Question

Plzzzz help me out...... ​

Answer 1

$\rm \longrightarrow I_{CD} = I_{GH} ( symmetry ) \\ \\ \rm \longrightarrow \: I_{AB} = I_{EF} = I\\ \rm\longrightarrow Now ,\: by \: using \: perpendicular \: axis \: theorem :- \\ \\\rm \longrightarrow I_{CD} + I_{GH} = 2 I_{AB} \\ \\ \rm \longrightarrow2 I_{CD} = 2 I_{AB} \\ \\ \rm \longrightarrow I_{CD} = I_{AB} \\ \\ \rm \longrightarrow I_{CD} = I \\ \\ \rm where \: \:I_{AB} = moment \: of \: inertia \: about \: AB \\\rm I_{CD} =moment \: of \: inertia \: about \: CD\\\rm I_{GH} = moment \: of \: inertia \: about \: GH\\\rm I_{EF} = moment \: of \: inertia \: about \: EF\\\rm I_{EF} = moment \: of \: inertia \: about \: EF$

Answer :-

• Option (A) I is correct

Answer 2

Answer:

$\rm \longrightarrow I_{CD} = I_{GH} ( symmetry ) \\ \\ \rm \longrightarrow \: I_{AB} = I_{EF} = I\\ \rm\longrightarrow Now ,\: by \: using \: perpendicular \: axis \: theorem :- \\ \\\rm \longrightarrow I_{CD} + I_{GH} = 2 I_{AB} \\ \\ \rm \longrightarrow2 I_{CD} = 2 I_{AB} \\ \\ \rm \longrightarrow I_{CD} = I_{AB} \\ \\ \rm \longrightarrow I_{CD} = I \\ \\ \rm where \: \:I_{AB} = moment \: of \: inertia \: about \: AB \\\rm I_{CD} =moment \: of \: inertia \: about \: CD\\\rm I_{GH} = moment \: of \: inertia \: about \: GH\\\rm I_{EF} = moment \: of \: inertia \: about \: EF\\\rm I_{EF} = moment \: of \: inertia \: about \: EF$

Answer :-

• Option (A) I is correct