QBManager

Heat Transfer

149581 The absolute temperature of a body $A$ is four times that of another body $B$ . For the two bodies, the difference in wavelengths, at which energy radiated is maximum is $3 μ$ . Then, the wavelength, at which the body $B$ radiates maximum energy, in micrometer, is:

1 2

2 2.5

3 4.00

4 4.5

Explanation:

C Given,
Absolute temperature of a body $A = 4 \times$ temperature of a body $B$
$∴ T_{A} = 4 T_{B}$
Wavelength of body B - Wavelength of body $A = 3 μ$
We know that,
Wein's displacement Law
$λ T =$ constant
$λ_{A} T_{A} = λ_{B} T_{B}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{4 T_{B}}$
$\frac{λ_{B}}{λ_{A}} = 4 \Rightarrow λ_{B} = 4 λ_{A}$
$∵ λ_{B} - λ_{A} = 3 μ m$
$∴ 4 λ_{A} - λ_{A} = 3 μ m$
$λ_{A} = 1 μ m$

AP EAMCET(Medical)-2004

Heat Transfer

149582 A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} K$ . The wavelength in nanometer at which its radiation becomes maximum is: $(b = 0.0029 mK)$

1 48

2 58

3 60

4 70

Explanation:

B Given,
Surface Temperature of particular star $= 5 \times 10^{4} K$
Wein's constant $b = 0.0029 mK$
We know that,
Wein's displacement law
$λ T =$ constant
$λ = \frac{b}{T}$
$λ = \frac{0.0029}{5 \times 10^{4}}$
$λ = 5.8 \times 10^{- 8} m$
$λ = 58 \times 10^{- 9} m$
$λ = 58 nm$

AP EAMCET(Medical)-2003

Heat Transfer

149583 The rate of emission of radiation of black body at temperature $27^{\circ} C$ is $E_{1}$ . If its temperature is increased to $327^{\circ} C$ the rate of emission of radiation is $E_{2}$ the relation between $E_{1}$ and $E_{2}$ is:

1

E_{2} = 24 E_{1}

2

E_{2} = 16 E_{1}

3

E_{2} = 8 E_{1}

4

E_{2} = 4 E_{1}

Explanation:

B Given,
$E_{1} = 27 + 273 = 300 K$
$E_{2} = 327 + 273 = 600 K$
We know that,
$E = σ T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{1}}{E_{2}} = {(\frac{300}{600})}^{4}$
$\frac{E_{1}}{E_{2}} = {(\frac{1}{2})}^{4}$
$\frac{E_{1}}{E_{2}} = \frac{1}{16}$
$E_{2} = 16 E_{1}$

AP EAMCET(Medical)-2002

Heat Transfer

149584 The temperature of a black body is increased by $50 %$ , then the percentage increase of radiations is approximately:

1

100 %

2

250 %

3

400 %

4

500 %

Explanation:

C Given,
Temperature of a black body $= T_{1}$ Increased temperature of black body $= \frac{3}{2} T_{1}$
We know that,
$E = σ T^{4} A$
$E_{1} = σ T_{1}^{4} A, E_{2} = σ T_{2}^{4} A$
Percentage increased in radiation $= \frac{E_{2} - E_{1}}{E_{1}} \times 100$
$= \frac{T_{2}^{4} - T_{1}^{4}}{T_{1}^{4}} \times 100$
$= \frac{{(\frac{3}{2} T_{1})}^{4} - T_{1}^{4}}{T_{1}^{4}} \times 100$
$= \frac{{(\frac{3}{2})}^{4} - 1}{1} \times 100$
$= \frac{\frac{81}{16} - 1}{1} \times 100$
$= \frac{65}{16} \times 100$
$= 406.25$
$%$ Increased in radiation $≃ 400 %$

AP EAMCET(Medical)-2001

Heat Transfer

149585 When the temperature of a black body increases, it is observed that the wavelength corresponding to maximum energy changes from $0.26 μ m$ to $0.13 μ m$ . The ratio of the emissive power of the body at the respective temperature is

