Related Mcqs:

1. Capacity (in tons/hr) of jaw/gyratory crusher is equal to (where, L = length of the receiving opening, cm S = greater width of the discharge opening, cm) ?

   A. 0.01 L.S
   B. 0.087 L.S
   C. L.S
   D. L.S/0.087

2. Gyratory crushers compared to the reciprocating jaw crushers__________________?

   A. Have greater capacity per unit of discharge area
   B. Crush intermittently
   C. Are less suitable for coarse materials
   D. Have less steady power consumption

3. The capacity of a gyratory crusher is ______________ that of a jaw crusher with the same gape, handling the same feed & for the same product size range.?

   A. Same as
   B. 2.5 times
   C. 5 times
   D. 10 times

4. Reciprocating screens are normally inclined at an angle of 5° with the horizontal and employ gyratory motion at feed end & reciprocating motion at the discharge end. They are not suitable for the screening of the ___________________?

   A. Light metal powder down upto 4 mesh size
   B. Dry chemicals
   C. Heavy tonnages of rocks or gravel
   D. Powdered food & granular materials

5. According to Taggart’s rule for selecting between a gyratory crusher and a jaw crusher; the later should be used, if the hourly tonnage to be crushed divided by the square of the gape expressed in cm is less than_______________?

   A. 0.00184
   B. 0.0184
   C. 0.184
   D. 1.84

6. Which of the following crushers can be considered as a combination of a jaw crusher and a roller crusher ?

   A. Rod mill
   B. Fluid energy mill
   C. Gyratory crusher
   D. Ball mill

7. Size reduction mechanism used in Jaw crushers is__________________?

   A. Attrition
   B. Compression
   C. Cutting
   D. Impact

8. Feed size of ≥ 25 cms can be accepted by _______________________?

   A. Ball mill
   B. Rod mill
   C. Fluid energy mill
   D. Jaw crusher

9. Theoretical capacity of crushing rolls in tons/hr is given by (where, V = peripheral velocity, m/sec. W = width of rolls, m Dr = distance between rolls ρ = density of material to be crushed, kg/m3 here, V =πND where, N = speed of the rolls in rotation per second (rps) D = diameter of rolls, m) ?

   A. 3.6 V.W.Dr.ρ
   B. 3.6 V.W. ρ
   C. 3.6 W.Dr.ρ
   D. 3.6 V.W.Dr/ρ

10. 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