Radiation MCQ Quiz - Objective Question with Answer for Radiation - Download Free PDF
Last updated on Jul 4, 2025
Latest Radiation MCQ Objective Questions
Radiation Question 1:
A metal object is heated inside a furnace, and a sensor placed above its surface records the thermal power radiated, denoted by P. The sensor’s display shows a value equal to x where x and thermal power has relation of 4x =(P / P0), where P0 is a fixed reference value. When the metal reaches a temperature of 327°C, the sensor reads a value of 1. Assuming that the metal's emissivity does not change with temperature, what will the sensor display when the surface temperature of the metal is increased to 2127°C?
Answer (Detailed Solution Below) 5
Radiation Question 1 Detailed Solution
Calculation:
The power radiated by a metal surface is given by the Stefan-Boltzmann law: P = e A T4, where e is the emissivity, A is the surface area, T is the absolute temperature in kelvin.
Convert temperatures to Kelvin: T1 = 327 + 273 = 600 K
T2 = 2127 + 273 = 2400 K
Then the radiated powers are:
P1 = e A (760)4
P2 = e A (3040)4
Sensor readings:
4x1 =(P1 / P0) ⇒x1 = log4(P1 / P0) = log4(e A (600)4 / P0)
4x2 =(P2 / P0) ⇒x2 = log4(P2 / P0) = log4(e A (2400)4 / P0)
Given: x1 = 1 Now dividing the two equations: x2 - x1 = log4((2400)4 / (600)4)
x2 = 1 + log4(44) = 1 + 4·log4(4) ⇒ x2 = 1 + 4 = 5
Radiation Question 2:
The Stefan-Boltzmann law is derived from:
Answer (Detailed Solution Below)
Radiation Question 2 Detailed Solution
Explanation:
The Stefan-Boltzmann Law
- The Stefan-Boltzmann law is a fundamental principle in thermal radiation, stating that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of its absolute temperature. Mathematically, the Stefan-Boltzmann law is expressed as:
E = σ × T4
Where:
- E: Total energy radiated per unit surface area (W/m2)
- σ: Stefan-Boltzmann constant (5.67 × 10-8 W/m2K4)
- T: Absolute temperature of the black body (K)
The Stefan-Boltzmann law is derived from Planck's Law, which describes the spectral distribution of electromagnetic radiation emitted by a black body in thermal equilibrium. By integrating Planck's law over all wavelengths, the Stefan-Boltzmann law can be obtained. This integration process effectively sums up the contributions of radiation from all wavelengths, yielding the total emissive power of the black body.
Planck's law:
- Planck's law describes the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature T.
- Planck’s law for the energy Eλ radiated per unit volume by a cavity of a blackbody in the wavelength interval λ to λ + Δλ can be written in terms of Planck’s constant (h), the speed of light (c = λ × v), the Boltzmann constant (k), and the absolute temperature (T):
Energy per unit volume per unit wavelength:
\({E_\lambda } = \frac{{8\pi hc}}{{{\lambda ^5}}} \times \frac{1}{{{e^{\frac{{hc}}{{kT\lambda }} - 1}}}}\)
Energy per unit volume per unit frequency:
\({E_\nu } = \frac{{8\pi h}}{{{c^3}}} \times \frac{{{\nu ^3}}}{{{e^{\frac{{hv}}{{kT}} - 1}}}}\)
So Planck’s distribution function:
\(E\left( {\omega ,T} \right) = \frac{1}{{{e^{\frac{{h\omega }}{\tau }}} - 1}}\)
Using planck’s law, when we plot Ebλ with λ, we get the curve as shown below.
As temperature increases, the peak of the curve shift towards a lower wavelength.
Radiation Question 3:
What is the primary mechanism by which thermal radiation transfers energy?
Answer (Detailed Solution Below)
Radiation Question 3 Detailed Solution
Explanation:
Thermal Radiation and Its Mechanism
- Thermal radiation is a mode of heat transfer that occurs through the emission of electromagnetic waves, primarily in the infrared spectrum, but it can also include visible light and other wavelengths. This form of energy transfer does not require a medium, meaning it can occur in a vacuum. The energy is emitted by all bodies that have a temperature above absolute zero, due to the thermal vibrations of their molecules and atoms. The amount and nature of the radiation depend on the temperature and the surface properties of the body.
Radiation as a heat transfer mechanism is governed by Stefan-Boltzmann’s law, which states:
Q = σ × A × T⁴
Where:
- Q = Heat transfer via radiation (W)
- σ = Stefan-Boltzmann constant (5.67 × 10⁻⁸ W/m²K⁴)
- A = Surface area of the body (m²)
- T = Absolute temperature of the body (K)
Thermal radiation is characterized by the following:
- Electromagnetic Waves: It is the primary mechanism by which thermal radiation occurs. These waves can travel through a vacuum, making radiation the dominant form of heat transfer in outer space.
