Correct Answer - C::D
As shown in the figure, the component of weight `Mgsintheta` tends to slide the poitn of contact (of the clinder with inclined plane) along its direction. The slicding friction acts in the opposite direction to this relative motion. Because of frictional force the cylinder rolls. Thus frictional force aids rotation but hinders translational motion.
Appling `F_("net")=ma` along the directon of inclined plane, we get
`Mgsintheta-f=Ma_(c)`
Where `a_(c)=` acceleration of centre of mass of the cylinder
`:. f=Mgsintheta-Ma_(c)`........i
But `a_(c)=(gsintheta)/(1+(I_(c))/(MR^(2)))=(gsintheta)/(1+(M(R^(2))/2)/(MR^(2)))`
`=2/3gsintheta`.........ii
From eqn i and ii
`f=(Mgsintheta)/3`
If `theta` is reduced, then friction force is reduced.
