Angle a has a weight of 3 and angle b has a weight of 4.Then weight of a+ b is
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Solution
weight of α + β=1(1/3)(1/4)=\(\frac{1}{(1/3)(1/4)}=12/7\)
Match List I and List II
A.Spectral resolution 1.Frequency of receiving radiations.
B.Temporal resolution 2.The wave length to in remote sensing which the remote sensing system is sensitive.
If the velocity of moving vehicles on a road is 24kmph,stopping distance is 19 metres and average length of vehicles is 6 metres, the basic capacity of lane, is –
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Solution
Basic Capacity =24×\(\frac{1000}{(19+6)}\)
= 960 \(\frac{veh.}{hour}\approx 1000\frac{veh.}{hour}\approx 1000\, veh/hour\)
The population equivalent of a city, given that the average sewage from the city is 100 × 106 lit/day and the average 5 day B.O.D is 320 mg/lit will be _______. Domestic Sewage
Qty. = 0.08 kg/person 1/day.
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Solution
4,00,000
Avg. Sewage = 100 × 106 lit/day
BOD5= 320 mg/lit
Total BOD5 Sewage = 320 mg/lit × 100 × 106 lit/day
= 32,000 kg/day
Population equivalent=\(\frac{32,000}{0.08}\)= 4,00,000
The following data pertain to a natural drain crossing an irrigation canal.
which of the following types of cross-drainage should bere commended in this case?
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Solution
As per given data,
Hence, this is a Syphon Aqueduct
A bridge has an expected life of 100 years and is designed for a flood magnitude of return period 100 years. Then the risk associated with this hydrologic deign will be _______%
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Solution
63.39
Risk = 1– qn
Where, q = 1 – p
and p = 1⁄T
Hence,Risk = 1-\(\left ( 1-\frac{1}{100} \right )^{100}\)
=1-(99⁄100)100
=1-(0.99)100
= 0.6339 = 63.39%
Characteristic length of a circular drainage pipe having depth(d) to diameter(D) ratio 0.5 will be
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Solution
Characteristic length =A⁄T
A=1⁄2(π⁄4D2)
T = D
LC=πD⁄8
or LC=\(\frac{\pi (2d)}{8}=\frac{\pi d}{4}\)
A Newtonian fluid fills the clearance between a shaft and a sleeve. When a force of 800N is applied to the shaft, parallel to the sleeve, the shaft attains a speed of 4.5cm/sec.If a force of 4KN is applied instead, the shaft will move with the speed of
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Solution
force = Fs=µ(V⁄T)πD×L
\(\frac{F_{s}}{V}=\frac{\pi DL\mu }{t}=constant\)
\(\frac{800}{4.5}=\frac{4000}{V}×V51\times1.5=\, 22.5cm/sec\)
A hollow hemispherical object of diameter D was immersed in water with its plane surface coinciding with the free surface .The Vertical component of force on the curved surface is given by FV =
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Solution
Vertical force =wt. of water contained in the hemisphere
Match list – I and list – II for different stages of flow in a pipeline and select the correct answer using the codes given below the list
List I | List II |
A. Laminar flow | 1. f = 64/Re |
B. Smooth turbulent flow (Re < 105) | 2. 1⁄√f = -0.8 + 2log(Re)√f |
C.Rough turbulent flow | \(3.\; f = \frac{0.3164}{\left ( R_{e} \right )^{1/4}}\) |
D. Smooth turbulent flow (Re > 106) | 4. 1⁄√f = 1.74 + 2log(Re⁄K) |
What is the Answer A B C D?