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# If the heart pushes 1cc of blood in one second under pressure $20000N{m^{-2}}$ the power of the heart is:A) $0.02W$B) $400W$C) $20mW$D) $0.2W$

Last updated date: 10th Sep 2024
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Hint: Power of any equipment is defined as the rate at which work is done or rate at which energy gets transferred from a place to another place or a form to another form.
$P = \dfrac{{\Delta W}}{{\Delta t}}$
Where $P$ is the power of the heart, $\Delta W$ is the work done by the heart and $\Delta t$ is the time taken by the heart.
In case of a system when the pressure is constant, the word done is;
$\Delta W = p\Delta V$
Where $\Delta W$ is the work done by the heart, $p$ is the pressure of the heart and $\Delta V$ is the volume of blood pumped every second.

Formulae used:
$P = \dfrac{{\Delta W}}{{\Delta t}}$
Where $P$ is the power of the heart, $\Delta W$ is the work done by the heart and $\Delta t$ is the time taken by the heart.
$\Delta W = p\Delta V$
Where $\Delta W$ is the work done by the heart, $p$ is the pressure of the heart and $\Delta V$ is the volume of blood pumped every second.

Complete step by step solution:
Power of any equipment is defined as the rate at which work is done or rate at which energy gets transferred from a place to another place or a form to another form.
It is given in the question that
$\Delta V = 1cc = {10^{ - 6}}{m^3}$
$p = 20000N{m^{ - 2}} = 2 \times {10^4}N{m^{ - 2}}$
$\Delta t = 1s$
In situations when power is to be calculated from pressure, we first confirm whether the pressure is constant or not and then use the appropriate formula for the work done. In this case;
$P = \dfrac{{\Delta W}}{{\Delta t}}$
$\Delta W = p\Delta V$
$\Rightarrow P = \dfrac{{p\Delta V}}{{\Delta t}}$
$\Rightarrow P = \dfrac{{2 \times {{10}^4}\left( {{{10}^{ - 6}}} \right)}}{1}$
$\Rightarrow P = 2 \times {10^{ - 2}}Nm{s^{ - 1}}$
$\Rightarrow P = 0.02W$

Thus the correct option is $\left( A \right), 0.02W.$

Note: It is important first check that all your units are in S.I. units, so as to not cause unnecessary mistakes. Also the final answer should also be in S.I. units. Note that the heart in this case acts as any motor pump would, using its power to cause a change in work done and hence a change in volume.