Exams › GATE › Technical › CE (Set No 3) Stage 2
8 questions with worked solutions.
Answer: c/(F_cγH)
Taylor's stability number is Sn = c_mobilized/(gamma H) with mobilized cohesion c/Fc, giving Sn = c/(Fc x gamma x H). That is option index 2, not the stored cFc/(gamma H).
Q2. The notation “SC” as per Indian Standard Soil Classification System refers to
Answer: Clayey sand
The notation "SC" in the Indian Standard Soil Classification System specifically denotes clayey sand, which is a soil type that contains a significant amount of sand mixed with clay, affecting its engineering properties.
Answer: √(k_z/kₓ)
To draw an isotropic flow net for anisotropic soil, the horizontal dimensions are scaled by the factor sqrt(k_z/k_x) (transformed x = x*sqrt(k_z/k_x)). The stored answer sqrt(kx/kz) is the reciprocal; the correct multiplying factor is sqrt(k_z/k_x).
Answer: I_L and I_N are always equal
Both the Lagrange polynomial and the Newton polynomial are methods of polynomial interpolation that yield the same interpolating polynomial when given the same set of data points. Therefore, their estimates at any point x* will be equal.
Answer: v_f k_j / 4
The maximum flow in the Greenshield's model occurs at half the free flow speed and half the jam density, leading to the formula for maximum flow being v_f k_j / 4, which represents the optimal balance between speed and density.
Answer: 19.93 minutes
The relationship between disinfection efficiency and time is often logarithmic, meaning that each additional log reduction in microorganisms requires a significantly longer time. Since achieving 50% disinfection takes 3 minutes, reaching 99% disinfection (which is a 2-log reduction) typically requires about 6 times that initial time, leading to approximately 19.93 minutes.
Answer: 2800 m
To achieve a scale of 1:8000, the height at which the aircraft flies must be calculated based on the focal length of the camera and the elevation of the terrain. The correct height of 2800 m ensures that the scale is maintained, allowing for accurate aerial photographs of the terrain.
Answer: 180
Standard axle = 80 kN, VDF = (P/80)^4. Buses: 80 x 2 axles of 40 kN = 160 x 0.0625 = 10. Trucks: 160 x [(40/80)^4 + (80/80)^4] = 160 x 1.0625 = 170. Total ESAL = 10 + 170 = 180.