1

\frac{16}{1}

2

\frac{4}{1}

3

\frac{1}{4}

4

\frac{1}{16}

Explanation:

D Given,
$λ_{1} = 0.26 μ m$
$λ_{2} = 0.13 μ m$
We know that,
Weins displacement law
$λ_{1} T_{1} = λ_{2} T_{2}$
$\frac{λ_{1}}{λ_{2}} = \frac{T_{2}}{T_{1}} = \frac{0.26}{0.13} = 2$
$T_{2} = 2 T_{1}$
Stefan's law,
$E \propto T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{{(2 T_{1})}^{4}} = \frac{1}{2^{4}} = \frac{1}{16}$

EAMCET-2002

Heat Transfer

149581 The absolute temperature of a body $A$ is four times that of another body $B$ . For the two bodies, the difference in wavelengths, at which energy radiated is maximum is $3 μ$ . Then, the wavelength, at which the body $B$ radiates maximum energy, in micrometer, is:

1 2

2 2.5

3 4.00

4 4.5

Explanation:

C Given,
Absolute temperature of a body $A = 4 \times$ temperature of a body $B$
$∴ T_{A} = 4 T_{B}$
Wavelength of body B - Wavelength of body $A = 3 μ$
We know that,
Wein's displacement Law
$λ T =$ constant
$λ_{A} T_{A} = λ_{B} T_{B}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{4 T_{B}}$
$\frac{λ_{B}}{λ_{A}} = 4 \Rightarrow λ_{B} = 4 λ_{A}$
$∵ λ_{B} - λ_{A} = 3 μ m$
$∴ 4 λ_{A} - λ_{A} = 3 μ m$
$λ_{A} = 1 μ m$

AP EAMCET(Medical)-2004

Heat Transfer

149582 A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} K$ . The wavelength in nanometer at which its radiation becomes maximum is: $(b = 0.0029 mK)$

1 48

2 58

3 60

4 70

Explanation:

B Given,
Surface Temperature of particular star $= 5 \times 10^{4} K$
Wein's constant $b = 0.0029 mK$
We know that,
Wein's displacement law
$λ T =$ constant
$λ = \frac{b}{T}$
$λ = \frac{0.0029}{5 \times 10^{4}}$
$λ = 5.8 \times 10^{- 8} m$
$λ = 58 \times 10^{- 9} m$
$λ = 58 nm$

AP EAMCET(Medical)-2003

Heat Transfer

149583 The rate of emission of radiation of black body at temperature $27^{\circ} C$ is $E_{1}$ . If its temperature is increased to $327^{\circ} C$ the rate of emission of radiation is $E_{2}$ the relation between $E_{1}$ and $E_{2}$ is:

1

E_{2} = 24 E_{1}

2

E_{2} = 16 E_{1}

3

E_{2} = 8 E_{1}

4

E_{2} = 4 E_{1}

Explanation:

B Given,
$E_{1} = 27 + 273 = 300 K$
$E_{2} = 327 + 273 = 600 K$
We know that,
$E = σ T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{1}}{E_{2}} = {(\frac{300}{600})}^{4}$
$\frac{E_{1}}{E_{2}} = {(\frac{1}{2})}^{4}$
$\frac{E_{1}}{E_{2}} = \frac{1}{16}$
$E_{2} = 16 E_{1}$

AP EAMCET(Medical)-2002

Heat Transfer

149584 The temperature of a black body is increased by $50 %$ , then the percentage increase of radiations is approximately:

1

100 %

2

250 %

3

400 %

4

500 %

Explanation:

