Question

Let \( R_1, R_2 \) and \( R_3 \) be the Reynolds numbers corresponding to the following three cases, respectively: 
Case 1: A 1 \(\mu\)m long bacterium swimming in water
Case 2: A 10 cm long fish swimming in water
Case 3: A 10 m long whale swimming in water
If each of them covers a distance equal to their respective body length in one second, which one of the following is CORRECT?

• A. \( R_1>R_2>R_3 \)
• B. \( R_1<R_2<R_3 \)
• C. \( R_3<R_1<R_2 \)
• D. \( R_2<R_1<R_3 \)

Hint:

The Reynolds number is directly proportional to the square of the characteristic length. Therefore, larger objects moving at the same speed have larger Reynolds numbers.

Answer 1

The Correct Option is B

Reynolds number (\( R \)) is a dimensionless quantity that describes the flow regime of a fluid, and is given by: \[ R = \frac{\rho v L}{\mu} \] where:
\( \rho \) is the density of the fluid,
\( v \) is the velocity of the object relative to the fluid,
\( L \) is the characteristic length of the object,
\( \mu \) is the dynamic viscosity of the fluid.
Step 1: Calculating Reynolds number for each case. Since all three cases describe an object moving at a velocity equal to its body length per second, the velocity \( v \) is equal to \( L \) (the characteristic length). Thus, Reynolds number simplifies to: \[ R = \frac{\rho L^2}{\mu} \] Step 2: Comparing the Reynolds numbers.
The Reynolds number is proportional to the square of the characteristic length of the object. Thus, the larger the object, the larger the Reynolds number.
In Case 1, the bacterium is 1 \(\mu\)m long, so \( R_1 \) is the smallest.
In Case 2, the fish is 10 cm long, so \( R_2 \) is larger than \( R_1 \).
In Case 3, the whale is 10 m long, so \( R_3 \) is the largest.
Conclusion: The correct answer is (B) \( R_1<R_2<R_3 \).