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Mechanical Operations - Section 1 (6)

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Mechanical Operations - Section 1

You can find the problems related to mechanical operation and can
solve these problems to test your aptitude ability.

41. 

Tabular bowl centrifuges as compared to disk bowl centrifuges

A. operate at higher speed.
B. employ bowl of larger diameter.
C. can not be operated under pressure/vacuum.
D. can't be used for separation of fine suspended solids from a liquid.

Answer: Option A

Explanation:

No answer description available for this question. Let us discuss.

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42. 

Which of the following is a pressure filter ?

A. Leaf filter (Moore filter).
B. Plate and flame filter.
C. Rotary drum filter.
D. Sand filter.

Answer: Option B

Explanation:

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43. 

Which of the following is not categorised as a "mechanical operation" ?

A. Agitation
B. Filtration
C. Size enlargement
D. Humidification

Answer: Option D

Explanation:

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44. 

For sizing of fine materials, the most suitable equipment is a

A. trommel
B. grizzly
C. shaking screen
D. vibrating screen

Answer: Option D

Explanation:

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45. 

In continuous filtration (at a constant pressure drop), filtrate flow rate varies inversely as the

A. square root of the velocity.
B. square of the viscosity.
C. filtration time only.
D. washing time only.

Answer: Option A

Explanation:

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46. 

Pick out the wrong statement.

A. Recycled coarse material to the grinder by a classifier is termed as circulating load.
B. Wear and tear in wet crushing is more than that in dry crushing of materials.
C. Size enlargement (opposite of size reduction) is not a mechanical operation.
D. A 'dust catcher' is simply an enlargement in a pipeline which permits the solids to settle down due to reduction in velocity of the dust laden gas.

Answer: Option C

Explanation:

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47. 

Power required to drive a ball mill with a particular ball load is proportional to (where, D = diameter of ball mill )

A. D
B. 1/D
C. D 2.5
D. 1/D 2.5

Answer: Option C

Explanation:

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48. 

If dp is the equivalent diameter of a non-spherical particle, Vp its volume and sp its surface area, then its sphericity is Φs is defined by

A. Φs = 6 Vp/dp sp
B. Φs = Vp/dp sp
C. Φs = 6 dp Sp/Vp
D. Φs = dp Sp/Vp

Answer: Option A

Explanation:

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