Total Emissive Power MCQ Quiz in मल्याळम - Objective Question with Answer for Total Emissive Power - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Apr 5, 2025
Latest Total Emissive Power MCQ Objective Questions
Top Total Emissive Power MCQ Objective Questions
Total Emissive Power Question 1:
The time period of rotation of the sun is \(25\) days and its radius is \(7 \times 10^8 \, m\). The Doppler shift for the light of wavelength \(6000 \, A^o\) emitted from the surface of the sun will be
Answer (Detailed Solution Below)
Total Emissive Power Question 1 Detailed Solution
\(\Delta \lambda = \lambda \dfrac{v}{c}\) and \(v = \dfrac{2\pi r}{T}\)
\(v = \dfrac{2\pi \times 7 \times 10^8}{25 \times 24 \times 3600}, \, c = 3 \times 10^8 \, m/s\)
\(\therefore \Delta \lambda = 0.04 \, \overset{o}{A}\)
Total Emissive Power Question 2:
If the temperature of solid surface changes from 27°C to 627°C, then its emissive power changes in the ratio of
Answer (Detailed Solution Below)
Total Emissive Power Question 2 Detailed Solution
Concept:
Stefan-Boltzmann Law
The thermal energy radiated by a black body per second per unit area is proportional to the fourth power of the absolute temperature and is given by:
E ∝ T4
E = σT4
σ = The Stefan – Boltzmann constant = 5.67 × 10-8 W m-2K-4
Calculation:
Given:
T1 = 27°C = 300 K, T2 = 627°C = 900 K
\(\frac{{{E_2}}}{{{E_1}}} = {\left( {\frac{{{T_2}}}{{{T_1}}}} \right)^4} = {\left( {\frac{{900}}{{300}}} \right)^4} =3^4 = 81\)
Total Emissive Power Question 3:
The intensity of solar radiaton is maximum at a wavelength of 0.49 μm. Assuming the Sun as a black body, what is the approximate total emissive power of sum? [Consider Wien's displacement constant = 2890 μm-K]
Answer (Detailed Solution Below)
Total Emissive Power Question 3 Detailed Solution
Concept:
Wein’s displacement law:
λmax.T = 2898 μmK
It states that the “product of absolute temperature and wavelength at which emissive power of a black body is a maximum, is constant”.
Calculation:
Given:
λmax = 0.49 microns = 0.49 × 10-6
λmax.T = 0.289 × 10-2
0.49 × 10-6 × T = 0.289 × 10-2
T = 5898 K
Emissive power, E = σT4 = 5.67 × 10-8 × (5898)4
E = 6.86 × 107 W/m2
Total Emissive Power Question 4:
A 100 W electric bulb has filament temperature of 3000° C. Assuming the filament to be black body, calculate the diameter of the wire if the length is 250 mm.
Answer (Detailed Solution Below)
Total Emissive Power Question 4 Detailed Solution
Concept:
Net radiation heat exchange between two infinitely long concentric cylindrical surfaces:
\({Q_{1 - 2}}=\frac{{\sigma \left( {T_1^4 - T_2^4} \right){A_1}}}{{\frac{1}{{{\epsilon_1}}} + \frac{{{A_1}}}{{{A_2}}}\left( {\frac{1}{{{\epsilon_2}}} - 1} \right)}}W\)
When a small body is kept in a large enclosure:
\({A_1} \ll {A_2} \Rightarrow \frac{{{A_1}}}{{{A_2}}} \approx 0\)
Q1 - 2 = σ ϵ1 A1 (T14 - T24)
For steady state condition of the filament:
Q = σ ϵf Af (Tf4 - T∞4)
Calculation:
Given:
Q = 100 W, T = 3000° C = 3273 K, L = 250 mm = 0.25 m
Af = π DL = 025 π D mm2
Now,
Q = σ ϵf Af Tf4
100 = 5.67 × 10-8 × 1 × 0.25 π D × (3273)4 = 5110.4 × 103D
D = 0.02 × 10-3 m
∴ D = 0.02 mm
Total Emissive Power Question 5:
A hole of area 1 cm2 is opened on the surface of a large spherical cavity whose inside temperature is maintained at 727°C. The value of Stefan-Boltzmann constant is 5.67 × 10-8 W/m2-K4. Assuming black body radiation, the rate at which the energy is emitted (in W) by the cavity through the hole, up to 3 digits after the decimal point, is _______.
