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Debanshi DattaEduRev Class 11 |
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Debanshi Datta
EduRev Class 11
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Discussed Questions
A cantilever of length (L) carries a distributed load whose intensity varies uniformly from zero at the fixed end to ‘w’ per unit length at the free end. The deflection at the free end will be- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A cantilever of length (L) carries a distributed load whose intensity varies uniformly from zero at the fixed end to ‘w’ per unit length at the free end. The deflection at the free end will be
a)
b)
c)
d)
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Neha Joshi answered • Oct 11, 2023 |
δ1 = 
δ2 = 
δ3 = δ1 − δ2 = 
= 
Calculate the rate of heat loss from a human body which may be considered as a vertical cylinder 30 cm in diameter, and 175 cm high while standing in a 30 km/hr wind at 15°C. The surface temperature of the human is 35°C At the mean film temperature,
The thermo-physical properties of air are
v = 15.33 x 10-6 m2/s
k = 0.0263 W/m - ° C
Pr = 0.7
Given, Nu = 0.13 (Gr x Pr)0.33
Correct answer is between '207,210'. Can you explain this answer?
Calculate the rate of heat loss from a human body which may be considered as a vertical cylinder 30 cm in diameter, and 175 cm high while standing in a 30 km/hr wind at 15°C. The surface temperature of the human is 35°C At the mean film temperature,
The thermo-physical properties of air are
v = 15.33 x 10-6 m2/s
k = 0.0263 W/m - ° C
Pr = 0.7
Given, Nu = 0.13 (Gr x Pr)0.33
The thermo-physical properties of air are
v = 15.33 x 10-6 m2/s
k = 0.0263 W/m - ° C
Pr = 0.7
Given, Nu = 0.13 (Gr x Pr)0.33
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Neha Joshi answered • Oct 11, 2023 |
Radiative heat transfer is given by,
A cast iron bracket, supporting the transmission shaft and the belt pulley, is fixed to the steel structure by means of four bolts as shown in Fig. There are two bolts atA and two bolts atB. The tensions in slack and tight sides of the belt are 5 kN and 10 kN respectively. The belt tensions act in a vertically downward direction. The distances are as follows,l1 = 50 mm, l2 = 150 mm, l = 200 mm The maximum permissible tensile stress in any bolt is 60 N/mm2 . Determine the size of the bolts (in mm2 )
- a) 145
- b) 170
- c) 150
- d) 135
Correct answer is option 'C'. Can you explain this answer?
A cast iron bracket, supporting the transmission shaft and the belt pulley, is fixed to the steel structure by means of four bolts as shown in Fig. There are two bolts at
A and two bolts at
B. The tensions in slack and tight sides of the belt are 5 kN and 10 kN respectively. The belt tensions act in a vertically downward direction. The distances are as follows,
l1 = 50 mm, l2 = 150 mm, l = 200 mm The maximum permissible tensile stress in any bolt is 60 N/mm2 . Determine the size of the bolts (in mm2 )
a)
145
b)
170
c)
150
d)
135
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Neha Joshi answered • Oct 11, 2023 |
Given,P = (5 + 10) kN,
l = 200 mm
t)max = 60 N/mm2Tensile stress due to the tendency of bracket to tilt about ‘C’
P1= Cl1, P2 = Cl2
Pl = 2P1l1 + 2P2l2 = 2(Cl1)l1 + 2(Cl2)l2
The maximum force will act on bolts denoted by 2
P2 = Cl2
Size of bolts, P2 = A(
t)max
t)max9000 = A(60)
∴A = 150 mm2
Slenderness ratio has dimensions of- a)cm
- b)cm-1
- c)dimensionless quantity
- d)cm2
Correct answer is option 'C'. Can you explain this answer?
Slenderness ratio has dimensions of
a)
cm
b)
cm-1
c)
dimensionless quantity
d)
cm2
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Neha Joshi answered • Oct 11, 2023 |
Slenderness ratio
A frictionless circular piston of area 10-2 m and mass 100 kg sinks into a cylindrical container of the same area filled with water of density 3 1000 kg/m as shown in the figure. The container has a hole of area 10-3 m2 at the bottom that is open to the atmosphere. Assuming there is no leakage from the edges of the piston and considering water to be incompressible, the magnitude of the piston velocity (in m/s) at the instant shown is __________ (correct to two decimal places).
