Question

A tank of oil has height of 4 m and density of 850 kg m\(^{-3} \). The gauge pressure at the bottom of the tank is (1 atm = 10\(^5\) Pa, Acceleration due to gravity = 10 ms\(^{-2} \))

• A. \( 34 \) kPa
• B. \( 384 \) kPa
• C. \( 284 \) kPa
• D. \( 200 \) kPa

Hint:

Gauge pressure at a depth in a fluid depends only on the density of the fluid, the acceleration due to gravity, and the depth. Use the formula \( P_{gauge} = \rho g h \). Ensure that all units are in the SI system before calculation. Convert the final answer to the required unit (kPa in this case).

Answer 1

The Correct Option is A

The gauge pressure at a certain depth \( h \) in a fluid of density \( \rho \) is given by the formula: $$ P_{gauge} = \rho g h $$ where \( g \) is the acceleration due to gravity.
Given: Height of the oil tank \( h = 4 \) m Density of the oil \( \rho = 850 \) kg m\(^{-3} \) Acceleration due to gravity \( g = 10 \) ms\(^{-2} \) Substitute the values into the formula: $$ P_{gauge} = (850 \text{ kg m}^{-3}) \times (10 \text{ ms}^{-2}) \times (4 \text{ m}) $$ $$ P_{gauge} = 8500 \times 4 \text{ kg m}^{-1} \text{ s}^{-2} $$ $$ P_{gauge} = 34000 \text{ Pa} $$ Convert the pressure from Pascal (Pa) to kilopascal (kPa): $$ P_{gauge} = 34000 \text{ Pa} \times \frac{1 \text{ kPa}}{1000 \text{ Pa}} = 34 \text{ kPa} $$ The gauge pressure at the bottom of the tank is 34 kPa.