Question

The human heart discharges $75c{m^3}$ of blood per beat against an average pressure of $10\;cm$ of Hg. Assuming that the pulse frequency is $75/\min$ , the power of the heart is (density of Hg = $13.6gm/c{m^3}$) :
(A) $1.275\;W$
(B) $12.5\;W$
(C) $0.125\;W$
(D) $125\;W$

Answer 1

Hint: - As we know that power is defined as the rate of the amount of work done and work is the product of force and distance and we also know that the force is the product of pressure and area. By using all these values we get the required formula and then substitute the values in this formula we will get the required power.
Formula used:
$power = \dfrac{{pressure \times area}}{{time}}$

Complete step-by-step solution:
As the power is the total amount of work done per unit time i.e.
$power = \dfrac{{work}}{{time}}$ ……….. $\left( 1 \right)$
As we also know that
$work = F \times r$
where $F$ is the force and $r$ is the distance
Putting the above value of force in the equation $\left( 1 \right)$ , we get
$power = \dfrac{{F \times r}}{t}$ ………. $\left( 2 \right)$
At present, we also know that pressure is the amount of force per unit area i.e.
$P = \dfrac{F}{A}$
$\Rightarrow F = P \times A$
On Substituting the above value in the equation $\left( 2 \right)$ , we get
$power = \dfrac{{P \times A \times r}}{t}$ ………… $\left( 3 \right)$
And we also know that volume
$V = A \times r$
On substituting the above value in the equation $\left( 3 \right)$ , we get
$power = \dfrac{{P \times V}}{t}$ ……………. $\left( 4 \right)$
Now, as it is given that volume
$V = 75c{m^3}$
Pulse frequency = $75/\min$
Therefore, the number of pulse in one second = $\dfrac{{75}}{{60}}$
Density of mercury $\rho = 13.6 \times {10^3}kg{m^{ - 3}}$
Pressure $P = 10cm$of Hg
$P = h\rho g = 0.1 \times 13.6 \times {10^3} \times 9.8$
On putting the above values in the equation$\left( 4 \right)$, we get
$Power = \dfrac{{0.1 \times 13.6 \times {{10}^3} \times 9.8 \times 75 \times 75 \times {{10}^{ - 6}}}}{{60}}$
$\Rightarrow power = \dfrac{{76.5}}{{60}} = 1.275W$

Hence, the correct option is (A) $1.275\;W$ .

Note: Human heart works like a hydraulic pump. We can determine the power of the heart through work-power relations. As we have work done by heart and rate of work done we can determine the power of the heart. $1$ mm of hg is a unit of pressure. It is called the column height of the mercury column with respect to a known pressure.