Physics problem on buoyant force, I need help

Question

If someone may help me with this attached Physics problem shown below. I've been having trouble and spending time on it, I can't figure it out. Thanks, much appreciated!

Answer 1

1. the weight of the solid is Mg = 0.25 × 9.8 = 2.45 N
2. the buoyant force by the pure water is F = 2.45 - 2 = 0.45 N
3. the buoyant force by the liquid is F' = 2.45 - 1.7 = 0.75 N
4. the buoyant force is calculated by the volume of the solid inside the liquid and the density of the liquid, assume the volume of the solid is V, the density of the water is d_w, and the density of the liquid is d_l,

F = V·d_w ⇒ V = F/ d_w ,

and

F' = V·d_l ⇒ V = F'/ d_l .

So we have

F/ d_w = F' / d_l

d_l = F' × d_w / F = d_w × F' / F = 0.75/0.45 * d_w = 1.667 d_w

if the density of water is 1 g/cm³, then the density of the liquid is 1.667 g/cm³.

Answer 2

The real weight of M= 250 g = 2.5 N ( taking g = 10) (using mass = 0.25 kg)

If apparent wt is 2N, then up thrust = 0.5 N which is also buoyant force =F

Now In another liquid L, the apparent weight become 1.7 N

New buoyant force is (2.5 - 1.7) N= 0.8 N = F'

As the volume of body immersed has not changed, more buoyant force by this liquid is more than water

Thus the density of this liquid is more than density of water

For a given volume of immersed body the buoyant force is proportional to density of liquid in which body is immersed

Using ratio proportion

Density of water/ Density of liquid = buoyant force in water/ buoyant force in liquid = 0.5/0.8

using density of water as 1000 kg/m³

Density of liquid = 8000/5 = 1600kg/m³