A page from level field book is as
R.L of Point B is _____________ m
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Solution
True diff. :-\(\frac{(1.75-1.48)+(1.60-1.30)}{2}\)
= 0.285
As point B is lower than A so 99.715 m
A dumpy level was set up at a position equidistant from two pegs A and B. The bubble was adjusted to its central position for each reading as it did not remain quite central when the telescope was moved from A to B. The readings on A and B were 1.481m and 1.591 m respectively. The instrument was then moved to D so that the distance between DB was about 5 times the distance DA and readings with the bubble central were 1.560 m and 1.655 m respectively. Collimation error of the instrument is __________ m.
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Solution
when bubble is central and dumpy level is equidistant,
True diff. = 1.591 – 1.481
= 0.110 m
True differance = (1.655 - 5e) – (1.560 -e)
0.110 = 0.095 – 4e
e = –0.004 m
In testing a dumpy level, following observations were recorded
Instrument location | Reading at A | Reading at B |
---|---|---|
Midway between A and B | 2.615 | 3.175 |
At 25 m from A and 75 m from B | 1.905 | 2.340 |
Then the tan of angle of inclination of the line of collimation will be ___________.
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Solution
True difference = 3.175 – 2.615
= 0.56 m
Apparent difference = 2.34 – 1.905 = 0.435 m
Reading at A= 1.905
Reading at B should be approx = 1.905 + 0.56 =2.465 m
As reading at Bis lesser so ‘α’ is downwards
Correct Reading at A =1.905 + 25 tan α
Correct Reading at B =2.340 + 75 tan α
Difference between A and B = 0.435 + 50 tan α
0.435 + 50 tanα= 0.56 tan α
= 0.0025
Match List I with List II and select a suitable answer by using the codes given below the Lists :-
List-I
A.Braking distance
B.Reaction time
C.Stopping sight distance
D.Overtaking sight distance
List-II
1.The minimum distance within which a vehicle moving with the design speed will overtake another vehicle moving in the same direction.
2.The minimum distance travelled between the instant of applying brakes and stoppage of the Vehicle.
3.The minimum distance within which a vehicle moving with design speed stops after reacting to the presence of obstruction.
4.The distance travelled during the reaction time.
A marshall specimen is prepared for bituminous concrete with a bitumen content of 4% by weight of total mix. The theoretical& measured unit weight of mix are 2.5 g/cm3 and 2.35 g/cm3 respectively. The bitumen has a specific gravity of 1.02. The percent voids in mineral aggregate filled with bitumen are __________
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Solution
\(VFB=\frac{V_{B}}{V_{V}+V_{B}}\times 100\)
\(=\frac{4/1.02}{2.55+4/1.02}\times 100\)
= 60.56%
The amount of methane produced from 2 kg of BOD stablised at STP will be _________ liters.Assume that the BOD contribution is due to glucose (C6 H12 O26)and anaerobic decomposition of glucose produces CO2 and CH4 only.
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Solution
\(\underset{180}{C_{6}H_{12}O_{6}}\rightarrow \underset{132}{3CO_{2}}+\underset{48}{3CH_{4}}\)
180 gm glucose produces→48 gm of CH3
2000 gm glucose will produce\(\frac{48}{180}\)×2000gm of CH4
1 mole of all gases at STP occupies 22.4 liters
CH4(16 gm)will occupy→22.4 liters
\(\frac{48}{180}\)×2000gm will occupy→\(\frac{22.4}{16}\times \frac{48}{180}\times 2000\)liter
=746.66 liters
Using Khosla theory, the value of exit gradient for a weir with a horizontal floor on a permeable foundation having width 15 m and depth of down stream sheet pile 2m will be______.Given the difference between upstream and down stream water level is 3 m.
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Solution
GE=\(\frac{H}{d}\frac{1}{\pi \sqrt{\lambda }}\)
where λ=\(\frac{1+\sqrt{1+\alpha ^{2}}}{2}\)
as α=b⁄d=15⁄2=7.5
\(\lambda =\frac{1+\sqrt{1+7.5^{2}}}{2}\)
= 4.283
GE=\(\frac{3}{2}\frac{1}{\pi \sqrt{4.283}}\)
GE= 0.230
In a watershed four rain gauges I,II, III and IV are installed.The depths of the normal annual rainfall at these stations are 50, 75, 80 and 120 cm respectively.The rain guage of station II went out of order during that particular year. The annual rainfall for that year, recorded at the remaining three stations was 80, 70, 50 cm. The rainfall at station II can be considered as _________ cm.
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Solution
Normal annual rainfall for station II is 75 cm.
75 × 0.9= 67.5 cm
75 × 1.1= 82.5 cm
Since, the neighbouring stations do not have their normal annual rainfall values in the range of 67.5 cm to 82.5 cm i.e. they differ by more than 10% hence, Normal Annual ratio method will be used
\(\frac{P_{2}}{N_{2}}=\frac{1}{m}\left ( \frac{P_{1}}{N_{1}}+\frac{P_{3}}{N_{3}}+\frac{P_{4}}{N_{4}} \right )\)
P2=\(\frac{75}{3}\left ( \frac{80}{50}+\frac{70}{80}+\frac{50}{120} \right )\)
P2= 72.29 cm
A hollow cylinder 1.0m long has internal and external diameters as 0.4m and 0.6m respectively. Both the ends of cylinder are open. If the weight of cylinder is 1000 N. Then
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Solution
GM = BM– BG
=1⁄V-\(\frac{H-h}{2}\)
π⁄4\(\left ( D_{O}^{2}-D_{I}^{2} \right )\gamma _{\omega }=100\)=100
\(h=\frac{1000}{(\frac{\pi }{4})(0.6^{2}-0.4^{2})9810}\)=0.649m
\(\bar{V}=\frac{\pi }{4}(D_{O}^{2}-D_{i}^{2})\, h=\frac{1000}{9810}\)=0.1019m3
\(GM=\frac{\frac{\pi }{64}(D_{O}^{4}-D_{i}^{4})}{0.1019}-\left ( \frac{1-0.65}{2} \right )\)
⇒0.05-0.175=-0.125m
Meta centric height is negative
Match the following pipe flow situations with their associated accelration.
A.Flow at constant rate 1.Zero accelration
passing through a bend.
B.Gradually changing flow 2.Local and convective both
through a straight pipe.
C.Gradually changing flow 3.Convective accelration
through a bend.
D.Flow at constant rate passing 4.Local acceleration
through a straight uniform
diameter pipe.
E.Flow at a constant rate passing through a pipe of gradually changing diameter