Quiz: Mechanical Engineering

Exam: GATE

Topic: Miscellaneous

Each question carries 2 mark

Negative marking: 1/3 mark

Time: 20 Minutes

Q1. The figure shown a pin-jointed plane truss loaded at the point M by hanging a mass of 100 kg. The member LN of the truss is subjected to a load of

(a) 0 Newton

(b) 490 Newtons in compression

(c) 981 Newtons in compression

(d) 981 Newtons in tension

Q2. A thin plate of uniform thickness is subject to pressure as shown in the figure below

Under the assumption of plane stress, which one of the following is correct?

(a) Normal stress is zero in the z-direction

(b) Normal stress is tensile in the z-direction

(c) Normal stress is compressive in the z-direction

(d) Normal stress varies in the z-direction

Q3. At a point in a stressed body the state of stress on two planes 45° apart is as shown below. Determine the two principal stresses in MPa.

(a) 8.242, 0.658

(b) 9.242, 0.758

(c) 9.242, 0.758

(d) 8.242, 0.758

Q4. A uniformly loaded propped cantilever beam and its free body diagram are shown below. The reactions are

(a) R_1=5qL/8,R_2=3qL/8,M=(qL^2)/8

(b) R_1=3qL/8,R_2=5qL/8,M=(qL^2)/8

(c) R_1=5qL/8,R_2=3qL/8,M=0

(d) R_1=3qL/8,R_2=5qL/8,M=0

Q5. The number of degrees of freedom of a planar linkage with 8 links and 9 simple revolute joints is

(a) 1

(b) 2

(c) 3

(d) 4

Q6. List – I (Gear type)

Worm gears

Cross helical gears

Bevel gears

Spur gears

List – II (Application)

Parallel shafts

Nonparallel, intersecting shafts

Nonparallel, nonintersecting shafts

Large speed ratios

Codes:

A B C D

(a) 2 3 1 4

(b) 4 1 3 2

(c) 4 3 2 1

(d) 3 1 2 4

Q7. A cantilever type gate hinged at Q is shown in the figure. P and R are the centers of gravity of the cantilever part and the counterweight respectively. The mass of the cantilever part is 75 Kg. the mass of the counterweight, for static balance, is

(a) 75 kg

(b) 150 kg

(c) 225 kg

(d) 300 kg

Q8. A Small steam whistle (Perfectly insulated and doing no shaft work) causes a drop of 0.8 kJ/kg in enthalpy of steam from entry to exit. If the kinetic energy of the steam at entry is negligible, the velocity of the steam at exit is

(a) 4 m/s

(b) 40 m/s

(c) 80 m/s

(d) 120 m/s

Q9. Proper gating design in metal casting

P. Influences the freezing range of the melt

Q. compensates the loss of fluidity of the melt

R. Facilitates top feeding of the melt

S. Avoids misrun

(a) P, R

(b) Q, S

(c) R, S

(d) R, S

Q10. A metal disc of 20 mm diameter is to be punched from a sheet of 2 mm thickness. The punch and the dia clearance is 3%. The required punch diameter is

(a) 19.88 mm

(b) 19.94 mm

(c) 20.06 mm

(d) 20.12 mm

Solutions

S1. Ans.(a)

Sol. Σf_H=0 & Σf_y=0

At joint ‘L’

∵ f_Lk -f_LM=0 (∵Σf_H=0 )

f_LN=0 (∵ Σf_y=0 )

Hence no force is cutting on the truss member L.N

S2. Ans.(a)

Sol. Thin plate of uniform thickness pertains to plane stress condition. So, stress out of plane would be zero.

S3. Ans.(b)

Sol.

2=((8+σ_y)/2)+((8-σ_y)/2) cos〖90°〗-3 Sin90°

σ_y=2MPa

σ_1,2=(8+2)/2±√(((8-2)/2)^2+3^2 )

=5±4.242

σ_1,2=9.242,+0.758 MPa

S4. Ans.(a)

Sol.

the given propped cantilever bam can be assumed to be consisting of two types of loads. First part is simple cantilever with UDL and the second part is cantilever beam with point load of R_2 at end

Balancing the forces, we get

R_1+R_2=qL—–(1)

Balancing the deflection at the end point as net deflection at the end is zero.

Deflection at B due to the UDL alone,

δ_B=(qL^4)/8EI

Deflection at B due to R_2 alone,

δ_b^’=(R_2 L^3)/3EI

∵δ_B=δ_B’

⇒(qL^4)/8EI=(R_2 L^3)/3EI⇒R_2=3qL/8

From (1),

R_1=qL-R_2

=qL-3qL/8=5qL/8

Moment,

M=R_2 L-qL×L\/2

=(3qL^2)/8-(qL^2)/2

= (qL^2)/8

S5. Ans.(c)

Sol. No. of links l = 8

No. of revolute joints j = 9

No. of higher pair, h = 0

Number of degree of freedom,

m = 3 (L-1)-2j-h

=3(8-1)-2×9-0

=3

S6. Ans.(c)

Sol. Worm gear – large speed ratio

Cross helical gears – Non- parallel, nonintersecting shafts

Bevel gear – Non parallel, interesting shafts

Spur gear – Parallel shafts.

S7. Ans.(d)

Sol.

Let counterweight at R is m kg

For static condition

ΣM_Q=0

m×0.5=75×2

m=300 kg

S8. Ans.(b)

Sol. Given data: = ∆h = 0.8 kJ/kg

⇒ h_1-h_2 = 0.8 kJ/kg

Applying steady flow energy equation

h_1=(V_1^2)/2+gz_1+q=h_2+V_2^2+g2_2+w

Assumption

q=0,w=0 V_1≈0 and change in KE and P.E neglet

∵h_1=h_2+(V_2^2)/2

h_1=h_2+(V_2^2)/2000

⇒h_1-h_2=(V_2^2)/2000

0.8×2000=V_2^2

V_2^2=1600

V_2=40 m\/sec

S9. Ans.(b)

Sol. Proper gating design can reduces pouring time, which makes up for loss of fluidity and avoids misrun defects.

S10. Ans.(a)

Sol. Given punched diameter; d = 20 mm

Shut thickness, t = 2 mm

Clearance; C = 3% of t

Punch diameter,

D=d-2C

=20-2×3/100×2

=19.58 mm