Explanation: The frost heave results in the lifting or swelling of the pavement surface. If the surface of the pavement has a differential rise due to frost heave, then the pavement will be damaged very badly. The differential rise happens because the centre of the pavement may rise higher than the sides or vice versa.

So, for the load 3629 kg, the design repetitions would be 250×\(\frac{15.31}{100}\)×365×20×8=2235260

Explanation: Formation width is the total width of the pavement, it is not an environmental factor. The environmental factors that affect the design of pavements can be related to the cutting and filling activities and the water table of the area. The height of embankment/filling, depth of cutting/excavation of the soil and the depth of water tale are considered.

\(P^{‘}=antilog(3.528)=3372.87 kg\)

Explanation: There are two ways to perform the capillary cut off to reduce the frost action. It can be done by inserting a layer of granular or bituminous material so that the rise of water can be stopped. Lowering the water table is not always possible. Laying pipes and use of geofibres are not employed as remedies for frost action.

Explanation: The equation relating the clear gap (d), centre to centre spacing (S) and radius of contact area (r) is given by S=(d+2a). So, S can be found out as S=70+2×150=370 mm.

Explanation: The formation of ice crystals does lead to the frost heave, but it is not formed on the pavement surface. The ice crystals are formed due to the freezing of water that is held in the subgrade soil.

Explanation: The strength evaluation of pavement materials is based on two things – CBR and elastic modulii. To find the elastic modulii, plate bearing test and triaxial compression tests are used. Triaxial compression test gives the resilient modulus of the subgrade soil.

Explanation: Equivalent wheel load is the equivalency of the load under consideration to the 2268 kg load. The number of repetitions of load 2268 kg i.e. 105000 is divided by the number of repetitions corresponding to the load under consideration i.e. 6500. The equivalent wheel load is hence obtained as 105000/6500 = 16.15. It is written as the closest power of 2, so the equivalent load factor would be 16.

Therefore, the design repetitions for the load 4082 kg would be 210×\(\frac{7.66}{100}\)×365×20×16=1878844.8

Explanation: Temperature is not a parameter affecting the moisture content of the subgrade, it is a factor affecting the design of pavements. The fluctuations in the water table, the type of pavement and shoulder, the capillary action and vapour movement influence the moisture content in the subgrade.

Explanation: The wheels of a modern bullock cart are steel wheels. When these wheels tread the pavement, they impose high stress on the pavement surface. While the stress is negligible in the bottom layers due to their small gross weight.

Explanation: The trailer portion carries the major load portion of the tractor-trailer unit. The axle provided is the tandem axle which consists of two axles on which two wheels are placed on one side of each axle. The tractor part consists of a single axle single wheel and single axle dual wheel assembly.

Explanation: Liquidity of a soil indicates the natural moisture content present in the soil. Soils having high plasticity are prone to a greater extent of swelling and shrinking than those with low plasticity and thus they tend to damage the pavement.

Reliability value is expressed as a percentage and it gives the probability that particular damage to the pavement would remain under permissible limits during its design period.