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Questions Related to energy production

Black body spectrum is

  1. continuous absorption

  2. line absorption

  3. continuous emission

  4. line emission


Correct Option: C
Explanation:

Black Body radiation involves emission of a number of wavelengths, which by definition is continuous emission.

Stefan-Boltzmanns Law for a perfect black body is represented by

  1. $\dfrac{dQ}{dt} = \sigma AT^2 $

  2. $\dfrac{dQ}{dt} = \sigma AT^3 $

  3. $\dfrac{dQ}{dt} = \sigma AT^4 $

  4. $ Q = \sigma AT^4 $


Correct Option: C
Explanation:
Stefan boltzmann's law, $\Rightarrow \cfrac { dQ }{ dt } =\sigma A\varepsilon { T }^{ 4 }$
( for black body, $\varepsilon=1$)
So, for black body stefan's boltzmann's law is $\Rightarrow \cfrac { dQ }{ dt } =\sigma A { T }^{ 4 }$
Stefan's boltzmann's law states that total energy radiated from a surface is proportional to the forth power of its absolute temperature.

Ferry's black body is accurately represented by 

  1. A fine hole in a double walled spherical cavity.

  2. A fine hole in a double walled spherical cavity, evacuated and painted black.

  3. A fine hole in a spherical cavity, evacuated and painted black.

  4. A fine hole in a black cavity.


Correct Option: B
Explanation:

Ferry's black body is accurately represented by a fine hole in a double walled spherical cavity, evacuated and painted black.

$\rightarrow$ Ferry designed the simplest black body. It is a double walled evacuated spherical cavity whose inner wall is blackened. The space between wall is evacuated to prevent heat loss by conduction and radiation. There is a fine hole in it. All the radiations incident upon this hole are absorbed by this black body. 

Ferry's black body is 

  1. a perfect reflector of heat.

  2. perfectly black.

  3. platinum black.

  4. almost perfectly black.


Correct Option: D
Explanation:
Ferry's black body is almost perfectly black.
As because emission of heat and absroption of heat takes place completely.

Black body radiation is

  1. White

  2. IR

  3. Black

  4. UV


Correct Option: A
Explanation:

a black body looks black because it absorbs light of all colors. so when it radiates energy as light, it will release all the colors which it had absorbed resulting in the emission of white light, because white light is the resultant of many component colors.

The original temperature of a black body is $727^\circ C$. Calculate temperature at which total radiant energy from this black body becomes double:

  1. $971K$

  2. $1189K$

  3. $2001K$

  4. $1458K$


Correct Option: B
Explanation:

$Rediant Energy = \sigma T^2$

$Energy = \sigma (1000)^4$
$E _2 = 2 E _1$
$Then$
$\sigma T _2 ^{4} = 2 \times \sigma (1000)^4$
$T _2 = 2^\frac{1}{4} \times1000$
$T _2 = 1189 K$

Temp. of black body is $3000K$ when black body cools. Then change in wavelength $\Delta \lambda=9$ micron corresponding to maximum energy density. Now temp. of black body is:

  1. $300K$

  2. $2700K$

  3. $270K$

  4. $1800K$


Correct Option: A

The rate of emission of radiation of a black body at 273$^{ \circ  }{ C }$ is E, then the rate of emission of radiation  of this body at 0$^{ \circ  }{ C }$ will be :-

  1. $\dfrac { E }{ 16 } $

  2. $\dfrac { E }{ 4 } $

  3. $\dfrac { E }{ 8 } $

  4. 0


Correct Option: A

For non black bodies, the range of values of emissivity $e$ is 

  1. $- 1 < e < 1 $

  2. $- 1 < e < 0 $

  3. $ 1 < e < 2 $

  4. $0 < e < 1 $


Correct Option: D

A blackbody does not

  1. emit radiation

  2. absorb radiation

  3. reflect radiation

  4. refract radiation


Correct Option: B