C Given,
Temperature of a black body $= T_{1}$ Increased temperature of black body $= \frac{3}{2} T_{1}$
We know that,
$E = σ T^{4} A$
$E_{1} = σ T_{1}^{4} A, E_{2} = σ T_{2}^{4} A$
Percentage increased in radiation $= \frac{E_{2} - E_{1}}{E_{1}} \times 100$
$= \frac{T_{2}^{4} - T_{1}^{4}}{T_{1}^{4}} \times 100$
$= \frac{{(\frac{3}{2} T_{1})}^{4} - T_{1}^{4}}{T_{1}^{4}} \times 100$
$= \frac{{(\frac{3}{2})}^{4} - 1}{1} \times 100$
$= \frac{\frac{81}{16} - 1}{1} \times 100$
$= \frac{65}{16} \times 100$
$= 406.25$
$%$ Increased in radiation $≃ 400 %$

AP EAMCET(Medical)-2001

Heat Transfer

149585 When the temperature of a black body increases, it is observed that the wavelength corresponding to maximum energy changes from $0.26 μ m$ to $0.13 μ m$ . The ratio of the emissive power of the body at the respective temperature is

1

\frac{16}{1}

2

\frac{4}{1}

3

\frac{1}{4}

4

\frac{1}{16}

Explanation:

D Given,
$λ_{1} = 0.26 μ m$
$λ_{2} = 0.13 μ m$
We know that,
Weins displacement law
$λ_{1} T_{1} = λ_{2} T_{2}$
$\frac{λ_{1}}{λ_{2}} = \frac{T_{2}}{T_{1}} = \frac{0.26}{0.13} = 2$
$T_{2} = 2 T_{1}$
Stefan's law,
$E \propto T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{{(2 T_{1})}^{4}} = \frac{1}{2^{4}} = \frac{1}{16}$

EAMCET-2002

Heat Transfer

149581 The absolute temperature of a body $A$ is four times that of another body $B$ . For the two bodies, the difference in wavelengths, at which energy radiated is maximum is $3 μ$ . Then, the wavelength, at which the body $B$ radiates maximum energy, in micrometer, is:

1 2

2 2.5

3 4.00

4 4.5

Explanation:

C Given,
Absolute temperature of a body $A = 4 \times$ temperature of a body $B$
$∴ T_{A} = 4 T_{B}$
Wavelength of body B - Wavelength of body $A = 3 μ$
We know that,
Wein's displacement Law
$λ T =$ constant
$λ_{A} T_{A} = λ_{B} T_{B}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{4 T_{B}}$
$\frac{λ_{B}}{λ_{A}} = 4 \Rightarrow λ_{B} = 4 λ_{A}$
$∵ λ_{B} - λ_{A} = 3 μ m$
$∴ 4 λ_{A} - λ_{A} = 3 μ m$
$λ_{A} = 1 μ m$

AP EAMCET(Medical)-2004

Heat Transfer

149582 A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} K$ . The wavelength in nanometer at which its radiation becomes maximum is: $(b = 0.0029 mK)$

1 48

2 58

3 60

4 70

Explanation:

B Given,
Surface Temperature of particular star $= 5 \times 10^{4} K$
Wein's constant $b = 0.0029 mK$
We know that,
Wein's displacement law
$λ T =$ constant
$λ = \frac{b}{T}$
$λ = \frac{0.0029}{5 \times 10^{4}}$
$λ = 5.8 \times 10^{- 8} m$
$λ = 58 \times 10^{- 9} m$
$λ = 58 nm$

AP EAMCET(Medical)-2003

Heat Transfer

149583 The rate of emission of radiation of black body at temperature $27^{\circ} C$ is $E_{1}$ . If its temperature is increased to $327^{\circ} C$ the rate of emission of radiation is $E_{2}$ the relation between $E_{1}$ and $E_{2}$ is:

1

E_{2} = 24 E_{1}

2

E_{2} = 16 E_{1}

3

E_{2} = 8 E_{1}

4

E_{2} = 4 E_{1}

Explanation:

B Given,
$E_{1} = 27 + 273 = 300 K$
$E_{2} = 327 + 273 = 600 K$
We know that,
$E = σ T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{1}}{E_{2}} = {(\frac{300}{600})}^{4}$
$\frac{E_{1}}{E_{2}} = {(\frac{1}{2})}^{4}$
$\frac{E_{1}}{E_{2}} = \frac{1}{16}$
$E_{2} = 16 E_{1}$

