Related Questions

The ratio of the hydraulic radius to the diameter of the channel, for maximum mean velocity of flow in a circular channel, in open channel flow is
A 0.3
B 0.9
C 0.03
D 0.66
The ratio of the depth of flow to the diameter of the channel for maximum discharge in a circular channel in open channel flow is
A 0.1
B 0.55
C 0.95
D 1.85
Like Thevenin's theorem, Norton's theorem provides a method of reducing a more complex circuit to a simpler, more manageable form for analysis.
A True
B False
If B and d are the bed width and depth of a channel in meters, the combined losses due to evaporation and seepage in cumecs per kilometer length of the channel, is
A $$\frac{1}{{50}} \times {\left( {{\text{B}} + {\text{d}}} \right)^{\frac{2}{3}}}$$
B $$\frac{1}{{100}} \times {\left( {{\text{B}} + {\text{d}}} \right)^{\frac{2}{3}}}$$
C $$\frac{1}{{150}} \times {\left( {{\text{B}} + {\text{d}}} \right)^{\frac{2}{3}}}$$
D $$\frac{1}{{200}} \times {\left( {{\text{B}} + {\text{d}}} \right)^{\frac{2}{3}}}$$
The critical depth for a channel is given by (where q = Unit discharge (discharge per unit width) through the channel)
A $${\left( {\frac{{\text{q}}}{{\text{g}}}} \right)^{\frac{1}{2}}}$$
B $${\left( {\frac{{{{\text{q}}^2}}}{{\text{g}}}} \right)^{\frac{1}{3}}}$$
C $${\left( {\frac{{{{\text{q}}^3}}}{{\text{g}}}} \right)^{\frac{1}{4}}}$$
D $${\left( {\frac{{{{\text{q}}^4}}}{{\text{g}}}} \right)^{\frac{1}{5}}}$$
একটি চ্যানেল দিয়ে প্রতি সেকেন্ডে সর্বোচ্চ 106 বিট ডেটা ট্রান্সফার হলে ঐ চ্যানেলে ব্যান্ডউইথ কত হবে? (If 106 bits data are transmitted at most per second through a channel, then what will be bandwidth of that channel?)
A 1MBps
B 1Mbps
C 10KBps
D 10Kbps
The major reason of hydrodynamic noise (i.e., noise resulting from liquid flow) is
A Pipe vibrations
B Cavitation
C Boundary layer separation
D Fluctuation in liquid flow
Aerodynamic noise resulting from turbulent gas flow is the most prevalent source of valve noise in fluid ilow control. It is caused due to
A Reynold stresses
B Shear forces
C Both A & B
D Neither A nor B
A rectangular channel has cross sectional area of 50 m2 If the channel section is to be most economical, calculate the depth. Take Breadth = 10m .
A 10 m
B 5 m
C 8 m
D 12 m
In a two lane channel, bottom width of channel is given by
A Manoeuvring lane + (2 × Bank clearance lane)
B (2 × Manoeuvring lane) + (2 × Bank clearance lane)
C (2 × Manoeuvring lane) + (2 × Bank clearance lane + ship clearance lane)
D Manoeuvring lane + (2 × Bank clearance lane) + ship clearance lane

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