
- ✓$8 L / 15$
- B$5 L / 6$
- C$(3 L / 4+R)$
- D$(7 L / 15+R)$

For system to be in equilibrium without toppling, following conditions must be fulfilled.
$(i)$ Centre of mass $C_1$ of sphere and upper block must lie inside the edge of lower block.
Taking origin of axes choosen at $C$, we have
$\frac{M}{2} \times y=M\left(\frac{L}{2}-y\right)$
$\Rightarrow \frac{y}{2}+y=\frac{L}{2} \text { or } y=\frac{L}{3}$
$(ii)$ Centre of mass of both of block and sphere must lie inside the edge of table.
So, again taking centre of mass $C_2$ as origin,
$\frac{3 M}{2}\left(x-\frac{L}{3}\right)+M\left(x-\frac{L}{3}-\frac{L}{2}\right)=0$
$\Rightarrow \frac{3 x}{2}-\frac{L}{2}+x-\frac{L}{3}-\frac{L}{2}=0$
$\Rightarrow \frac{5 x}{2}=\frac{4 L}{3}$
$\Rightarrow x =\frac{8 L}{15}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

$\frac{\text{E}}{2}$
$2\text{E}$
$\text{E}$
$\frac{\text{E}}{4}$