AP EAMCET(Medical)-2002

Heat Transfer

149584 The temperature of a black body is increased by $50 %$ , then the percentage increase of radiations is approximately:

1

100 %

2

250 %

3

400 %

4

500 %

Explanation:

C Given,
Temperature of a black body $= T_{1}$ Increased temperature of black body $= \frac{3}{2} T_{1}$
We know that,
$E = σ T^{4} A$
$E_{1} = σ T_{1}^{4} A, E_{2} = σ T_{2}^{4} A$
Percentage increased in radiation $= \frac{E_{2} - E_{1}}{E_{1}} \times 100$
$= \frac{T_{2}^{4} - T_{1}^{4}}{T_{1}^{4}} \times 100$
$= \frac{{(\frac{3}{2} T_{1})}^{4} - T_{1}^{4}}{T_{1}^{4}} \times 100$
$= \frac{{(\frac{3}{2})}^{4} - 1}{1} \times 100$
$= \frac{\frac{81}{16} - 1}{1} \times 100$
$= \frac{65}{16} \times 100$
$= 406.25$
$%$ Increased in radiation $≃ 400 %$

AP EAMCET(Medical)-2001

Heat Transfer

149585 When the temperature of a black body increases, it is observed that the wavelength corresponding to maximum energy changes from $0.26 μ m$ to $0.13 μ m$ . The ratio of the emissive power of the body at the respective temperature is

1

\frac{16}{1}

2

\frac{4}{1}

3

\frac{1}{4}

4

\frac{1}{16}

Explanation:

D Given,
$λ_{1} = 0.26 μ m$
$λ_{2} = 0.13 μ m$
We know that,
Weins displacement law
$λ_{1} T_{1} = λ_{2} T_{2}$
$\frac{λ_{1}}{λ_{2}} = \frac{T_{2}}{T_{1}} = \frac{0.26}{0.13} = 2$
$T_{2} = 2 T_{1}$
Stefan's law,
$E \propto T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{{(2 T_{1})}^{4}} = \frac{1}{2^{4}} = \frac{1}{16}$

EAMCET-2002

Heat Transfer

149581 The absolute temperature of a body $A$ is four times that of another body $B$ . For the two bodies, the difference in wavelengths, at which energy radiated is maximum is $3 μ$ . Then, the wavelength, at which the body $B$ radiates maximum energy, in micrometer, is:

1 2

2 2.5

3 4.00

4 4.5

Explanation:

C Given,
Absolute temperature of a body $A = 4 \times$ temperature of a body $B$
$∴ T_{A} = 4 T_{B}$
Wavelength of body B - Wavelength of body $A = 3 μ$
We know that,
Wein's displacement Law
$λ T =$ constant
$λ_{A} T_{A} = λ_{B} T_{B}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{4 T_{B}}$
$\frac{λ_{B}}{λ_{A}} = 4 \Rightarrow λ_{B} = 4 λ_{A}$
$∵ λ_{B} - λ_{A} = 3 μ m$
$∴ 4 λ_{A} - λ_{A} = 3 μ m$
$λ_{A} = 1 μ m$

AP EAMCET(Medical)-2004

Heat Transfer

149582 A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} K$ . The wavelength in nanometer at which its radiation becomes maximum is: $(b = 0.0029 mK)$

1 48

2 58

3 60

4 70

Explanation:

B Given,
Surface Temperature of particular star $= 5 \times 10^{4} K$
Wein's constant $b = 0.0029 mK$
We know that,
Wein's displacement law
$λ T =$ constant
$λ = \frac{b}{T}$
$λ = \frac{0.0029}{5 \times 10^{4}}$
$λ = 5.8 \times 10^{- 8} m$
$λ = 58 \times 10^{- 9} m$
$λ = 58 nm$

