Select Page
Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

# Multiple choice question for engineering

## Set 1

1. The solid angle is defined by a region by the rays of a sphere, and is measured as
a) An/r 2
b) An/r
c) An/r 3
d) An/r 4
Where,
An is projection of incident surface normal to line of propagation.

Answer: a [Reason:] Solid angle is represented by α.

2. The plane angle is defined by a region by the rays of a circle, and is measured as
a) 3 L/ r
b) 2 L/ r
c) L/ r
d) 4 L / r
Where,
L is arc of length and r is radius of circle.

Answer: c [Reason:] It is the ratio of arc of length on the circle to the radius of the circle.

3. When the incident surface is a sphere, the projection of surface normal to the line of propagation is the silhouette disk of the sphere which is a circle of the diameter of
a) Parabola
b) Sphere
c) Triangle
d) Hyperbola

Answer: b [Reason:] It must be a circle of diameter of a sphere.

4. If I n denotes the normal intensity and I α represents the intensity at angle α, then
a) I α = 2 I n cos α
b) I α = 3 I n cos α
c) I α = 4 I n cos α
d) I α = I n cos α

Answer: d [Reason:] The intensity of radiation in a direction from the normal is proportional to cosine of the angle.

5. The intensity of normal radiation I n is how much times the emissive power?
a) 1/π
b) 2/ π
c) 3/ π
d) 4/ π

Answer: a [Reason:] I n = σ T 4/ π and E = σ T 4.

6. A small surface emits diffusively, and measurements indicate that the total intensity associated with emission in the normal direction I n = 6500 W/square m sr. The emitted radiation is intercepted by three surfaces. Mark calculations for intensity associated with emission

a) 3500 W/m2 sr
b) 4500 W/m2 sr
c) 5500 W/m2 sr
d) 6500 W/m2 sr

Answer: d [Reason:] For a diffusion emitter, the intensity of the emitted radiation is independent of direction.

7. Consider a deep-space probe constructed as 1 m diameter polished aluminum sphere. Estimate the equilibrium temperature that the probe reaches if the solar energy received is 300 W/m2. For solar radiation, absorptivity of aluminum is 0.3 and the average emissivity appropriate for aluminum at low temperature is 0.04
a) 415.67 K
b) 315.67 K
c) 215.67 K
d) 115.67 K

Answer: b [Reason:] Q in = α q A P = 70.7 W. Q out = E σ b A T 4.

8. The total emissive power of the emitter with area d A and temperature T is given by
a) E = 2 σ T 4 d A
b) E = 3 σ T 4 d A
c) E = σ T 4 d A
d) E = ½ σ T 4 d A

Answer: c [Reason:] E = I n π d A.

9. A black body of 0.2 m2 area has an effective temperature of 800 K. Calculate the intensity of normal radiations
a) 1234.65 W/m2 sr
b) 7396.28 W/m2 sr
c) 3476.74 W/m2 sr
d) 8739.43 W/m2 sr

Answer: b [Reason:] In = α T 4/π = 7396.28 W/m2 sr.

10. The energy radiated out decreases with increases in α and becomes zero at an angle of
a) 45
b) 30
c) 0
d) 90

Answer: d [Reason:] I α = I n cos α. So at 90 degree it becomes zero.

## Set 2

1. If radiant energy E B emitted by the black surface strikes the non-black surface. If non-black surface has absorptivity α, it will absorb how much radiations?
a) α E B
b) 2 α E B
c) 3 α E B
d) 4 α E B

Answer: a [Reason:] The remainder (1 – α) will be reflected back for full absorption at the black surface.

2. If two surfaces are at the same temperature, then the conditions correspond to mobile thermal equilibrium for which the resultant interchange of heat is zero are
a) 3 E – α E B = 0
b) 2 E – α E B = 0
c) E – α E B = 0
d) ½ E – α E B = 0

Answer: c [Reason:] The remainder (1 – α) will be reflected back for full absorption at the black surface.

3. The absorptivity of black body equals to
a) 2
b) 1
c) 3
d) 4

Answer: b [Reason:] The absorptivity of black body equals to unity.

4. A diathermanous body
a) Shines as a result of incident radiation
b) Gets heated up a result of absorption of incident radiation
c) Allows all the incident radiation to pass through it
d) Partly absorbs and partly reflects the incident radiation

Answer: c [Reason:] It behaves like a body that allows all the incident radiation to pass through it.

5. Choose the false statement
a) Snow is nearly black to thermal radiation
b) Absorption of radiation occurs in a very thin layer of material n ear the surface
c) Transmissivity varies with wavelength of incident radiation
d) Most of the engineering materials have rough surfaces, and these rough surfaces give regular (specular) reflections

Answer: d [Reason:] Rough surfaces give diffused reflections. Reflections from highly polished and smooth surfaces have regular characteristics.

