Question 1 |
For a 2^{\circ} curve on a high speed Broad Gauge (BG) rail section, the maximum sanctioned speed is 100 km/h and the equilibrium speed is 80 km/h. Consider dynamic gauge of BG rail as 1750 mm. The degree of curve is defined as the angle subtended at its center by a 30.5 m arc. The cant deficiency for the curve (in mm,round off to integer) is ________________
45 | |
65 | |
57 | |
28 |
Question 1 Explanation:
\begin{aligned} \text { Length of curve }&=\text { Radius } \times \text { Degree of curve }\\ \frac{30.5 \times 180}{2^{\circ} \times \pi}&=R\\ R &=873.76 \mathrm{~m} \\ C_{d} &=C_{t h}-C_{a c t} \\ C_{\mathrm{act}} &=\frac{G V^{2}}{127 R}=\frac{1750 \times 100^{2}}{127 \times 873.76}=157.70 \mathrm{~mm} \\ C_{\mathrm{ev}} &=\frac{G V^{2}}{127 R}=\frac{1750 \times 80^{2}}{127 \times 873.76}=100.930 \mathrm{~mm} \\ C_{\mathrm{def}} &=C_{\mathrm{act}}-C_{t h}=157.70-100.930=56.77 \mathrm{~mm} \\ &=57 \mathrm{~mm} \end{aligned}
Question 2 |
The stopping sight distance (SSD) for a level highway is 140 m for the design speed of 90 km/h. The acceleration due to gravity and deceleration rate are 9.81 \mathrm{~m} / \mathrm{s^2} and 3.5 \mathrm{~m} / \mathrm{s^2}, respectively. The perception/reaction time (in s,round off to two decimal places) used in the SSD calculation is ______________
6.14 | |
5.02 | |
2.02 | |
4.12 |
Question 2 Explanation:
\begin{aligned} \mathrm{SSD} &=140 \mathrm{~m} \\ V &=90 \mathrm{kmph} \\ a &=3.5 \mathrm{~m} / \mathrm{s}^{2} \\ \mathrm{SSD} &=V t_{R}+\frac{V^{2}}{2 g f} \\ a &=g f \\ 140 &=\left(\frac{5}{18} \times 90 \times t_{R}\right)+\frac{\left(\frac{5}{18} \times 90\right)^{2}}{2 \times 3.5} \\ t_{R} &=2.028 \mathrm{~seconds} \end{aligned}
Question 3 |
In a three-phase signal system design for a four-leg intersection, the critical flow ratios for each phase are 0.18, 0.32, and 0.22. The total loss time in each of the phases is 2s. As per Webster's formula, the optimal cycle length (in s, round off to the nearest integer) is _______________
28 | |
46 | |
50 | |
58 |
Question 3 Explanation:
\begin{aligned} L &=2 \times 3=6 \text { seconds } \\ n &=3 \\ y &=(0.18+0.32+0.22) \\ C_{0} &=\left(\frac{1.5 L+5}{1-y}\right)=\left(\frac{1.5 \times 6+5}{1-(0.18+0.32+0.22)}\right) \\ &=50 \text { seconds } \end{aligned}
Question 4 |
The softening point of bitumen has the same unit as that of
distance | |
temperature | |
time | |
viscosity |
Question 4 Explanation:
Softening point is the temperature at which bitumen becomes soft and starts flowing.
Question 5 |
Relationship between traffic speed and density is described using a negatively sloped straight line. If v_{f} is the free flow speed then the speed at which the maximum flow occurs is
0 | |
\frac{v_{f}}{4} | |
\frac{v_{f}}{2} | |
v_{f} |
Question 5 Explanation:
Speed at maximum flow =\frac{V_{f}}{2}
Question 6 |
The longitudinal section of a runway provides the following data:
\begin{array}{|c|c|} \hline \text { End-to-end runway (m) } & \text { Gradient (\%) } \\ \hline 0 \text { to 300 } & +1.2 \\ \hline 300 \text { to } 600 & -0.7 \\ \hline 600 \text { to } 1100 & +0.6 \\ \hline 1100 \text { to } 1400 & -0.8 \\ \hline 1400 \text { to } 1700 & -1.0 \\ \hline \end{array}
The effective gradient of the runway (in %, round off to two decimal places) is _____________
\begin{array}{|c|c|} \hline \text { End-to-end runway (m) } & \text { Gradient (\%) } \\ \hline 0 \text { to 300 } & +1.2 \\ \hline 300 \text { to } 600 & -0.7 \\ \hline 600 \text { to } 1100 & +0.6 \\ \hline 1100 \text { to } 1400 & -0.8 \\ \hline 1400 \text { to } 1700 & -1.0 \\ \hline \end{array}
The effective gradient of the runway (in %, round off to two decimal places) is _____________
0.12 | |
0.32 | |
0.54 | |
0.69 |
