Answer
$\rho = 600~kg/m^3$
Work Step by Step
We can find the volume of the displaced water:
$W_{app} = Weight-F_b$
$F_b = Weight-W_{app}$
$F_b = 30~N-20~N$
$F_b = 10~N$
$\rho_w~V~g = 10~N$
$V = \frac{10~N}{\rho_w~g}$
$V = \frac{10~N}{(1000~kg/m^3)(9.8~m/s^2)}$
$V = 0.00102~m^3$
Thus the volume of the object is $~~0.00102~m^3$
We can find the density of the other liquid:
$W_{app} = Weight-F_b$
$F_b = Weight-W_{app}$
$F_b = 30~N-24~N$
$F_b = 6.0~N$
$\rho~V~g = 10~N$
$\rho = \frac{6.0~N}{(0.00102~m^3)(9.8~m/s^2)}$
$\rho = 600~kg/m^3$
