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 $3 \mu$. 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 $\mathrm{A}=4 \times$ temperature of a body $\mathrm{B}$ $\therefore \quad \mathrm{T}_{\mathrm{A}}=4 \mathrm{~T}_{\mathrm{B}}$ Wavelength of body B - Wavelength of body $A=3 \mu$ We know that, Wein's displacement Law $\lambda \mathrm{T}=$ constant $\lambda_{\mathrm{A}} \mathrm{T}_{\mathrm{A}}=\lambda_{\mathrm{B}} \mathrm{T}_{\mathrm{B}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{4 \mathrm{~T}_{\mathrm{B}}}$ $\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}}=4 \Rightarrow \lambda_{\mathrm{B}}=4 \lambda_{\mathrm{A}}$ $\because \quad \lambda_{\mathrm{B}}-\lambda_{\mathrm{A}}=3 \mu \mathrm{m}$ $\therefore \quad 4 \lambda_{\mathrm{A}}-\lambda_{\mathrm{A}}=3 \mu \mathrm{m}$ $\lambda_{\mathrm{A}}=1 \mu \mathrm{m}$
AP EAMCET(Medical)-2004
Heat Transfer
149582
A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} \mathrm{~K}$. The wavelength in nanometer at which its radiation becomes maximum is: $(b=0.0029 \mathrm{mK})$
1 48
2 58
3 60
4 70
Explanation:
B Given, Surface Temperature of particular star $=5 \times 10^{4} \mathrm{~K}$ Wein's constant $b=0.0029 \mathrm{mK}$ We know that, Wein's displacement law $\lambda \mathrm{T}=$ constant $\lambda=\frac{\mathrm{b}}{\mathrm{T}}$ $\lambda=\frac{0.0029}{5 \times 10^{4}}$ $\lambda=5.8 \times 10^{-8} \mathrm{~m}$ $\lambda=58 \times 10^{-9} \mathrm{~m}$ $\lambda=58 \mathrm{~nm}$
AP EAMCET(Medical)-2003
Heat Transfer
149583
The rate of emission of radiation of black body at temperature $27^{\circ} \mathrm{C}$ is $\mathrm{E}_{1}$. If its temperature is increased to $327^{\circ} \mathrm{C}$ the rate of emission of radiation is $E_{2}$ the relation between $E_{1}$ and $E_{2}$ is:
1 $\mathrm{E}_{2}=24 \mathrm{E}_{1}$
2 $\mathrm{E}_{2}=16 \mathrm{E}_{1}$
3 $\mathrm{E}_{2}=8 \mathrm{E}_{1}$
4 $\mathrm{E}_{2}=4 \mathrm{E}_{1}$
Explanation:
B Given, $\mathrm{E}_{1}=27+273=300 \mathrm{~K}$ $\mathrm{E}_{2}=327+273=600 \mathrm{~K}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{300}{600}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{1}{16}$ $\mathrm{E}_{2}=16 \mathrm{E}_{1}$
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 $=\mathrm{T}_{1}$ Increased temperature of black body $=\frac{3}{2} \mathrm{~T}_{1}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4} \mathrm{~A}$ $\mathrm{E}_{1}=\sigma \mathrm{T}_{1}^{4} \mathrm{~A}, \quad \mathrm{E}_{2}=\sigma \mathrm{T}_{2}^{4} \mathrm{~A}$ Percentage increased in radiation $=\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{E}_{1}} \times 100$ $=\frac{\mathrm{T}_{2}^{4}-\mathrm{T}_{1}^{4}}{\mathrm{~T}_{1}^{4}} \times 100$ $=\frac{\left(\frac{3}{2} \mathrm{~T}_{1}\right)^{4}-\mathrm{T}_{1}^{4}}{\mathrm{~T}_{1}^{4}} \times 100$ $=\frac{\left(\frac{3}{2}\right)^{4}-1}{1} \times 100$ $=\frac{\frac{81}{16}-1}{1} \times 100$ $=\frac{65}{16} \times 100$ $=406.25$ $\%$ Increased in radiation $\simeq 400 \%$
