**T***he* volume of the rectangular tank = (L x B x H)**=(30 x16 x14)****Ullage pipe projecting above its top = 0.5m****RD of oil = 0.78****The length of the pipe line from tank to the ship = 10km****= 10000m****The diameter of the pipe = (d)****= 40cm****Hence, radius = (40 /2 )****= 20cm**

**Volume of the tank = (L x B x H)****= (30m x16m x14m)**** =6720m ^{3}**

**Area = (L x B)**** = (30 x 16)**** = 480m ^{2}**

**The total ullage =0.78m**

**So ,ullage inside the tank = (0.78 – 0.50)****= 0.28m**

**If ullage inside the tank = 0.25m,The depth of oil to be loaded is**** = (14 -0.28) m**** =13.72m**

**Volume of the oil = (L x B x D)****=( 30x 16x 13.72)m****= 6585.6 m ^{3}**

**Mass of the oil to be load = (volume x density)****= (6585.6 x 0.78)****= 5136.768 t**

**Now , volume of oil in the pipe = (πr ^{2}h)**

**=(3.1416 x 0.2 x 0.2 x 10000)**

**=1256.64m**^{3}**Depth of oil when it poured in to the tank ****= (volume /area)****= (1256.64 /480)****= 2.618m**

**Depth of the tank without the pipe line oil ****= (13.72 -2.618)**** = 11.102m**

**Thus the ullage will be = (14 -11.102)****= 2.898m**

**Exact ullage where pump should be shut = (2.898 + 0.5)m**

** = 3.398m.**

**Exact ullage where pump should be shut = (2.898 + 0.5)m**

**= 3.398m.**