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Perfectly black body - class-XI

Description: perfectly black body
Number of Questions: 24
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Tags: physics heat transfer thermal properties energy production
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The coefficient of transmission for an ideal black body is :

  1. infinity

  2. zero

  3. 1

  4. more than one


Correct Option: B
Explanation:

Ideal black body absorbs all the radiation, no transmission, no reflection.

Out of the following, which body is not an ideal black body?

  1. Wein's black body

  2. Ferry's black body

  3. coal

  4. sun


Correct Option: C
Explanation:

wein's and ferry's black bodies are well defined in there papers and sun can also be considered as black body because it is very good emitter. but coal is just a normal carbon compound and neither is it a good absorber or emmiter

Read the following statements carefully
(A) Black body radiation is white
(B) Emissivity of a body is equal to its absorptive power
Mark correct option:

  1. Statement (A) is correct

  2. Statement (B) is correct

  3. Both are correct

  4. Both are wrong


Correct Option: C
Explanation:
  • (A) The combination of radiation of all the visible wavelengths makes white light. Hence black  body radiation is white

(B) Emissivity of body is equal to it's absorptive power, it is not less then absorptive power or greater than absorptive power
  • Hence both the statements are true
  • Hence option C is the right answer
           

The absorptivity of Lamp black is:

  1. 0.91

  2. 0.98

  3. 1.00

  4. 0.99


Correct Option: C
Explanation:

Black body is a body which absorbs and emit all types of radiations completely. Since we know that lamp black is perfect black body so its absorptivity is $1.00$

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

An ideal black body is a :

  1. lump of charcoal heated to a high temperature

  2. metal coated with a black dye

  3. glass surface coated with coal tar

  4. hollow enclosure blackened inside and having a small hole


Correct Option: D
Explanation:

Hollow enclosure blackened inside and having a small hole is a very good example of a black body.
Suppose once light enters inside it.
It may be absorbed or it may be reflected.
Since it is blackened from inside, there is a high probability that it will be absorbed.
Now if it is reflected, it will suffer multiple reflections and it is very unlikely that it will come out of the hole because the aperture of hole is too small. Moreover, with each reflection, more and more fraction of it will be absorbed. So, it will serve as a good black body.

Which of the following is more close to a black body?

  1. Black board paint

  2. Green leaves

  3. Black holes

  4. Red roses


Correct Option: C
Explanation:

we know that $\alpha +\rho +\tau =1$
${\alpha}= absorptivity$
${\rho}=reflectivity$
${\tau}=transmitivity$
so for black body ${\rho}\  and \ {\tau} \ will\  be \ zero$
so ${\alpha}=1$ so black hole has also ${\alpha}=1$ which is equivalent to black body.

Hence we can consider black holes as black body.

Initially a black body at absolute temperature $T$ is kept inside a closed chamber at absolute temperature $T _{o}$. Now the chamber is slightly opened to allow sun rays to enter. It is observed that temperatures $T$ and $T _{o}$ remains constant.Which of the following statement is/are true?

  1. The rate of emission of energy from the black body remains the same

  2. The rate of emission of energy from the black body increases

  3. The rate of absorption of energy by the black body increases.

  4. The energy radiated by the black body equals the energy absorbed by it


Correct Option: A,D
Explanation:

It is given that the absolute temperatures of both the black body and the surroundings are constant with time, even after sunlight(radiation) is incident on it.

  • When a body absorbs radiation, its temperature increases
  • When a body emits radiation, its temperature decreases
Also the sun, being a source of infinite radiation(very large source of radiation).
We infer from this that the incident radiation should be of constant magnitude.
And if the temperature of the black body is a constant, that means it's emission and absorption of radiation are matched and equal. The absorption is of constant magnitude, because the sun's radiation is of constant value. Hence the emission is of constant value also and is equal to the absorption. The options follow.

An ideal black body at room temperature is thrown in a furnace. It is observed that

  1. initially it is darkest body and at later time the brightest

  2. it is darkest body at all the times

  3. it cannot be distinguished at all the times

  4. initially it is the darkest body and at later times it cannot be distinguished


Correct Option: D
Explanation:

Initially the black body is in bright environment. So, it appears dark and finally when its temperature becomes equal to temperature of furnace, it becomes invisible

A spherical body of area A and emissivity $e = 0.6$ is kept inside a perfectly black body. Total heat radiated by the body at temperature $T$ 

  1. $ 0.8\ e\sigma AT^4$

  2. $ 0.4\ e\sigma AT^4$

  3. $ 0.6\ e\sigma AT^4$

  4. $ 1.0\ e\sigma AT^4$


Correct Option: D
Explanation:
According to Stefan's Boltzman law, the thermal energy radiated by a black body radiator per second per unit area is proportional to fourth power of the absolute temperature and is given by
$\dfrac{P}{A} = \sigma T^4$ ..............(1)
For the hot bodies other than black body radiator equation (1) becomes,
$\dfrac{P}{A} = e \sigma T^4$
$P = e \sigma A T^4$ .................(2)
where, $e$ is the emissivity of the body.
Now, when such hot body is kept inside a perfectly black body, the total thermal radiation is the sum of emitted radiations (in open) and the part of incident radiations reflected from the walls of the perfectly black body. This will give black body radiations, hence the total radiations emitted by the body will be,
$P = 1.0 e \sigma A T^4$.

Emissivity of a perfect black body is

  1. always $0$.

  2. always $1$.

  3. between $0$ and $1$.

  4. always $>1$.


Correct Option: B
Explanation:
Emissivity of a perfect black body is always 1.
The best absorber is defined as the object which can absorb all the electromagnetic radiations falling upon it. The black body is not only a perfect absorber but it is also the best in emitting radiation. Also, a black bosy in thermal equlibrium has emissivity, $\epsilon=1$

Radiation coming from the hole of a Ferry's Black body is called 

  1. black radiation.

  2. cavity radiation.

  3. Ferry's radiation.

  4. None of these


Correct Option: B
Explanation:
$\rightarrow$ Radiation coming from the hole of ferry's black body is called cavity radiation because this black body is double walled evacuated spherical cavity whose inner wall is blackened

Which of the following is an example of a black body radiation?

  1. The cooling of earth at night

  2. Solar radiations

  3. Heat currents from a black surface

  4. None of these


Correct Option: A,B
Explanation:
The Earth is heated by the flux of energy it receives from sun and cools down by emitting infrared radiations to space. So we can say that cooling of earth is example of the black body radiation.
Sun has no solid surface, any radiation that hits the Sun is scattered and absorbed until it is completely lost. This makes the Sun very close to black bodies as far as absorption is concerned and as a result it would be reasonable to suppose that it is black body radiators.
So, the radiation of sun is the black body radiation.

The Wien's displacement law for a black body is
($T$ is the absolute temperature in $K$
$b$
 is a constant of proportionality 
$e$ is the emissivity of the black body)

  1. $\lambda _{max} T = b$

  2. $\lambda _{max} T = e$

  3. $\lambda _{max} b = T$

  4. None of these


Correct Option: A
Explanation:
According to wein's displacement law there is inverse relation between $\lambda _{max}$ of radiation emitted by black body and its temperature (absolute)
$\lambda _{max}\; \alpha \; \cfrac{1}{T} \Rightarrow =b\cfrac{1}{T} \Rightarrow \lambda _{max} T=b$
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