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 $0.26 \mu \mathrm{m}$ to $0.13 \mu \mathrm{m}$. 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, $\lambda_{1}=0.26 \mu \mathrm{m}$ $\lambda_{2}=0.13 \mu \mathrm{m}$ We know that, Weins displacement law $\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{0.26}{0.13}=2$ $\mathrm{~T}_{2}=2 \mathrm{~T}_{1}$ Stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\left(2 \mathrm{~T}_{1}\right)^{4}}=\frac{1}{2^{4}}=\frac{1}{16}$
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 $3 \mu$. 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 $\mathrm{A}=4 \times$ temperature of a body $\mathrm{B}$ $\therefore \quad \mathrm{T}_{\mathrm{A}}=4 \mathrm{~T}_{\mathrm{B}}$ Wavelength of body B - Wavelength of body $A=3 \mu$ We know that, Wein's displacement Law $\lambda \mathrm{T}=$ constant $\lambda_{\mathrm{A}} \mathrm{T}_{\mathrm{A}}=\lambda_{\mathrm{B}} \mathrm{T}_{\mathrm{B}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{4 \mathrm{~T}_{\mathrm{B}}}$ $\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}}=4 \Rightarrow \lambda_{\mathrm{B}}=4 \lambda_{\mathrm{A}}$ $\because \quad \lambda_{\mathrm{B}}-\lambda_{\mathrm{A}}=3 \mu \mathrm{m}$ $\therefore \quad 4 \lambda_{\mathrm{A}}-\lambda_{\mathrm{A}}=3 \mu \mathrm{m}$ $\lambda_{\mathrm{A}}=1 \mu \mathrm{m}$
AP EAMCET(Medical)-2004
Heat Transfer
149582
A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} \mathrm{~K}$. The wavelength in nanometer at which its radiation becomes maximum is: $(b=0.0029 \mathrm{mK})$
1 48
2 58
3 60
4 70
Explanation:
B Given, Surface Temperature of particular star $=5 \times 10^{4} \mathrm{~K}$ Wein's constant $b=0.0029 \mathrm{mK}$ We know that, Wein's displacement law $\lambda \mathrm{T}=$ constant $\lambda=\frac{\mathrm{b}}{\mathrm{T}}$ $\lambda=\frac{0.0029}{5 \times 10^{4}}$ $\lambda=5.8 \times 10^{-8} \mathrm{~m}$ $\lambda=58 \times 10^{-9} \mathrm{~m}$ $\lambda=58 \mathrm{~nm}$
AP EAMCET(Medical)-2003
Heat Transfer
149583
The rate of emission of radiation of black body at temperature $27^{\circ} \mathrm{C}$ is $\mathrm{E}_{1}$. If its temperature is increased to $327^{\circ} \mathrm{C}$ the rate of emission of radiation is $E_{2}$ the relation between $E_{1}$ and $E_{2}$ is:
1 $\mathrm{E}_{2}=24 \mathrm{E}_{1}$
2 $\mathrm{E}_{2}=16 \mathrm{E}_{1}$
3 $\mathrm{E}_{2}=8 \mathrm{E}_{1}$
4 $\mathrm{E}_{2}=4 \mathrm{E}_{1}$
Explanation:
B Given, $\mathrm{E}_{1}=27+273=300 \mathrm{~K}$ $\mathrm{E}_{2}=327+273=600 \mathrm{~K}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{300}{600}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{1}{16}$ $\mathrm{E}_{2}=16 \mathrm{E}_{1}$
