# Mechanical engineering Questions and solutions set - 4

Mechanical engineering Questions and solutions set - 4
Question:-
Option (A)
8 min
Option (B)
12 min
Option (C)
16 min
Option (D)
20 min
Correct Option:
B
Question Solution:
No. of stroke required = 30mm/(0.3mm/stroke)=100stroke
Distance travelled in one stroke = 20 + 600 + 20 = 640 mm
time taken in forward direction = 640(mm)/(8000mm/min)=0.08min = 4.8 sec.
time taken in reverse direction =(4.8/2)=2.4sec
(\ Speed ratio =(Time taken in forward dis)/(Time taken in backward dis )=2/1)
\ Time taken for one stroke = 4.8 + 2.4 = 7.2 Sec.
\ time taken for 100 stroke = 7.2×100 = 720 sec = 12 min
Question:-
Option (A)
0.03972 MPa
Option (B)
-0.03972 MPa
Option (C)
-0.07945 MPa
Option (D)
0.07945 MPa
Correct Option:
C
Question Solution:
a = 12.5×10-6/°C
E = 200 G Pa
rise in temperature = ΔT = 20°C
Under free expansion (dl) = alΔT
= 12.5×10-6×0.5×20
1.25×10-4 m
Temperature stress due to permitted expansion d
stemp=(E(dl-d)/l)
Force =(E(dl-d)/l)x A =Kd
Þ200x109[(1.25x10-4-d)/0.5]x(p/4)x0.012=50x103x d
Þ1.25x10-4-d=1.59x10-3d
Þ d=1.248x10-4m
\ stemp=(200x109(1.25-1.248)x10-4/0.5)
=80000 N/M2 (compressive)
=-0.08 MPa
Question:-
Option (A)
50
Option (B)
16
Option (C)
29
Option (D)
38
Correct Option:
C
Question Solution:
Profit (at P) = 2×0 + 5×10 = 50 (\ x1 and x2 > 0)
Profit (at Q) = 2×8 + 5×0 = 16 max. profit = 29
Profit (at R) = 2×7 + 5×3 = 29
Question:-
Option (A)
Slot I
Option (B)
Slot II
Option (C)
Slot III
Option (D)
Any of three
Correct Option:
B
Question Solution:
Case I
Q=Ö(2x2000x20)/0.16x 2=500(<1000)
Ordering cost =(2000/500)x20=80
Average inventory cost =(500+0)/2x0.16x2=80
Total cost =2000x2+80+80
=4160
Case II
Q=Ö(2x2000x20)/0.16x1.9=513(<1000)
Ordering cost =(2000/1000)x20=40
Average inventory cost =(1000+0)/2x0.16x1.9 = 152
Total cost =2000x1.9+40+152=Rs.3992

Case III
Q=Ö (2x2000x20)/0.16x1.86=518(<1000)
Ordering cost =(2000/1000)x20=20
Average inventory cost =(2000+0)/2x0.16x1.86=297.6
Total cost =2000x1.86+20+297.6=Rs. 4037.6
\ slot II is optimum
Question:-
Option (A)
710 m/s
Option (B)
692.5 m/s
Option (C)
671.3 m/s
Option (D)
667.9 m/s
Correct Option:
B
Question Solution: h1 = 3000 KJ/kg
h2 = 2762 kg/kg
v1 = 60 m/s
v2 = ?
Applying steady flow energy equation
h1+(v21/2000)=h2+(v22/2000)

Þ 3000+(602/2000) =2762+ (v22/2000)
Þ v2=692.53 m/s
Question:-
Option (A)
30 KW
Option (B)
15 KW
Option (C)
12 KW
Option (D)
6 KW
Correct Option:
C
Question Solution:
Q0 = 3600×2× [27 – (-3)] + 3600 × 230 + 3600 ×0.5×[-3-(-23)]
= 1080000 KJ / 10 hr
= 30 KW
(COP)Carnot =250/(300-250)=5
(COP)actual=(5/2)=2.5
\(Q0/W)=2.5Þ W=(30/2.5)=12 KW 