- Emissivity: The ability of a material to emit energy as thermal radiation is determined by its emissivity, which ranges from 0 (perfect reflector) to 1 (perfect emitter or blackbody).
- Temperature Dependence: The intensity and wavelength distribution of the emitted radiation depend on the temperature of the object.
Radiation Question 4:
Two identical plates P and Q, radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures TP and TQ, respectively, with TQ < TP, as shown in Fig. 1. The radiated power transferred per unit area from P to Q is 𝑊0. Subsequently, two more plates, identical to P and Q, are introduced between P and Q, as shown in Fig. 2. Assume that heat transfer takes place only between adjacent plates. If the power transferred per unit area in the direction from P to Q (Fig. 2) in the steady state is 𝑊𝑆, then the ratio \(\frac{W_{0}}{W_{S}}\) is _______
Answer (Detailed Solution Below) 3.00
Radiation Question 4 Detailed Solution
Calculation:
From Fig. 1:
W0 = σ A (TP4 - TQ4)
From Fig. 2 (two intermediate screens inserted):
WS = σ (TP4 - T14) = σ (T14 - T24) = σ (T24 - TQ4)
⇒ Each step has the same energy flow, so:
3WS = σ (TP4 - TQ4) = W0
⇒ ½W0 / WS = 3
Final Answer: 3.00
Radiation Question 5:
Emissivity factor for the energy emitted by a grey body is given by
(where, E = Energy emitted by a grey body per m2 per unit time and EB = Energy emitted by a perfect black body per m2 per unit time)
Answer (Detailed Solution Below)
Radiation Question 5 Detailed Solution
Explanation:
Emissivity Factor for Energy Emitted by a Grey Body
- Emissivity is a measure of the efficiency of a surface in emitting thermal radiation compared to a perfect black body. A black body is an idealized physical object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence, and emits the maximum amount of energy at any given temperature. A grey body, however, emits less energy than a black body at the same temperature.
Mathematically, the emissivity factor (ε) is defined as the ratio of the energy emitted by a grey body per unit area per unit time (E) to the energy emitted by a perfect black body per unit area per unit time (EB) at the same temperature:
\( \varepsilon = \frac{E}{E_B} \)
- \( E \) = Emitted energy by grey body (W/m²)
- \( E_B \) = Emitted energy by black body (W/m²)
Key Points:
- For a perfect black body, emissivity is equal to 1 because it emits the maximum possible energy at any given temperature.
- For real-world objects (grey bodies), emissivity is less than 1, as they emit less energy compared to a black body.
- Emissivity is a dimensionless quantity and typically ranges between 0 and 1.
Physical Significance:
The emissivity factor is crucial in determining the thermal radiation properties of materials. It plays a significant role in applications such as:
- Thermal engineering, where heat transfer calculations involve radiation.
- Designing heat exchangers and radiative cooling systems.
- Understanding and modeling the behavior of materials in high-temperature environments.
- Developing thermal imaging systems and sensors.
Top Radiation MCQ Objective Questions
Radiation thermal resistance may be written as [where F, A, σ are shape factor, Area and Stefan-Boltzmann constant respectively]
Answer (Detailed Solution Below)
Radiation Question 6 Detailed Solution
Download Solution PDFExplanation:
Net radiation heat exchange between two bodies is given by:
Q̇ = AF × σ × (T14 - T24)
where F, A, σ are shape factor, Area and Stefan-Boltzmann constant respectively
Now explanding (T14 - T24)
Q̇ = AF × σ × ((T1)2)2 - (T2)2)2)
Q̇ = AF × σ × (T12 - T22) × (T12 + T22)
Q̇ = AF × σ × (T1 - T2)(T1 + T2) × (T12 + T22)
\(\dot Q =\frac{T_1-T_2}{\frac{1}{\sigma \times AF \times(T_1+T_2)\times(T_1^2+T_2^2)}}\)
Comparing with the electrical analogy \(i=\frac VR\)
We will get thermal resistance as \(\frac{1}{{FA\sigma \left( {{T_1} + {T_2}} \right)\left( {T_1^2 + T_2^2} \right)}}\)
For an opaque plane surface the radiosity, irradiation and emissive power are respectively 16, 24 and 12 W/m2. Determine the emissivity of surface.
