A tank 4m x 3m x 2m containing an oil of specific gravity 0.83 is moved with an acceleration g/2 m sec2. The ratio of the pressures at its bottom when it is moving vertically up and down, is

ANS:B - 3

When the tank is moving vertically up or down with an acceleration, the pressure at the bottom of the tank is affected by the acceleration of the tank and the gravitational force acting on the fluid inside. Let's denote:

• Pu​ as the pressure at the bottom when the tank is moving upward,
• Pd​ as the pressure at the bottom when the tank is moving downward,
• ρ as the density of the oil,
• g as the acceleration due to gravity.

When the tank is moving upward, the effective acceleration acting on the fluid is g+g/2=3g/2​. When the tank is moving downward, the effective acceleration acting on the fluid is g−g/2=g/2​. Using the hydrostatic pressure equation: P=ρgh The pressure at the bottom of the tank is directly proportional to the density of the fluid and the height of the fluid column above the point where pressure is measured. The height of the fluid column is the same whether the tank is moving up or down. So, the ratio of pressures is: pu/pd=pg(height of fluid column)/pg(height of fluid column)=(3/2)g/(1/2)g=3​​ Therefore, the ratio of pressures at the bottom when the tank is moving vertically up and down is 3.