Question

The gravitational force exerted by an object is given by F 5 mg where F is the force in newtons, m is the mass in kilograms, and g is the acceleration due to gravity, 9.81 m/s². Calculate the force exerted per unit of area by a column of mercury (density 5 13.59 g/cm³) that is 76.0 cm high. How high would a column of water (density 5 1.00 g/cm³) have to be to exert the same force?

Answer 1

Answer:

Part a)

$P = 1.013 \times 10^5 Pa$

Part b)

$H = 10.3 m$

Explanation:

As we know that force due to gravity is given as

$F = mg$

now for force per unit area is given as

$P = \frac{F}{A}$

$P = \frac{\rho Ah g}{A}$

$P = \rho g H$

so for mercury we will have

$P = (13.59 \times 10^3) (9.81)(0.76)$

$P = 1.013 \times 10^5 Pa$

Part b)

Now if water also apply same force per unit area

then we will have

$P = \rho g H$

$1.013 \times 10^5 = (1000) 9.81 H$

$H = 10.3 m$