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 $=\mathrm{T}_{1}$ Increased temperature of black body $=\frac{3}{2} \mathrm{~T}_{1}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4} \mathrm{~A}$ $\mathrm{E}_{1}=\sigma \mathrm{T}_{1}^{4} \mathrm{~A}, \quad \mathrm{E}_{2}=\sigma \mathrm{T}_{2}^{4} \mathrm{~A}$ Percentage increased in radiation $=\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{E}_{1}} \times 100$ $=\frac{\mathrm{T}_{2}^{4}-\mathrm{T}_{1}^{4}}{\mathrm{~T}_{1}^{4}} \times 100$ $=\frac{\left(\frac{3}{2} \mathrm{~T}_{1}\right)^{4}-\mathrm{T}_{1}^{4}}{\mathrm{~T}_{1}^{4}} \times 100$ $=\frac{\left(\frac{3}{2}\right)^{4}-1}{1} \times 100$ $=\frac{\frac{81}{16}-1}{1} \times 100$ $=\frac{65}{16} \times 100$ $=406.25$ $\%$ Increased in radiation $\simeq 400 \%$
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 $0.26 \mu \mathrm{m}$ to $0.13 \mu \mathrm{m}$. 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, $\lambda_{1}=0.26 \mu \mathrm{m}$ $\lambda_{2}=0.13 \mu \mathrm{m}$ We know that, Weins displacement law $\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{0.26}{0.13}=2$ $\mathrm{~T}_{2}=2 \mathrm{~T}_{1}$ Stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\left(2 \mathrm{~T}_{1}\right)^{4}}=\frac{1}{2^{4}}=\frac{1}{16}$
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 $3 \mu$. 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 $\mathrm{A}=4 \times$ temperature of a body $\mathrm{B}$ $\therefore \quad \mathrm{T}_{\mathrm{A}}=4 \mathrm{~T}_{\mathrm{B}}$ Wavelength of body B - Wavelength of body $A=3 \mu$ We know that, Wein's displacement Law $\lambda \mathrm{T}=$ constant $\lambda_{\mathrm{A}} \mathrm{T}_{\mathrm{A}}=\lambda_{\mathrm{B}} \mathrm{T}_{\mathrm{B}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{4 \mathrm{~T}_{\mathrm{B}}}$ $\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}}=4 \Rightarrow \lambda_{\mathrm{B}}=4 \lambda_{\mathrm{A}}$ $\because \quad \lambda_{\mathrm{B}}-\lambda_{\mathrm{A}}=3 \mu \mathrm{m}$ $\therefore \quad 4 \lambda_{\mathrm{A}}-\lambda_{\mathrm{A}}=3 \mu \mathrm{m}$ $\lambda_{\mathrm{A}}=1 \mu \mathrm{m}$
AP EAMCET(Medical)-2004
Heat Transfer
149582
A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} \mathrm{~K}$. The wavelength in nanometer at which its radiation becomes maximum is: $(b=0.0029 \mathrm{mK})$
1 48
2 58
3 60
4 70
Explanation:
B Given, Surface Temperature of particular star $=5 \times 10^{4} \mathrm{~K}$ Wein's constant $b=0.0029 \mathrm{mK}$ We know that, Wein's displacement law $\lambda \mathrm{T}=$ constant $\lambda=\frac{\mathrm{b}}{\mathrm{T}}$ $\lambda=\frac{0.0029}{5 \times 10^{4}}$ $\lambda=5.8 \times 10^{-8} \mathrm{~m}$ $\lambda=58 \times 10^{-9} \mathrm{~m}$ $\lambda=58 \mathrm{~nm}$
AP EAMCET(Medical)-2003
Heat Transfer
149583
The rate of emission of radiation of black body at temperature $27^{\circ} \mathrm{C}$ is $\mathrm{E}_{1}$. If its temperature is increased to $327^{\circ} \mathrm{C}$ the rate of emission of radiation is $E_{2}$ the relation between $E_{1}$ and $E_{2}$ is:
1 $\mathrm{E}_{2}=24 \mathrm{E}_{1}$