6. The emissivity and the absorptivity of a real surface are equal for radiation with identical temperature and wavelength. This law is referred to as
a) Kirchhoff’s law
b) Lambert’s law
c) Planck’s law
d) Wein’s displacement law

Answer: a [Reason:] Emissivity and absorptivity are related by Kirchhoff’s law.

7. With an increase in wavelength, the monochromatic emissive power of a black body
a) Increases
b) Decreases
c) Decreases, reaches a minimum and then increases
d) Increases, reaches a maximum and then decreases

Answer: d [Reason:] It firstly increases to its maximum value and then decreases to zero.

8. The temperature of a solid surface changes from 27 K to 627 K. The emissive power changes would then confirm to the ratio
a) 6:1
b) 9:1
c) 81:1
d) 27:1

Answer: c [Reason:] E2/E1 = (T2/T1)4 = 81.

9. If the temperature of a hot body is increased by 50%, the amount of radiations emitted by it would increase by nearly
a) 200%
b) 500%
c) 50%
d) 100%

Answer: b [Reason:] E2/E1 = (T2/T1)4 = 5.06.

10. Consider two surfaces, one absolutely black and the other non-black. These surfaces are arranged parallel to each other and so close that the radiation of one falls totally on the other. Choose the correct option

a) 1 denotes the radiant energy E emitted by the non-black surface impinges on the black surface
b) 1 denotes the radiant energy E emitted by the black surface impinges on the non-black surface
c) 2 and 3 denotes the quantity α E b
d) 4 denotes the quantity (1 – α) E b

Answer: a [Reason:] The radiant energy E b emitted by the black surface strikes the non-black surface. If the non-black surface has absorptivity α, it will absorb α E b radiations.

## Set 3

1. For laminar film condensation on a vertical plate, the velocity distribution at a distance δ from the top edge is given by
a) p g (δ y – y/2)/σ
b) p g (δ y – y 2)/σ
c) p g (δ y – y 2/2)/σ
d) p g (δ – y 2/2)/σ

Answer: c [Reason:] An equation for the velocity distribution as a function of some distance from the wall surface can be set up by considering the equilibrium between the gravity and viscous forces on an elementary volume of the liquid film.

2. For laminar film condensation on a vertical plate, the film thickness is given by
a) [4 k δ (t sat – t s) x/p 2 g h f g] 0.25
b) [4 k δ (t sat – t s) x/p 2 g h f g] 0.5
c) [4 k δ (t sat – t s) x/p 2 g h f g]
d) [4 k δ (t sat – t s) x/p 2 g h f g] 1.5

Answer: a [Reason:] The film thickness increases as the fourth root of the distance down the surface.

3. For laminar film condensation on a vertical plate, the mass flow rate of the condensate per unit depth of the film at any position x is given by
a) p g δ 3/ 3
b) p g δ 3/ 3 σ
c) p g δ 2/ 3 σ
d) p g δ 2/ 3

Answer: b [Reason:] Mass flow rate = mean flow velocity * flow area * density.

4. For laminar film condensation on a vertical plate, the gravitational acceleration g is replaced by
a) 4 g sin α
b) 3 g sin α
c) 2 g sin α
d) g sin α
Where, α is the inclination angle with the horizontal

Answer: d [Reason:] It is replaced by sin α.

5. For laminar film condensation on a vertical plate, the local heat transfer coefficient at the lower edge of the plate is given by
a) [k 3 p 2 g h f g /4 δ l (t sat – t s)] 0.25
b) [k 3 p 2 g h f g /4 δ l (t sat – t s)] 0.5
c) [k 3 p 2 g h f g /4 δ l (t sat – t s)] 0.1
d) [k 3 p 2 g h f g /4 δ l (t sat – t s)] 0.125

Answer: a [Reason:] The rate of condensation heat transfer is higher at the upper end of the plate than at the lower end.

6. Mark the wrong statement with respect to laminar flow condensation on a vertical plate
a) The rate of condensation heat transfer is maximum at the upper edge of the plate and progressively decreases as the lower edge is approached
b) The average heat transfer coefficient is two third of the local heat transfer coefficient at the lower edge of the plate
c) At a definite point on the heat transfer, the film coefficient is directly proportional to thermal conductivity and inversely proportional to thickness of film at the point
d) The film thickness increases as the fourth root of the distance down the upper edge

Answer: b [Reason:] The average heat transfer coefficient is 4/3 of the local heat transfer coefficient at the lower edge of the plate.

