The human ciculatory system can be thought of as a closed system of interconnecting pipes through which fluid is continuously circulated by two pumps the two pumps the right and left verticles of the heart, work as simple two-stroke force pumps. The muscles of the heart regulate the force by contracting and relaxing. the contraction (systole) lasts about 0.2s and a complete systole/diastole (contraction/relaxation) cycle lasts about 0.8s. For flood pressures and speeds in the normal range. the volume flow rate of blood through a blood vessel is directly proportional tot he pressure difference over a length of the vessel and to the fourth power of the radius of the vessel. The total mechanical energy per unit volume of blood just as it leaves the heart is `E//V=rhogh+P+rhov^(2)`
Q. During intense exercise the volume of blood pumped per second by an athlete,s heart increases by a factor of 7, and his blood pressure increases by 20%/by what factor does the power output of the heart increase during exercise?

Options

A. 1.2
B. 3.5
C. 7
D. 8.4

Answer

To solve the problem, we need to determine the factor by which the power output of the heart increases during intense exercise. We will use the relationship between power, blood pressure, and the volume of blood pumped.

### Step-by-Step Solution:

1. **Understanding Power Output**:
The power output \( P \) of the heart can be expressed as:
\[ P = \text{Blood Pressure} \times \text{Volume Flow Rate} \]
where the volume flow rate is the amount of blood pumped per second.

2. **Given Information**:
- The volume of blood pumped per second increases by a factor of \( 7 \).
- The blood pressure increases by \( 20\% \).

3. **Expressing the Increase in Blood Pressure**:
An increase of \( 20\% \) in blood pressure can be expressed as:
\[ \text{New Blood Pressure} = \text{Old Blood Pressure} \times (1 + 0.20) = \text{Old Blood Pressure} \times 1.2 \]

4. **Calculating the New Power Output**:
Let \( P_0 \) be the original power output. The new power output \( P_1 \) can be calculated as:
\[ P_1 = (\text{New Blood Pressure}) \times (\text{New Volume Flow Rate}) \]
Substituting the expressions we have:
\[ P_1 = (1.2 \times \text{Old Blood Pressure}) \times (7 \times \text{Old Volume Flow Rate}) \]

5. **Expressing the New Power Output in Terms of Old Power Output**:
Since \( P_0 = \text{Old Blood Pressure} \times \text{Old Volume Flow Rate} \), we can rewrite \( P_1 \):
\[ P_1 = 1.2 \times 7 \times P_0 = 8.4 \times P_0 \]

6. **Finding the Factor of Increase in Power Output**:
The factor by which the power output increases is:
\[ \text{Factor of Increase} = \frac{P_1}{P_0} = 8.4 \]

### Conclusion:
The power output of the heart increases by a factor of \( 8.4 \) during intense exercise.