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Multiple choice question for engineering

Set 1

1. How many types of fluid flow are characterized in the realms of fluid mechanics?
a) 1
b) 2
c) 3
d) 4

View Answer

Answer: b [Reason:] There are two types of flow i.e. laminar and turbulent flow.

2. In which fluid flow, the motion of fluid particles is irregular?
a) Turbulent
b) Laminar
c) One dimensional
d) Two dimensional

View Answer

Answer: a [Reason:] It proceeds along erratic and unpredictable paths.

3. Following are the characteristics of turbulent flow
(i) Eddying
(ii) Sinuous
(iii) Rectilinear
Identify the correct option
a) 2 and 3
b) 1 and 3
c) 1, 2 and 3
d) 1 and 2

View Answer

Answer: d [Reason:] Obviously a turbulent flow is eddying and sinuous rather than rectilinear in character.

4. The nature of fluid flow is governed by following parameters
(i) Mean flow velocity
(ii) Density of fluid
(iii) Dynamic viscosity of the fluid
Identify the correct statements
a) 1 and 3
b) 1 only
c) 1, 2 and 3
d) 2 and 3

View Answer

Answer: c [Reason:] Osborne Reynolds, an English scientist confirmed the existence of these two regimes experimentally.

5. The value of convective coefficient of air in case of free convection is
a) 3-7 W/m2 K
b) 3-4 W/m2 K
c) 8-9 W/m2 K
d) 9-9.5 W/m2 K

View Answer

Answer: a [Reason:] In case of air, the value of convective coefficient is small i.e. 3 W/m2 K to 7 W/m2 K. This is due to presence of some moisture in air.

6. The fluid particles move in flat or curved un-mixing layers or streams and follow a smooth continuous path. This type of flow is known as
a) Steady flow
b) Stream flow
c) Turbulent flow
d) Laminar flow

View Answer

Answer: d [Reason:] The paths of fluid movement are well defined so it is laminar flow.

7. The characteristic dimension d in the relation R E = V d p/δ is the equivalent diameter and is defined as how many times the cross-sectional flow area divided by wetted perimeter
a) 7
b) 4
c) 1
d) 6

View Answer

Answer: b [Reason:] It is defined as four times the equivalent diameter and is defined as how many times the cross-sectional flow area divided by wetted perimeter.

8. For a duct of rectangular cross-section with length l and breadth b, the value of d e is
a) l b / l + b
b) 2 l b
c) 2 l b / l + b
d) 4 l b / l + b

View Answer

Answer: c [Reason:] d e = 4 (l) (b)/2 l + 2 b.

9. In many flow situations, the duct can be
(i) Circular
(ii) Rectangle
(iii) Trapezoidal
(iv) Annulus
Identify the correct option
a) 1 and 2
b) 1, 2, 3 and 4
c) 1, 2 and 3
d) 3 and 4

View Answer

Answer: b [Reason:] In many flow situations, the duct is not circular but is rectangle, trapezoidal or even annulus formed by a tube within another tube.

10. If an annulus has an inner diameter of d 1 and an outer diameter of d 2 then the equivalent diameter is
a) 2 d 2 – d 1
b) d 2 – 2 d 1
c) d 1 – d 2
d) d 2 – d 1

View Answer

Answer: d [Reason:] 4 (π/4) (d 22 – d 12)/π (d 1 + d 2).

Set 2

1. Glycerin at 10 degree Celsius flows past a flat plate at 20 m/s. Workout the velocity components at a point P(x, y) in the fluid flow where
x = 2 m from the leading edge of the plate
y = 5 cm from the plate surface
For glycerin at 10 degree Celsius, kinematic viscosity = 2.79 * 10 -3 m2/s
a) 15.92 m/s and 0.0952 m/s
b) 16.92 m/s and 0.0952 m/s
c) 17.92 m/s and 0.0752 m/s
d) 18.92 m/s and 0.0752 m/s

View Answer

Answer: b [Reason:] u/U INFINITY = 0L.846 and v/U INFINITY (Re) ½ = 0.57.

