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# Multiple choice question for engineering

## Set 1

1. Which fin yields the maximum heat flow per unit weight?
a) Straight triangular fin
b) Curved triangular fin
c) Parabolic fin
d) Hyperbolic fin

Answer: a [Reason:] In straight triangular fin, there is maximum heat flow.

2. Heat dissipation by every segment of the fin is
a) Sometimes same
b) Same
c) Not same
d) Sometimes same or sometimes not same

Answer: c [Reason:] It is always different as fins are not uniform with respect to cross-sectional area.

3. “If a fin of a constant cross section is used, there would be wastage of material”. Chose the correct option
a) True
b) False

Answer: a [Reason:] Cross section must vary to utilize the material.

4. Which one is true regarding parabolic fin?
a) It dissipates the minimum amount of heat at minimum material cost
b) It dissipates the minimum amount of heat at maximum material cost
c) It dissipates the maximum amount of heat at maximum material cost
d) It dissipates the maximum amount of heat at minimum material cost

Answer: d [Reason:] In this case a parabolic fin is of great practical importance.

5. For parabolic fin, the curve follows which law?
a) y = C/x2
b) y = C x4
c) y = C x2
d) y = C x1/2

Answer: c [Reason:] Equation of parabola is y = 4 x2 or x = 4 y2.

6. The correction length for cylindrical fin is
a) L C = L + d/4
b) L C = 2 L + d/4
c) L C = 3 L + d/4
d) L C = 4 L + d/4
Where, d is the diameter

Answer: a [Reason:] Area = π d2/4.

7. Provision of fins on a given heat transfer surface will be more effective if there is
a) Fewer but thick fins
b) Large number of thick fins
c) Fewer but thin fins
d) Large number of thin fins

Answer: d [Reason:] Increase in ratio of perimeter P to be cross sectional area A C brings about improvement in the effectiveness of fins.

8. The heat dissipation at any section of parabolic fin is given by
a) (t2 – t1) (b) (δ)
b) k (t2 – t1) (b) (δ)
c) k (t2 – t1) (δ)
d) k (t2 – t1) (b)

Answer: b [Reason:] Q = q x (A X) = k (t2 – t1) (b) (δ).

9. An air cooled cylindrical wall is to be fitted with triangular fins of 3 cm thickness at base and 12 cm in height. The fins are made from stainless steel with density 8000 kg/m3 and thermal conductivity 17.5 W/m K. The wall temperature is 600 degree Celsius and the fin is exposed to an environment with t a = 30 degree Celsius and h = 20 W/m2 K. What is the temperature distribution along the fin?
a) t = 10 + 250 I 0 [6.056 (x) 1/2].
b) t = 20 + 250 I 0 [6.056 (x) 1/2].
c) t = 30 + 250 I 0 [6.056 (x) 1/2].
d) t = 40 + 250 I 0 [6.056 (x) 1/2].

Answer: c [Reason:] α/α 0 = t – t 0/t 0 – t a = I 0 [2 B (x) ½]/ I 0 [2 B (l) ½]. Here B = (2 h l/k δ) ½ = 3.028.

10. Consider the above problem, make calculations for the rate of heat flow per unit mass of fin material used
a) 126.53 W/kg
b) 154.76 W/kg
c) 134.87 W/kg
d) 165.46 W/kg

Answer: a [Reason:] Q = b (2 h k δ) ½ α 0 I 1 [2 B (L) ½/ I 0 [2 B (L) ½ = 1822 W. Mass of fin per meter width = 14.4 kg. Therefore rate of heat flow per unit mass = 1822/14.4 = 126.53 W/kg.

## Set 2

1. The distribution of radiant energy is non uniform with respect to
a) Wavelength
b) Direction
c) Both wavelength and direction
d) Time

Answer: c [Reason:] It should be non-uniform with respect to both wavelength and direction.

2. “Spatial distribution is also known as directional distribution”. Choose the correct option
a) True
b) False

Answer: a [Reason:] Spatial or directional both are same.

3. Of the radiant energy 350W/m2 incident upon a surface 250W/m2 is absorbed, 60W/m2 is reflected and the remainder is transmitted through the surface. Workout the value for absorptivity for the surface material
a) 0.123
b) 0.714
c) 0.684
d) 0.386

Answer: b [Reason:] Absorptivity = 250/350 = 0.714.

