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.

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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.

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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.

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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.

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138. The method of the slices is applicable to
A. homogenous soils
B. stratified soils
C. saturated soils
D. non-uniform slopes

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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

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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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