7. A plate condenser of dimensions l * b has been designed to be kept with side l in the vertical position. However due to oversight during erection and installation, it was fixed with side b vertical. How would this affect the heat transfer? Assume laminar conditions and same thermos-physical properties and take b = l/2
a) The condenser should be installed with shorter side horizontal
b) The condenser should be installed with longer side horizontal
c) The condenser should be installed with longer side vertical
d) The condenser should be installed with shorter side vertical

Answer: d [Reason:] h 1 = 0.943 [k 3 p 2 g h f g/δ l (t sat – t s)] 0.25, h2 = 0.943 [k 3 p 2 g h f g/δ b (t sat – t s)] 0.25. So, h 1/h 2 = 0.8409.

8. Determine the length of a 25 cm outer diameter tube if the condensate formed on the surface of the tube is to be same whether it is kept vertical or horizontal
a) 61.5 cm
b) 71.5 cm
c) 81.5 cm
d) 91.5 cm

Answer: b [Reason:] h v = 0.943 [k 3 p 2 g h f g/δ l (t sat – t s)] 0.25, h H = 0.943 [k 3 p 2 g h f g/δ d (t sat – t s)] 0.25, l/d = 2.86.

9. The critical Reynolds number for transition from laminar to turbulent film condensation is
a) 2000
b) 1900
c) 1800
d) 1700

Answer: c [Reason:] This should be 1800 for perfect transition from laminar to turbulent film condensation.

10. Which of the following is a wrong statement in the context of convective heat transfer coefficient in laminar film condensation?
The heat transfer coefficient varies as
a) – ¼ power of acceleration due to gravity
b) ½ power of density of liquid
c) ¼ power of enthalpy of evaporation
d) – ½ power of dynamic viscosity

Answer: a [Reason:] h = 0.943 [k 3 p 2 g h f g/δ l (t sat – t s)] 0.25. Further the heat transfer coefficient varies as ¼ power of acceleration due to gravity.

## Set 4

1. The Nusselt number is related to Reynolds number in laminar and turbulent flows respectively as
a) R e-1/2 and R e0.8
b) R e1/2 and R e0.8
c) R e-1/2 and R e-0.8
d) R e1/2 and R e-0.8

Answer: b [Reason:] Nusselt number = h l/k and Reynolds number = p V l/µ.

2. A hot plate of 15 cm2 area maintained at 200 degree celsius is exposed to still air at 30 degree Celsius temperature. When the smaller side of the plate is held vertical, convective heat transfer rate is 15 per cent higher than bigger side of the plate is held vertical. Find size of the plate. The appropriate correlation for the convection coefficient is

Nu = 0.60 (Gr Pr) 0.25
a) l = 5.123 cm and b = 2.928 cm
b) l = 6.123 cm and b = 3.928 cm
c) l = 7.123 cm and b = 4.928 cm
d) l = 8.123 cm and b = 5.928 cm

Answer: a [Reason:] t f = 200 + 30/2 = 115 degree celsius. Pr = µ c/k, Gr = b3 p 2 β g d t/µ 2 and Nu = h b/k.

3. Which of the following is true for laminar flow?
a) 104 < G r P r < 107
b) 104 < G r P r < 108
c) 104 < G r P r < 109
d) 104 < G r P r < 1010

Answer: c [Reason:] The product G r P r is often referred to as Rayleigh number, and its value sets the criterion of laminar character of flow.

4. A horizontal heated plate at 200 degree Celsius and facing upwards has been placed in still air at 20 degree Celsius. If the plate measures 1.25 m by 1 m, make calculations for the heat loss by natural convection. The convective film coefficient for free convection is given by the following empirical relation
h = 0.32 (α) 0.25 W/m2 K
Where α is the mean film temperature in degrees kelvin
a) 6006 W
b) 5006 W
c) 4006 W
d) 3006 W

Answer: a [Reason:] α = 273 + 200 + 20/2 = 383 degree Celsius. Therefore, rate of heat loss = h A d t = 3006 W.

5. For laminar flow, Reynolds number must not be less than
a) 1000
b) 2000
c) 20000
d) 40000

Answer: d [Reason:] It should not be less than 40000.

6. For laminar flow, Prandtl number must be more than
a) 0.05
b) 0.2
c) 0.6
d) 0.3

Answer: c [Reason:] It must be more than 0.6. It is indicative of the relative ability of the fluid to diffuse momentum and internal energy by molecular mechanisms.

