Question

The heart of a man pumps 5 litres of blood through the arteries per minute at a pressure of $150mm$ of mercury. If the density of mercury be $13.6\times {{10}^{3}}kg{{m}^{-3}}$ and $g=10m{{s}^{-2}}$, then the power of heart in watt is:
$\begin{align}
  & \text{A}\text{. }1.50 \\
 & \text{B}\text{. }1.70 \\
 & \text{C}\text{. }2.35 \\
 & \text{D}\text{. }3.0 \\
\end{align}$

Answer 1

- Hint: The pump law gives us the relationship between flow rate of a fluid and its pumping velocity. For calculating the power generated by pumping, we can use the expression that gives the pumping power in the form of pressure rise across the pump, and the pumping velocity.

Formula used:
$\text{Power of heart}=\dfrac{PV}{t}$
Where,
$P$ is the pressure at which the blood pumps
$V$ is the volume of the pumped blood
$t$ is the time

Complete step-by-step solution
Centrifugal pumps obey a law known as Pump law, which states that the flow rate or capacity is directly proportional to the pumping speed and the power required by the pump is directly proportional to the cube of pumping speed.
\[m=\rho Av\]
Where,
$m$ is the mass flow rate in $Kg{{s}^{-{{1}^{{}}}}}$
$\rho $ is the density in $Kg{{m}^{-3}}$
$A$ is the area in ${{m}^{2}}$
$v$ is the velocity in $m{{s}^{-1}}$
Flow rate is described as the volume of fluid passing by some location or a particular point through an area during a period of time.
Expression for flow rate:
$Q=Vt$
Where,
$Q$ is the flow rate
$V$ is the volume
$t$ is the elapsed time
Flow rate is defined as the volume of fluid per unit time flowing past a point through a fixed area.
The volume of blood pumped by man’s heart is,
$V=5\text{ litres}$
That is,
$\begin{align}
  & V=5\times {{10}^{-3}}{{m}^{3}} \\
 & \left( \because 1\text{ litre = }{{10}^{-3}}{{m}^{3}} \right) \\
\end{align}$
The time in which this volume of blood pumps,
$t=1\min =60\sec $
The pressure at which the blood pumps,
$P=150mm\text{ of Hg = }0.15m\text{ of Hg}$
That is,
$\begin{align}
  & P=\left( 0.15m \right)\times \left( 13.6\times {{10}^{3}}Kg{{m}^{3}} \right)\times \left( 10m{{s}^{-2}} \right) \\
 & P=0.15\times 13.6\times {{10}^{3}}\times 10 \\
 & P=20.4\times {{10}^{3}}N{{m}^{-2}} \\
\end{align}$
Now,
Power of the heart is given as,
$\text{Power of heart}=\dfrac{PV}{t}$
Where,
$P$ is the pressure at which the blood pumps
$V$ is the volume of the pumped blood
$t$ is the time
Putting values,
$\begin{align}
  & P=20.4\times {{10}^{3}}N{{m}^{-2}} \\
 & V=5\times {{10}^{-3}}{{m}^{3}} \\
 & t=60\sec \\
\end{align}$
We get,
\[\begin{align}
  & \text{Power of heart = }\dfrac{PV}{t}=\dfrac{20.4\times {{10}^{3}}\times 5\times {{10}^{-3}}}{60} \\
 & \text{Power of heart = }\dfrac{80}{60}=1.70W
\end{align}\]
The power of the heart is $1.70W$
Hence, the correct option is B.

Note: Students should not get confused between the terms, pressure and the flow rate. Glow rate refers to the amount of fluid coming out of a fixture in a certain amount of time, while pressure refers to the amount of force that is put on the fluid to make it move from one place to another or the amount of force that the fluid exerts when coming out of the fixture.