AP EAMCET(Medical)-2003

Heat Transfer

149583 The rate of emission of radiation of black body at temperature $27^{\circ} C$ is $E_{1}$ . If its temperature is increased to $327^{\circ} C$ the rate of emission of radiation is $E_{2}$ the relation between $E_{1}$ and $E_{2}$ is:

1

E_{2} = 24 E_{1}

2

E_{2} = 16 E_{1}

3

E_{2} = 8 E_{1}

4

E_{2} = 4 E_{1}

Explanation:

B Given,
$E_{1} = 27 + 273 = 300 K$
$E_{2} = 327 + 273 = 600 K$
We know that,
$E = σ T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{1}}{E_{2}} = {(\frac{300}{600})}^{4}$
$\frac{E_{1}}{E_{2}} = {(\frac{1}{2})}^{4}$
$\frac{E_{1}}{E_{2}} = \frac{1}{16}$
$E_{2} = 16 E_{1}$

AP EAMCET(Medical)-2002

Heat Transfer

149584 The temperature of a black body is increased by $50 %$ , then the percentage increase of radiations is approximately:

1

100 %

2

250 %

3

400 %

4

500 %

Explanation:

C Given,
Temperature of a black body $= T_{1}$ Increased temperature of black body $= \frac{3}{2} T_{1}$
We know that,
$E = σ T^{4} A$
$E_{1} = σ T_{1}^{4} A, E_{2} = σ T_{2}^{4} A$
Percentage increased in radiation $= \frac{E_{2} - E_{1}}{E_{1}} \times 100$
$= \frac{T_{2}^{4} - T_{1}^{4}}{T_{1}^{4}} \times 100$
$= \frac{{(\frac{3}{2} T_{1})}^{4} - T_{1}^{4}}{T_{1}^{4}} \times 100$
$= \frac{{(\frac{3}{2})}^{4} - 1}{1} \times 100$
$= \frac{\frac{81}{16} - 1}{1} \times 100$
$= \frac{65}{16} \times 100$
$= 406.25$
$%$ Increased in radiation $≃ 400 %$

AP EAMCET(Medical)-2001

Heat Transfer

149585 When the temperature of a black body increases, it is observed that the wavelength corresponding to maximum energy changes from $0.26 μ m$ to $0.13 μ m$ . The ratio of the emissive power of the body at the respective temperature is

1

\frac{16}{1}

2

\frac{4}{1}

3

\frac{1}{4}

4

\frac{1}{16}

Explanation:

D Given,
$λ_{1} = 0.26 μ m$
$λ_{2} = 0.13 μ m$
We know that,
Weins displacement law
$λ_{1} T_{1} = λ_{2} T_{2}$
$\frac{λ_{1}}{λ_{2}} = \frac{T_{2}}{T_{1}} = \frac{0.26}{0.13} = 2$
$T_{2} = 2 T_{1}$
Stefan's law,
$E \propto T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{{(2 T_{1})}^{4}} = \frac{1}{2^{4}} = \frac{1}{16}$

EAMCET-2002

Heat Transfer

149581 The absolute temperature of a body $A$ is four times that of another body $B$ . For the two bodies, the difference in wavelengths, at which energy radiated is maximum is $3 μ$ . Then, the wavelength, at which the body $B$ radiates maximum energy, in micrometer, is:

1 2

2 2.5

3 4.00

4 4.5

Explanation:

C Given,
Absolute temperature of a body $A = 4 \times$ temperature of a body $B$
$∴ T_{A} = 4 T_{B}$
Wavelength of body B - Wavelength of body $A = 3 μ$
We know that,
Wein's displacement Law
$λ T =$ constant
$λ_{A} T_{A} = λ_{B} T_{B}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{T_{A}}$
$\frac{λ_{A}}{λ_{B}} = \frac{T_{B}}{4 T_{B}}$
$\frac{λ_{B}}{λ_{A}} = 4 \Rightarrow λ_{B} = 4 λ_{A}$
$∵ λ_{B} - λ_{A} = 3 μ m$
$∴ 4 λ_{A} - λ_{A} = 3 μ m$
$λ_{A} = 1 μ m$

