Question

A container of 100 cm^2 cross sectional area is having a liquid of mass 5 kg .What is the pressure at bottom of the container

Answer 1

Pressure applied at any point is equal to Force per area
I.e. P = F/A
therefore Pressure at bottom of container is
Force applied by mass of liquid upon area
P = mg/ A
P = 5×10/100×10^-4. ( area is in cm^2)
P = .5 ×10^4
P = 5000

Answer 2

Answer:

The pressure acting at bottom of the container is 5kPa.

Explanation:

The pressure is defined as the physical force exerted on an object. It is the force acting per unit area. i.e.

$P=\frac{F}{A}$           (1)

Where,

P=pressure acting on an area

F=force exerted on a particular area

A=area over which the force acts

From the question we have,

A=100cm²=100×10⁻⁴=10⁻²m²           (1cm=10⁻²m)

m=5kg

And the force acting on the bottom of the container is given as,

$F=mg$          (2)       (g=acceleration due to gravity=10m/s²)

By substituting the required values in equation (2) we get;

$F=5\times 10=50N$    (3)

By putting the value of F and A in equation (1) we get;

$P=\frac{50}{10^-^2}=5000Pa$

$P=5kPa$

Hence, the pressure acting at bottom of the container is 5kPa.