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, the Reynolds number will be
A. 300
B. 337.5
C. 600
D. 675
Answer: Option D
Join The Discussion
Comments ( 3 )
Related Questions on Hydraulics and Fluid Mechanics
A. 22.5 m/sec.
B. 33 m/sec.
C. 40 m/sec.
D. 90 m/sec.
A. the weight of the body
B. more than the weight of the body
C. less than the weight of the body
D. weight of the fluid displaced by the body
The difference of pressure between the inside and outside of a liquid drop is
A. $${\text{p}} = {\text{T}} \times {\text{r}}$$
B. $${\text{p}} = \frac{{\text{T}}}{{\text{r}}}$$
C. $${\text{p}} = \frac{{\text{T}}}{{2{\text{r}}}}$$
D. $${\text{p}} = \frac{{2{\text{T}}}}{{\text{r}}}$$
A. cannot be subjected to shear forces
B. always expands until it fills any container
C. has the same shear stress.at a point regardless of its motion
D. cannot remain at rest under action of any shear force
I think answer should be 6750
Explaination -
# Given -
r = 15 cm = 0.15 m
v = 2 m/s
μ = 0.8 poise
ρ = 0.9×1000 = 900 kg/m^3
# Solution -
Reynold's number is calculated by -
R = 2.ρ.v.r / μ
R = 2 × 900 × 2 × 0.15 / 0.8
R = 675
Therefore, Reynolds number iz 675
How plz Explain,,,?