[amp_mcq option1=”300″ option2=”337.5″ option3=”600″ option4=”675″ correct=”option1″]
The correct answer is $\boxed{\text{B)} 337.5}$.
The Reynolds number is a dimensionless number that is used to characterize the flow of a fluid around a solid object. It is defined as follows:
$$Re = \frac{\rho v D}{\mu}$$
where:
- $\rho$ is the density of the fluid,
- $v$ is the velocity of the object relative to the fluid,
- $D$ is the characteristic length of the object, and
- $\mu$ is the dynamic viscosity of the fluid.
In this case, the density of the fluid is $\rho = 0.9 \times 1000 \text{ kg}/\text{m}^3 = 900 \text{ kg}/\text{m}^3$, the velocity of the object is $v = 2 \text{ m}/\text{s}$, the characteristic length of the object is the radius of the sphere, $D = 0.15 \text{ m}$, and the dynamic viscosity of the fluid is $\mu = 0.8 \text{ poise} = 0.8 \times 10^{-3} \text{ Pa} \cdot \text{s}$. Substituting these values into the definition of the Reynolds number, we get:
$$Re = \frac{(900 \text{ kg}/\text{m}^3)(2 \text{ m}/\text{s})(0.15 \text{ m})}{(0.8 \times 10^{-3} \text{ Pa} \cdot \text{s})} = 337.5$$
Therefore, the Reynolds number for a sphere of radius 15 cm moving with a uniform velocity of 2 m/sec through a liquid of specific gravity 0.9 and dynamic viscosity 0.8 poise is 337.5.