aptitude and reasoing

REASONING

21. How many cubes will have exactly two face painted only in Red ?

a) 10 b) 12 c) 14 d) 8

22. How many cubes will have at least three sides painted ?

a) 8 b) 6 c) 3 d) 2

23. How many cubes will have no face painted at all ?

a) 1 b) 2 c) 3 d) 4

Direction (Qs. 24 to 28) : Read the following information carefully and answer the question below it.

A dice with it’s face numbered 1 to 6, is shown in three different position X, Y and Z

24. Which number lies at the bottom face of dice X ?

a) 1 b) 2 c) 3 d) 4

25. Which number lies at the bottom face of dice Y ?

a) 6 b) 5 c) 2 d) 1

26. Which number lies opposite of 6 ?

a) 1 b) 2 c) 4 d) 5

27. Which numbers are hidden behind the numbers 6 and 5 in the dice ?

a) 1 & 4 b) 1 & 3 c) 4 & 3 d) 1 & 2

28. Which of the hidden number adjacent to 5 in die X are common to the hidden numbers adjacent to 5 in

die Z ?

a) 1 & 4 b) 2 c) 6 d) None of these

Direction (Qs. 29 to 32) : Read the following information carefully and answer the question below it.

Six dice with their upper faces erased are as shown :

The sum of the number of dots on the opposite faces is 7.

29. If the dice (i), (ii), (iii) have even number of dots on their bottom faces, than what would be the total

number of dots on the top faces ?

a) 14 b) 7 c) 21 d) 12

30. If dice (i), (ii), (iii) have even number of dots on their bottom faces and the dice (iv), (v), (vi) have odd

number of dots on their top faces then what would be the difference in the total number of top face

between these two sets ?

a) 0 b) 1 c) 2 d) 3

31. If odd numbered dice have odd number on their bottom faces what would be the total number of dots

on the tope faces of these dice ?

a) 4 b) 6 c) 10 d) 12

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REASONING

32.) If even numbered dice have even number of dots on their top faces what would be the total number of

dots on the top faces of these dice ?

a) 18 b) 14 c) 12 d) 10

Practice Questions Ans :
1-b, 2-c, 3-d, 4-a, 5-c, 6-c, 7-d, 8-d, 9-c, 10-b, 11-b. 12-d, 13-a, 14-a, 15-b, 16-a, 17-c, 18-c, 19-c, 20-c, 21-b,
22-a, 23-a, 24-b, 25-c, 26-a, 27-b, 28-d, 29-b, 30-c, 31-d, 32-a

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REASONING
20. MISCELLANEOUS

Class Work :

Statement : 1.) A single-point cutting tool with 120 rake angle is used to machine a steel-piece.

The depth of cut, i.e. uncut thickness is 0.81 mm. The chip thickness under orthogonal machining

condition is 1.8 mm. The shear angle is approximately

a) 220 b) 260 c) 560 d) 760

2.) In orthogonal turning of low carbon steel pipe with principal cutting edge angle of 900, the main

cutting force is 1000 N and the feed force is 800 N. The shear angle is 250 and orthogonal rake

angle is zero. Employing Merchant’s theory, the ratio of friction force to normal force acting on the

cutting too is

a) 1.56 b) 1.25 c) 0.80 d) 0.64

3.) In orthogonal turning of medium carbon steel, the specific machining energy is 2.0 J/mm3. The

cutting velocity, feed and depth of cut are 120 m/min, 0.2 mm/rev and 2 mm respectively. The

main cutting force in N is

a) 40 b) 80 c) 400 d) 800

4.) In orthogonal turning of a low carbon steel bar of diameter 150 mm with uncoated carbide

tool, the cutting velocity is 90 m/min. The feed is 0.24 mm/rev and the depth of cut is 2 mm. The

chip thickness obtained is 0.48 mm. If the orthogonal rake angle is zero and the principal cutting

edge angle is 900, the shear angle in degree is

a) 20.56 b) 26.56 c) 30.56 d) 36.56

5.) Details pertaining to an orthogonal metal cutting process are given below :

Chip thickness ratio 0.4

Underformed thickness 0.6 mm

Bake angle +100

Cutting speed 2.5 m/s

Mean thickness of 25 microns

Primary shear zone

The shear strain rate in s-1 during the process is

a) 0.1781 x 105 b) 0.7754 x 105 c) 1.0104 x 105 d) 4.397 x 105

Common data for below questions

A cylinder is turned on a lathe with orthogonal machining principle. Spindle rotates at 200 rpm.

The axial feed rate is 0.25 mm per revolution. Depth of cut is 0.4 mm. The rake angle is 100. In the

analysis it is found that the shear angle is 27.750.

6A) The thickness of the produced chip is

a) 0.511 mm b) 0.528 mm c) 0.818 mm d) 0.846 mm

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REASONING

6B.) In the above problem, the coefficient of friction at the chip tool interface obtained using

Earnest and Merchant theory is

a) 0.18 b) 0.6 c) 0.71 d) 0.98

Common data for below questions :

In an orthogonal machining operation :

Uncut thickness = 0.5 mm, Cutting speed = 20 m/min, Rake angle = 150 , Width of cut =5 mm,

Chip thickness = 0.7 mm, Thrust force = 200 N, Cutting force = 1200 N, Assume Merchant’s theory.

7A.) The values of shear angle and shear strain respectively, are

a) 30.30 and 1.98 b) 30.30 and 4.23 c) 40.20 and 2.97 d) 40.20 and 1.65

7B.) The coefficient of friction at the tool-chip interface is

a) 0.23 b) 0.46 c) 0.85 d) 0.95

8.) Following data correspond to an orthogonal Turning of a 100 mm diameter rod on a lathe. Rake

angle : +150; Uncut chip thickness 0.5 mm, nominal chip thickness after the cut 1.25 mm. The

shear angle (in degrees) for this process is ___________ (correct to two decimal places).

9.) During the turning of a 20 mm diameter steel bar at a spindle speed of 400 rpm, a tool life of 20

minute is obtained. When the same bar is turned at 200 rpm, the tool life becomes 60 minute.

Assume that Taylor’s tool life equation is valid. When the bar is turned at 300 rpm, the tool life (in

minute) is approximately

a) 25 b) 32 c) 40 d) 50

10.) The tool life equation for HSS tool is VT0.14f0.7d0.4 = Constant. The tool life (T) of 30 min is

obtained using the following cutting conditions :

V = 45 m/min, f = 0.35 mm, d = 2.0 mm

If speed (V), feed (f) and depth of cut (d) are increased individually by 25%, the tool life (in min) is

a) 0.15 b) 1.06 c) 22.50 d) 30.0

11.) Orthogonal turning of a mild steel tube with a tool of rake angle 100 carried out at a feed of

0.14 mm/rev. If the thickness of the chip produced is 0.28 mm, the values of shear angle and shear

strain will be respectively

a) 28020’ and 2.19 b) 22020’ and 3.53 c) 24030’ and 3.53 d) 37020’ and 5.19

Linked data for below questions

In orthogonal turning of a bar of 100 mm diameter with a feed of 0.25 mm/rev, depth of cut of 4

mm and cutting velocity of 90 m/min, it is observed that the main (tangential) cutting force is

perpendicular to the friction force acting at the chip-tool interface. The main (tangential) cutting

force is 1500 N.

12A.) The orthogonal rake angle of the cutting tool in degree is

a) Zero b) 3.58 c) 5 d) 7.16

12B.) The normal force acting at the chip-tool interface

a) 1000 b) 1500 c) 2000 d) 2500

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