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Heat Transfer

149555 If $T$ is the temperature of a body then the rate at which energy is radiated from the body is proportional to

1

T

2

T^{2}

3

T^{3}

4

T^{4}

Explanation:

D Stefan Boltzmann law stated that the point of radiation emitted by a black body per unit area is directly proportional to the fourth power of the temperature.
$Q =\in σ A (T)^{4}$
$Q \propto T^{4}$

TS EAMCET 29.09.2020

Heat Transfer

149567 Temperature of two stars in ratio $3 : 2$ If wavelength of maximum intensity of first body is $400 \AA$ What is corresponding wavelength of second body?

1

9000 \AA

2

6000 \AA

3

2000 \AA

4

8000 \AA

Explanation:

B According to Wien's displacement law
$λ_{m} T =$ constant
$∴ \frac{{(λ_{m})}_{1}}{{(λ_{m})}_{2}} = \frac{T_{2}}{T_{1}}$
$\frac{T_{1}}{T_{2}} = \frac{3}{2}, {(λ_{m})}_{1} = 4000 \AA$
$∴ {(λ_{m})}_{2} = \frac{T_{1}}{T_{2}} \times λ_{m 1} = \frac{3}{2} \times 4000 = 6000 \AA$

DCE-2007

Heat Transfer

149592 Which one among the following radiations carries maximum energy?

1 Ultraviolet rays

2 Gamma-rays

3 X-rays

4 Infra-red rays

NDA (I) 2008

Heat Transfer

149597 The temperature of a blackbody radiation enclosed in a container of volume $V$ is increased from $100^{\circ} C$ to $1000^{\circ} C$ . The heat required in the process is

1

4.79 \times 10^{- 4} cal

2

9.21 \times 10^{- 5} cal

3

2.17 \times 10^{- 4} cal

4

7.54 \times 10^{- 4} cal

5 None of these

Explanation:

E According to Stefan's law,
$\frac{P}{A} = σ T^{4} (∵ σ =$ Stefan's constant $)$
Here, area of the body is not given, so information is insufficient.

WB JEE 2016

Heat Transfer

149555 If $T$ is the temperature of a body then the rate at which energy is radiated from the body is proportional to

1

T

2

T^{2}

3

T^{3}

4

T^{4}

Explanation:

D Stefan Boltzmann law stated that the point of radiation emitted by a black body per unit area is directly proportional to the fourth power of the temperature.
$Q =\in σ A (T)^{4}$
$Q \propto T^{4}$

TS EAMCET 29.09.2020

Heat Transfer

149567 Temperature of two stars in ratio $3 : 2$ If wavelength of maximum intensity of first body is $400 \AA$ What is corresponding wavelength of second body?

1

9000 \AA

2

6000 \AA

3

2000 \AA

4

8000 \AA

Explanation:

B According to Wien's displacement law
$λ_{m} T =$ constant
$∴ \frac{{(λ_{m})}_{1}}{{(λ_{m})}_{2}} = \frac{T_{2}}{T_{1}}$
$\frac{T_{1}}{T_{2}} = \frac{3}{2}, {(λ_{m})}_{1} = 4000 \AA$
$∴ {(λ_{m})}_{2} = \frac{T_{1}}{T_{2}} \times λ_{m 1} = \frac{3}{2} \times 4000 = 6000 \AA$

DCE-2007

Heat Transfer

149592 Which one among the following radiations carries maximum energy?

1 Ultraviolet rays

2 Gamma-rays

3 X-rays

4 Infra-red rays

NDA (I) 2008

Heat Transfer

149597 The temperature of a blackbody radiation enclosed in a container of volume $V$ is increased from $100^{\circ} C$ to $1000^{\circ} C$ . The heat required in the process is

1

4.79 \times 10^{- 4} cal

2

9.21 \times 10^{- 5} cal

3

2.17 \times 10^{- 4} cal

4

7.54 \times 10^{- 4} cal

5 None of these

Explanation:

E According to Stefan's law,
$\frac{P}{A} = σ T^{4} (∵ σ =$ Stefan's constant $)$
Here, area of the body is not given, so information is insufficient.

