Answer:
![\Delta p = 90.7 kPa](https://img.qammunity.org/2020/formulas/physics/high-school/ysg51hv5gqaocy5f1a5qbm3gxzpsx5ntho.png)
Step-by-step explanation:
specific gravity of oil is
![= (\rho_(oil))/(\rho_w)](https://img.qammunity.org/2020/formulas/physics/high-school/g98gc8bc0p132q7r7u4b3bwh52mov6rm3x.png)
![\rho_(oil) = 0.85*1000 = 850 kg/m3](https://img.qammunity.org/2020/formulas/physics/high-school/sff85zbi2zqvdxfs9jrm9sxtgngmdl3p25.png)
we know that
change in pressure for oil is given as
![\Delta p = \rho gh](https://img.qammunity.org/2020/formulas/physics/high-school/km7r5lph6t4znvd3wl3yiaxpgnehqaiwdz.png)
here density and h is for oil
![\Delta p = 850*5 *9.81 = 41,692.5 kPa](https://img.qammunity.org/2020/formulas/physics/high-school/x9987juziy4qkchili6b0ll8j4tp6potyc.png)
change in pressure for WATER is given as
![\Delta p = \rho gh](https://img.qammunity.org/2020/formulas/physics/high-school/km7r5lph6t4znvd3wl3yiaxpgnehqaiwdz.png)
here density is for water and h is for water
![\Delta p = 1000*5 *9.81 = 49,050 kPa](https://img.qammunity.org/2020/formulas/physics/high-school/yt6euizoqyub1knfumndpudc0p4225c6pn.png)
pressure change due to both is given as
![\Delta p = 41692.3 + 49050 = 90742.5 N/m2](https://img.qammunity.org/2020/formulas/physics/high-school/i6xo7bddlur2u4t1igz2jx83vkjlgox8o5.png)
![\Delta p = 90.7 kPa](https://img.qammunity.org/2020/formulas/physics/high-school/ysg51hv5gqaocy5f1a5qbm3gxzpsx5ntho.png)