If a liquid of density ρ and coefficient of viscosity η flow with a velocity v through a pipe of diameter D, then the Reynold’s number is X.
If the velocity of the liquid flowing through the pipe increases to 2v and the diameter of the pipe is reduced to D/4 (keeping all the other parameters the same), the new Reynold’s number is Y.
What will be the value of X : Y?
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A.
Sat Dec 30 1899 01:04:00 GMT+0000 (Coordinated Universal Time)
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B.
Sat Dec 30 1899 04:01:00 GMT+0000 (Coordinated Universal Time)
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C.
Sat Dec 30 1899 01:02:00 GMT+0000 (Coordinated Universal Time)
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D.
Sat Dec 30 1899 02:01:00 GMT+0000 (Coordinated Universal Time)
Correct Answer:
D. Sat Dec 30 1899 02:01:00 GMT+0000 (Coordinated Universal Time)
Explanation:
The Reynolds number is defined by the formula Re = (?vD) / ?. Initially, the Reynolds number X is equal to (?vD) / ?. When the velocity is doubled to 2v and the diameter is reduced to D/4, the new Reynolds number Y becomes (? * 2v * D/4) / ?, which simplifies to (?vD) / (2?) or X/2. To find the ratio X : Y, we compare X to X/2, which simplifies to 2:1. Therefore, the correct ratio is 2:01.
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