Which of the following statements are correct
1.For water at 100°C at sea level, the vapour pressure is equal to atmospheric pressure.
2.Viscosity of a fluid is the property exhibited by it both in static and in dynamic conditions.
3.Real fluids have lower viscosity then ideal fluids.
4.Dynamic viscosity is the force per unit velocity gradient.
5.Vapour pressure of a liquid is independent of the externally exerted pressure.
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Solution
Dynamic viscosity is the property of fluid in motion in which one layer of fluid exerts viscous force on the other layer.Ideal fluids have no viscosity and surface tension and they are in compressible.Dynamic viscosity is the stress per unit velocity gradient.
The Leq for fluctuating noise level for 100 minutes will be____dB.
Given 40 dB for 20 minutes
60 dB for 30 minutes
50 dB for 15 minutes
55 dB for 35 minutes
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Solution
Leq=10 log\(\sum_{i=1}^{4}10\frac{L_{i}}{10}\times t_{i}\)
⇒\( 10log\left ( \left ( 10^{(40/10)}\times \frac{20}{100} \right )+\left ( 10^{60/10}\times \frac{30}{100} \right ) \\ +\left ( 10^{50/10}\times \frac{15}{100} \right )+\left ( 10^{55/10}\times \frac{35}{100} \right ) \right )\)
⇒10 log((104× 0.2) + (106× 0.3) + (105+ 0.15) + (105.5× 0.35))
⇒10 log (427679.71)
⇒56.311 dB
If the wind velocity at a height of 10m measured as 60 km/hr the velocity at a height of 1 m above the ground can be expected to be in Km/hr.
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Solution
nαch1/7
60 = c(10)1/7....(i)
n= c(1)1/7....(ii)
dividing (ii) by(i)
\(\frac{V}{60}=\frac{1}{(10)^{1/7}}\)
v= 43.18 km/hr.
At a hydel power plant, two turbo generators of capacity 25,000 kw each have been installed. It the load on the plant varies from a minimum of 10,000 KW to a maximum of 40,000 KW then the load factor will be ______%
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Solution
Load factor :\(\frac{Avg.load\, over\, certain\, period}{Peak\, load\, during\, that\, period}\)
\(\Rightarrow \frac{(10,000+40,000)}{\frac{2}{40,000}}\)
\(\Rightarrow \frac{25,000}{40,000}\)
In percentage = \(\frac{25}{40}\)×100=62.5%
A four lane undivided road is at present carrying a traffic of 1485 commercial vehicle per day. It needs to be strengthened for the growing traffic needs. The vehicle damage factor has been found to be 2.8. The rate of growth of traffic is 5% per annum. The period of construction is 5 years. The pavement is to be designed for 20 years after completion.
Lateral distribution factor is 0.4. The cumulative standard axles to be used in design, is _______ msa.
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Solution
P = 1485 cvpd
A = 1485 × (1 + 0.05)5
= 1895.278 cvpd.
∴Design standard axles
=\(\frac{365A[(1+r^{n})-I]}{r}\times LDF\times VDF\)
=\(\frac{365\times 1895.278[1.05^{20}-1]}{.05}\times 0.4\times 2.8\)
= 25619158.829 =25.62 msa
In Marshall method of mix design, the course aggregates,fine aggregates, filler and Bitumen, having respective sp.the Gravities of 2.6, 2.7, 2.65 and 1.02 mixed in the ratio of 0.55: 0.346 : 0.48 : 0.056 respectively. The theoretical sp.gravity of the mix would be ________
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Solution
Gt=\(=\frac{\gamma t}{\gamma w}=\frac{x}{0.575\frac{x}{\gamma w}\times \gamma w}\)
= 1.74
The live load for a sloping roof with slope 15° where access is not provided to roof is taken as
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Solution
⇒0.75 – 0.2(15 – 10)
⇒0.65 KN/m2
A rectangular beam of 300 mm width, 500 mm effective depth which is subjected to ultimate moment of 50KNm, ultimate shear force of 50 KN and Torsional moments of 40 KNm.Consider M-20 grade concrete and Fe 415 grade steel,effective cover of 35 mm. Keeping above data in mind the Area of compression steel required will be ______ mm2
.
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Solution
Mut=\(\frac{T_{u}}{1.7}\left ( 1+\frac{D}{b} \right )\)
=\(\frac{40}{1.7}\left ( 1+\frac{535}{300} \right )\)
= 65.49 KNm
Mut > Mu
Me1= 50 + 65.49 = 115.49 KNm
Me2= 65.49 – 50 = 15.49 KNm
Moment corresponding compression steel = 15.49KNm
\(A_{sc}=\frac{M_{e2}}{0.87f_{y}(d-d')}\)
\(\Rightarrow \frac{15.49\times 10^{6}}{0.87\times 415(500-35)}\)
= 92.27 mm2
Match List I (Type of member) and List II (Basic span to effective depth) and select the correct answer :
List I List II
A. Simply supported one way slab 1.26
B.Continuous one way slab 2.20
C.Simply supported two way slab 3.35
D.Continuous two way slab 4.40
In a system two weightless rigid bars AB and BC of length ‘2a’ each having hinge supports at the ends A and C,respectively are connected to each other at B by a frictionless hinge (Internal hinge). The rotation at the hinge is restrained by a rotational spring of the stiffness K and system assumes a straight line configuration ABC.The rotation at the supports due to vertical load P acting at joint B.
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Solution
work done by load = strain energy stored in the spring
1⁄2×P×δ=1⁄2k(2θ)2
1⁄2×P×θa2=1⁄2K4θ2
Pθa = 2K θ2
θ=\(\frac{Pa}{2k}\)