Question

The minimum pressure required to force the blood of density $1.04gc{m}^{-3}$ from the heart to the top of the head (vertical distance = $60cm$) is

Answer 1

Step 1: Given data & Formula used

In a body, the velocity of blood is the same everywhere, thus, $v_1=v_2$.

Let $\rho =1.04gc{m}^{-3}$ and $g=9.8m{s}^{-2}=980cm{s}^{-2}$

The difference in height between the top and the bottom point will be: $h_2-h_1=60cm$.

Formula used:

According to Bernoulli's theorem, $P_1-P_2=\rho g(h_2-h_1)+\frac{1}{2}\rho (v_2^2-v_1^2)$

Where, $P_1\to$Pressure at top

$P_2\to$ Pressure at bottom

$\rho \to$ density of blood

$g\to$ acceleration due to gravity

$h_1\to$ height at the bottom point

$h_2\to$ height at top point

$v_1\to$ velocity at bottom point

$v_2\to$ velocity at top point

Step 2: Substituting the given values in the equation and obtaining the result

Now, on substituting the values in the formula, we get:

$P_2-P_1=1.04\times 980\times 60dynec{m}^{-2}$

$\Rightarrow P_2-P_1=6.1\times {10}^{4}dynec{m}^{-2}$

Minimum pressure required to force the blood of density $1.04gc{m}^{-3}$ from the heart to the top of the head is $6.1\times {10}^{4}dynec{m}^{-2}$.