
The heart of a man pumps $4000 c.c.$ of blood through the arteries per minute at a pressure of $130 mm of Hg$. If the density of mercury is $13.6 × 10^3 kg{m}^{-3}$, what is the power of the heart?
A) 0.115 HP
B) 1.15 HP
C) 1.15 watt
D) 2.30 watt
Answer
568.8k+ views
Hint
As we know that power is the work/time and work is the product of force and distance and we also know that force is the product of pressure and area. By using all these values we get
$power = \dfrac{{pressure \times area}}{{time}}$
and then substitute the values in this formula we will get required power.
Step by step solution
As power is defined as the amount of work done per unit time i.e.
$ \Rightarrow power = \dfrac{{work}}{{time}}$………………….. (1)
As we also know that
$work = F \times r$
Where, $F$ is the force and $r$ is the distance
Put this value in equation (1), we get
$ \Rightarrow power = \dfrac{{F \times r}}{t}$……………….. (2)
Now we also that pressure is defined as the force per unit area i.e.
$P = \dfrac{F}{A}$
$ \Rightarrow F = P \times A$
Substitute this value in equation (2), we get
$ \Rightarrow power = \dfrac{{P \times A \times r}}{t}$ ……………… (3)
And we also know that volume
$V = A \times r$
On substituting this value in equation in (3), we get
$ \Rightarrow power = \dfrac{{P \times V}}{t}$……………………… (4)
Now, as it is given that volume
$V = 4000 c.c. $
Time $t = 1 min =60 sec$
Density of mercury $\rho = 13.6 \times {10^3}kg{m^{ - 3}}$
Pressure P = 130mm of Hg
$ \Rightarrow P = h\rho g = 0.13 \times 13.6 \times {10^3} \times 9.8$
Put these values in equation (4), we get
$
\Rightarrow power = \dfrac{{0.13 \times 13.6 \times {{10}^3} \times 9.8 \times 4000 \times {{10}^{ - 6}}}}{{60}} \\
\Rightarrow power = \dfrac{{69.3}}{{60}} = 1.15W \\
$
Hence, option (C) is correct.
Note
Human heart works like a hydraulic pump. We can calculate the power of the heart by using work-power relation. As we have work done by heart and rate of work done we can calculate power of heart. mm of Hg is a unit of pressure. It is defined as the column height of mercury column with respect to given pressure.
As we know that power is the work/time and work is the product of force and distance and we also know that force is the product of pressure and area. By using all these values we get
$power = \dfrac{{pressure \times area}}{{time}}$
and then substitute the values in this formula we will get required power.
Step by step solution
As power is defined as the amount of work done per unit time i.e.
$ \Rightarrow power = \dfrac{{work}}{{time}}$………………….. (1)
As we also know that
$work = F \times r$
Where, $F$ is the force and $r$ is the distance
Put this value in equation (1), we get
$ \Rightarrow power = \dfrac{{F \times r}}{t}$……………….. (2)
Now we also that pressure is defined as the force per unit area i.e.
$P = \dfrac{F}{A}$
$ \Rightarrow F = P \times A$
Substitute this value in equation (2), we get
$ \Rightarrow power = \dfrac{{P \times A \times r}}{t}$ ……………… (3)
And we also know that volume
$V = A \times r$
On substituting this value in equation in (3), we get
$ \Rightarrow power = \dfrac{{P \times V}}{t}$……………………… (4)
Now, as it is given that volume
$V = 4000 c.c. $
Time $t = 1 min =60 sec$
Density of mercury $\rho = 13.6 \times {10^3}kg{m^{ - 3}}$
Pressure P = 130mm of Hg
$ \Rightarrow P = h\rho g = 0.13 \times 13.6 \times {10^3} \times 9.8$
Put these values in equation (4), we get
$
\Rightarrow power = \dfrac{{0.13 \times 13.6 \times {{10}^3} \times 9.8 \times 4000 \times {{10}^{ - 6}}}}{{60}} \\
\Rightarrow power = \dfrac{{69.3}}{{60}} = 1.15W \\
$
Hence, option (C) is correct.
Note
Human heart works like a hydraulic pump. We can calculate the power of the heart by using work-power relation. As we have work done by heart and rate of work done we can calculate power of heart. mm of Hg is a unit of pressure. It is defined as the column height of mercury column with respect to given pressure.
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