A given arteriole has a resistance of 2. What would the new resistance...
Resistance in a tube is equal to

, where L is the length of the tube, n is the viscosity, and r is the radius.
Therefore, the resistance is proportional to 1/r
4
= The new resistance would be 1/8
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A given arteriole has a resistance of 2. What would the new resistance...
Given Information:
- Resistance of the arteriole = 2
- The radius of the arteriole doubles
To find:
- The new resistance of the arteriole after the radius doubles
Formula:
- Resistance (R) is inversely proportional to the fourth power of the radius (r) of the vessel.
- Mathematically, R ∝ 1/r^4
Solution:
1. Initial Resistance:
- According to the given information, the initial resistance of the arteriole is 2.
2. Relationship between Resistance and Radius:
- The resistance of a vessel is inversely proportional to the fourth power of its radius.
- When the radius of a vessel doubles, the resistance will change accordingly.
3. Doubling the Radius:
- Let's assume the initial radius of the arteriole is 'r'.
- When the radius doubles, the new radius becomes '2r'.
4. Relationship between Initial and New Resistance:
- According to the formula, the initial resistance is inversely proportional to the fourth power of the initial radius: R ∝ 1/r^4.
- The new resistance will be inversely proportional to the fourth power of the new radius: R' ∝ 1/(2r)^4.
5. Calculating the New Resistance:
- Substituting the new radius (2r) into the formula, we get: R' ∝ 1/(2r)^4 = 1/16r^4.
- The new resistance is 1/16 times the initial resistance.
- Therefore, the new resistance of the arteriole after the radius doubles is 1/16 of the initial resistance.
6. Answer:
- The new resistance of the arteriole after the radius doubles is 1/8 (which is equivalent to 1/16).