Answer (Detailed Solution Below)
Radiation Question 7 Detailed Solution
Download Solution PDFConcept:
Irradiation (G): The thermal radiation falling or incident upon a surface per unit time and per unit area is called irradiation
Radiosity (J): The total radiation energy leaving a surface per unit time per unit area is called radiosity.
Emissive Power (E): It is radiation energy leaving a surface.
J = Emitted Energy + Reflected part of incident energy
J = E + ρ G
For an opaque surface transmissivity (τ) = 0,
For any surface α + ρ + τ = 1
Where α = absorptivity, ρ = reflectivity, τ = transmissivity
Consider a small real body in thermal equilibrium with its surrounding blackbody cavity then by Kirchhoff's Law of radiation
⇒ absorptivity = emissivity ⇒ α = ϵ
∴ ρ + ϵ = 1, ρ = 1 - ϵ
J = E + (1 - ϵ) G
Calculation:
Given:
radiosity (J) = 16 W/m2, irradiation (G) = 24 W/m2,
Emissive power (E) = 12 W/m2
J = E + (1 - ϵ) G
16 = 12 + (1 - ϵ) 24
\(\epsilon= 1 - \frac{4}{{24}} = 0.83\)
J = E + ρG
J = εEb + ρG
Eb = Emissive power of a perfect black body
The peak wavelength of radiation emitted by a black body at a temperature of 2000 K is 1.45 μm. If the peak wavelength of emitted radiation changes to 2.90 μm, then the temperature (in K) of the black body is
Answer (Detailed Solution Below)
Radiation Question 8 Detailed Solution
Download Solution PDFConcept:
From Wein's displacement law
\({\lambda _{max}}T = 2898\;\mu m - k\left( {constant} \right)\)
Thus, \({\lambda _{peak}}T = \lambda _{peak}'T'\)
Calculation:
Given,
Black body λpeak = 1.45 μm at 2000 K.
Now, \(\lambda _{peak}' = 2.90\;\mu m\)
1.45 × 2000 = 2.90 × T'
T' = 1000 KA solid sphere of radius r1 = 20 mm is placed concentrically inside a hollow sphere of radius r2 = 30 mm as shown in figure.
The view factor F21 for radiation heat transfer is
Answer (Detailed Solution Below)
Radiation Question 9 Detailed Solution
Download Solution PDFConcept:
F11 + F12 = 1
But F11 = 0
∴ F12 = 1
Now from reciprocating law
A1 F12 = A2 F21
Calculation:
\(\Rightarrow 4\pi r_1^2 \times I = 4\pi r_2^2 \times {F_{21}}\)
\(\Rightarrow {F_{21}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = {\left( {\frac{2}{3}} \right)^2} = \frac{4}{9}\)Solar radiation of 1000 W/m2 is incident on a grey opaque surface with an emissivity of 0.4 and emissive power (black body) of 400 W/m2. The radiosity of the surface will be:
Answer (Detailed Solution Below)
Radiation Question 10 Detailed Solution
Download Solution PDFConcept:
Irradiation (G): Total radiation incident upon a surface per unit time per unit area.
Radiosity (J): Total radiation leaving a surface per unit time per unit area.
Radiosity comprises the original emittance from the surface plus the reflected portion of any radiation incident upon it.
J = E + ρG
J = εEb + ρG
Eb = Emissive power of a perfect black body
α + ρ + τ = 1
For opaque body: τ = 0 ⇒ α + ρ = 1 ⇒ ρ = 1 - α = 1 - ε
J = εEb + (1 - ε)G = E + (1 - ε)G
Calculation:
Given:
G = 1000 W/m2, Eb = 400 W/m2, ϵ = 0.4
J = 400 × 0.4 + (1 - 0.4)1000 = 760 W/m2According to Stefan-Boltzman law the radiation energy emitted byby a black body is directly proportional to (where T is the absolute temperature Of the body)
Answer (Detailed Solution Below)
Radiation Question 11 Detailed Solution
Download Solution PDFExplanation:
Stephen's Boltzmann law:
According to Stefan’s law, the radiant energy emitted by a black body is directly proportional to the fourth power of its absolute temperature.
Q̇ = єσAT4
Where Q̇ is Radiate energy, σ is the Stefan-Boltzmann Constant, T is the absolute temperature in Kelvin, є is Emissivity of the material, and A is the area of the emitting body.
- Stefan's Law is used to accurately find the temperature Sun, Stars, and the earth.
- A black body is an ideal body that absorbs or emits all types of electromagnetic radiation.
From above it clear that the amount of radiation emitted by a perfectly black body is proportional to the fourth power of temperature on an ideal gas scale. Thus, option 3 is correct.