2 $\mathrm{E}_{2}=16 \mathrm{E}_{1}$
3 $\mathrm{E}_{2}=8 \mathrm{E}_{1}$
4 $\mathrm{E}_{2}=4 \mathrm{E}_{1}$
Explanation:
B Given, $\mathrm{E}_{1}=27+273=300 \mathrm{~K}$ $\mathrm{E}_{2}=327+273=600 \mathrm{~K}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{300}{600}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{1}{16}$ $\mathrm{E}_{2}=16 \mathrm{E}_{1}$
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 $=\mathrm{T}_{1}$ Increased temperature of black body $=\frac{3}{2} \mathrm{~T}_{1}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4} \mathrm{~A}$ $\mathrm{E}_{1}=\sigma \mathrm{T}_{1}^{4} \mathrm{~A}, \quad \mathrm{E}_{2}=\sigma \mathrm{T}_{2}^{4} \mathrm{~A}$ Percentage increased in radiation $=\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{E}_{1}} \times 100$ $=\frac{\mathrm{T}_{2}^{4}-\mathrm{T}_{1}^{4}}{\mathrm{~T}_{1}^{4}} \times 100$ $=\frac{\left(\frac{3}{2} \mathrm{~T}_{1}\right)^{4}-\mathrm{T}_{1}^{4}}{\mathrm{~T}_{1}^{4}} \times 100$ $=\frac{\left(\frac{3}{2}\right)^{4}-1}{1} \times 100$ $=\frac{\frac{81}{16}-1}{1} \times 100$ $=\frac{65}{16} \times 100$ $=406.25$ $\%$ Increased in radiation $\simeq 400 \%$
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 $0.26 \mu \mathrm{m}$ to $0.13 \mu \mathrm{m}$. 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, $\lambda_{1}=0.26 \mu \mathrm{m}$ $\lambda_{2}=0.13 \mu \mathrm{m}$ We know that, Weins displacement law $\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{0.26}{0.13}=2$ $\mathrm{~T}_{2}=2 \mathrm{~T}_{1}$ Stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\left(2 \mathrm{~T}_{1}\right)^{4}}=\frac{1}{2^{4}}=\frac{1}{16}$
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 $3 \mu$. 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 $\mathrm{A}=4 \times$ temperature of a body $\mathrm{B}$ $\therefore \quad \mathrm{T}_{\mathrm{A}}=4 \mathrm{~T}_{\mathrm{B}}$ Wavelength of body B - Wavelength of body $A=3 \mu$ We know that, Wein's displacement Law $\lambda \mathrm{T}=$ constant $\lambda_{\mathrm{A}} \mathrm{T}_{\mathrm{A}}=\lambda_{\mathrm{B}} \mathrm{T}_{\mathrm{B}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{4 \mathrm{~T}_{\mathrm{B}}}$ $\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}}=4 \Rightarrow \lambda_{\mathrm{B}}=4 \lambda_{\mathrm{A}}$ $\because \quad \lambda_{\mathrm{B}}-\lambda_{\mathrm{A}}=3 \mu \mathrm{m}$ $\therefore \quad 4 \lambda_{\mathrm{A}}-\lambda_{\mathrm{A}}=3 \mu \mathrm{m}$ $\lambda_{\mathrm{A}}=1 \mu \mathrm{m}$
AP EAMCET(Medical)-2004
Heat Transfer
149582
A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} \mathrm{~K}$. The wavelength in nanometer at which its radiation becomes maximum is: $(b=0.0029 \mathrm{mK})$
1 48
2 58
3 60
4 70
Explanation:
