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Home RRB JE Stage-II RRB JE Stage II : Mechanical Engineering Quiz | 17th August

RRB JE Stage II : Mechanical Engineering Quiz | 17th 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. From design point of view, spherical vessel are preffered over cylindrical pressure vessel because they
(a)Are cost effective in fabrication
(b)Have uniform higher circumferential stress
(c)Uniform lower circumferential stress
(d)Have a large volume for the same quantity of material used

S1. Ans.(c)
Sol. In spherical vessel, the circumferential stresss and the longitudinal stress is equal and less than circumferential stress of cylindrical vessel

Q2. The stretch in a steel rod of circular section, having a length L subjected to a tensie load and tapering uniformly from a diameter d1, at one end to a diameter d2, at the other end, is given by
(a) PL/4E d1d2
(b PLπ /E d1d2
(c) PLπ /4E d1d2
(d)4PL/π E d1d2

S2. Ans.(d)
Sol. Deflection of circular tapering rod subjected to tensile load P. is
δ L= 4PL/ πd1d2E

Q3. For a composite bar consisting of a bar enclosed inside a tube of another material and when compressed under a load Was a whole through rigid plates at the end of the bar. The equation of compatibility is given by (suffixes 1 and 2 refer to bar and tube respectively)
(a) W1+W2= W
(b) W1+W2 = constant
(c) W1/A1E1=W/A2E2
(d) W1/A1E2=W2/A2E1

S3. Ans.(c)
Sol. (Strain)1=(×Strain)2
Therefore :=W1/A1×E1=W2/A2×E2

Q4. A tapering bar (diameter of end sections being, d, and d2) have the same length and are subjected the same and a bar of uniform cross-section’d axial pull. Both the bars will have the same extension if ‘d’ is equal to
(a) (d1+d2)/2
(b) √d1d2
(c) (√d1d2)/2
(d) (d1d2)/2

Q5. A slender bar of 100 mm2 cross-section is subjected to loading as shown in the figure below. If the modulus of elasticity is taken as 200 * 109 Pa, then the elongation produced in the bar will be

(a) 10mm 200KN 200KN
(c) 1mm 100KN
(d) nil
0.5m 1.0m 0.5m

Q6. If permissible stress in plates of joint through a pin as shown in the figure is 200 MPa, then the width w will be

(a) 15mm
(b) 20mm

S6. Ans.(a)
Sol. Axo=F
(W- 10) x 2 x 200 2000
Therefore W-10=5
therefore W=15 mm

Q7. In the arrangement as shown in the figure, the stepped steel bar ABC is loaded by a load P. The material has Young’s modulus E= 200 GPa and the two portions AB and BC have area of cross section 1 cm2 and 2 cm2respectively. The magnitude of load P required to fill up the gap of 0.75 mm is

(a) 10kN
(c) 20kN

S7. Ans.(a)
Sol. -[{(P/200)×1000}+{(P/100)×1000}]

Q8. The maximum shearing stress induced in the beam section at any layer at any position along the beam length (shown in the figure) is equal to

(a) 30 kgf/cm2 200mm
(b) 40 kgf/cm2 1m
(c) 50 kgf/cm2
(d) 60 kgf/cm2

S8. Ans.(a)
Sol. For rectangular section
Max shear stress
= 30 kgf/cm2

Q9. If Poisson’s ratio of a material is 0.5, then the elastic modulus for the material is
(a) three times its shear modulus
(b) four times its shear modulus
(c) equal to its shear modulus
(d) indeterminate

S9. Ans.(a)
Sol. E=2G(1+μ)
Put μ=0.5
Therefore E= 3G

Q10. The number of independent elastic constants required to express the stress-strain relationship for a linearly elastic isotropic material is
(a) one
(b) two
(c) three
(d) four

S10. Ans.(b)
Sol. By Hookes law
E and μ, young modulus and poison’s ratio

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