# Mechanical engineering Questions and solutions set - 5

Mechanical engineering Questions and solutions set - 5
Question:-
Option (A)
50×106 Nmm
Option (B)
12.5×106 Nmm
Option (C)
100×106 Nmm
Option (D)
25×106 Nmm
Correct Option:
A
Question Solution:
Total load = (1/2)x2x37.5
= 37.5 KN. Line of action of this total load will be at (2/3)x2=(4/3) m from fixed end.
Bending moment at fixed end
= 37.5 ×103 ×(4/3)x103
= 50 × 106 Nmm
Question:-
Option (A)
there is no energy or mass transfer across the boundary
Option (B)
there is no mass transfer, but energy transfer exists
Option (C)
there is no energy transfer, but mass transfer exists.
Option (D)
both energy and mass transfer take place across the boundary, but the mass transfer is controlled by values.
Correct Option:
B
Question Solution: Question:-
Option (A)
16 (2/3) KJ
Option (B)
33 (1/3) KJ
Option (C)
37 (1/3) KJ
Option (D)
66 (2/3) KJ
Correct Option:
A
Question Solution:
h=(Work done)/(heat Supplied)
Þ heat supplied = (50KJ)/(0.75)=66.67 KJ
\ Heat engine is 75% efficient
\ heat rejected is 25% of heat supplied
\ Heat rejected = 66.67 × 0.25
= 16.67 KJ
Question:-
Option (A)
0 KJ
Option (B)
1 KJ
Option (C)
2 KJ
Option (D)
3 KJ
Correct Option:
C
Question Solution:
Net pressure action on the piston = 300 KPa - 100 KPa
= 200 KPa
(which is constant throughout the expansion process)
and change in volume = dV = 0.01 m3
\ W.D in expansion process = pdv
={200x103(N/m2}x(0.01m3)
= 2 × 103 Nm
= 2 KJ
Question:-
Option (A)
(pl)/((1-2m)E)
Option (B)
(pl(1-2m))/E
Option (C)
(plm)/E
Option (D)
(pl)/( mE)
Correct Option:
B
Question Solution:
sx=(P/A); sy=0; sz=0
Îx=(sx/E) Îy=-m(sx/E) and Îz=-m(sx/E)
Îv=Îx+Îv+Îy=(sx/E)-2m(sx/E)=( sx/E)(1-2m)
Þ Îv =(dv/v)=(P/AE)(1-2m)
Þ dv=(Pv/AE)(1-2m) Þ dv=(pl/E)(1-2m)(because v=Al)
Question:-
Option (A)
8.7
Option (B)
13.9
Option (C)
17.3
Option (D)
27.3
Correct Option:
C
Question Solution:
Characteristics length (Lc) = (V/As)={((4/3)pr3)/(4pr2)}=(r/3)
= ((2.5x10-3)/3)=8.333x10-4m
Biot Number (Bi) =(hLc/K)
=((250x8.33x10-4)/400)=5.208x10-4
Thermal diffusivity (a)=(K/rcp)=(400)/(9000x385)=1.1544x10-4
Fourier Number (F0) =(at/Lc2)={(1.1544x10-4x t)/(8.333x10-4)2}=166.247 t
Now
(q/qi)=(T-Tsurr)/(Ti-Tsurr)=e-Bi x F0
Þ(T-300)/(500-300)=e-5.208x10-4x166.247t
Þ ((T-300)/200)=e-0.0866t
Þ en{(T-300)/200}=-0.0866t
Þ en(T-300)-en200=-0.0866t
Differentiate w.r.t t
(1/(T-300))(dT/dt)=-0.0866 Þ (dT/dt)at=500 =-0.0866 x (500-300) =-17.32 K/S