4. Thermal radiation strikes a surface which has a reflectivity of 0.55 and transmissivity of 0.032. A quantity known as flux is found out to be 95 W/m 2. Determine the rate of incident flux.
a) 123.34 W/m2
b) 333.37 W/m2
c) 122.27 W/m2
d) 227.27 W/m2

Answer: d [Reason:] Incident flux = 95/1 – 0.032 – 0.55 = 227.27 W/m2.

5. What is the wavelength for visible light?
a) 3.9 * 10 -1 to 7.8 * 10 -1 micron meter
b) 4.9 * 10 -1 to 7.8 * 10 -1 micron meter
c) 5.9 * 10 -1 to 7.8 * 10 -1 micron meter
d) 6.9 * 10 -1 to 7.8 * 10 -1 micron meter

Answer: a [Reason:] This is the maximum and minimum wavelength for visible light.

6. Radiant energy with an intensity of 800 W/m2 strikes a flat plate normally. The absorptivity is thrice the reflectivity and twice the transmissivity. Determine the rate of absorption
a) 236.40 W/m2
b) 336.40 W/m2
c) 436.40 W/m2
d) 536.40 W/m2

Answer: c [Reason:] Rate of absorption = 0.5455 * 800 = 436.40 W/m2.

7. A thin metal plate of 4 cm diameter is suspended in atmospheric air whose temperature is 290 K. This plate attains a temperature of 295 K when one of its face receives radiant energy from a heat source at the rate of 2 W. If heat transfer coefficient on both surfaces of the plate is stated to be 87.5 W/m 2 K, workout the reflectivity of the plate
a) 0.35
b) 0.45
c) 0.55
d) 0.65

Answer: b [Reason:] Heat loss by convection from both sides of the plates = 2 h A d t = 1.1 W. Energy lost by reflection = 2.0 – 1.1 = 0.9 W.

8. On clear night there is radiation from earth’s surface to the space. On such a night, the water particles on the plant leaves radiate to the sky whose temperature may be taken as 200 K. The water particles receive heat by convection from the surrounding air, the convection heat transfer coefficient has a value of 30 W/m2 K. If the water should not freeze, make calculations for the air temperature
a) 280.474 K
b) 345.645 K
c) 123.456 K
d) 874.387 K

Answer: a [Reason:] Heat radiated to sky = Heat received by convection. So, temperature of air = 224.22/30 + 273 = 280.474 K.

9. “Isothermal furnaces, with small apertures, approximate a black body and are frequently used to calibrate heat flux gauges”. Chose the correct answer
a) True
b) False

Answer: a [Reason:] A small hole leading into a cavity thus acts very nearly as a black body.

10. “A surface element emits radiation in all directions, the intensity of radiation is however same in different directions. Choose the correct answer
a) True
b) False

Answer: b [Reason:] It must be different in different directions.

## Set 3

1. Clamp coupling employ bolts.
a) True
b) False

Answer: a [Reason:] The two halves of sleeve are clamped together by bolts.

2. Entire torque is transmitted by friction in the clamp coupling.
a) True
b) False

Answer: b [Reason:] Some torque is transferred by friction and some by key.

3. Does shaft has to be shifted i axial direction to remove clamp coupling.
a) Yes
b) No

Answer: b [Reason:] Clamp coupling unlike muff coupling is easily removed without shifting the shaft in axial direction.

4. If shaft diameter is 40mm, calculate the diameter of sleeves in clamp coupling.
a) 100mm
b) 80mm
c) 60mm
d) 40mm

5. If shaft diameter is 40mm, calculate the length of sleeves in clamp coupling.
a) 80mm
b) 140mm
c) 100mm
d) 120mm

6. If 8 bolts are emplaced in a clamp coupling with shaft diameter 80mm d, calculate the tensile force on each bolt if coefficient of friction is 0.3 and torque transmitted is 4000N-m.
a) 51234.4N
b) 45968.3N
c) 41666.7N
d) None of the listed

7. If 8 bolts are emplaced in a clamp coupling with shaft diameter 80mm d, calculate the diameter of each bolt if coefficient of friction is 0.3 and torque transmitted is 4000N-m. Permissible tensile stress is 80N/mm².
a) 27mm
b) 25mm
c) 23mm
d) 21mm

8. Power is transmitted by only in rigid flange coupling.
a) True
b) False

Answer: b [Reason:] Power is transmitted by key as well as bolts.

