A. Ball mill
B. Rod mill
C. Fluid energy mill
D. Jaw crusher
Related Mcqs:
- As per Taggart’s formula, the capacity (kg/hr) of Jaw & Gyratory crushers (for gapes of 10 to 60 cms) is equal to (where, L = Length of feed opening, cms S = Maximum width of discharge opening, cms) ?
A. LS
B. 93 LS
C. 250 LS
D. √(LS) - Out of the following size reduction equipments, the maximum feed size can be accepted by the__________________?
A. Tube mill
B. Ball mill
C. Jaw crusher
D. Jet pulveriser - Work index is the gross energy (kWh/tonne of feed) necessary to reduce a very large feed to such a size that 80% of product particles will pass through a 0.1 mm screen. The value of work index determined for wet grinding should be multiplied with ______________ to get the same for dry grinding?
A. 1.0
B. 0.5
C. 1.34
D. 4.34 - Which of the following equations is Rittinger’s crushing law? (Where P = power required by the machine, m = feed rate, k = a constant, D̅ sa & D̅ sb = volume surface mean diameter of feed & product respectively) ?
A. P/m = K/ √Dp
B. P/m = K . ln D̅ sa/D̅ sb
C. P/m = K . (1/ D̅ sb – 1/D̅ sa)
D. None of these - According to Bond crushing law, the work required to form particle of size ‘D’ from very large feed is (where (S/V)p and (S/V)f are surface to volume ratio of the product and feed respectively) ?
A. (S/V)p
B. √(S/V)p
C. (S/V)2p
D. (S/V)f - The equivalent diameter of channel of a constant non-circular cross-section of 3 cm by 6 cm will be _______________ cms?
A. 20
B. 12
C. 8
D. 2 - Xanthates are used in the froth floatation process as a/an _______________________?
A. Conditioner
B. Frother
C. Collector
D. Activator - Close circuit grinding by a ball mill with air sweeping employs a _______________________?
A. Classifier
B. Cyclone separator between mill & classifier
C. Both A. & B.
D. Neither A. nor B. - In a size reduction crushing operation, feed size is 300 to 1500 mm while the product size is 100 to 300 mm. This is a case of the ________________ crushing?
A. Secondary
B. Fine
C. Primary
D. Ultrafine - The basic filtration equation is given as dt/dV = (μ/A ΔP). [(α .CV/A) + Rm], where, V is volume of the filtrate; A is the filtration area, a is specific cake resistance, μ is viscosity of the filtrate, and C is the concentration of the solids in the feed slurry. In a 20 minutes constant rate filtration, 5 m3 of filtrate was obtained. If this is followed by a constant pressure filtration, how much more time in minutes, it will take for another 5 m3 of filtrate to be produced? Neglect filter medium resistance, Rm; assume incompressible cake ?
A. 10
B. 20
C. 25
D. 30