A cubical block of wood of edge 3cm floats in water. The lower surface of cube just touches free ends of vertical spring fixed at bottom of the pot. Then the maximum weight that can be put on the block without wetting it is [Density of wood $=800{\text{kgm}}^{-3}$ spring constant of the spring$=\text{5}0{\text{Nm}}^{-\text{1}}$ , Take$\text{g}=\text{1}0{\text{ms}}^{-\text{2}}$ ]

Given that a cubical block.

$\rho =800{\text{kgm}}^{-3}$

$\text{S}=3\text{cm}=3\times {10}^{-2}\text{m}$

$\text{V}=27\times {10}^{-6}{\text{m}}^{\text{3}}$

is floating on water just touching the spring.

Let a weight = Mg is placed on it.

Such that block just completely immersed pushing the spring.

We can have

$\text{Mg}+\text{Vdg}={\text{Vd}}_{\text{2}}\text{g}+\text{k}x$

$\Rightarrow$ $\text{Mg}+\text{27} \times \text{1}0 \times \text{8}00 \times \text{1}{0}^{-6}=\text{27}\times \text{1}{0}^{-\text{6}}\times \text{1}000\times \text{1}0+\text{5}0\times 0.0\text{6}$

$\Rightarrow$ $\text{Mg}+\text{216}\times \text{1}{0}^{-\text{3}}=\text{27}0\times \text{1}{0}^{-\text{3}}+\text{3}00\times \text{1}{0}^{-\text{3}}$

$\Rightarrow$ $\text{Mg}+\text{216}\times \text{1}{0}^{-\text{3}} =\text{57}0\times \text{1}{0}^{-\text{3}}$

$\Rightarrow$ $\text{Mg}=\left(\text{57}0-\text{216}\right)\text{1}{0}^{-\text{3}}$

$\Rightarrow$ $=\text{354}\times \text{1}{0}^{-\text{3}} \text{N}$

        $\therefore \text{Mg}=0.\text{354N}$