Question 1 |
A parabolic vertical crest curve connects two road segments with grades +1.0% and -2.0%. If a 200 m stopping sight distance is needed for a driver at a height of 1.2 m to avoid an obstacle of height 0.15 m, then the minimum curve length
should be ______ m. (round off to the nearest integer)
241 | |
365 | |
115 | |
273 |
Question 1 Explanation:
Given that, n_1=+1% and n_2=-2%
n=n_1-n_2=3%
SSD = 200 m
and h_1= 1.2 m and h_2= 0.15 m
as given n_1 up gradient, and n_2 - down gradient.
So curve is summit curve.
Assume L > SSD
\begin{aligned} L &=\frac{NS^2}{2(\sqrt{n_1}+\sqrt{n_2})^2} \\ &= \frac{3}{100} \times \frac{(200)^2}{2 \times (\sqrt{1.2}+\sqrt{0.15})^2}\\ &=272.91>200 \\ L&=272.91m \end{aligned}
n=n_1-n_2=3%
SSD = 200 m
and h_1= 1.2 m and h_2= 0.15 m
as given n_1 up gradient, and n_2 - down gradient.
So curve is summit curve.
Assume L > SSD
\begin{aligned} L &=\frac{NS^2}{2(\sqrt{n_1}+\sqrt{n_2})^2} \\ &= \frac{3}{100} \times \frac{(200)^2}{2 \times (\sqrt{1.2}+\sqrt{0.15})^2}\\ &=272.91>200 \\ L&=272.91m \end{aligned}
Question 2 |
The base length of the runway at the mean sea level (MSL) is 1500 m. If the
runway is located at an altitude of 300 m above the MSL, the actual length
(in m) of the runway to be provided is ____________. (round off to the nearest
integer)
1245 | |
2354 | |
1605 | |
1248 |
Question 2 Explanation:
Correction for elevation = It should increase at a
rate of 7% per 300 m rise in elevation from MSL.
Given that
Basic runway length as MSL =1500m
Elevation =300m
Correction =\frac{7}{100}\times \frac{300}{300}\times 1500=105m
The actual length of runway= 1500+150=1605m
Given that
Basic runway length as MSL =1500m
Elevation =300m
Correction =\frac{7}{100}\times \frac{300}{300}\times 1500=105m
The actual length of runway= 1500+150=1605m
Question 3 |
The vehicle count obtained in every 10 minute interval of a traffic volume survey done in peak one hour is given below.
\begin{array}{|c|c|}\hline \text{Time Interval}& \text{Vehicle Count} \\ \text{(in minutes)}& \\ \hline 0-10& 10\\ \hline 10-20 &11\\ \hline 20-30&12\\ \hline 30-40&15\\ \hline 40-50 & 13\\ \hline 50-60 &11\\ \hline \end{array}
The peak hour factor (PHF) for 10 minute sub-interval is __________. (round off to one decimal place)
\begin{array}{|c|c|}\hline \text{Time Interval}& \text{Vehicle Count} \\ \text{(in minutes)}& \\ \hline 0-10& 10\\ \hline 10-20 &11\\ \hline 20-30&12\\ \hline 30-40&15\\ \hline 40-50 & 13\\ \hline 50-60 &11\\ \hline \end{array}
The peak hour factor (PHF) for 10 minute sub-interval is __________. (round off to one decimal place)
0.2 | |
0.4 | |
0.8 | |
0.1 |
Question 3 Explanation:
\begin{aligned}
PHF&=\frac{\text{Peak flow during 1hour}}{6 \times \text{Peak flow during 10 minutes}}\\
&=\frac{10+11+12+15+13+11}{6 \times 15}\\
&=\frac{72}{6 \times 15}\\
&=0.8
\end{aligned}
Question 4 |
A horizontal curve is to be designed in a region with limited space. Which of
the following measure(s) can be used to decrease the radius of curvature?
Decrease the design speed. | |
Increase the superelevation. | |
Increase the design speed. | |
Restrict vehicles with higher weight from using the facility. |
Question 4 Explanation:
e+f=\frac{V^2}{127R}
R=\frac{V^2}{127(e+f)}
R=\frac{V^2}{127(e+f)}
Question 5 |
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 5 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 6 |
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 6 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 7 |
A highway designed for 80 km/h speed has a horizontal curve section with radius 250 m. If the design lateral friction is assumed to develop fully, the required super elevation is
0.02 | |
0.05 | |
0.07 | |
0.09 |
Question 7 Explanation:
\begin{aligned} V &=80 \mathrm{kmph}, R=250 \mathrm{~m} \\ e+f &=\frac{V^{2}}{127 R} \\ e+0.15 &=\frac{80^{2}}{127 \times 250} \\ e &=0.051 \end{aligned}
Question 8 |
The shape of the most commonly designed highway vertical curve is
circular (single radius) | |
circular (multiple radii) | |
parabolic | |
spiral |
Question 8 Explanation:
The ideal vertical curve is a 2^{\circ} parabola.
Question 9 |
The design speed of a two-lane two-way road is 60 km/h and the longitudinal coefficient
of friction is 0.36. The reaction time of a driver is 2.5 seconds. Consider acceleration
due to gravity as 9.8 m/s^2. The intermediate sight distance (in m, round off to the nearest
integer) required for the load is ________
124 | |
246 | |
186 | |
162 |
Question 9 Explanation:
Given : f = 0.36; v = 60 km; g = 9.8 m/s^2;\; t_R=2.5s
\begin{aligned} SSD&= \left ( 0.278Vt_R+\frac{V^2}{254f} \right )\\ &= 0.278 \times 60 \times 2.5 +\frac{60^2}{254 \times 0.36}\\ &= 41.7+39.37=81m\\ ISD &=2 \times SSD\\ &=81 \times 2=162m \end{aligned}
\begin{aligned} SSD&= \left ( 0.278Vt_R+\frac{V^2}{254f} \right )\\ &= 0.278 \times 60 \times 2.5 +\frac{60^2}{254 \times 0.36}\\ &= 41.7+39.37=81m\\ ISD &=2 \times SSD\\ &=81 \times 2=162m \end{aligned}
Question 10 |
In an urban area, a median is provided to separate the opposing streams of traffic. As
per IRC : 86-1983, the desirable minimum width (in m, expressed as integer) of the
median, is __________.
1.2 | |
5 | |
7.5 | |
2.6 |
Question 10 Explanation:
As per IRC : 86-1983
Desirable minimum width of median in urban roads = 5 m
Desirable minimum width of median in urban roads = 5 m
There are 10 questions to complete.
Very helpful. Thanku so much🔥
Please check the solution of question 28
M= (R-d)-(R-d) cos a/2 For two lane
You are wrong bro !
m= R-(R-d)cos(a/2)
solution is correct.