Question

.a block of wood of 1.5kg mass floats on water with 68 percent of its volume submerged. a lead block is placed on the wood and the wood is the fully submerged with the lead entirely out of the water. find the mass of the lead block?

Answer 1

M=1.5 kg

V_in = 68%

$M = ρ_{water} V_{in} \\ V_{in} = \frac{1.5}{1000} =1.5 \times 10^{-3} \;m^3 \\ V_o = \frac{V_{in}}{0.68} \\ V_o = 2.206 \times 10^{-3} \; m^3$

When the additional mass of lead is placed on the top, the remaining 32% of volume also goes into the water.

$m_{block}=1000 \times 0.32 \times 2.206 \times 10^{-3} = 0.706 \; kg$