Answer
The specific gravity of the alcohol is $0.785$
Work Step by Step
We can find the buoyant force exerted on the aluminum cylinder:
$F_b = 1.03~N - 0.730~N = 0.30~N$
According to Archimedes' principle, the buoyant force is equal to the weight of the alcohol that is displaced. We can find the mass of the alcohol that is displaced:
$mg = F_b$
$m = \frac{F_b}{g}$
$m = \frac{0.30~N}{9.80~m/s^2}$
$m = 0.0306~kg$
We can find the density of the alcohol:
$\rho = \frac{m}{V} = \frac{0.0306~kg}{3.90\times 10^{-5}~m^3} = 785~kg/m^3$
We can find the specific gravity:
$\frac{785~kg/m^3}{1000~kg/m^3} = 0.785$
The specific gravity of the alcohol is $0.785$.
