Two identical uniform rectangular blocks (with longest side $\mathrm{L}$ ) and a solid sphere of radius $\mathrm{R}$ are to be balanced at the edge of a heavy table such that the centre of the sphere remains at the maximum possible horizontal distance from the vertical edge of the table without toppling as indicated in the figure.
If the mass of each block is $\mathrm{M}$ and of the sphere is $\mathrm{M/2}$, then the maximum distance $\mathrm{x}$ that can be achieved is

• Correct answer
  A

  $8 \mathrm{L} / 15$
• B

  $5 \mathrm{L} / 6$
• C

  $(3 \mathrm{L} / 4+\mathrm{R})$
• D

  $(7 \mathrm{L} / 15+\mathrm{R})$

Solution

$2+\mathrm{S}$ System lie above edge of 1 . $\begin{array}{l} \frac{M}{2} y-M\left(\frac{L}{2}-y\right)=0 \\ \frac{y}{2}+y=\frac{L}{2} \\ y=\frac{L}{3} \end{array}$ Now $1+2+\mathrm{S}$ centre of mass wili lie above the table $\frac{3 M}{2}\left(x-\frac{L}{3}\right)+M\left(x-\frac{L}{3}-\frac{L}{2}\right)=0$ $\frac{3 x}{2}-\frac{L}{2}+x-\frac{L}{3}-\frac{L}{2}=0$ $\frac{5 x}{2}=\frac{4 L}{3} \quad \quad x=\frac{8 L}{15}$

Hence, the correct answer is option: 'A'.