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Asked: 2020-05-06T01:20:27+05:30 In:

GATE MT 2020 Q3. Select the correct spectra (shown on a log-log scale on the figures) for emission from a gray surface and a black body, both maintained at 1000 K.

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14%A ( 7 voters )
14%B ( 7 voters )
18%C ( 9 voters )
54%D ( 27 voters )
Based On 50 Votes

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GATE MT 2020 Q3. Select the correct spectra (shown on a log-log scale on the figures) for emission from a gray surface and a black body, both maintained at 1000 K.
GATE MT 2020 Q3. Select the correct spectra (shown on a log-log scale on the figures) for emission from a gray surface and a black body, both maintained at 1000 K.
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    4 Answers

    1. Gold B.TECH, NIT DURGAPUR
      This answer was edited.

      As the emissivity of black body is equal to one and gray body emissivity is less than one so the emissive power of black body is higher than that of gray body

      As the emissivity of black body is equal to one and gray body emissivity is less than one so the emissive power of black body is higher than that of gray body

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    2. Iron
      This answer was edited.

      Black body absorbs 100% of the radiations and hence its emmisivity is 1. Grey body absorbs the waves partially and hence its emmisivity is less than 1. So emmisive power of blackbody is greater than grey body

      Black body absorbs 100% of the radiations and hence its emmisivity is 1. Grey body absorbs the waves partially and hence its emmisivity is less than 1. So emmisive power of blackbody is greater than grey body

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    3. Best Answer
      Gold official account of Askmemetallurgy

      As per Kirchoff's law: Since no body is perfect emitter, its emittance is less than that of the black body M(b). Thus, emissivity of real (gray M(g)) body is defined by E(g)= M(g)/M(b) Black body : E=1 white body E=0 Gray body: E<1 Selective radiator E'<1 Correct answer is option D.  

      As per Kirchoff’s law:

      Since no body is perfect emitter, its emittance is less than that of the black body M(b). Thus, emissivity of real (gray M(g)) body is defined by E(g)= M(g)/M(b)

      Black body : E=1

      white body E=0

      Gray body: E<1

      Selective radiator E'<1

      Correct answer is option D.

       

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    4. Platinum⭐ GATE Metallurgy Coaching

      Option D

      Option D

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