Question

A cylindrical block of wood of mass $2\text{kg}$ is floating in water with its axis vertical. It is depressed by a distance $x$ and net force on it is given by $F=kx$. Then, value of $k$ in $\text{N/m}$ is
[Use $\pi =\frac{22}{7}$]

Answer 1

According to question,



From initial case -
On applying condition of equilibrium along vertical
$\Rightarrow F_b-Mg=0$
$\Rightarrow F_b=Mg$
$\Rightarrow {\rho }_{l}gV=2\times 10=20$
[$V=$ volume of water being displaced]
$\Rightarrow {\rho }_{l}gAt=20...\left(1\right)$
$[V=\text{Area}(A)\times \text{depth}(t\left)\right]$
From final case -
Net force along the vertical direction is given by,
$F=F_b^{\prime }-Mg...\left(2\right)$
$\Rightarrow F={\rho }_{l}g{V}^{\prime }-Mg$
where, $V^{\prime }=A(t+x)$ is the volume of water displaced.
$\because [V^{\prime }=\text{Surface Area (A)}\times \text{depth}(t+x\left)\right]$
$\Rightarrow F={\rho }_{l}gA(t+x)-Mg$
$\Rightarrow F={\rho }_{l}gAt+{\rho }_{l}gAx-Mg$
$\Rightarrow F={\rho }_{l}gAt+(1000\times 10\times \pi (0.07{)}^2\times x)-(2\times 10)$
Substituting $\left({\rho }_{l}gAt\right)$ from Eq.$\left(1\right)$,
$\Rightarrow F=154x$
Also according to question, net force on cylindrical block can be written as, $F=kx$
$\Rightarrow kx=154x$
$\Rightarrow k=154\text{N/m}$
Hence the value of $k$ is $154$.