2. A plate 0.3 m long is placed at zero angle of incidence in a stream of 15 degree Celsius water moving at 1 m/s. Find out the stream wise velocity component at the mid-point of the boundary layer. For water at 15 degree Celsius
p = 998.9 kg /m3
µ = 415.85 * 10 -2 kg/hr m
a) 0.736 m/s
b) 0.636 m/s
c) 0.536 m/s
d) 0.436 m/s

View Answer

Answer: a [Reason:] n = y (U INFINITY/v x) ½ = 2.5.

3. Air at 25 degree Celsius flows over a flat surface with a sharp leading edge at 1.5 m/s. Find the boundary layer thickness at 0.5 from the leading edge. For air at 25 degree Celsius, kinematic viscosity = 15.53* 10n -6 m2/s
a) 4.1376 cm
b) 3.1376 cm
c) 2.1376 cm
d) 1.1376 cm

View Answer

Answer: d [Reason:] δ = 5 x/ (Re) ½ = 1.1376 m.

4. Local skin friction coefficient is given by
a) 0.646/ (Re) 1/2
b) 1.646/ (Re) 1/2
c) 2.646/ (Re) 1/2
d) 3.646/ (Re) 1/2

View Answer

Answer: a [Reason:] It should be 0.646/ (Re) ½.

5. A plate 0.3 m long is placed at zero angle of incidence in a stream of 15 degree Celsius water moving at 1 m/s. Find out the maximum boundary layer thickness. For water at 15 degree Celsius. For water at 15 degree Celsius
p = 998.9 kg /m3
µ = 415.85 * 10 -2 kg/hr m
a) 4.945 m
b) 3.945 m
c) 2.945 m
d) 1.945 m

View Answer

Answer: c [Reason:] Re = l p U INFINITY/µ.

6. Shear stress at the middle of the plate is given by
a) T W = 0.964 p U INFINITY 2/2 (Re) 1/2
b) T W = 0.864 p U INFINITY 2/2 (Re) 1/2
c) T W = 0.764 p U INFINITY 2/2 (Re) 1/2
d) T W = 0.664 p U INFINITY 2/2 (Re) 1/2.

View Answer

Answer: d [Reason:] T W = 3 µ U INFINITY/2 δ = 0.664 p U INFINITY 2/2 (Re) ½.

7. Boundary layer thickness is given by
a) δ = 5.64 x/ (Re) ½
b) δ = 5.64 x/ (Re) ½
c) δ = 6.64 x/ (Re) ½
d) δ = 7.74 x/ (Re) ½

View Answer

Answer: a [Reason:] δ/x = (140 * 2/13) ½ (µ/x p U INFINITY) ½.

8. Air at 25 degree Celsius flows over a flat surface with a sharp leading edge at 1.5 m/s. Find the value of Reynolds number. For air at 25 degree Celsius, kinematic viscosity = 15.53* 10n -6 m2/s
a) 38694
b) 12846
c) 48294
d) 76386

View Answer

Answer: c [Reason:] Re = x U INFINITY/v = 48294.

9. A plate 0.3 m long is placed at zero angle of incidence in a stream of 15 degree Celsius water moving at 1 m/s. Find out the maximum value of the normal component of velocity at thr trailing edge of the plate. For water at 15 degree Celsius
p = 998.9 kg /m3
µ = 415.85 * 10 -2 kg/hr m
a) 1.6885 * 10 -2 m/s
b) 1.6885 * 10 -3 m/s
c) 1.6885 * 10 -4 m/s
d) 1.6885 * 10 -5 m/s

View Answer

Answer: b [Reason:] v/U INFINITY (Re) ½ =0.860.

Set 3

1. The relationship (Wavelength) MAX T = constant, between the temperature of a black body and the wavelength at which maximum value of monochromatic emissive power occurs is known as
a) Planck’s law
b) Kirchhoff’s law
c) Lambert’s law
d) Wein’s law

View Answer

Answer: d [Reason:] This is the Wien’s law. From the spectral distribution of black body emissive power, it is apparent that the wavelength associated with maximum rate of emission depends up on the absolute temperature of the radiating surface.

2. A body at 500 K cools by radiating heat to ambient atmosphere maintained at 300 K. When the body has cooled to 400 K, the cooling rate as a percentage of original rate is about
a) 31.1
b) 41.5
c) 50.3
d) 80.4

View Answer

Answer: a [Reason:] Q2/Q1 = (400)4 – (300)4/ (500)4 – (300)4 = 0.32.

