If velocity is zero over $${\frac{1}{3}^{{\text{rd}}}}$$ of a cross-section and is uniform over remaining $${\frac{2}{3}^{{\text{rd}}}}$$ of the cross-section, then the correction factor for kinetic energy is
A. $$\frac{4}{3}$$
B. $$\frac{3}{2}$$
C. $$\frac{9}{4}$$
D. $$\frac{{27}}{8}$$
Answer: Option C
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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
Ua=(2/3).Umax
KE correction Fctr= (umax³ .(2/3 A))/(ua³)(A)
= 9/4
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