134. The shearing force acting along the slice of a curved surface of slippage, causes the soil to slide

A. down at the centre

B. down at the toe

C. upward at the centre

D. none of these.

135. Failure of a slope occurs only when total shear force is

A. equal to total shearing strength

B. greater than total shearing strength

C. less than total shearing strength

D. none of these.

136. If S, L and R are the arc length, long chord and radius of the sliding circle then the perpendicular distance of the line of the resultant cohesive force, is given by

A.a=SR/L

B. a= LS/R

C. a=LR/S

D. none of these.

137. If the failure of a finite slope occurs through the toe, it is known as

A. slope failure

B. face failure

C. base failure

D. toe failure.

138. The method of the slices is applicable to

A. homogenous soils

B. stratified soils

C. saturated soils

D. non-uniform slopes

139. In a purely cohesive soil, the critical centre lies at the intersection of

A. perpendicular bisector of slope and the locus of the centre

B. perpendicular drawn at 1/3rd slope from toe and the locus of the centre

C. perpendicular drawn at 2/3rd slope from toe and the locus of the centre

D. directional angles

140. If the cohesive force, (c), is 1.5 t/m2, the density (y) of the soil is 2.0 t/m3, factor of safety (F) is 1.5 and stability factor (SO is 0.05, the safe height of the slope, is

A. 5 metrets

B. 8 metres

C. 10 metres

D. 12 metres.

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