3. Two spheres A and B of same material have radius 1 m and 4 m, and temperatures 4000 K and 2000 K respectively. Then the energy radiated by sphere A is
a) Greater than that of sphere B
b) Less than that of sphere B
c) Equal to that of sphere B
d) Two times that of sphere B

View Answer

Answer: c [Reason:] E A/E B = (1)2 (4000)4/ (4)2 (2000)4 = 1.

4. A small body has a total emissive power of 4.5 kW/m2. Determine its surface temperature of maximum emission
a) 530.77 K
b) 345.65 K
c) 236.54 K
d) 367.8 K

View Answer

Answer: a [Reason:] E = σ T4. So, T = 530.77 K.

5. A small black body has a total emissive power of 4.5 k W/m2. In which range of the spectrum does this wavelength fall?
a) Thermal region
b) Cosmic region
c) Visible region
d) Infrared region

View Answer

Answer: d [Reason:] (Wavelength) T = 2.8908 * 10 -3. This must be the wavelength of infrared region.

6. The sun emits maximum radiation of 0.52 micron meter. Assuming the sun to be a black body, Calculate the surface temperature of the sun
a) 2345 K
b) 5573 K
c) 9847 K
d) 6492 K

View Answer

Answer: b [Reason:] T = 2.8908 * 10 -3/0.52 * 10 -6 = 5573 K.

7. Consider the previous problem, determine the maximum monochromatic emissive power of the sun’s surface
a) 4.908 * 10 13 W/m2
b) 5.908 * 10 13 W/m2
c) 6.908 * 10 13 W/m2
d) 7.908 * 10 13 W/m2

View Answer

Answer: c [Reason:] (E) MAX = 1.285 * 10 -5 T5 = 6.908 * 10 13 W/m2.

8. A furnace emits radiation at 2000 K. Treating it as a black body radiation, calculate the wavelength at which emission is maximum
a) 1.449 * 10 -6 m
b) 2.449 * 10 -6 m
c) 3.449 * 10 -6 m
d) 4.449 * 10 -6 m

View Answer

Answer: a [Reason:] (Wavelength) T = 2.8908 * 10 -3. So, wavelength = 1.449 * 10 -6 m.

9. Four identical pieces of copper painted with different colors of paints were heated to the same temperature and then left in the environment to cool. Which of the following paint will give fast cooling?
a) White
b) Rough
c) Black
d) Shining

View Answer

Answer: c [Reason:] The emissivity of black paint is highest i.e. unity. Consequently, the emitted radiant energy will be maximum when painted black.

10. A surface for which emissivity is constant at all temperatures and throughout the entire range of wavelength is called
a) Opaque
b) Grey
c) Specular
d) Diathermanous

View Answer

Answer: b [Reason:] Foe grey surface, emissivity is constant. Grey surface radiates much more because of constant wavelength throughout its entire surface.

Set 4

1. Two black discs each of diameter 50 cm are placed parallel to each other concentrically at a distance of one meter. The discs are maintained at 1000 K and 500 K. Calculate the heat flow between the discs when no other surface is present
a) 317.27 W
b) 417.27 W
c) 517.27 W
d) 617.27 W

View Answer

Answer: b [Reason:] Q = F 12 A 1 σ B (T 1 4 – T 2 4).

2. Two black discs each of diameter 50 cm are placed parallel to each other concentrically at a distance of one meter. The discs are maintained at 1000 K and 500 K. Calculate the heat flow between the discs when the disks are connected by a cylindrical black no-flux surface
a) 2417.68 W
b) 3417.68 W
c) 4417.68 W
d) 5417.68 W

View Answer

Answer: d [Reason:] Q = F 12 A 1 σ B (T 1 4 – T 2 4).

3. Heat exchange between two black surfaces enclosed by an insulated surface is given by
a) Q 12 = A 1 σ b (T 14 – T24) [A 2 – A 1 F 122/A 1 + A 2 – 2 A 1 F 12].
b) Q 12 = 2 A 1 σ b (T 14 – T24) [A 2 – A 1 F 122/A 1 + A 2 – 2 A 1 F 12].
c) Q 12 = 3 A 1 σ b (T 14 – T24) [A 2 – A 1 F 122/A 1 + A 2 – 2 A 1 F 12].
d) Q 12 = 4 A 1 σ b (T 14 – T24) [A 2 – A 1 F 122/A 1 + A 2 – 2 A 1 F 12].

