Quiz: Mechanical Engineering

Exam: GATE

Topic: Miscellaneous

Each question carries 2 mark

Negative marking: 1/3 mark

Time: 20 Minutes

Q1. In a heat engine operating in a cycle between a source temperature of 606 °C and a sink temperature of 20°C, what will be the least rate of heat rejection per kW net output of the engine?

(a) 0.50 kW

(b) 0.667 kW

(c) 1.5 kW

(d) 0.0341 kW

Q2. Two reversible engines operate between thermal reservoirs at 1200 K, T_2 K and 300 K such that 1st engine receives heat from 1200 K reservoir and rejects heat to thermal reservoir at T_2 K, while the 2nd engine receives heat from thermal reservoir at T_2 K and rejects heat to the thermal reservoir at 300 K. the efficiency of both the engines is equal. What is the value of temperature T_2?

(a) 400 K

(b) 500 K

(c) 600 K

(d) 700 K

Q3. What is minimum amount of work input to a refrigerator which convert 1 kg of water at 20 °C into ice at -5 °C while max. COP of refrigerator is 10?

(a) 40 kJ

(b) 50 kJ

(c) 42.91 kJ

(d) 52.91 kJ

Q4. Two steel truss members, AC and BC, each having cross sectional are of 100 mm², are subjected to a horizontal force F as shown in figure. All the joints are hinged.

If F = 1 kN, the magnitude of the vertical reaction force developed at the point B in kN is

(a) 0.63

(b) 0.32

(c) 1.26

(d) 1.46

Q5. A natural feed journal bearing of diameter 50 mm and length 50 mm operating at 20 revolution/sec. carries a load of 2.0 kN. The lubricant used has a viscosity of 20 MPa/s. the radial clearance is 50 μm. The Sommerfeld number for the bearings is

(a) 0.062

(b) 0.125

(c) 0.250

(d) 0.785

Q6. The part of a gating system which regulates the rate of pouring of molten metal is

(a) pouring basin

(b) runner

(c) choke

(d) ingate

Q7. A concentrated load P acts on a simply supported beam of span L at a distance L/3 from the left support. The bending moment at the point of application of the load is given by

(a) PL/3

(b) (2 PL)/3

(c) PL/9

(d) (2 PL)/9

Q8. A strip with cross – sectional area 150 mm × 4.5 mm is being rolled with 20% reduction of area using 450 mm diameter rollers. The angel subtended by the deformation zone at the roll centre is (in radians)

(a) 0.01

(b) 0.02

(c) 0.03

(d) 0.06

Q9. The inlet angle of runner blades of a Francis turbine is 90°. The blades are so shaped that the tangential component of velocity at blade outlet is zero. The flow velocity remains constant throughout the blade passage and is equal to half of the blade velocity at runner inlet. The blade efficiency of the runner is

(a) 25%

(b) 50%

(c) 80%

(d) 89%

Q10. The atomic packing factor for a material with body centered cubic structure is

(a) 0.68

(b) 0.53

(c) 0.89

(d) 0.87

Solutions

S1. Ans.(a)

Sol. T_2 = 20 + 273 = 293 K

T_1 = 606 + 273 = 879

η=W/Q_1 =1-T_2/T_1

Q_1=1/(1-(293/879) )=1.5 kW

S2. Ans.(c)

Sol. T_1= 1200 K

T_3 = 300 K

η=W/Q_1 =1-T_2/T_1

Q_1=1/(1-(293/879) )=1.5 kW

S3. Ans.(c)

Sol. Q = m_w C_pW (0-20)+m_i C_pi (-0.5-0)-325

= -1 × 4.18 (20) – 1 × 2.09(5) – 335

= -429.05 kJ

∵W_min=Q/(COP)_max =429.05/10=42.905

= 42.91 kJ

S4. Ans.(a)

Sol. using lame’s theorem

T_1/(Sin120°)= T_1/(Sin135°)=F/(Sin105°)

T_1 = 0.8965 F

T_2 = 0.732 F

Vertical reaction at B.

R_B=T_2 Cos30°=0.732 cos30°

R_B=0.634 kN

S5. Ans.(b)

Sol. Sommerfeld number

S=(r/c)^2×μN/P

Where

r is radius of journal

μ is viscosity of lubricant

N is no. of revolution per second

P is bearing pressure on projected area

C is radial clearance

There fore

p^’=P/(d×l)=2000/(50×50)=0.8 N\/〖mm〗^2

S=(25/(50×10^(-3) ))^2×(20×20×10^(-3))/(0.8×10^6 )

=0.125

S6. Ans.(a)

Sol. pouring basing regulates the rate of molten metal, maintains the required rate of liquid metal flow and also reduces turbulence at the sprue entrance.

S7. Ans.(d)

Sol. T_1=2P/3

T_2=P\/3

At point of application

BM=2P/3×L/3=(2 PL)/9

S8. Ans.(d)

Sol. Bite angle = tan^(-1) [√R∆h/(R-∆h\/2)]

From equation, we have 20% reduction in cross-section axis

150×4.5×0.8=150×h

h=3.6 mm

∆h=4.5-3.6=0.9 mm, Given R=450/2=225

θ=tan^(-1) (√(225×0.9)/(225-0.9/2))=3.62°

θ=0.06 rad

S9. Ans.(d)

Sol. Given data:

B_i=90°

V_Buo=0

V_fi=V_f0=(ui^2)/gH∵V_wi=ui

By energy balance equation,

ρQgH=ρQV_wi u_0+(ρQV_0^2)/2

gH=ui^2+(V_f0^2)/2=ui^2+(ui^2)/8

gH =9/8 ui^2

∵η_h=(ui^2)/(9/8 ui^2 )=8/9=0.8888

η_H=88.88 %≈89%

S10. Ans.(a)

Sol. Atomic packing factor =

(Volume occupied by Atoms in cell)/(Volume of unit cell)

=(N_avg×4/3 πR^3)/a^3 |

for bcc structure,a=4r/√3 and N_avg=2

by putting these value,APF=0.68