Answer:
The minimum mass of the lead for the combination to submerge is 155 g.
Explanation:
let M be the mass of the wood.
let m be the minimum mass of lead to be added for the combination to submerge.
let ρ1 be the density of the liquid.
let ρ2 be the density of the wood.
let ρ3 be the density of lead.
let g be the gravitational acceleration.
For the combination to submerge, the weight of the wood combined with the weight of lead should at least be equal to the buyant force, that is:
weight of wood and lead = buyant force
g×(M+m) = g×(ρ1)×(M/ρ2 + m/ρ3)
M+m = (ρ1)×(M/ρ2 + m/ρ3)
m - ρ1×m/ρ3 = (ρ1)×(M/ρ2) - M
m(1 - ρ1/ ρ3) = M(ρ1/ρ2 -1)
m = [M(ρ1/ρ2 -1)]/[(1 - ρ1/ ρ3)]
= [(300)(1.1/0.75 -1)]/[(1 - 1.1/ 11.3)]
= 155 g
Therefore, the minimum mass of the lead for the combination to submerge is 155 g.