View Answer

Answer: a [Reason:] This is the net heat exchange between two black surfaces enclosed by an insulated surface.

4. Heat exchange between two gray surfaces enclosed by an adiabatic surface is given by
a) Q 12 = A (T 14 – T 24) / [1/E 1 + 1/E 2 – 2 + 2/1 + F 12].
b) Q 12 = A σ b (T 14 – T 24) / [1/E 1 + 1/E 2 + 2/1 + F 12].
c) Q 12 = A σ b (T 14 – T 24) / [1/E 1 + 1/E 2 – 2 + 2/1 + F 12].
d) Q 12 = A σ b (T 14 – T 24) / [1/E 1 + 1/E 2 – 2].

View Answer

Answer: c [Reason:] This is the net heat exchange between two gray surfaces enclosed by an adiabatic surface.

5. A blind cylindrical hole of 2 cm diameter and 3 cm length is drilled into a metal slab having emissivity 0.7. If the metal slab is maintained at 650 K, make calculations for the radiation heat escape from the hole
heat-transfer-questions-campus-interviews-q5
a) 7 W
b) 3 W
c) 1 W
d) 9 W

View Answer

Answer: b [Reason:] Q = E 1 A 1 σ b T 14 [1 – F 11/1 – (1 – E 1) F 11].

6. A cavity in the shape of a frustum of a cone has diameter 30 cm and 60 cm and the height is 80 cm. If the cavity is maintained at temperature of 800 K, determine the heat loss from the cavity when the smaller diameter is at the bottom
a) 6577 W
b) 2367 W
c) 8794 W
d) 3675 W

View Answer

Answer: a [Reason:] Q = E 1 A 1 σ b T 14 [1 – F 11/1 – (1 – E 1) F 11].

7. Consider the above problem, find how this heat loss would be affected if the cavity is positioned with bigger diameter at the base
a) 75.06 % (increase)
b) 55.06 % (decrease)
c) 65.06 % (increase)
d) 75.06 % (decrease)

View Answer

Answer: d [Reason:] Percentage change in heat flow = 6577 – 1640/6577 = 0.7506.

8. A conical cavity of base diameter 15 cm and height 20 cm has inside surface temperature 650 K. If emissivity of each surface is 0.85, determine the net radiative heat transfer from the cavity
a) 168.3 W
b) 158.3 W
c) 148.3 W
d) 138.3 W

View Answer

Answer: a [Reason:] Q = E 1 A 1 σ b T 14 [1 – F 11/1 – (1 – E 1) F 11]. Here, F 11 = 0.649 and A 1 = 0.0503 m2.

9. A cylindrical cavity of base diameter 15 cm and height 20 cm has inside surface temperature 650 K. If emissivity of each surface is 0.85, determine the net radiative heat transfer from the cavity
a) 194 W
b) 184 W
c) 174 W
d) 164 W

View Answer

Answer: c [Reason:] Q = E 1 A 1 σ b T 14 [1 – F 11/1 – (1 – E 1) F 11]. Here, F 11 = 0.842 and A 1 = 0.11186 m2.

10. What is the unit of coefficient of radiant heat transfer?
a) W/K
b) W/m2 K
c) W/m2
d) W/m K

View Answer

Answer: b [Reason:] Its value can be calculated from the heat flux equation for any configuration.

Set 5

1. The total radiant energy leaving a surface per unit time per unit surface area is represented by
a) Radiation
b) Radiosity
c) Irradiation
d) Interchange factor

View Answer

Answer: b [Reason:] It comprises the original emittance from the surface plus the reflected portion of any radiation incident upon it.

2. Determine the radiant heat flux between two closely spaced, black parallel plates radiating only to each other if their temperatures are 850 K and 425 K. The plates have an area of 4 m2
a) .040
b) .030
c) .020
d) .010

View Answer

Answer: d [Reason:] Q 12 = F 12 A 1 σ b (T 14 – T 24) = .010.

