Question

A solid rectangular block 0.100m×0.080m ×0.060m and float freely in a liquid of density 13546 kg/m³ if the depth of liquid is 0.45m up the block largest side find the density of the block

Answer 1

I assume the block floats with its largest face (0.100 m × 0.080 m) horizontal, and that the submerged depth along that face is 0.045 m (0.45 m would be larger than the block height 0.060 m, so I take it as a typo).

Given:
- block dims: 0.100 × 0.080 × 0.060 m → V_block = 0.100·0.080·0.060 = 4.80×10^−4 m^3
- submerged depth h = 0.045 m so displaced volume V_disp = base area·h = (0.100·0.080)·0.045 = 3.60×10^−4 m^3
- liquid density ρ_liq = 13546 kg/m^3

Equilibrium (buoyant force = weight) gives ρ_block · V_block = ρ_liq · V_disp, so
ρ_block = ρ_liq · (V_disp / V_block) = 13546 · (3.60×10^−4 / 4.80×10^−4) = 13546 · 0.75 ≈ 1.016×10^4 kg/m^3.

Answer: ρ_block ≈ 1.02 × 10^4 kg/m^3 (≈ 10,160 kg/m^3).

If you actually meant 0.45 m (not 0.045 m): since 0.45 m > block height, the block would be fully submerged — the problem as stated would be inconsistent unless the block density equals the liquid density (13546 kg/m^3) for neutral buoyancy. Confirm which depth was intended if you want me to recheck.