Mechanics — Hard JEE physics MCQ
A uniform wooden stick of mass $1.6 \mathrm{~kg}$ and length $l$ rests in an inclined manner on a smooth, vertical wall of height $h(<l)$ such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of $30^{\circ}$ with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio $h / l$ and the frictional force $f$ at the bottom of the stick are
$\left(g=10 \mathrm{~ms} \mathrm{~s}^{2}\right)$
- A. $\frac{h}{l}=\frac{\sqrt{3}}{16}, f=\frac{16 \sqrt{3}}{3} \mathrm{~N}$
- B. $\frac{h}{l}=\frac{3}{16}, f=\frac{16 \sqrt{3}}{3} \mathrm{~N}$
- C. $\frac{h}{l}=\frac{3 \sqrt{3}}{16}, f=\frac{8 \sqrt{3}}{3} \mathrm{~N}$
- D. $\frac{h}{l}=\frac{3 \sqrt{3}}{16}, f=\frac{16 \sqrt{3}}{3} \mathrm{~N}$
Solution
The correct option is **D**.
