b
(b)
Let $m$ be the mass of each coin. The centre of mass on $N$ coins kept on lid is
$CM =\frac{40 m \times 0+N m \times 9}{(40+N) m}=\frac{9 N}{N+40} \,cm$
The geometric centre of submerged part after keeping $N$ coins will be
$GC =\frac{3(40+N)}{40} \,cm$
For equilibrium, $CM = GC$
$\Rightarrow \quad \frac{9 N}{N+40}=\frac{(40+N) 3}{40}$
$\Rightarrow 3 N^2-480 N+4800=0$
This gives, $N=10.72$, which is closed to $10$ .