- Home
- Standard 11
- Physics
A wide bottom cylindrical massless plastic container of height $9 \,cm$ has $40$ identical coins inside it and is floating on water with $3 \,cm$ inside the water. If we start putting more of such coins on its lid, it is observed that after $N$ coins are put, its equilibrium changes from stable to unstable. Equilibrium in floating is stable if the geometric centre of the submerged portion is above the centre of the mass of the object). The value of $N$ is closed to
$6$
$10$
$16$
$24$
Solution
(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 $10$ .
