Q1.An isolated 3-h rainfall event on a small catchment produces a hydrograph peak and point of
inflection on the falling limb of the hydrograph at 7 hours and 8.5 hours respectively, after
the start of the rainfall. Assuming, no losses and no base flow contribution, the time of
concentration (in hours) for this catchment is approximately
Answer : Option DExplaination / Solution:
For small catchment, time of concentration is equal to lag time of peak flow.
Q2.The Muskingum model of routing a flood through a stream reach is expressed as
O2 = K0I2 + K1I1 + K2O1, where K0, K1 and K2 are the routing coefficients for the concerned
reach, I1 and I2 are the inflows to reach, and O1 and O2 are the are the outflows from the reach
corresponding to time steps 1 and 2 respectively. The sum of K0, K1 and K2 of the model is
Answer : Option DExplaination / Solution: No Explaination.
Q4.A road is being designed for a speed of 110 km/hr on a horizontal curve with a super
elevation of 8%. If the coefficient of side friction is 0.10, the minimum radius of the curve (in
m) required for safe vehicular movement is
Q5.X is 1 km northeast of Y. Y is 1 km southeast of Z. W is 1 km west of Z. P is 1 km south of
W. Q is 1 km east of P. What is the distance between X and Q in km?
Q7.On a section of a highway the speed-density relationship is linear and is given by v = [80-(2/3)k]; where v is in km/h and k is in veh/km. The capacity (in veh/h) of this section of the highway would be
Q8.The survey carried out to delineate natural features, such as hills, rivers, forests and manmade
features, such as towns, villages, buildings, roads, transmission lines and canals is
classified as
Answer : Option DExplaination / Solution:
Topographic survey is carried out to delineate natural features.
Q10.The chainage of the intersection point of two straights is 1585.60 m and the angle of
intersection is 140o . If the radius of a circular curve is 600.00 m, the tangent distance (in m)
and length of the curve (in m), respectively are
Answer : Option CExplaination / Solution:
Length of curve, L = (π/180)× ∆×R
= (π/180)×40×600 = 418.88m
Tangent distance, T = R tan(∆/2)
= 600tan(40/2) = 218.38m
Total Question/Mark :
Scored Mark :
Mark for Correct Answer : 1
Mark for Wrong Answer : -0.5
Mark for Left Answer : 0