9. If shaft diameter is 60mm, how many bolts are recommended for rigid flange coupling?
a) 2
b) 3
c) 4
d) 5

Answer: c [Reason:] If shaft diameter is between 40mm and 100mm, 4 bolts are used.

10. Determine the diameter of the bolts used in rigid flange coupling if transmitted torque is 270N-m, pitch circle diameter=125mm and four bolts are emplace in the coupling. Permissible shear stress in the bolts is 70N/mm².
a) 3.8mm
b) 3.6mm
c) 4.4mm
d) 4mm

11. The hub is treated as a solid shaft while calculating torsional shear stress in the hub.
a) True
b) False

Answer: b [Reason:] Hub is treated as a hollow shaft.

12. Find the shear stress in a flange at the junction of hub in rigid flanged coupling if torsional moment is 2980N-m and diameter of hub being 125mm. Also the thickness of flange is 25mm.
a) 6.77N/mm²
b) 10.24N/mm²
c) 4.84N/mm²
d) 4.22N/mm²

13. If shaft diameter is 30mm and number of pins emplaced are 6, then the diameter of the pin will be?
a) 6.4mm
b) 5.6mm
c) 6.1mm
d) 5.9mm

14. Calculate the force acting on the each pin in flexible coupling if torque transmitted is 397N-m and PCD=120mm with number of pins 6.
a) 1400.3N
b) 1102.8N
c) 1320.3N
d) None of the listed

## Set 4

Answer the question 1-7 with respect to figure 1
1. The two welds shown in the figure experience equal resisting force?
a) True
b) False

Answer: b [Reason:] As the welded joints are unsymetrically loaded hence resisting forces are different in the two welds.

2. The moment of forces about the CG is ?
a) Zero
b) Infinite
c) P₁y₁+P₂y₂
d) None of the listed

Answer: a [Reason:] External force passes through CG hence moment is zero.

3. Which equations can be used to find the resisting force in the two welds?
a) P=P₁+P₂
b) P₁+P₂=0
c) Both P=P₁+P₂ and P₁y₁=P₂y₂
d) P₁y₁=P₂y₂

Answer: c [Reason:] One equation is of equilibrium and the other conservation of moment about CG.

4. For the following welded joint l₁y₁=l₂y₂.
a) True
b) False

Answer: a [Reason:] P₁y₁=P₂y₂ and P₁=0.707hl₁τ and P₂=0.707hl₂τ.

5. Find the total length of weld required to with stand the load of 100kN. Permissible shear stress in weld is 70kN/mm².
a) None of the listed
b) 101.03mm
c) 202.06mm
d) 30309mm

6. Find the length of weld 2 required to with stand the load of 100kN. Permissible shear stress in weld is 70kN/mm².
a) None of the listed
b) 102.22mm
c) 132.4mm
d) 70.16mm

Answer: c [Reason:] By finding CG y₂=103.9mm, therefore l₁y₁=l₂y₂ and l₁+l₂=202.06mm.

7. Find the length of weld 1 required to with stand the load of 100kN. Permissible shear stress in weld is 70kN/mm².
a) None of the listed
b) 102.22mm
c) 132.4mm
d) 70.16mm

Answer: d [Reason:] By finding CG y₂=103.9mm, therefore l₁y₁=l₂y₂ and l₁+l₂=202.06mm.

Figure 2

Answer the following questions with respect to figure 2

8. The following welded joint is an example of
a) Eccentric Load in plane of welds
b) Axial load in plane of welds
c) Axial load in plane perpendicular to plane of welds
d) None of the mentioned

Answer: a [Reason:] The figure clearly depicts.

9. What kind of stresses does the welded joint undergo?
a) Torsional shear stress
b) Direct shear stress
c) Direct and torsional shear stress
d) None of the listed

Answer: c [Reason:] After shifting force to CG, we have a force and a moment about CG.

10. While considering moment of inertia for calculating torsional shear stress, J=I(xx) + I(yy), which of the following can be neglected in context with the following figure?
a) I(xx)
b) I(yy)
c) Both I(xx) and I(yy)
d) None of the listed

Answer: a [Reason:] I(xx) is negligible as compared to I(yy) as I(xx)=ltᵌ/12 and t is very less as compared to l so I(xx) is neglected in comparison to I(yy).