B Given, Surface Temperature of particular star $=5 \times 10^{4} \mathrm{~K}$ Wein's constant $b=0.0029 \mathrm{mK}$ We know that, Wein's displacement law $\lambda \mathrm{T}=$ constant $\lambda=\frac{\mathrm{b}}{\mathrm{T}}$ $\lambda=\frac{0.0029}{5 \times 10^{4}}$ $\lambda=5.8 \times 10^{-8} \mathrm{~m}$ $\lambda=58 \times 10^{-9} \mathrm{~m}$ $\lambda=58 \mathrm{~nm}$
AP EAMCET(Medical)-2003
Heat Transfer
149583
The rate of emission of radiation of black body at temperature $27^{\circ} \mathrm{C}$ is $\mathrm{E}_{1}$. If its temperature is increased to $327^{\circ} \mathrm{C}$ the rate of emission of radiation is $E_{2}$ the relation between $E_{1}$ and $E_{2}$ is:
1 $\mathrm{E}_{2}=24 \mathrm{E}_{1}$
2 $\mathrm{E}_{2}=16 \mathrm{E}_{1}$
3 $\mathrm{E}_{2}=8 \mathrm{E}_{1}$
4 $\mathrm{E}_{2}=4 \mathrm{E}_{1}$
Explanation:
B Given, $\mathrm{E}_{1}=27+273=300 \mathrm{~K}$ $\mathrm{E}_{2}=327+273=600 \mathrm{~K}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{300}{600}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{1}{16}$ $\mathrm{E}_{2}=16 \mathrm{E}_{1}$
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 $=\mathrm{T}_{1}$ Increased temperature of black body $=\frac{3}{2} \mathrm{~T}_{1}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4} \mathrm{~A}$ $\mathrm{E}_{1}=\sigma \mathrm{T}_{1}^{4} \mathrm{~A}, \quad \mathrm{E}_{2}=\sigma \mathrm{T}_{2}^{4} \mathrm{~A}$ Percentage increased in radiation $=\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{E}_{1}} \times 100$ $=\frac{\mathrm{T}_{2}^{4}-\mathrm{T}_{1}^{4}}{\mathrm{~T}_{1}^{4}} \times 100$ $=\frac{\left(\frac{3}{2} \mathrm{~T}_{1}\right)^{4}-\mathrm{T}_{1}^{4}}{\mathrm{~T}_{1}^{4}} \times 100$ $=\frac{\left(\frac{3}{2}\right)^{4}-1}{1} \times 100$ $=\frac{\frac{81}{16}-1}{1} \times 100$ $=\frac{65}{16} \times 100$ $=406.25$ $\%$ Increased in radiation $\simeq 400 \%$
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 $0.26 \mu \mathrm{m}$ to $0.13 \mu \mathrm{m}$. 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, $\lambda_{1}=0.26 \mu \mathrm{m}$ $\lambda_{2}=0.13 \mu \mathrm{m}$ We know that, Weins displacement law $\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{0.26}{0.13}=2$ $\mathrm{~T}_{2}=2 \mathrm{~T}_{1}$ Stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\left(2 \mathrm{~T}_{1}\right)^{4}}=\frac{1}{2^{4}}=\frac{1}{16}$
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 $3 \mu$. 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 $\mathrm{A}=4 \times$ temperature of a body $\mathrm{B}$ $\therefore \quad \mathrm{T}_{\mathrm{A}}=4 \mathrm{~T}_{\mathrm{B}}$ Wavelength of body B - Wavelength of body $A=3 \mu$ We know that, Wein's displacement Law $\lambda \mathrm{T}=$ constant $\lambda_{\mathrm{A}} \mathrm{T}_{\mathrm{A}}=\lambda_{\mathrm{B}} \mathrm{T}_{\mathrm{B}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}$ $\frac{\lambda_{\mathrm{A}}}{\lambda_{\mathrm{B}}}=\frac{\mathrm{T}_{\mathrm{B}}}{4 \mathrm{~T}_{\mathrm{B}}}$ $\frac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{A}}}=4 \Rightarrow \lambda_{\mathrm{B}}=4 \lambda_{\mathrm{A}}$ $\because \quad \lambda_{\mathrm{B}}-\lambda_{\mathrm{A}}=3 \mu \mathrm{m}$ $\therefore \quad 4 \lambda_{\mathrm{A}}-\lambda_{\mathrm{A}}=3 \mu \mathrm{m}$ $\lambda_{\mathrm{A}}=1 \mu \mathrm{m}$