AP EAMCET(Medical)-2004

Heat Transfer

149582 A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} K$ . The wavelength in nanometer at which its radiation becomes maximum is: $(b = 0.0029 mK)$

1 48

2 58

3 60

4 70

Explanation:

B Given,
Surface Temperature of particular star $= 5 \times 10^{4} K$
Wein's constant $b = 0.0029 mK$
We know that,
Wein's displacement law
$λ T =$ constant
$λ = \frac{b}{T}$
$λ = \frac{0.0029}{5 \times 10^{4}}$
$λ = 5.8 \times 10^{- 8} m$
$λ = 58 \times 10^{- 9} m$
$λ = 58 nm$

AP EAMCET(Medical)-2003

Heat Transfer

149583 The rate of emission of radiation of black body at temperature $27^{\circ} C$ is $E_{1}$ . If its temperature is increased to $327^{\circ} C$ the rate of emission of radiation is $E_{2}$ the relation between $E_{1}$ and $E_{2}$ is:

1

E_{2} = 24 E_{1}

2

E_{2} = 16 E_{1}

3

E_{2} = 8 E_{1}

4

E_{2} = 4 E_{1}

Explanation:

B Given,
$E_{1} = 27 + 273 = 300 K$
$E_{2} = 327 + 273 = 600 K$
We know that,
$E = σ T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{1}}{E_{2}} = {(\frac{300}{600})}^{4}$
$\frac{E_{1}}{E_{2}} = {(\frac{1}{2})}^{4}$
$\frac{E_{1}}{E_{2}} = \frac{1}{16}$
$E_{2} = 16 E_{1}$

AP EAMCET(Medical)-2002

Heat Transfer

149584 The temperature of a black body is increased by $50 %$ , then the percentage increase of radiations is approximately:

1

100 %

2

250 %

3

400 %

4

500 %

Explanation:

C Given,
Temperature of a black body $= T_{1}$ Increased temperature of black body $= \frac{3}{2} T_{1}$
We know that,
$E = σ T^{4} A$
$E_{1} = σ T_{1}^{4} A, E_{2} = σ T_{2}^{4} A$
Percentage increased in radiation $= \frac{E_{2} - E_{1}}{E_{1}} \times 100$
$= \frac{T_{2}^{4} - T_{1}^{4}}{T_{1}^{4}} \times 100$
$= \frac{{(\frac{3}{2} T_{1})}^{4} - T_{1}^{4}}{T_{1}^{4}} \times 100$
$= \frac{{(\frac{3}{2})}^{4} - 1}{1} \times 100$
$= \frac{\frac{81}{16} - 1}{1} \times 100$
$= \frac{65}{16} \times 100$
$= 406.25$
$%$ Increased in radiation $≃ 400 %$

AP EAMCET(Medical)-2001

Heat Transfer

149585 When the temperature of a black body increases, it is observed that the wavelength corresponding to maximum energy changes from $0.26 μ m$ to $0.13 μ m$ . The ratio of the emissive power of the body at the respective temperature is

1

\frac{16}{1}

2

\frac{4}{1}

3

\frac{1}{4}

4

\frac{1}{16}

Explanation:

D Given,
$λ_{1} = 0.26 μ m$
$λ_{2} = 0.13 μ m$
We know that,
Weins displacement law
$λ_{1} T_{1} = λ_{2} T_{2}$
$\frac{λ_{1}}{λ_{2}} = \frac{T_{2}}{T_{1}} = \frac{0.26}{0.13} = 2$
$T_{2} = 2 T_{1}$
Stefan's law,
$E \propto T^{4}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{{(2 T_{1})}^{4}} = \frac{1}{2^{4}} = \frac{1}{16}$

EAMCET-2002

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