Question

In standing person of $180 cm$ height, the blood pressure at feet level is higher than at head level (due to hydrostatic component) by (assume gravitational acceleration $g = 9.81 m s^{-2}$ and blood density $d = 1.04 g cm^{-3}$ )
a) $183.6 hPa$
b) $18.36 Pa$
c) $1836 Pa$
d) $1020 Pa$

Answer 1

Hint: Height of a person, density of blood and gravitational constant are given. We have to find blood pressure. Pressure is given by the product of density, acceleration due to gravity and height. First, we convert all the values of a given variable into standard units.

Complete step-by-step solution:
Given: height of a person, $h = 180 cm = 1.8 m$
Density of blood, $d = 1.04 g cm^{-3} = 1.04 \times 10^{3} Kg m^{-3}$
$d = 1040 Kg m^{-3}$
gravitational acceleration, $g = 9.81 m s^{-2}$
Blood Pressure is given by: $ P = dgh$
where, d is the density of blood.
g is the gravitational acceleration.
h is the height of the person.
Put all the values of d, g and h in $P = dgh$.
$P = 1040 \times 9.81 \times 1.8$
It gives,
$P = 18364.3 Pa$
In (a) $183.6 hPa$, hPa is hecto pascals which equals to $100 Pa$.
So, $183.6 hPa$ means $18360 Pa$.
The blood pressure is $18364.3 Pa$.
Option (a) is correct.

Note: The pressure applied by an immobile fluid or hydrostatic pressure is the pressure in a balanced system that depends only upon the height of the fluid, the density of the fluid, and the acceleration due to gravity. Its units are the alike as the pressure. As the depth rises, the pressure applied by the fluid becomes high.