Question 6 Explanation:
Assuming RL of start of runway as datum i.e., RL = 0 m)
\begin{aligned} \text { Effective gradient } &=\left[\frac{\text { Maximum difference in reduced level }}{\text { Total runway length }}\right] \\ &=\left[\frac{4.5-(-0.9)}{1700} \times 100\right] \% \\ &=0.3176 \% \simeq 0.32 \% \end{aligned}
\begin{aligned} \text { Effective gradient } &=\left[\frac{\text { Maximum difference in reduced level }}{\text { Total runway length }}\right] \\ &=\left[\frac{4.5-(-0.9)}{1700} \times 100\right] \% \\ &=0.3176 \% \simeq 0.32 \% \end{aligned}
Question 7 |
Spot speeds of vehicles observed at a point on a highway are 40, 55, 60, 65 and 80 km/h. The space-mean speed (in km/h, round off to two decimal places) of the observed vehicles is ______________
56.99 | |
25.36 | |
47.85 | |
96.36 |
Question 7 Explanation:
\begin{aligned} V_{1}=40, V_{2}=55, V_{3}=60, & V_{4}=65, V_{5}=80 \\ V_{S}=&\left(\frac{n}{\frac{1}{V_{1}}+\frac{1}{V_{2}}+\ldots+\frac{1}{V_{n}}}\right)=\frac{1}{40}+\frac{1}{55}+\frac{1}{60}+\frac{1}{65}+\frac{1}{80} \\ =& 56.99 \mathrm{kmph} \end{aligned}
Question 8 |
Vehicular arrival at an isolated intersection follows the Poisson distribution. The mean vehicular arrival rate is 2 vehicle per minute. The probability (round off to two decimal places) that at least 2 vehicles will arrive in any given 1-minute interval is _____
0.59 | |
0.47 | |
0.24 | |
0.89 |
Question 8 Explanation:
P(n,t)=\frac{(\lambda t)^n e^{-\lambda t}}{n!}
\lambda =2 \;v/m
t = 1 \text{ minute}
probability of at least 2 vehicles arriving in 1 ? minute is given by,
1-(P(0,1)+P(1,1))
\begin{aligned} &= 1-\left \{ \frac{(2 \times 1)^0 e^{-2(1)}}{0!}+\frac{(2 \times 1)^1 e^{-2(1)}}{1!} \right \}\\ &=1-\left \{ \frac{1(e^{-2})}{1}+ \frac{2(e^{-2})}{1} \right \} \\ &= 1-(e^{-2}+2e^{-2})\\ &=1-3e^{-2}=0.59 \end{aligned}
\lambda =2 \;v/m
t = 1 \text{ minute}
probability of at least 2 vehicles arriving in 1 ? minute is given by,
1-(P(0,1)+P(1,1))
\begin{aligned} &= 1-\left \{ \frac{(2 \times 1)^0 e^{-2(1)}}{0!}+\frac{(2 \times 1)^1 e^{-2(1)}}{1!} \right \}\\ &=1-\left \{ \frac{1(e^{-2})}{1}+ \frac{2(e^{-2})}{1} \right \} \\ &= 1-(e^{-2}+2e^{-2})\\ &=1-3e^{-2}=0.59 \end{aligned}
Question 9 |
On a road, the speed - density relationship of a traffic stream is given by u=70 - 0.7k (where speed, u, is in km/h and density, k is in veh/km). At the capacity condition, the average time headway will be
0.5s | |
1.0s | |
1.6s | |
2.1s |
Question 9 Explanation:
\begin{aligned} \mathrm{u}&=70-0.07 \mathrm{k} \\ \mathrm{u}&=70\left[1-\frac{\mathrm{k}}{\frac{70}{0.7}}\right]\\ V_{f} &=70 \mathrm{kmph} \\ \mathrm{k}_{\mathrm{j}} &=\frac{70}{0.7}=100 \mathrm{veh} / \mathrm{km} \\ \mathrm{q}_{\max } &=\frac{1}{4} V_{f} k_{j}=\left(\frac{1}{4} \times 70 \times 100\right)=1750 \mathrm{veh} / \mathrm{hr} \\ \mathrm{q}_{\max } &=\frac{3600}{h_{i}} \\ 1750 &=\frac{300}{h_{i}}\\ & \text{Average time headway}\\ h_{i}&=\frac{3600}{1750}=2.057=2.1 \mathrm{sec} \end{aligned}
Question 10 |
A signalized intersection operates in two phases. The lost time is 3 seconds per phase. The maximum ratios of approach flow to saturation flow for the two phases are 0.37 and 0.40. The optimum cycle length using the Webster's method (in seconds, round off to one decimal place) is ______________
44.8 | |
92.6 | |
25.8 | |
60.9 |
Question 10 Explanation:
\begin{aligned} \mathrm{n} &=2, \mathrm{~L}=3 \times 2=6 \mathrm{sec} \\ \mathrm{y}_{1} &=0.37, \mathrm{y}_{2}=0.40 \\ \mathrm{Y} &=\mathrm{y}_{1}+\mathrm{y}_{2} \\ &=0.37+0.40 \\ &=0.77 \\ \mathrm{C}_{0} &=\left(\frac{1.5 \mathrm{~L}+5}{1-\mathrm{Y}}\right)=\frac{1.5 \times 6+5}{1-0.77} \\ &=60.87 \mathrm{sec} \simeq 60.9 \mathrm{sec} \end{aligned}
There are 10 questions to complete.