WB JEE 2016

Heat Transfer

149555 If $T$ is the temperature of a body then the rate at which energy is radiated from the body is proportional to

1

T

2

T^{2}

3

T^{3}

4

T^{4}

Explanation:

D Stefan Boltzmann law stated that the point of radiation emitted by a black body per unit area is directly proportional to the fourth power of the temperature.
$Q =\in σ A (T)^{4}$
$Q \propto T^{4}$

TS EAMCET 29.09.2020

Heat Transfer

149567 Temperature of two stars in ratio $3 : 2$ If wavelength of maximum intensity of first body is $400 \AA$ What is corresponding wavelength of second body?

1

9000 \AA

2

6000 \AA

3

2000 \AA

4

8000 \AA

Explanation:

B According to Wien's displacement law
$λ_{m} T =$ constant
$∴ \frac{{(λ_{m})}_{1}}{{(λ_{m})}_{2}} = \frac{T_{2}}{T_{1}}$
$\frac{T_{1}}{T_{2}} = \frac{3}{2}, {(λ_{m})}_{1} = 4000 \AA$
$∴ {(λ_{m})}_{2} = \frac{T_{1}}{T_{2}} \times λ_{m 1} = \frac{3}{2} \times 4000 = 6000 \AA$

DCE-2007

Heat Transfer

149592 Which one among the following radiations carries maximum energy?

1 Ultraviolet rays

2 Gamma-rays

3 X-rays

4 Infra-red rays

NDA (I) 2008

Heat Transfer

149597 The temperature of a blackbody radiation enclosed in a container of volume $V$ is increased from $100^{\circ} C$ to $1000^{\circ} C$ . The heat required in the process is

1

4.79 \times 10^{- 4} cal

2

9.21 \times 10^{- 5} cal

3

2.17 \times 10^{- 4} cal

4

7.54 \times 10^{- 4} cal

5 None of these

Explanation:

E According to Stefan's law,
$\frac{P}{A} = σ T^{4} (∵ σ =$ Stefan's constant $)$
Here, area of the body is not given, so information is insufficient.

WB JEE 2016

Heat Transfer

149555 If $T$ is the temperature of a body then the rate at which energy is radiated from the body is proportional to

1

T

2

T^{2}

3

T^{3}

4

T^{4}

Explanation:

D Stefan Boltzmann law stated that the point of radiation emitted by a black body per unit area is directly proportional to the fourth power of the temperature.
$Q =\in σ A (T)^{4}$
$Q \propto T^{4}$

TS EAMCET 29.09.2020

Heat Transfer

149567 Temperature of two stars in ratio $3 : 2$ If wavelength of maximum intensity of first body is $400 \AA$ What is corresponding wavelength of second body?

1

9000 \AA

2

6000 \AA

3

2000 \AA

4

8000 \AA

Explanation:

B According to Wien's displacement law
$λ_{m} T =$ constant
$∴ \frac{{(λ_{m})}_{1}}{{(λ_{m})}_{2}} = \frac{T_{2}}{T_{1}}$
$\frac{T_{1}}{T_{2}} = \frac{3}{2}, {(λ_{m})}_{1} = 4000 \AA$
$∴ {(λ_{m})}_{2} = \frac{T_{1}}{T_{2}} \times λ_{m 1} = \frac{3}{2} \times 4000 = 6000 \AA$

DCE-2007

Heat Transfer

149592 Which one among the following radiations carries maximum energy?

1 Ultraviolet rays

2 Gamma-rays

3 X-rays

4 Infra-red rays

NDA (I) 2008

Heat Transfer

149597 The temperature of a blackbody radiation enclosed in a container of volume $V$ is increased from $100^{\circ} C$ to $1000^{\circ} C$ . The heat required in the process is

1

4.79 \times 10^{- 4} cal

2

9.21 \times 10^{- 5} cal

3

2.17 \times 10^{- 4} cal

4

7.54 \times 10^{- 4} cal

5 None of these

Explanation:

E According to Stefan's law,
$\frac{P}{A} = σ T^{4} (∵ σ =$ Stefan's constant $)$
Here, area of the body is not given, so information is insufficient.

WB JEE 2016

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