Q1: If the atmospheric pressure on the surface of an oil tank (sp. gr. 0.8) is 0.1 kg/cm^2, the pressure at a depth of 2.5 m, is

ANS:C - 3 metre of water

Given, Atmospheric pressure P, oil = 0.1kg/cm² = 10000N/m².

{How I converted 0.1kg/cm² to N/m².
For this, we should know.
1kg = 10N & 1m = 100Cm.
So, (0.1x10)N÷(100-¹x100-¹)m².
= 10000N/m²}.

Density of oil = Specific gravity of oil x Density of water.
=>0.8x1000 = 800kg/m³.

The pressure at 2.5 m depth of oil surface will be:

P = Poil+{Density of oil (p)x Acc due to gravity(g) x depth of oil surface(h)}.
P = 10000N/m²+(800x10x2.5)N/m²,
P = 30000N/m² or Pa.

Now, converting to water head:

P = {Density of water(p) x Acc due to gravity(g) x depth of water(h)}
=>30000N/m²=1000x10xh
h = 3 meters of the water column. Sure, let me explain in more detail.

The pressure exerted by a fluid increases with depth. In this case, we have an oil tank with a specific gravity of 0.8 and an atmospheric pressure of 0.1 kg/cm2 on the surface. To find the pressure at a depth of 2.5 m, we can use the formula:

Pressure = Atmospheric pressure + (Density of fluid x Acceleration due to gravity x Depth).

Since the specific gravity of oil is less than 1, we can use the formula:
Pressure = Atmospheric pressure + (Specific gravity x Atmospheric pressure x Depth).

Plugging in the values, we get:
Pressure = 0.1 kg/cm2 + (0.8 x 0.1 kg/cm2 x 2.5 m)
Simplifying the equation, we find that the pressure at a depth of 2.5 m is equivalent to the pressure exerted by 3 m of water.