7. Air at atmospheric pressure and 20 degree Celsius flows with 6 m/s velocity through main trunck duct of air conditioning system. The duct is rectangular in cross-section and measures 40 cm by 80 cm. Determine heat loss per meter length of duct corresponding to unit temperature difference. The relevant thermos-physical properties of air are
v = 15 * 10 -6 m 2/s
α = 7.7 * 10 -2 m2/hr
k = 0.026 W/m degree
a) 32.768 W
b) 42.768 W
c) 52.768 W
d) 62.768 W

Answer: b [Reason:] Q = h A d t, Area = 2 (a + b) (1) = 2.4 m2, h = 365.34 k/l.

8. The local film coefficient for laminar flow past a flat plate may be obtained from the correlation
a) Nu = 0.332 (Re) 0.5 (Pr) 0.33
b) Nu = 0.332 (Re) 0.5 (Pr) 0.43
c) Nu = 0.332 (Re) 0.5 (Pr) 0.53
d) Nu = 0.332 (Re) 0.5 (Pr) 0.63

Answer: a [Reason:] Fluid properties are evaluated at mean film temperature.

9. For a plate of length l, an average value of Nusselt number is given by
a) Nu = 0.664 (Re) 0.5 (Pr) 0.54
b) Nu = 0.664 (Re) 0.5 (Pr) 0.74
c) Nu = 0.664 (Re) 0.5 (Pr) 0.27
d) Nu = 0.664 (Re) 0.5 (Pr) 0.33

Answer: d [Reason:] For a plate of length l, an average value of Nusselt number or convection coefficient may be obtained by integration.

10. The correlation for liquid metal is given by
a) Nu = 0.465 (Re) (Pr)
b) Nu = 0.565 (Re) (Pr)
c) Nu = 0.665 (Re) (Pr)
d) Nu = 0.765 (Re) (Pr)

Answer: b [Reason:] It is valid for Pr less than or equal to 0.05.

## Set 5

1. According to lumped system analysis, solid possesses thermal conductivity that is
a) Infinitely large
b) Infinitely small
c) Moderate
d) 50% small

Answer: a [Reason:] Solutions to the many of the transient heat flow problems are obtained by the lumped system parameter analysis.

2. The temperature and rate of heat conduction are undoubtedly dependent on
a) Time coordinates
b) Space coordinates
c) Mass coordinates
d) Both time and space coordinates

Answer: d [Reason:] It should depend on both time coordinates and space coordinates.

3. Glass spheres of 2 mm radius and at 500 degree Celsius are to be cooled by exposing them to an air stream at 25 degree Celsius. Find the minimum time required for cooling to a temperature of 60 degree Celsius. Assume the following property values
Density = 2250 kg/m3
Specific heat = 850 J/kg K
Conductivity = 1.5 W/m K
a) 13.78 seconds
b) 14.78 seconds
c) 15.78 seconds
d) 16.78 seconds

Answer: b [Reason:] t – t a/t I – t a = exponential (- h A T/p V c).

4. Which is true regarding lumped system analysis?
(i) Conductive resistance = 0
(ii) Convective resistance = 0
(iii) Thermal conductivity = 0
(iv) Thermal conductivity = infinity
Identify the correct statements
a) 1 and 2
b) 1, 2 and 4
c) 1 and 4
d) 2 and 4

Answer: c [Reason:] Solids have infinite thermal conductivities. It implies that internal conductance resistance is very low.

5. Which of the following is an example of lump system analysis?
a) Heating or cooling of fine thermocouple wire due to change in ambient temperature
b) Heating of an ingot in an furnace
c) Cooling of bars
d) Cooling of metal billets in steel works

Answer: a [Reason:] Others are the examples of non-periodic variation.

6. What is the criterion for the applicability of lump system analysis?
a) Mean length
b) Normal length
c) Characteristics length
d) Mass no

Answer: c [Reason:] The first set in establishing a criteria for the applicability of lump system analysis is to define a characteristics length.

7. What is the value of characteristics length for cylinder?
a) R/5
b) R/4
c) R/3
d) R/2

Answer: d [Reason:] π R2 L/2 π R L = R/2.

8. During heat treatment, cylindrical pieces of 25 mm diameter, 30 mm height and at 30 degree Celsius are placed in a furnace at 750 degree Celsius with convective coefficient 80 W/m2 degree. Find the value of biot number if thermal conductivity is 40 W/m degree
a) 0.0082
b) 0.0072
c) 0.0062
d) 0.0052

Answer: a [Reason:] For a cylindrical piece, the characteristic linear dimension is, l = volume/surface area = .00441 m. So, biot number = hl/k = -.00882.

9. What is the value of characteristics length for sphere?
a) R/2
b) R/3
c) R/4
d) R/5

Answer: b [Reason:] 4/3 π R3/4 π R2 = R/3.

10. What is the value of characteristics length for cube?
a) L/3
b) L/4
c) L/5
d) L/6