For the given uniform square lamina ab sub d , whose centre is -Turito

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Physics-

For the given uniform square lamina $A B C D$ , whose centre is $O$

Physics-General

$I_{A C} = I_{E F}$
$\sqrt{2} I_{A C} = I_{E F}$
$I_{A D} = {4 I}_{E F}$
$I_{A C} = \sqrt{2} I_{E F}$

Answer:The correct answer is: $I subscript A D end subscript equals blank 4 I subscript E F end subscript$ Let the each side of square lamina is $d$ .
So, $I subscript E F end subscript equals I subscript G H end subscript$ (due to symmetry)
And $I subscript A C end subscript equals I subscript B D end subscript$ (due to symmetry)
Now, according to theorem of perpendicular axis,

$I subscript A C end subscript plus I subscript B D end subscript equals I subscript 0 end subscript$
$rightwards double arrow 2 I subscript A C end subscript equals I subscript 0 end subscript$ (i)
and $I subscript E F end subscript plus I subscript G H end subscript equals I subscript 0 end subscript$
$rightwards double arrow 2 I subscript E F end subscript equals I subscript 0 end subscript$ (ii)
From Eqs. (i) and (ii), we get
$I subscript A C end subscript equals I subscript E F end subscript$
$therefore I subscript A D end subscript equals I subscript E F end subscript plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction$
$equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction plus fraction numerator m d to the power of 2 end exponent over denominator 4 end fraction open parentheses a s I subscript E F end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 12 end fraction close parentheses$
So, $I subscript A D end subscript equals fraction numerator m d to the power of 2 end exponent over denominator 3 end fraction equals 4 I subscript E F end subscript$

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