Answer
$SG\approx 1.53$
Work Step by Step
Calculate the pressure due to oil:
$P_{oil}=\gamma_{oil}\cdot h_{oil}=8500\cdot 5.0=42,500\text{ Pa}$
Calculate the pressure due to unknown liquid:
$P_{unknown}=P_{total}-P_{oil}=65,000-42,500=22,500\text{ Pa}$
Calculate the specific weight of the unknown liquid:
$\gamma_{unknown}=\frac{P_{unknown}}{h_{unknown}}=\frac{22,500}{1.5}=15,000\text{ N/m}^3$
Determine specific gravity:
$SG=\frac{\gamma_{unknown}}{\gamma_{water}}=\frac{15,000}{9810}\approx 1.53$
