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RRB JE Stage II : Mechanical Engineering Quiz | 21st 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.

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Q1. Consider the following:
1.Relative size
.2 Crystal structure
3. Chemical affinity
4. Valency
Which of these factors govern relative solubility of two metals in each other in the solid state?
(a) 1,2 and 3 only
(b) 2,3 and 4 only
(c) 1,2 and 4 only
(d) 1, 2,3 and 4

S1. Ans.(d)
Sol. Factors govern relative solubility of two metals are:
1. Relative size
2. Valency
3. Chemical affinity
4. Crystal structure

Q2. Surface imperfections which separate two onentations that are mirror image of one another
is called
(a) Stacking fault
(b) Grain boundary
(c) Tilt boundary
(d) Twinned boundary

S2. Ans.(d)
Sol. When the surface imperfection orientations on one side are mirror image of opposite side such grain boundary defeats are called twin boundary defects.

Q3. Solid material chemical bonds are
(a) lonic, molecular and fusion
(b) Covalent, fusion and fission
(c) lonic, covalent and molecular
(d) Fission, molecular and ionic

S3. Ans.(c)
Sol. Solid material chemical bonds are ionic Bond, covalent bond and molecular bond. lonic bond is the strong electrostatic attraction between cation and anion. Covalent bonds are one in which there is a sharing of one or more electrons from the adjacent atoms.

Q4. Match List-I with List-II and select the correct answer using the code given below the Lists
A. Alpha iron
B. Zinc
C Glass
D. Copper
2. BCC
3. HCP
4. Amorphous
(a) 1 4 3 2
(b) 2 4 3 1
(c) 1 3 4 2
(d) 2 3 4 1

S4. Ans.(d)
Sol. Examples of :
(a) BCC – Alpha iron, Delta iron, Chromium,Tungsten.
(b) FCC – Copper, Aluminium, Argon, Nickel.
(c) HCP- Beryllium, Cadmium, Coba ,Magnesium.
(d) Glass has Amorphous Structure
BCC-Body centered cubic
FCC-Face centered cubic
HCP-Hexagonal close-packed

Q5. Consider the following statements regarding material:
1.In micro structure external forms of the crystal may reflect the internal symmetry of atom
2.Electronic structure is arrangement of electrons in different orbit of atom
3.A crystal is formed by a non repeating cell in uni-direction
4. Nuclear structure can be studied by Mossbauer studies
Which of these statements are correct?
(a) 1, 2 and 3
(b) 1, 2 and 4
(c) 1, 3 and 4
(d) 2, 3 and 4

S5. Ans.(b)
Sol. (i) The atoms in a metal crystal are arranged in a regular, repeated three-dimensional pattern.
(ii) A crystal is a solid where the atoms form a periodic arrangement.
(iii) Mossbauer effect involves resonant and recoilless emission and absorption of gamma radiation by atomic nuclei found in a solid. It is also known as recoilless nuclear resonance fluorescence.

Q6. Copper has FCC structure, its atomic radius is 1.28 Å, and atomic mass is 63.5. The density of copper will be
(a) 8.9 x 10³ kgmm³
(b) 8.9 x 10³ kg/cm³
(c) 8.9x103kg/m3
(d) 8.9 x 10³ g/mm³

S6. Ans.(c)
Sol. Density(g/m³)
=Mass x Number of atoms in a unit cell/Volume x Avagadro’s number
Volume of unit cell= a³ =(2√2r)³
= 47.46 x 〖10〗^(-30) m³
Density=(4x 63.5)/(47.46x〖10〗^(-3)x6.023 x 10²³)
=8.9 x 10⁶ g/m³
=8.9 x 10³ kg/m³

Q7. A plane intersects the coordinate axes at x=2/3 ;y=1/3;and z =1/2
What is the Miller index of this plane?
(a) 932
(c) 423
(b) 432
(d) 364

S7. Ans.(d)
Sol. Miller indices is defined as rationalized reciprocal of fractional intercepts taking along three crystallographic direction. Intercepts are[2/3,1/3,1/2] Reciprocal of Intercepts[3/2,3,2] Converting to whole number by multiplying by 2.
Miller indices = (3 6 4)

Q8. If the atomic radius of aluminium is r, what is its unit cell volume?
(a) (2r/√2)^3
(b) (4r/√2)^3
(c) (2r/√3)^3
(d) (4r/√3)^3

S8. Ans.(b)
Sol. Aluminium-For FCC crystal structure
FCC crystal structure
√2a= 4r
a =4r/√2

Q9. The following terms relate to floating bodies: Centre of gravity…G, Metacentre…M, Weight W ,Buoyant force.FB of floating body…,
Match List-l (Condition) with List-Il (Result) and select the correct answer using the codes given
below the lists:
A. G is above M
B. G and M Coincide
C. G is below M
1. Stable equilibrium
2. Unstable equilibrium
3. Floating body
4. Neutral equilibrium
(a) 1 3 2 4
(b) 3 1 4 2
(c) 2 3 4 1
(d) 2 4 1 3

S9. Ans.(d)
Sol. For completely submerged body stable equilibrium exist when centre of gravity is below centre of buoyancy. Unstable equilibrium exist when centre of gravity is above the centre of buoyancy and neutral equilibrium exists when centre of gravity exists with centre of buoyancy (B).A floating body will be in stable equilibrium when metacenter is above centre of gravity, in unstable equilibrium when metacenter is below centre of gravity. And if centre of gravity coincide with metacenter, body will be in neutral equilibrium. The metacentric height depends on the volume of liquid displaced and the distance between the metacentre and the centre of gravity

Q10. Consider the following statements: The metacentric height of a floating body depends
1. directly on the shape of its water-line area
2.on the volume of liquid displaced by the body
3.on the distance between the metacentre and the centre of gravity
4.on the second moment of water-line area
Which of these statements are correct?
(a) 1 and 2
(b) 2 and 3
(c) 3 and 4
(d) 1 and 4

S10. Ans.(b)
Sol. The metacentric height depends on the volume of liquid displaced and the distance between the metacentre and the centre of gravity.

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