Answer (Detailed Solution Below) 5.6 - 5.8
Total Emissive Power Question 5 Detailed Solution
Heat energy emitted by a body is given according to Stefan – Boltzmann low.
E = ϵ.A.σT4
Where ϵ is the emissivity
A = area
σ = Stefan-Boltzmann constant
T = Temperature in k.
Here ϵ = 1 (for black body)
∴ E = 1 × 10-4 × 5.67 × 10-8 × (1000)4 = 5.670Total Emissive Power Question 6:
A giant star emits radiant energy at a rate 104 times greater than Sun, whereas its surface temperature is only half (2900 K) that of Sun. The Sun’s radius is 7 × 108 m. Considering ϵ = 1 for both, the radius of the star will be ______ × 1010 m.
Answer (Detailed Solution Below) 27.5 - 28.5
Total Emissive Power Question 6 Detailed Solution
\(\begin{array}{l} \dot Q = \epsilon A\sigma {T^4}\\ {{\dot Q}_B} = {10^4}{{\dot Q}_S}\\ {A_B}T_B^4 = {10^4}{A_S}T_S^4\\ R_B^2\;T_B^4 = {10^4}R_S^2\;T_S^4\\ {\rm{As\;}}{T_B} = \frac{{{T_S}}}{2}\\ {\left( {\frac{{{R_B}}}{{{R_S}}}} \right)^2} = {\left( {\frac{{{T_S}}}{{{T_B}}}} \right)^4} \times {10^4} = {2^4} \times {10^4}\\ \frac{{{R_B}}}{{{R_S}}} = 4 \times {10^2} = 400 \end{array}\)
RB = 400 × 7 × 108 = 28 × 1010 mTotal Emissive Power Question 7:
A surface of area 0.5m2, emissivity 0.8 and temperature 1500C is placed in a large, evacuated chamber whose walls are maintained at 250C. What is the net rate of radiation exchange between the surface and the chamber walls (in W)?
Answer (Detailed Solution Below)
Total Emissive Power Question 7 Detailed Solution
The net rate at which radiation is transferred from the surface to the chamber walls is:
\(Q = A\sigma \left( {T_s^4 - T_{surr}^4} \right) = 0.8 \times 0.5 \times 5.67 \times {10^{ - 8}}\left( {{{423}^4} - {{298}^4}} \right) = 547.25\;W\)
Total Emissive Power Question 8:
The heat transfer coefficient including convection radiation is 30 W/m2°C for the outer surface of the pipe in a large enclosure. Assume pipe surface is black and wall of pipe surface and enclosure are 200°C and 100°C respectively. The convection heat transfer coefficient will be
Answer (Detailed Solution Below)
Total Emissive Power Question 8 Detailed Solution
T1 = 200 + 273 = 473 K
T2 = 100 + 272 = 373 K
\({Q_{12\;rad}} = {A_1}{\epsilon_1}\sigma \left( {T_1^4 - T_2^4} \right)\)
\(= 1 \times 1 \times 5.67 \times {10^{ - 8}}\left( {{{473}^4} - {{373}^4}} \right)\)
= 1740.56 W/m2°C
\({Q_{1 - 2\;rad}} = {A_1} \cdot {h_{rad}}\left( {{T_1} - {T_2}} \right)\)
1740.56 = 1 × hrad(473 - 373)
hrad = 17.4 W/m2°C
hconv = 30 - hrad = 12.6 W/m2°C