- At O.
- At a distance less than $\frac{\text{l}}2{}$ from O.
- At a distance $\frac{\text{l}}2{}$ from O.
- At a distance larger than $\frac{\text{l}}2{}$ from O.
Explanation:
It is given that there is no force along x-axis.
COM of rod will remain and will not shift along x-axis (horizontal direction).
Force gravity is acting along y-axis (vertical direction). So, COM will shift along the y-axis by $\frac{\text{l}}2{}$ distance and COM of horizontal rod is at a distance $\frac{\text{l}}2{}$ from one end.
Therefore, lower end of the rod will remain at a distance $\frac{\text{l}}2{}$ from O.
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$\frac{4(2+\text{b})}{3(3+\text{b})}$
$\frac{3(3+\text{b})}{4(2+\text{b})}$
$\frac{4(3+\text{b})}{3(2+\text{b})}$