\[\begin {array} {1 1} (a)\;\sqrt {\frac{3g}{l}}. \sin \theta/2 & \quad (b)\;\sqrt {\frac{6g}{l}}. \sin \theta/2 \\ (c)\;\sqrt {\frac{3g}{l}}. \cos \theta/2 & \quad (d)\;\sqrt {\frac{6g}{l}}. \cos \theta/2 \end {array}\]

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I dint understand how did you write that height mgl/2(1-cos@) plzz explain by diagram .

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Since the reaction force at the height is not doing any work, total mechanical energy is conserved.

$\Delta PE= mg l/2 (1- \cos \theta)$

$\Delta K = \large\frac{1}{2} I_0. w^2$

$\qquad=\large\frac{1}{2} \frac{ml^2}{3}.w^2$

$\large\frac{1}{2} \large\frac{ml^2}{3}$$ w^2= mgl$$\;.\ \sin ^2 \theta/2$

$w= \sqrt {\large\frac{6g}{l}} \sin \theta/2$

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