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Intensity of radiation at a surface in perpendicular direction is equalto: [IES-2005; 2007]
- a)Product of emissivity of surface and 1/π
- b)Product of emissivity of surface and π
- c)Product of emissive power of surface and 1/ π
- d)Product of emissive power of surface and π
Correct answer is option 'C'. Can you explain this answer?
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Intensity of radiation at a surface in the perpendicular direction can be defined as the amount of radiant energy that is incident on a unit area of the surface per unit time. It is a measure of the rate at which energy is being transferred to the surface through radiation.
The intensity of radiation at a surface depends on various factors, including the emissivity and emissive power of the surface. Emissivity is a measure of how well a surface emits radiation compared to an ideal blackbody, while emissive power is the total amount of radiation emitted by a surface per unit area.
Let's break down the given options to understand why option 'C' is the correct answer:
a) Product of emissivity of surface and 1/b:
This option suggests that the intensity of radiation is proportional to the emissivity of the surface and inversely proportional to 'b'. However, 'b' is not defined in the question or any other context, so this option is not valid.
b) Product of emissivity of surface and c:
This option suggests that the intensity of radiation is proportional to the emissivity of the surface and a constant 'c'. However, 'c' is also not defined in the question or any other context, so this option is not valid.
c) Product of emissive power of surface and 1:
This option suggests that the intensity of radiation is proportional to the emissive power of the surface. This is the correct answer because the emissive power represents the total amount of radiation emitted by the surface per unit area. In the perpendicular direction, the intensity is equal to the emissive power.
d) Product of emissive power of surface and:
This option is not complete and does not provide any meaningful information, so it is not a valid answer.
In conclusion, the intensity of radiation at a surface in the perpendicular direction is equal to the product of the emissive power of the surface. This means that the more radiation a surface emits per unit area, the higher the intensity of radiation at that surface.
The intensity of radiation at a surface depends on various factors, including the emissivity and emissive power of the surface. Emissivity is a measure of how well a surface emits radiation compared to an ideal blackbody, while emissive power is the total amount of radiation emitted by a surface per unit area.
Let's break down the given options to understand why option 'C' is the correct answer:
a) Product of emissivity of surface and 1/b:
This option suggests that the intensity of radiation is proportional to the emissivity of the surface and inversely proportional to 'b'. However, 'b' is not defined in the question or any other context, so this option is not valid.
b) Product of emissivity of surface and c:
This option suggests that the intensity of radiation is proportional to the emissivity of the surface and a constant 'c'. However, 'c' is also not defined in the question or any other context, so this option is not valid.
c) Product of emissive power of surface and 1:
This option suggests that the intensity of radiation is proportional to the emissive power of the surface. This is the correct answer because the emissive power represents the total amount of radiation emitted by the surface per unit area. In the perpendicular direction, the intensity is equal to the emissive power.
d) Product of emissive power of surface and:
This option is not complete and does not provide any meaningful information, so it is not a valid answer.
In conclusion, the intensity of radiation at a surface in the perpendicular direction is equal to the product of the emissive power of the surface. This means that the more radiation a surface emits per unit area, the higher the intensity of radiation at that surface.
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Intensity of radiation at a surface in perpendicular direction is equalto: [IES-2005; 2007]a)Product of emissivity of surface and 1/πb)Product of emissivity of surface and πc)Product of emissive power of surface and 1/ πd)Product of emissive power of surface and πCorrect answer is option 'C'. Can you explain this answer?
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Intensity of radiation at a surface in perpendicular direction is equalto: [IES-2005; 2007]a)Product of emissivity of surface and 1/πb)Product of emissivity of surface and πc)Product of emissive power of surface and 1/ πd)Product of emissive power of surface and πCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Intensity of radiation at a surface in perpendicular direction is equalto: [IES-2005; 2007]a)Product of emissivity of surface and 1/πb)Product of emissivity of surface and πc)Product of emissive power of surface and 1/ πd)Product of emissive power of surface and πCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Intensity of radiation at a surface in perpendicular direction is equalto: [IES-2005; 2007]a)Product of emissivity of surface and 1/πb)Product of emissivity of surface and πc)Product of emissive power of surface and 1/ πd)Product of emissive power of surface and πCorrect answer is option 'C'. Can you explain this answer?.
Intensity of radiation at a surface in perpendicular direction is equalto: [IES-2005; 2007]a)Product of emissivity of surface and 1/πb)Product of emissivity of surface and πc)Product of emissive power of surface and 1/ πd)Product of emissive power of surface and πCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about Intensity of radiation at a surface in perpendicular direction is equalto: [IES-2005; 2007]a)Product of emissivity of surface and 1/πb)Product of emissivity of surface and πc)Product of emissive power of surface and 1/ πd)Product of emissive power of surface and πCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Intensity of radiation at a surface in perpendicular direction is equalto: [IES-2005; 2007]a)Product of emissivity of surface and 1/πb)Product of emissivity of surface and πc)Product of emissive power of surface and 1/ πd)Product of emissive power of surface and πCorrect answer is option 'C'. Can you explain this answer?.
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