If ruling gradient is I in 20 and there is also a horizontal curve of radius 76 m, then the compensated grade should be
A. 3 %
B. 4%
C. 5 %
D. 6%
Answer: Option B
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Highway facilities are designed for
A. annual average hourly volume
B. annual average daily traffic
C. thirtieth highest hourly volume
D. peak hourly volume of the year
The provision of traffic signals at intersections
A. reduces right angled and rear end collisions
B. increases right angled and rear end collisions
C. reduces right angled collisions but may increase rear end collisions
D. reduces rear end collisions but may increase right angled collisions
In CBR test the value of CBR is calculated at
A. 2.5 mm penetration only
B. 5.0 mm penetration only
C. 7.5 mm penetration only
D. both 2.5 mm and 5.0 mm penetrations
If aggregate impact value is 20 to 30 percent, then it is classified as
A. exceptionally strong
B. strong
C. satisfactory for road surfacing
D. unsuitable for road surfacing
Sol
Here,
Ruling gradient= 1 in 20, 1/20=0.05, 0.05×100%=5%
Radius of curve(R)=76m
Then,
Grade compensation=30+R/R %
=30+76/76
=1.39%
Max. amount of grade compensation=75/R %
=75/76
=0.98%
Take min. Value,
Grade compensation=0.98%
Compensated grade=Ruling gradient -Grade compensation
=5%-0.98%
=4.02% it's near to 4%
The answer is Option B
Maximum value of { 30 +R /R , or 75/R
= (30+76) /76 , or 75/76
= 1.39, or .98
Maximum value = 1.39
Here, rolling gradient= 1÷20 = 0.05 × 100% = 5%
So, 5% - 1.39% = 3.61%
Since, minimum grade compensation= 4% , maximum= 6.7%
So, here results came as 3.61% (should consider minimum grade compensation of 4%)
So, answer is = 4%
Rulling gradient=1:20=5 %
Grade Compensation(%)=(30+76)/76=1.39
Maximum limit of grade compensation=75/76
=0.986
Compensated Grade=5-(.986)=4.014>4
Therefore compensatwd grade will be 4 %