}^2} + {{60}^2}} \right] + M{{\left[ {50} \right]}^2}}}$$ = $${1 \over 4}$$
IO' = IO + Md2 {parallel axis theorem}
= $${M \over {12}}\left[ {{{80}^2} + {{60}^2}} \right]$$ + M[50]2
$$ \therefore $$ $${{{I_O}} \over {{I_{O'}}}}$$ = $${{{M \over {12}}\left[ {{{80}^2} + {{60}^2}} \right]} \over {{M \over {12}}\left[ {{{80}^2} + {{60}^2}} \right] + M{{\left[ {50} \right]}^2}}}$$ = $${1 \over 4}$$
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