A cube (density 0.5 gm/ cc) of side 10 cm floats in water placed in a cylindrical beaker of base area 1500 cm². When a mass m is placed on the wooden block the water level in the beaker rises by 2 mm. mass  $m=x\times {10}^{-2}gm$. Then x= .

Let the cube dips further by y cm and water level rises by 2 mm.
 

Then equating the volumes (/// volume = \\\ volume in figure)
⇒ Volume of water raised = volume of extra depth of wood
  $\Rightarrow 100y=(1500-100)\frac{2}{10}=1400\times \frac{2}{10}=280$
   y = 2.8 cm
   Extra upthrust
 ${\rho }_{waer}\times (2.8+0.2)\times 100g=mg$
   m = 300 gm