If radiant energy EB emitted by the black surface strikes the non-black surface and if the non-black surface has absorptivity α, then it will absorb how much radiation?
Answer (Detailed Solution Below)
Radiation Question 12 Detailed Solution
Download Solution PDFExplanation:
Let E be the total emissive power of the body and α be the absorptivity of the body.
α be the absorptivity of the body.
Emissivity (ϵ) is the ratio of emissive power of a non-black body to the black body.
ϵ = E/Eb
Kirchhoff’s law also holds for monochromatic radiation, for which
\(\frac{{{E_{\lambda 1}}}}{{{\alpha _{\lambda 1}}}} = \frac{{{E_{\lambda 2}}}}{{{\alpha _{\lambda 2}}}} = \frac{{{E_{b\lambda }}}}{{{\alpha _{b\lambda }}}} = \frac{{{E_{b\lambda }}}}{1}\)
∵ Absorptivity of a black body is one.
\(\therefore \frac{{{E_\lambda }}}{{{\alpha _\lambda }}} = {E_{b\lambda }} \Rightarrow {\alpha _\lambda } = \frac{{{E_\lambda }}}{{{E_{b\lambda }}}} = {\epsilon_\lambda }\)
Therefore, the monochromatic emissivity of a black body is equal to the monochromatic absorptivity at the same wavelength.
Eλ = αλEbλ
E = α EB
Important Points
According to Kirchhoff’s law, the ratio of total emissive power to absorptivity is constant for all bodies which are in thermal equilibrium with the surroundings. Therefore, it means that the emissivity of a body is equal to its absorptivity.
The rate of energy emission from unit surface area through unit solid angle, along a normal to the surface, is known as
Answer (Detailed Solution Below)
Radiation Question 13 Detailed Solution
Download Solution PDFExplanation:
Intensity of radiation:
The radiation intensity for emitted radiation Ie(θ, ϕ) is defined as the rate at which radiation energy dQ̇e is emitted in the (θ, ϕ) direction per unit area normal to this direction and per unit solid angle about this direction.
\({I_e}\left( {\theta ,\phi } \right)\; = \;\frac{{d{{\dot Q}_e}}}{{dA\cos \theta .d\omega }}\; = \;\frac{{d{{\dot Q}_e}}}{{dA\cos \theta \sin \theta \;d\theta d\phi }}\;\left( {\frac{W}{{{m^2}}}.sr} \right)\)
Emissivity (ϵ):
The emissivity of a surface represents the ratio of the radiation emitted by the surface at a given temperature to the radiation emitted by a blackbody at the same temperature.
For black body surface, ϵ = 1.
Irradiation (G):
The radiation flux incident on a surface from all directions is called irradiation G.
Absorptivity, reflectivity and transmissivity:
The fraction of irradiation absorbed by the surface is called the absorptivity α, the fraction reflected by the surface is called the reflectivity ρ, and the fraction transmitted is called the transmissivity τ.
\(\alpha + \rho + \tau \; = \;1\)
Thermal radiation extends over the range of
Answer (Detailed Solution Below)
Radiation Question 14 Detailed Solution
Download Solution PDFConcept:
- Thermal radiation is the transmission of heat in the form of radiant energy from one body to another body by the thermal motion of charged particles of matter.
- The mechanism of radiation is divided into 3 phases
- Conversion of thermal energy of the hot source into electromagnetic waves
- Passage of wave motion through intervening medium
- Transformation of waves into heat
- Thermal radiation is limited to wavelengths ranging from 0.1 to 100 µm of the electromagnetic spectrum.
- It includes some portion of UV radiation (0.1 to 0.4 µm), entire visible radiation (0.4 to 0.7 µm), and entire infrared radiation (0.7 to 100 µm)
Absorptivity of the grey body
Answer (Detailed Solution Below)
Radiation Question 15 Detailed Solution
Download Solution PDFExplanation:
When thermal radiation falls onto an object,
- The radiation will be absorbed by the surface of the object, causing its temperature to change
- The radiation will be reflected from the surface of the body, causing no temperature change
- The radiation will pass completely through the object, causing no temperature change
Absorptivity (α) is a measure of how much of the radiation is absorbed by the body
Reflectivity (ρ) is a measure of how much is radiation is reflected
Transmissivity (τ) is a measure of how much radiation passes through the object.
Each of these parameters is a number that ranges from 0 to 1.
Absorptivity may be a function of wavelength and/or direction, and is related to the emissivity of the region by Kirchhoff's law. The absorptivity is identically equal to unity for black bodies and is independent of temperature and wavelength for gray bodies.