Correct answer is '1.456'. Can you explain this answer?
A frictionless circular piston of area 10-2 m and mass 100 kg sinks into a cylindrical container of the same area filled with water of density 3 1000 kg/m as shown in the figure. The container has a hole of area 10-3 m2 at the bottom that is open to the atmosphere. Assuming there is no leakage from the edges of the piston and considering water to be incompressible, the magnitude of the piston velocity (in m/s) at the instant shown is __________ (correct to two decimal places).
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Neha Joshi answered • Oct 11, 2023 |
Total gauge pressure at top 
Total gauge pressure at bottom = 0
Total gauge pressure at bottom = 0
A double fillet welded joint with parallel fillet weld of length L and leg B is subjected to a tensile force P. Assuming uniform stress distribution, the shear stress in the weld is given by- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
A double fillet welded joint with parallel fillet weld of length L and leg B is subjected to a tensile force P. Assuming uniform stress distribution, the shear stress in the weld is given by
a)
b)
c)
d)
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Neha Joshi answered • Oct 11, 2023 |
Where
Le = Effective length
Le = 2L(Double fillet weld)
.
A circular plate 1 m in diameter is submerged vertically in water such its upper edge is 8 m below the free surface of water. The total hydrostatic pressure force on one side of plate is[1988]- a)6.7 kN
- b)65.4 kN
- c)45.0 kN
- d)77.0 kN
Correct answer is option 'B'. Can you explain this answer?
A circular plate 1 m in diameter is submerged vertically in water such its upper edge is 8 m below the free surface of water. The total hydrostatic pressure force on one side of plate is
[1988]
a)
6.7 kN
b)
65.4 kN
c)
45.0 kN
d)
77.0 kN
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Neha Joshi answered • Oct 11, 2023 |
Total hydrostatic pressure force, F = ρghA
F = 1000 × 9.81 × 8.5 × 0.785 = 65458 N
or
F = 65.45 KN
Total strain energy stored in a simply supported beam of span, 'L' and flexural rigidity 'EI 'subjected to a concentrated load 'W' at the centre is equal to- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Total strain energy stored in a simply supported beam of span, 'L' and flexural rigidity 'EI 'subjected to a concentrated load 'W' at the centre is equal to
a)
b)
c)
d)
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Neha Joshi answered • Oct 11, 2023 |
Strain energy
=
= 
= 
Alternative method: In a funny way you may use Castigliano’s theorem,
δ = 
We know that δ = 
for simply supported beam in concentrated load at mid span.
for simply supported beam in concentrated load at mid span.Then δ = 
or
orU =
partially integrating
partially integratingwith respect to W we ge U = 
A truss consists of horizontal members (AC, CD, DB and EF) and vertical members (CE and DF) having length l each. The-members AE, DE and BF are inclined at 45° to the horizontal For the uniformly distributed load p per unit length on the members EF of the truss shown in figure given below, the force in the member CD is
[2003]- a)pl/2
- b)pl
- c)0
- d)2pl/3
Correct answer is option 'A'. Can you explain this answer?
A truss consists of horizontal members (AC, CD, DB and EF) and vertical members (CE and DF) having length l each. The-members AE, DE and BF are inclined at 45° to the horizontal For the uniformly distributed load p per unit length on the members EF of the truss shown in figure given below, the force in the member CD is
[2003]
a)
pl/2
b)
pl
c)
0
d)
2pl/3
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Neha Joshi answered • Oct 11, 2023 |
Total load on EF member = pl
where l = length of EF
For horizontal equilibrium
RA + RB = Pl
Taking moment about A, we have
Considering a point A.
For horizontal equilibrium FAC + FAE cos 45º = 0
For vertical equilibrium
RA – FAE sin 45º = 0
–ve sign shows that the force on member AC is opposite of assumed. Now considering a point C For horizontal equilibrium
where l = length of EF
For horizontal equilibrium
RA + RB = Pl
Taking moment about A, we have
Considering a point A.
For horizontal equilibrium FAC + FAE cos 45º = 0
For vertical equilibrium
RA – FAE sin 45º = 0
–ve sign shows that the force on member AC is opposite of assumed. Now considering a point C For horizontal equilibrium
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