When a cube floats, the fraction submerged (f) equals the ratio of its density to the liquid's density: f = ρ_cube/ρ_liq. Initially, f_i = 1/n.

After a temperature change θ, densities change due to cubical expansion. The new density is ρ' = ρ/(1+γθ). The new submerged fraction f' becomes:

$\frac{{\rho }_{\text{cube}}}{{\rho }_{\text{liq}}}\times \frac{1+{\gamma }_{\text{ℓ}}\theta }{1+{\gamma }_{\text{m}}\theta }$

Since γθ is small, we approximate 1/(1+γ_mθ) ≈ (1-γ_mθ). Substituting f_i = 1/n gives:

$\frac{1}{n}\times (1+{\gamma }_{\text{ℓ}}\theta )(1-{\gamma }_{\text{m}}\theta )\approx \frac{1}{n}[1+({\gamma }_{\text{ℓ}}-{\gamma }_{\text{m}}\left)\theta \right]$

Final Answer: $\frac{[1+({\gamma }_{\text{ℓ}}-{\gamma }_{\text{m}}\left)\theta \right]}{n}$