# Mechanical Operations Questions and Answers – Energy for Size Reduction

This set of Mechanical Operations Questions and Answers for Freshers focuses on “Energy for Size Reduction”.

1. The limiting diameter, Dp is ____
a) d + r
b) 0.5R + r
c) 0.04R + d
d) R +r

Explanation: The limiting size Dp of particle that can be nipped by the rolls are calculated by
DP = 0.04R + d.

2. The maximum size of the product is _____
a) 2d
b) 4d
c) 5d
d) 2.5d

Explanation: The maximum size of the product is equal to 2d, the particle depends on the pacing between the rolls.

3. Roll crushers, are set to a reduction ratio of ____
a) 4:1
b) 3
c) Either 3 or 4:1
d) 1:3

Explanation: To operate more efficiently the reduction ratio is set to 3 or 4:1, the maximum particle size is 1/3 or ¼.

4. What is the forces exerted by the rolls?
a) 100 to 500 N/cm
b) 8700 to 70,000 N/cm
c) 500 to 1000 N/m
d) 7 to 300 N/cm

Explanation: The force exerted by the roll is from 8700 to 70,000 N/cm of the roll width, to allow material to pass without damaging the quality.

5. Which of the following is defines the theoretical capacity?
a) C = 50 uwd
b) C = 10 uwd
c) C = 5 uwd
d) C = 4.5 uwd

Explanation: The theoretical capacity is defined as C = 50 uwd, where u is the peripheral speed of rolls in fps, and w is the width.
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6. Which of the following is the capacity Q in tons?
a) 0.0068 ND wds
b) 1.20 ND wds
c) 5 ND wds
d) 10 N D

Explanation: If N is the speed of the roll in rpm and D roll diameter in inches and thus Q = 0.068 ND wds, which gives results in tons per hour.

7. The feed size of the roll crusher is _____
a) 2/3 to 3 inches
b) ½ to 3 inches
c) ¼ inches
d) 8 inches

Explanation: The feed size of the roll crusher is in between ½ to 3 inches and the product ranges between ½ to 20 inches.

8. Calculate the limiting size, if R is 2.5 cm and the half width is given as 10 cm?
a) 5.1
b) 0.1
c) 0.5
d) 10.1

Explanation: As Dp = 0.04R + d, 0.04*25 + 10 = 10.1 cm.

9. Calculate capacity, if peripheral speed is 100 fps, width is 10 ft. and d is 5 ft?
a) 250,000
b) 100,000
c) 45,000
d) 50,000

Explanation: As C= 50 uwd so, C = 50*100*10*5 = 250,000 ft3/hr.

10. Calculate Q, if Speed is 250 rpm and D is 10 inches and wds factor is given as 500?
a) 11500 T/hr
b) 8500 T/hr
c) 9000 T/hr
d) 5000 T/hr

Explanation: As Q = 0.068*250*10*500 = 8500 T/hr.

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