A 600 mm square bearing plate settles by 15 mm in plate load test on a cohesionless soil under an intensity of loading of 0.2 N/ram². The settlement of a prototype shallow footing 1 m square under the same intensity of loading is

For Cohesive soil or clay,
(i) Sf/Sp = Bf/Bp
But for Cohesionless soil or sand,
(ii) Sf/Sp = [ Bf (Bp+0.3) / Bp (Bf+0.3) ]^2
Here, it is cohesionless soil, so eq. (ii) will be apply
=》 Sf/0.015 = [ 1 (0.6 +0.3) / 0.6 (1 + 0.3) ]^2

=》 Sf / 0.015 = [ 0.9 / (0.6* 1.3) ]^2

=》 Sf/0.015 = 1.33

=》 Sf = 0.01995 m = 19.95 mm
Which is in between 15mm to 25 mm..
So, Answer B is correct.

Sf=sp(bf(bp+30)÷bp(bf+30)^2 15(100(60+30)÷60(100+30). 9000÷7800 =1.15. 1.15^2 = 1.3225×15=19.83mm. Settlement of plate sp=15mm Width of plate Bf=1m=100cm. Width of plate Bp=600mm=60cm.

Pp=Sf × Wp/Wf
15= Sf × 600/1000
Sf= 25mm....exact

Please someone describe this