11. Calculate the direct shear stress in the welds by ignoring torsional shear stress if P=8kN and thickness of weld is 5mm.
a) None of the listed
b) 12.33N/mm
c) 13.33N/mm
d) 14.33N/mm

12. Determine the torsional shear stress in the welds if P=8kN and thickness of the welds is assumed t.
a) None of the listed
b) 456.34/t
c) 543.13/t
d) 589.31/t

Answer: c [Reason:] M=8000×150,r(farthest point)=√25²+25²,J=2x[tlᵌ/12 + Ar₁²] ; τ=Mr/J.

13. Find the thickness of the weld if P=8kN and permissible shear stress in the welds is 100N/mm²
a) 4mm
b) 5mm
c) 6mm
d) 7mm

Answer: c [Reason:] Direct shear stress and torsional shear stress act at an angle of 45’,135’,225’ and 315’. Maximum stress will be at the point where they act at 45’. Direct=8000/120t or 66.67/t and torsional =543.13/t. Hence net shear=592.1/t =100.

## Set 5

1. The various cutting tool materials used are:
a) high speed steels
b) cast cobalt alloys
c) carbides
d) all of the mentioned

Answer: d [Reason:] Materials used for cutting tools are a) high speed steels b) cast cobalt alloys c) carbides d) coated tools e) alumina-based ceramics f) cubic boron nitride g) silicon nitride based ceramics h) diamond i) whisker reinforced materials and nano materials

2. High speed steels are suitable for making
a) high positive rake angle tools
b) interrupted cuts
c) machine tools with ow stiffness that are subject to vibration
d) all of the mentioned

Answer: d [Reason:] Because of their toughness (hence high resistance to fracture), high-speed steels are suitable especially for (a) high positive rake-angle tools (i.e., those with small included angles), (b) interrupted cuts, (c) machine tools with low stiffness that are subject to vibration and chatter, and (d) complex and single-piece tools, such as drills, reamers, taps, and gear cutters.

3. ___________ improves toughness, wear resistance, and high temperature strength.
a) Chromium
c) Tungsten
d) None of the mentioned

Answer: a [Reason:] Chromium improves toughness, wear resistance, and high-temperature strength.Vanadium improves toughness, abrasion resistance, and hot hardness. Tungsten and cobalt have similar effects, namely, improved strength and hot hardness. Molybdenum improves wear resistance, toughness, and high-temperature strength and hardness.

4. _____________ contains nickel molybdenum matrix.
a) Chromium
b) Titanium carbide
c) Tungsten
d) None of the mentioned

Answer: b [Reason:] Titanium carbide (TiC) consists of a nickel-molybdenum matrix. It has higher Wear resistance than tungsten carbide but is not as tough. Titanium carbide is suitable for machining hard materials (mainly steels and cast irons) and for cutting at speeds higher than those appropriate for tungsten carbide.

5. Coating materials used are
a) titaniun nitride
b) titanium carbide
c) titanium carbonitride
d) all of the mentioned

Answer: d [Reason:] Commonly used coating materials are titanium nitride (TiN), titanium carbide (TiC), titanium carbonitride (TiCN), and aluminum oxide (Al2O3).

6. Characteristics of coated cutting tools are:
a) high hardness
b) chemical stability
c) low thermal conductivity
d) all of the mentioned

Answer: d [Reason:] Coatings for cutting tools and dies should have the following general characteristics: ° High hardness at elevated temperatures, to resist wear. ° Chemical stability and inertness to the workpiece material, to reduce wear. ° Low thermal conductivity, to prevent temperature rise in the substrate. ° Compatibility and good bonding to the substrate, to prevent flaking or spalling. ° Little or no porosity in the coating, to maintain its integrity and strength.

7. Ceramic tools are fixed ti tool body by
a) soldering
b) brazing
c) welding
d) clamping

Answer: b [Reason:] The ceramic tools are fixed to a tool body by brazing. These tools have greater tool life than carbide tools.

8. The carbide tools operating at very low cutting speeds
a) reduces tool life
b) increases tool life
c) have no effect on tool life
d) spoils the work piece

Answer: a [Reason:] The carbide tools operating at very low cutting speeds below 30m/min reduces tool life.

9. High speed steel tools retain their hardness upto a temperature of
a) 2500
b) 3500
c) 5000
d) 9000