\[(a)\;Mg(\sqrt 2 +1)\quad (b)\;Mg \sqrt 2 \quad (c)\;\frac{mg}{\sqrt 2}\quad (d)\;Mg(\sqrt 2-1)\]

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Work done by trension -Work done by force =Work done by gravitational force

$0+F \times AB =Mg \times AC$

$F=Mg \bigg(\large\frac{AC}{AB}\bigg)$

$AB= l\sin 45$

$\qquad=\large\frac{l}{\sqrt 2}$

$AC=OC-OA$

$\qquad=l-l \cos 45$

$\qquad=Mg \bigg(\large\frac{1-\frac{1}{2}}{\Large\frac{1}{\sqrt 2}}\bigg)$

$AC=OC-OA$

$\qquad=l-l \cos 45$ (l-length of string)

Therefore $F=Mg(\sqrt 2-1)$

Hence d is the correct answer.

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