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An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum
diameters of the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the maximum
velocities of fluid in this pipe is
diameters of the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the maximum
velocities of fluid in this pipe is
- a)9/16
- b)(v3)/2
- c)81/256
- d)3/4
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
An ideal fluid flows (laminar flow) through a pipe of non-uniform diam...
Given information:
- Fluid flows through a pipe of non-uniform diameter.
- Maximum diameter of the pipe = 6.4 cm
- Minimum diameter of the pipe = 4.8 cm
To find:
The ratio of the minimum and maximum velocities of fluid in this pipe.
Solution:
1. Velocity of fluid in a pipe is inversely proportional to the cross-sectional area of the pipe. When the diameter of the pipe decreases, the cross-sectional area decreases, and hence the velocity increases.
2. The cross-sectional area of a pipe is given by the formula: A = πr^2, where r is the radius of the pipe.
3. Let the maximum velocity of the fluid be V_max and the corresponding radius be r_max.
4. According to the given information, the maximum diameter of the pipe is 6.4 cm. Therefore, the maximum radius is r_max = 6.4/2 = 3.2 cm = 0.032 m.
5. Similarly, let the minimum velocity of the fluid be V_min and the corresponding radius be r_min.
6. According to the given information, the minimum diameter of the pipe is 4.8 cm. Therefore, the minimum radius is r_min = 4.8/2 = 2.4 cm = 0.024 m.
7. The ratio of the velocities is given by: V_min/V_max = A_max/A_min, where A_max and A_min are the cross-sectional areas of the pipe at maximum and minimum diameters, respectively.
8. Substituting the values of the radii into the formula for cross-sectional area, we get: A_max = π(0.032)^2 and A_min = π(0.024)^2.
9. Simplifying the ratio, we have: V_min/V_max = (π(0.024)^2)/(π(0.032)^2) = (0.024/0.032)^2 = (3/4)^2 = 9/16.
10. Therefore, the ratio of the minimum and maximum velocities of the fluid in the pipe is 9/16.
Hence, the correct answer is option 'A' (9/16).
- Fluid flows through a pipe of non-uniform diameter.
- Maximum diameter of the pipe = 6.4 cm
- Minimum diameter of the pipe = 4.8 cm
To find:
The ratio of the minimum and maximum velocities of fluid in this pipe.
Solution:
1. Velocity of fluid in a pipe is inversely proportional to the cross-sectional area of the pipe. When the diameter of the pipe decreases, the cross-sectional area decreases, and hence the velocity increases.
2. The cross-sectional area of a pipe is given by the formula: A = πr^2, where r is the radius of the pipe.
3. Let the maximum velocity of the fluid be V_max and the corresponding radius be r_max.
4. According to the given information, the maximum diameter of the pipe is 6.4 cm. Therefore, the maximum radius is r_max = 6.4/2 = 3.2 cm = 0.032 m.
5. Similarly, let the minimum velocity of the fluid be V_min and the corresponding radius be r_min.
6. According to the given information, the minimum diameter of the pipe is 4.8 cm. Therefore, the minimum radius is r_min = 4.8/2 = 2.4 cm = 0.024 m.
7. The ratio of the velocities is given by: V_min/V_max = A_max/A_min, where A_max and A_min are the cross-sectional areas of the pipe at maximum and minimum diameters, respectively.
8. Substituting the values of the radii into the formula for cross-sectional area, we get: A_max = π(0.032)^2 and A_min = π(0.024)^2.
9. Simplifying the ratio, we have: V_min/V_max = (π(0.024)^2)/(π(0.032)^2) = (0.024/0.032)^2 = (3/4)^2 = 9/16.
10. Therefore, the ratio of the minimum and maximum velocities of the fluid in the pipe is 9/16.
Hence, the correct answer is option 'A' (9/16).
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An ideal fluid flows (laminar flow) through a pipe of non-uniform diam...
A1V1 =A2V2 where A =area and V= velocity
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An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimumdiameters of the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the maximumvelocities of fluid in this pipe isa)9/16b)(v3)/2c)81/256d)3/4Correct answer is option 'A'. Can you explain this answer?
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