3. What is the value of grey body factor for concentric cylinders?
a) 3/ [1 – e 1/e 1 + 1 + 1 – e 2/e 2 (A 1/A 2)].
b) 4/ [1 – e 1/e 1 + 1 + 1 – e 2/e 2 (A 1/A 2)].
c) 1/ [1 – e 1/e 1 + 1 + 1 – e 2/e 2 (A 1/A 2)].
d) 2/ [1 – e 1/e 1 + 1 + 1 – e 2/e 2 (A 1/A 2)].

View Answer

Answer: c [Reason:] Here, F 12 = 1.

4. The net heat exchange between the two grey surfaces may be written as
a) (Q 12) NET = E b 1 – E b 2/ (1 – e 1/A 1 e 1 + 1/A 1 F 12 + 1 – e 2/A 2 e 2)
b) (Q 12) NET = 2 E b 1 – E b 2/ (1 – e 1/A 1 e 1 + 1/A 1 F 12 + 1 – e 2/A 2 e 2)
c) (Q 12) NET = E b 1 – 2 E b 2/ (1 – e 1/A 1 e 1 + 1/A 1 F 12 + 1 – e 2/A 2 e 2)
d) (Q 12) NET = 2 E b 1 – 3 E b 2/ (1 – e 1/A 1 e 1 + 1/A 1 F 12 + 1 – e 2/A 2 e 2)

View Answer

Answer: a [Reason:] This equation gives the electrical network corresponding to surface resistances of two radiating bodies.

5. The net rate at which the radiation leaves the surface is given by
a) e (E b J)/1 – 4 e
b) e (E b J)/1 – 3 e
c) e (E b J)/1 – 2 e
d) e (E b J)/1 – e

View Answer

Answer: d [Reason:] The net rate at which the radiation leaves the surface is given by the difference between its radiosity and the incoming irradiation.

6. A ring (E = 0.85) of 8 cm inner and 16 cm outer diameter is placed in a horizontal plane. A small element (E = 0.7) of 1 cm2 is placed concentrically 8 cm vertically below the center of the ring. The temperature of the ring is 800 K and that of small area is 400 K. Find the radiant heat gain by the small ring
a) – 10.59 J/hour
b) – 11.59 J/hour
c) – 12.59 J/hour
d) – 13.59 J/hour

View Answer

Answer: b [Reason:] Q 12 = (F g) 12 A 1 σ b (T 14 – T 24) = A 1 σ b (T 14 – T 24)/ (I/E 1 – 1) + 1/F 12 + (I/E 2 – 1) A 2/A 1.

7. Two opposed, parallel, infinite planes are maintained at 420 K and 480 K. Calculate the net heat flux between these planes if one has an emissivity of 0.8 and other an emissivity of 0.7
a) 534.86 W/m2
b) 634.86 W/m2
c) 734.86 W/m2
d) 834.86 W/m2

View Answer

Answer: c [Reason:] Q 12 = (F g) 12 A 1 σ b (T 14 – T 24) and (F g) 12 = 1/ (I/E 1 – 1) + 1/F 12 + (I/E 2 – 1) A 2/A 1.

8. Consider the above problem, if temperature difference is doubled by raising the temperature 480 K to 540 K, then how this heat flux will be affected?
a) 1803.55 W/m2
b) 1703.55 W/m2
c) 1603.55 W/m2
d) 1503.55 W/m2

View Answer

Answer: a [Reason:] Q 2 = 0.59 (5.67 * 10 -8) (540 4 – 420 4).

9. The total radiant energy incident upon a surface per unit time per unit area is known as
a) Shape factor
b) Radiosity
c) Radiation
d) Irradiation

View Answer

Answer: d [Reason:] Some of it may be reflected to become a part of the radiosity of the surface.

10. Which one of the following is true for opaque non-black surface?
a) J = E +2 p G
b) J = E + p G
c) J = 2 E + p G
d) J = ½ E + p G

View Answer

Answer: b [Reason:] For an opaque non-black surface of constant radiation characteristics, the total radiant energy leaving the surface is the sum of its original emittance and the energy reflected by it out of the irradiation impinging on it.

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