AP EAMCET(Medical)-2004
Heat Transfer
149582
A particular star (assuming it as a black body) has a surface temperature of about $5 \times 10^{4} \mathrm{~K}$. The wavelength in nanometer at which its radiation becomes maximum is: $(b=0.0029 \mathrm{mK})$
1 48
2 58
3 60
4 70
Explanation:
B Given, Surface Temperature of particular star $=5 \times 10^{4} \mathrm{~K}$ Wein's constant $b=0.0029 \mathrm{mK}$ We know that, Wein's displacement law $\lambda \mathrm{T}=$ constant $\lambda=\frac{\mathrm{b}}{\mathrm{T}}$ $\lambda=\frac{0.0029}{5 \times 10^{4}}$ $\lambda=5.8 \times 10^{-8} \mathrm{~m}$ $\lambda=58 \times 10^{-9} \mathrm{~m}$ $\lambda=58 \mathrm{~nm}$
AP EAMCET(Medical)-2003
Heat Transfer
149583
The rate of emission of radiation of black body at temperature $27^{\circ} \mathrm{C}$ is $\mathrm{E}_{1}$. If its temperature is increased to $327^{\circ} \mathrm{C}$ the rate of emission of radiation is $E_{2}$ the relation between $E_{1}$ and $E_{2}$ is:
1 $\mathrm{E}_{2}=24 \mathrm{E}_{1}$
2 $\mathrm{E}_{2}=16 \mathrm{E}_{1}$
3 $\mathrm{E}_{2}=8 \mathrm{E}_{1}$
4 $\mathrm{E}_{2}=4 \mathrm{E}_{1}$
Explanation:
B Given, $\mathrm{E}_{1}=27+273=300 \mathrm{~K}$ $\mathrm{E}_{2}=327+273=600 \mathrm{~K}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{300}{600}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{1}{16}$ $\mathrm{E}_{2}=16 \mathrm{E}_{1}$
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 $=\mathrm{T}_{1}$ Increased temperature of black body $=\frac{3}{2} \mathrm{~T}_{1}$ We know that, $\mathrm{E}=\sigma \mathrm{T}^{4} \mathrm{~A}$ $\mathrm{E}_{1}=\sigma \mathrm{T}_{1}^{4} \mathrm{~A}, \quad \mathrm{E}_{2}=\sigma \mathrm{T}_{2}^{4} \mathrm{~A}$ Percentage increased in radiation $=\frac{\mathrm{E}_{2}-\mathrm{E}_{1}}{\mathrm{E}_{1}} \times 100$ $=\frac{\mathrm{T}_{2}^{4}-\mathrm{T}_{1}^{4}}{\mathrm{~T}_{1}^{4}} \times 100$ $=\frac{\left(\frac{3}{2} \mathrm{~T}_{1}\right)^{4}-\mathrm{T}_{1}^{4}}{\mathrm{~T}_{1}^{4}} \times 100$ $=\frac{\left(\frac{3}{2}\right)^{4}-1}{1} \times 100$ $=\frac{\frac{81}{16}-1}{1} \times 100$ $=\frac{65}{16} \times 100$ $=406.25$ $\%$ Increased in radiation $\simeq 400 \%$
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 $0.26 \mu \mathrm{m}$ to $0.13 \mu \mathrm{m}$. 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, $\lambda_{1}=0.26 \mu \mathrm{m}$ $\lambda_{2}=0.13 \mu \mathrm{m}$ We know that, Weins displacement law $\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}=\frac{0.26}{0.13}=2$ $\mathrm{~T}_{2}=2 \mathrm{~T}_{1}$ Stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\left(2 \mathrm{~T}_{1}\right)^{4}}=\frac{1}{2^{4}}=\frac{1}{16}$
