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# Quiz: Mechanical Engineering 8th May

Quiz: Mechanical Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 mark
Negative marking: 1/3 mark
Time: 20 Minutes
Q1. The value of β and μ, for an ideal gas is
(a) β=1/ρ and μ_J=0
(b) β=1/T and μ_J=0
(c) β=1/T and μ_J<0
(d) β=1/ρ and μ_J<0
Where, β is volume expansivity and μ_J is joule kelvin coefficient

Q2. Consider the following statements with regard to joule – kelvin expansion and choose the correct one.
(a) Inversion curve is the locus of all points at which μJ is zero.

(b) inversion curve is the locus of all point having maximum value of μ_j

(c) Inversion curve is the locus of all point having minimum value of μ_J

(d) None of these

Q3. At triple point, (l = latent heat)

(a) l_sublimation=l_vaporization+l_fusion

(b) l_sublimation=l_vaporization -l_fusion
(c) l_vaporization=l_sublimaton+l_fusion
(d) l_fusion=l_vaporization+l_sublimation

Q4. Which one of the following represents the condensation of a mixture saturated liquid and saturated vapour on the enthalpy – entropy diagram?
(a) A horizontal line
(b) An inclined line of constant slope
(c) A vertical line
(d) A curved line

Q5. A cubic casting of 50 mm side undergoes volumetric solidification shrinkage and volumetric solid contraction of 4% and 6% respectively. No riser is used. Assume uniform cooling in all directions. The side of the cube after solidification and contraction is
(a) 48.32 mm
(b) 49.90 mm
(c) 49.94 mm
(d) 49.96 mm

Q6. What is the approximate % change in the life, t, of the tool with zero rake angle used in orthogonal cutting when its clearance angle, α, is changed from 10 to 7 deg?
(Hints; Flank wear rate is proportional to cot α)
(a) 30% increase
(b) 30% decrease
(c) 70% increase
(d) 70% increase

Q7. A 20° full depth involute spur pinon of 4 mm module and 21 teeth is to transmit 15 kW at 960 rpm. Its face width is 25 mm.
The tangential force transmitted (in N) is
(a) 3552
(b) 2611
(c) 1776
(d) 1305

Q8. The ratio of tension on the tight side to that on the slack in a flat belt drive is:
(a) proportional to the product of coefficient of friction and lap angle
(b) an exponential function of the product of coefficient of friction and lap angle
(c) proportional to the lap angle
(d) proportional to the coefficient of friction

Q9. Two streams of liquid metal which are not hot enough to fuse properly result into casting defect known as
(a) cold shut
(b) swell
(c) sand wash
(d) scab

Q10. For resistance spot welding of 1.5 mm thick steel sheets, the current required is of the order
(a) 10 Amp
(b) 100 Amp
(c) 1000 Amp
(d) 10,000 Amp

Solutions

S1. Ans.(b)
Sol. μ_j (∂T/∂p)_h=1/C_P [(∂V/∂T)_p-V]
Volume expansivity, β = 1/V (∂V/∂T)_p
For an ideal gas,
μ_j=V/C_P [T/V (∂V/∂T)_p-1]=V/C_P [βT-1]=0
And β=1/T

S2. Ans.(a)
Sol. inversion curve is the locus of all points at which μ_j is zero.

S3. Ans.(a)
Sol. At triple point,
l_sublimation=l_vapourization+l_fusion

S4. Ans.(b)
Sol. An inclined line of constant slope represents the condensation of a mixture of saturated liquid and saturated vapour on the enthalpy – entropy diagram (Mollier diagram).

S5. Ans.(a)
Sol. Side of cube after solidification = ∛(0.96×(50)^3 )=49.32 mm
Slide of cube after contractors = ∛(0.94×(49.32)^3 )=48.32 mm
= 48.32 mm

S6. Ans.(b)
Sol. Tool life, T ∝1/(flank wear)
T ∝1/cotα
T ∝ Tanα
T_2/T_1 =(tanα_2)/(tanα_1 )=(tan7°)/(tan10°)=0.7
T_2=0.7 T_1
Tool life decreased by 30%

S7. Ans.(a)
Sol. Module: (m) = 4 mm; number of teeth, T = 21
Pressure angle, ϕ = 20°; face width, b = 25 mm
Full depth involute spur pinion
Transmission of power,
P=15 kW at 960 rpm
Power = Torque × angular speed
P=T×ω=F_t×r×ω
P=F_t×r×ω
⇒15×1000=f_t×42×10^(-3)×2π×960/60
⇒F_t=3552.56N

S8. Ans.(b)
Sol. The ratio of tension on the tight side (T_1 ) to that of back side (T_2) in a flat belt drive is given by:
T_1/T_2 =e^(μθ)min
Here, μ = coefficient of friction
θ = Angle of contact or lap angle

S9. Ans.(a)
Sol. When two streams of liquid metal came closer to each other from opposite side and their temperature is not enough to fuse properly. so, an effect is produced i.e. cold shut. Another reason of this defect is the lack of fluidity.

S10. Ans.(d)
Sol. for resistance spot welding, current required
3937 (t_1+t_2 )A
=3937 (1.5+1.5)
=11811 A
≈10000A

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