RRB JE Stage II : Mechanical Engineering Quiz | 26th August

Dear aspirants,

RRB JE Stage II exam is scheduled to be conducted from 28th August to 1st September 2019. The overall merit for the selection will be made on the basis of the marks scored in stage II, so candidates need to prepare thoroughly for the second stage. To help the aspirants, Adda247 has come up with a study plan in which we will provide you daily quizzes of all technical subjects. The below quiz will be helpful for the candidates who opted Mechanical as the exam group in RRB JE stage II.

Q1. Consider the following statements:
Addition of silicon to cast iron
1. Promotes graphite module formation
2. Promotes graphite flake formation
3. Increases the fluidity of the molten metal
4. Improves the ductility of cast iron
Which of these statements are correct?
(a) 1 and 4
(c) 1 and 3
(b) 2 and 3
(d) 3 and 4

S1. Ans.(b)
Sol. Silicon is strong graphitizer and promotes graphitization (ie decomposition of cementite to iron and graphite Graphitizer improves the fluidity of iron).

Q2. Eutectoid reaction occurs at
(a) 600 C
(c) 114°C
(b) 727°C
(d) 1493°C3

S2. Ans.(c)
In eutectoid reaction, the austenite transforms into a phase mixture of ferrite (containing 0.76% C) and cementite. This phase mixture is known as pearlite. The eutectoid reaction occurs at a constant temperature. This is known as eutectoid temperature and it is 727°C.

Q3. Match List-I (Name of material) with List-lI (% Carbon range) and select the correct answer
A. Hypo-eutectoid steel
B. Hyper-eutectoid steel
C. Hypo-eutectic cast iron
D. Hyper-eutectic cast iron
(1). 4.3-6.67
(2) 2.0-4.3
(3) 0.8-2.0
 A B C D
(a) 4 3 2 1
(b) 1 3 4 2
(c) 4 1 2 3
(d) 1 2 3 4

S3. Ans.(a)
Sol. Hypoeutectoid steel 0.008-0.76
Hyperutectoid steel 0.76 – 2.1
Hypoeutectic steel 2.1-4.3
Hypereutectic steel 4.36.64

Q4. Which one of the following sets of constituents is in equilibrium cooling of expected
hypereutectoid steel from austenitic state?
a) Ferrite and pearlite
(b) Cementite and pearlite
(c) Ferrite and bainite
(d) Cementite and martensite

S4. Ans.(b)
Sol. Hypereutectoid steel when cooled in equilibrium will result in pro-eutectoid cementite and pearlite whereas hypoeutectoid steel when cooled in equilibrium will result in pro-eutectoid ferrite and pearlite.

Q5. A given steel test specimen is studied unden metallurgical microscope. Magnification used 100 X. In that different phases are observed. One of them is Fe3C. The observed phase Fe3C is also known as
(a) ferrite
(c) cementite
(b) austenite
(d) martensite

Q6. Assertion (A): Carbon would form an interstitial solid solution with iron.
Reason (R): The atomic radius of iron is smaller than that of carbon.
(a) both A and R are true and R is the correct explanation of A
(b) both A and R are true but R is not a correct explanation of A
(c) A is true but R is false
(d) A is false but R is true

S6. Ans.(c)
Sol. Carbon in present as interstitial impurity in iron matrix. The size of iron atom is larger than that of carbon.

Q7. Heating the hypoeutectoid steels to 30°C above the upper critical temperature line, soaking at the temperature and then cooling slowly to room temperature to form a perite and ferrite structure is known as
(a) Hardening
(b) Normalising
(c) Tempering
(d) Annealing

S7. Ans.(d)
Sol. Slow cooling signifies cooling in furnace so it refer annealing process

Q8. In a eutectic system, two elements are completely
(a) Insoluble in solid and liquid state
(b) Soluble liquid state
(c) Soluble in solid state
(d) Insoluble in liquid state

Q9. Assertion (A) Carbon forms interstitial solid solution when added to iron.
:Reason (R) : The atomic radius of carbon atom is much smaller than that of iron.
a) both A and R are true and R is the correct explanation of A
(b) both A and R are true but R is not a correct explanation of A
(c) A is true but R is false
(d) A is false but R is true


Q10. Pearlite consists of
(a) 6.67% C and 93.33% ferrite
b) 12% Fe and 87% cementite
(c) 13% C and 87% ferrite
(d)13% cementite and 87% ferrite

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