Civil Engineering - UPSC Civil Service Exam Questions - Discussion

Angularity number = 67-(100w/CG),
W=Weight of aggregate in kg,
G= Specific Gravity,
C= Volume (in Kg in termed water filled that volume),
Angularity index= (3 Angularity number/20)+1.

Angularity number = 67-(100w/CG).
= 67-(100 * 5.2/1000 * 3 * 2.65)
= 1.59.

W=Weight of aggregate in kg,
G= Specific Gravity,
C= Volume (in Kg in termed water filled that volume),
Angularity index= (3 Angularity number/20)+1.
= ( 3* 1.59/20)+1.
= 1.238 (correct answer).

The weight of actual aggregate to be filled= 2.65*1000= 2650 kg per 1-metre cube.
Per .003 m^3 volume.
Req aggregate = 2650*.003 = 7.95 kg,
The weight of aggregate filled = 5.2 kg,
Flakiness index = (7.95 - 5.2)/7.95 = .34.

Angularity number = 67%-% of solid value.

%of solid value = [(weight of aggregate)÷(G*weight of water having equal volume)]*100
%of solid value = (5.2*100)÷(2.56*0.003*1000)
= 1.59.

ANGULARITY INDEX={(3÷20)*angularity number)+1.
= [(3÷20)*1.59]+1.
= 1.2385 .

Angularity No. = 67-(100w/CG).
= 67-[ (100*5.2)÷(2.65*1000*0.003)],
Ang. No. = 1.591.

Then calculated the angularity index:
Ang. Index= 3 * angularity no./20 + 1.
= 3*1.591/20 + 1.
Ang.index = 1.23865.

Angularity number = 67-(100w/CG).
= 67-(100 * 5.2/1000 * 3 * 2.65)
Ang.No.= 1.591
Angularity Index= (3 Angularity number/20)+1.
= ( 3* 1.59/20)+1.
Ang.index= 1.23865.

Specific gravity = 2.65.
v = .003
wt = 5.2kg
Vs = 5.2/(2.65*10^3)=0.0019
Vv = V-Vs
Vv = 0.003-0.0019 = 0.0011.
Angularity index = Vv/V.
=.0011/.003 = .37 (.35 approx)
So, option B is correct.