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# Quiz: Mechanical Engineering 5th May

Quiz: Mechanical Engineering
Exam: GATE
Topic: Miscellaneous

Each question carries 2 mark
Negative marking: 1/3 mark
Time: 20 Minutes

Q1. Steam flows at a rate of 10 kg/s through a supersonic ratio & velocity. Throat diameter is 50 mm. Density ratio & velocity ratio with reference to throat & exit are respectively 2.45 and 0.8. what is diameter at the exit?
(a) 122.5 mm
(b) 58 mm
(c) 70 mm
(d) 62.5 mm
Q2. A reciprocating water jacketed NH3 compressor installed in cold storage delivers 3 kg/min of NH3 with enthalpy of 1670 kJ/kg. At the suction line the enthalpy of NH3 is 1465 kJ/kg. if power input of the compressor is 16 kW what will be heat transfer rate to water jacket if ∆KE and ∆PE are neglected?
(a) -5.75 kW
(b) 5.75 kW
(c) -10.75 kW
(d) 10.75 kW

Q3. For the given fluctuating fatigue load, the values of stress amplitude and stress ratio are respectively

(a) 100 MPa and 5
(b) 250 MPa and 5
(c) 100 MPa and 0.20
(d) 250 MPa and 0.20

Q4. A cylindrical pin of 25_(+0.010)to(+0.020) mm diameter is electroplated. Plating thickness is 〖2.0〗^(+0.005) mm. Neglecting the gauge tolerance, the diameter (in mm, up to 3 decimal points accuracy) of the GO ring gauge to inspect the plated pin is____________
(a) 29.030
(b) 25.020
(c) 27.030
(d) 23.030

Q5. The number of customers arriving at a railway reservation counter is Poisson distributed with an arrival rate of eight customers per hour. The reservation clerk at this counter takes six minutes per customer on an average with an exponentially distributed service time. The average number of the customers in the queue will be
(a) 3
(b) 3.2
(c) 4
(d) 4.2

Q6. A metric thread of pitch 2 mm and thread angle 60° is inspected for its pitch diameter using 3- wire method. The diameter of the best size wire in mm is
(a) 0.866
(b) 1.000
(c) 1.154
(d) 2.000

Q7. A gas is stored in a cylindrical tank of inner radius 7m and wall thickness 50 mm. the gauge pressure of the gas is 2 MPa. The maximum shear stress (in MPa) in the wall is
(a) 35
(b) 70
(c) 140
(d) 280

Q8. A cylindrical uranium fuel rod of radius 5 mm in a nuclear is generating heat at the rate of 4 × 10^7 W/m³. the rod is cooled by a liquid (convective heat transfer coefficient 1000 W/m²K) at 25°C. at steady state, the surface temperature (in K) of the rod is
(a) 308
(b) 398
(c) 418
(d) 448

Q9. If the elongation factor during rolling of an ingot is 1.22. The minimum number of passes needed to produce a section 250×250 mm from an ingot of 750×750 mm are
(a) 8
(b) 9
(c) 10
(d) 12

Q10. For A four bar linkage in toggle position, the value of mechanical advantage is
(a) 0.0
(b) 0.5
(c) 1.0
(d) ∞

Solutions

S1. Ans.(c)
Sol. W=(AV_1)/v_1 =(A_2 V_1)/v_2
∵ A_1 V_1 ρ_1=A_2 V_2 ρ_2
∵ A_2=A_1 (V_1/V_2 )(ρ_1/ρ_2 )
∵ π/4 (d)^2=π/4 (50)^2 (2.45)×0.8
d = 70 mm
S2. Ans.(a)
Sol. h_1+q=h_2+W
∵ 1670 + q = 1465 + ((16×60)/3)
q = -115 kJ/kg = -115 × 3/60 kW
= 5.75 kW

S3. Ans.(c)
Sol. From the given figure we can see that
σ_max=250 MPa
σ_min=50 MPa
We know that,
Stress Amplitude = (σ_max-σ_min)/2
= =(250-50)/2=100MPa.
Stress Ratio =σ_min/σ_max
=50/250=0.20

S4. Ans.(a)
Sol. Cylindrical pin is
= 25_(+0.010)to(+ 0020) mm
We know that Go ring gauge work for pin on upper limit
i.e. = 25 + 0.020
= 25.020 mm
After electroplating, thickness will be added in twice because of both side addition of plating
=25.020+2×[2.005]
=29.030mm

S5. Ans.(b)
Sol. Arrival rate (λ) = 8 per hour
Service rate (μ) = 60/6 = 10 per hour
Average no. of customer in the queue
L_q=λ^2/(μ(μ-λ) )=(8)^2/10(10-8) =3.2

S6. Ans.(c)
Sol.
Pitch; P = 2 mm
Thread Angle: 2θ = 60°
θ = 30°
for best size wire
dW=P/2 secθ
=2/2 sec30°=1.154 mm

S7. Ans.(c)
Sol. Maximum shear stress in the wall
=σ_1/2=pd/4t=(2×14×1000)/(4×50)=140 MPa

S8. Ans.(b)
Sol. Given data; r = 5 mm = 0.005 m, d = 2r = 0.010 m, q_G=4×10^7 W\/m^2
h_0=1000 W\/m^2 k,T_n=25°C
∵ for steady state heat transfer,
Rate of heat generation in the rod = Rate of convection heat transfer from rod surface to the fluid.
q_G×Volume of rod=h_w A (T_S-T_0 )
q_G×π/4 d^2 l=h_0×πdl(T_S-T_0 )
(dq_G)/4=h(T_S-T_0 )⇒(0.010×4×10^7)/4=1000 (T_S-25 )
or,T_S-25=100
T_S=100+25=125°C
=(125+273)
=398K

S9. Ans.(c)
Sol. Elongation factor =A_0/A_1 =1.2
Let no. of passes be r, in order to get desired section of 250×250 mm^2 from ingot of 750×750 mm^2
Here, we have =(750×750)/(1.22)^n =250×250
(1.22)^n=(750×750)/(250×250)=9 ∴ =11.04

S10. Ans.(d)
Sol.

ω_4 of the output link DC become zero at the extreme positions. The extreme positions of the linkage are known as “Toggle position”.
∴ mechanical advantage =ω_input/ω_output
∵ ω_output=0
∴ Mechanical advantage = ∞

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