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(D) Let the mass of the rod be $M$ and its length be $l$.Weight of the rod is $W = Mg$.Initially,for equilibrium,the force $F$ exerted by each man is

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$\frac{W}{4}$

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(D) Let the mass of the rod be $M$ and its length be $l$.Weight of the rod is $W = Mg$.Initially,for equilibrium,the force $F$ exerted by each man is $F + F = Mg$,so $F = \frac{Mg}{2} = \frac{W}{2}$.When one man lets go,the rod starts rotating about the other end (let it be $A$) due to the torque provided by its weight acting at the center of mass.The torque about end $A$ is $	au = Mg \frac{l}{2}$.The moment of inertia of the rod about end $A$ is $I = \frac{Ml^2}{3}$.Using $	au = I\alpha$,we have $Mg \frac{l}{2} = \left( \frac{Ml^2}{3} ight) \alpha$.Solving for angular acceleration,$\alpha = \frac{3g}{2l}$.The linear acceleration of the center of mass is $a = \frac{l}{2} \alpha = \frac{l}{2} \left( \frac{3g}{2l} ight) = \frac{3g}{4}$.Applying Newton's second law for the center of mass,$Mg - F' = Ma$,where $F'$ is the new force exerted by the remaining man.$F' = Mg - Ma = Mg - M \left( \frac{3g}{4} ight) = Mg \left( 1 - \frac{3}{4} ight) = \frac{Mg}{4} = \frac{W}{4}$.

$A$ horizontal heavy uniform bar of weight $W$ is supported at its ends by two men. At the instant one of the men lets go of his end of the rod,the force on the other man's hand changes to:

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