Question

a body weighs 160 grams in air,130 grams in water and 136 grams in oil.what is the specific gravity of oil?

Answer 1

Let the density of the body =  p kg/m³
                          its volume = v m³
weight of body in a medium = (density of body-density of medium)×v×g

Case:1
weight in air = 160g = 0.16kg × g m/s²   (g=acceleration due to gravity)
density of air ≈ 0
Thus
0.16×g = (p-0)vg
⇒v = 0.16/p

Case:2
weight in water = 130g = 0.13kg × g m/s²
density of water = w kg/m³
Thus 
0.13 × g = (p - w)vg
⇒p-w = 0.13/v
⇒p-w = 0.13/(0.16/p)
⇒p-w = 0.13p/0.16
⇒p-w = 13p/16
⇒p - 13p/16 -w = 0
⇒3p/16 = w
⇒p = 16w/3

Case:3
weight in oil = 136g = 0.136kg × g m/s²
density of oil = d kg/m³
Thus 
0.136 × g = (p - d)vg
⇒0.136 = (p-d)v
⇒0.136 = (p-d)(0.16/p)

⇒0.136 = (0.16)$\frac{p-d}{p}$

⇒ 0.136/0.16 = 1-$\frac{d}{p}$

⇒0.85 = 1 -$\frac{d}{16w/3}$

⇒0.85 = 1 - $\frac{3}{16} \frac{d}{w}$

But accrding to defination of specific gravity, specific gravity of oil = $\frac{d}{w}$

⇒ $\frac{3}{16} \frac{d}{w}$ = 1 - 0.85

⇒ $\frac{3}{16} \frac{d}{w}$ = 0.15 

⇒$\frac{d}{w}$ = $\frac{16}{3}$ ×0.15

⇒$\frac{d}{w}$ = $\frac{16}{3} * \frac{15}{100}$

⇒$\frac{d}{w}$ = 0.8

Answer 2

specific gravity of a medium = density of the medium / density of water 
         = ρ = ρ_{medium} / ρ_{water}

$= \frac{weight\ in\ air - weight\ in\ medium}{weight\ in\ air - weight\ in\ water}\\ \\= \frac{W_{air}-W_{oil}}{W_{air}-W_{water}} \\ \\= \frac{160-136}{160-130} =\frac{24}{30} = 0.80$