Advertisements

*Depth of the tank = 10m**L, B, H, of triangle = (12 x 12 x 12**RD = 0.8*

*Allowing, 3% of the volume of the oil loaded for expansion.**Area of the triangle = 1/2 (base x height)*

*In triangle PQR, RQ in the height which divide the base at same length.**So In triangle PSR*

*SR ^{2} =PR^{2} –PS^{2}*

*= (12*

^{2}– 5^{2})*= (144 -25)*

*=119*

*So, SR = 10.9m**So area of the triangle = 1/2 (10×10.9)**=54.5m ^{2}*

*Volume of the tank = (10 x54.5)**= 545m ^{3}*

*Now, volume of the oil = (Total volume -free space)*

**l**et ‘V’ be the volume of the oil*545 = (V + 3/100 x V)**545 = (V + 3V/ 100)**545 = (103V /100)**V = (545 x 100) /103**V = 529.126m ^{3}*

*Mass of the oil = (density x volume)*

= 0.8 x 529.126*=423.3 t*