UPPSC Lec'21 ME: Daily Practices Quiz. 08-Nov-2021

Quiz: Mechanical Engineering
Exam: UPPSC AE/Lecturer
Topic: Miscellaneous

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 8 Minutes

Q1. The percentage area of the curve in normal distribution curve for ± 3 σ limits?
(a) 99.73%
(b) 95.45%
(c) 68.26%
(d) None of the above

Q2. In case of R-chart, the upper and lower control limits are respectively
(a) A_2 R ̅ and A_3 R ̅
(b) D_3 R ̅ and D_4 R ̅
(c) A_3 R ̅ and A_2 R ̅
(d) D_4 R ̅ and D_3 R ̅

Q3. The control chart, which is used to measure “variability of variability” within the samples?
(a) X ̅ chart
(b) R chart
(c) P chart
(d) C chart

Q4. Bolts are designed on the basis of
(a) direct tensile stress
(b) direct shear stress
(c) direct compressive stress
(d) direct bearing stress

Q5. In case of trapezoidal thread, the frictional torque at mean radius while raising load is where,
W = load, d_m = mean diameter, θ= angle of friction, α = Helix angle
(a) (Wd_m)/2×(μsecθ+tanα)/(1-μsecθ.tanα)
(b) (Wd_m)/2×(μsecθ+tanα)/(1+μsecθ.tanα)
(c) (Wd_m)/2×(μsecθ-tanα)/(1-μsecθ.tanα)
(d) (Wd_m)/2×(μsecθ-tanα)/(1+μsecθ.tanα)

Q6. What is the maximum efficiency of a screw jack?
(a) (1-cosφ)/(1+cosφ)
(b) (1+sinφ)/(1-sinφ)
(c) (1-tanφ)/(1+tanφ)
(d) (1-sinφ)/(1+sinφ)

Solutions
S1. Ans. (a)
Sol. for ± 3 σ limits, it is 99.73%
for ± 2 σ limits, it is 95.45%
for ± σ limits, it is 68.26%

S2. Ans. (b)
Sol. Upper control limit is D_4 R ̅
And lower control limit is D_3 R ̅
The value of D_4 and D_3 depends upon sample size.

S3. Ans. (b)
Sol. R chart is used to measure “variability of variability.”

S4. Ans. (a)
Sol. Bolts are designed on the basis of direct tensile stress.

S5. Ans. (a)
Sol. The frictional torque at mean radius in case of trapezoidal thread while raising load is
T=(Wd_m)/2×(μsecθ+tanα)/(1-μsecθ.tanα)

S6. Ans. (d)
Sol. Efficiency of screw jack is
ɳ=tanα/(tan⁡(φ+α))
where, φ & α are friction angle and helix angle repspectively.
It is maximum when the helix angle i.e α=π/4-φ/2
And the maximum efficiency is
ɳ_max=(1-sinφ)/(1+sinφ)