Question

a 25 cm. thick block of ice floating on fresh water can support an 80 kg man standing on it, what is the smallest area of the ice block? (sp gr. of ice=0.917)

Answer 1

Weight of the ice is $W_{ic}=g(\rho_{ice}V)\implies W_{ic}=g(917V)=917gV\: \text{N}$ ,Weight of the man is $W_m=80g\:\text{N}$ ,thus net weight $W_{total}=W_{ic}+W_m=917gV+80g\:\text{N}$ .

Now, Buoyant force is $F_b=\rho Vg$ ,Hence according to the problem

$$F_b=W_{total}\\ \implies 917V+80=1000V\implies V=1.27m^3$$

As, thickness is $25cm=0.25m$ ,thus required area is $1.27/0.25=5.08m^2$