A cubic block of side 0.5m float on water with 30% of its volume under water. what is the maximum weight (in kg) that can be put on the block without fully submerging it under water ?___________{Take density of water 10³kg/m³}_____________$\times$10^-1

Given that the cubical block's side = 0.5 metres 
Water has a density of $={10}^{3}kg/m^3$

We know that 

$\mathrm{Density}=\frac{\mathrm{mass}}{\mathrm{volume}}\Rightarrow \mathrm{mass}=\mathrm{volume}\times \mathrm{density}$

When a body floats, its weight causes an upward thrust.

${50}^3\times \frac{30}{100}\times 1\times \mathrm{g}={\mathrm{M}}_{\mathrm{cube}}\mathrm{g\_\_\_\_\_\_}\left(1\right)$

Let m mass should be placed, then

${50}^3\times 1\times \mathrm{g}=({\mathrm{M}}_{\mathrm{Cube}}+\mathrm{m})\mathrm{g\_\_\_\_\_\_}\left(2\right)$

Subtracting equation 1 from 2,

$\left({\left(50\right)}^3\times 1\times \mathrm{g}\right)-({\left(50\right)}^3\times \frac{30}{100}\times 1\times \mathrm{g})=({\mathrm{M}}_{\mathrm{cube}}+\mathrm{m})\mathrm{g}-{\mathrm{M}}_{\mathrm{cube}}\mathrm{g}$${50}^3\mathrm{g}-({\left(50\right)}^3\times 0.3\times \mathrm{g})={\mathrm{M}}_{\mathrm{cube}}\mathrm{g}+\mathrm{mg}-{\mathrm{M}}_{\mathrm{cube}}\mathrm{g}$

$\Rightarrow \mathrm{mg}={50}^3\times \mathrm{g}(1-0.3)\Rightarrow \mathrm{mg}=125\times {10}^3\times 0.7\mathrm{g}$

$\Rightarrow \mathrm{m}=875\times {10}^{-1}\mathrm{kg}$

Hence the correct answer is 875.