149520
Temperatures of two stars are in the ratio 3:2. If wavelength of maximum intensity of first star is $4500 \AA$, the corresponding wavelength for second star is
1 $2250 \AA$
2 $3000 \AA$
3 $9000 \AA$
4 $6750 \AA$
Explanation:
D Given that, $\mathrm{T}_{1}: \mathrm{T}_{2}=3: 2$ $\lambda_{1}=4500 \AA, \lambda_{2}=$ ? According to Wein's displacement law, $\lambda \mathrm{T}=$ constant $\lambda_{1} T_{1}=\lambda_{2} T_{2}$ $\lambda_{2}=\lambda_{1} \times \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\lambda_{2}=\lambda_{1} \times \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\therefore \quad 4500 \times \frac{3}{2}$ $6750 \AA$
MHT-CET 2020
Heat Transfer
149521
An iron nail changes its colour from red to orange red and then to bluish white, when heated strongly in flames. This change of colour can be explained on the basis of
1 Newton's law of cooling
2 Kirchoff's law
3 Wein's displacement law
4 Stefin's law of radiation
Explanation:
C According to Wein's displacement law, $\lambda_{\max } \mathrm{T}=\mathrm{b}$ Where, $\lambda_{\text {max }}=$ wavelength $\mathrm{T}=\text { Temperature }$ $\lambda_{\max } \mathrm{T}=\mathrm{b}$ $\lambda=\frac{\mathrm{b}}{\mathrm{T}}$ $\lambda \propto \frac{1}{\mathrm{~T}} \quad \text { (Let, } \mathrm{b}=\text { constant) }$ It states that wavelength changes with temperature. That's why change of colour can be explained by Wein's displacement law.
MHT-CET 2019
Heat Transfer
149522
The original temperature of a black body is $727^{\circ} \mathrm{C}$. The temperature to which the black body must be raised so as to double the total radiant energy is
149523
The maximum wavelength of radiation emitted by a star is $289.8 \mathrm{~nm}$. Then intensity of radiation for the star is (Given : Stefan's constant $=5.67 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-4}$, Wien's constant, $b=\mathbf{2 8 9 8} \mu \mathrm{mK}$ )
1 $5.67 \times 10^{-12} \mathrm{Wm}^{-2}$
2 $5.67 \times 10^{8} \mathrm{Wm}^{-2}$
3 $10.67 \times 10^{14} \mathrm{Wm}^{-2}$
4 $10.67 \times 10^{7} \mathrm{Wm}^{-2}$
Explanation:
B Given that, $\lambda=289.8 \mathrm{~nm}$ Wein's constant $b=2898 \mu \mathrm{mK}$ Applying Wein's displacement law, $\lambda \times \mathrm{T}=\mathrm{b}$ $\mathrm{T}=\frac{\mathrm{b}}{\lambda}$ $\mathrm{T}=\frac{2898 \mu \mathrm{mk}}{289.8 \mathrm{~nm}}$ $\mathrm{~T}=10^{4} \mathrm{~K}$ According to Stefan's Boltzmann law Radiation intensity $\mathrm{R}=\sigma \mathrm{T}^{4}$ $=5.67 \times 10^{-8} \times\left(10^{4}\right)^{4}$ $=5.67 \times 10^{8} \mathrm{Wm}^{-2}$
149520
Temperatures of two stars are in the ratio 3:2. If wavelength of maximum intensity of first star is $4500 \AA$, the corresponding wavelength for second star is
1 $2250 \AA$
2 $3000 \AA$
3 $9000 \AA$
4 $6750 \AA$
Explanation:
D Given that, $\mathrm{T}_{1}: \mathrm{T}_{2}=3: 2$ $\lambda_{1}=4500 \AA, \lambda_{2}=$ ? According to Wein's displacement law, $\lambda \mathrm{T}=$ constant $\lambda_{1} T_{1}=\lambda_{2} T_{2}$ $\lambda_{2}=\lambda_{1} \times \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\lambda_{2}=\lambda_{1} \times \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\therefore \quad 4500 \times \frac{3}{2}$ $6750 \AA$
MHT-CET 2020
Heat Transfer
149521
An iron nail changes its colour from red to orange red and then to bluish white, when heated strongly in flames. This change of colour can be explained on the basis of
1 Newton's law of cooling
2 Kirchoff's law
3 Wein's displacement law
4 Stefin's law of radiation
Explanation:
C According to Wein's displacement law, $\lambda_{\max } \mathrm{T}=\mathrm{b}$ Where, $\lambda_{\text {max }}=$ wavelength $\mathrm{T}=\text { Temperature }$ $\lambda_{\max } \mathrm{T}=\mathrm{b}$ $\lambda=\frac{\mathrm{b}}{\mathrm{T}}$ $\lambda \propto \frac{1}{\mathrm{~T}} \quad \text { (Let, } \mathrm{b}=\text { constant) }$ It states that wavelength changes with temperature. That's why change of colour can be explained by Wein's displacement law.
MHT-CET 2019
Heat Transfer
149522
The original temperature of a black body is $727^{\circ} \mathrm{C}$. The temperature to which the black body must be raised so as to double the total radiant energy is
149523
The maximum wavelength of radiation emitted by a star is $289.8 \mathrm{~nm}$. Then intensity of radiation for the star is (Given : Stefan's constant $=5.67 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-4}$, Wien's constant, $b=\mathbf{2 8 9 8} \mu \mathrm{mK}$ )
1 $5.67 \times 10^{-12} \mathrm{Wm}^{-2}$
2 $5.67 \times 10^{8} \mathrm{Wm}^{-2}$
3 $10.67 \times 10^{14} \mathrm{Wm}^{-2}$
4 $10.67 \times 10^{7} \mathrm{Wm}^{-2}$
Explanation:
B Given that, $\lambda=289.8 \mathrm{~nm}$ Wein's constant $b=2898 \mu \mathrm{mK}$ Applying Wein's displacement law, $\lambda \times \mathrm{T}=\mathrm{b}$ $\mathrm{T}=\frac{\mathrm{b}}{\lambda}$ $\mathrm{T}=\frac{2898 \mu \mathrm{mk}}{289.8 \mathrm{~nm}}$ $\mathrm{~T}=10^{4} \mathrm{~K}$ According to Stefan's Boltzmann law Radiation intensity $\mathrm{R}=\sigma \mathrm{T}^{4}$ $=5.67 \times 10^{-8} \times\left(10^{4}\right)^{4}$ $=5.67 \times 10^{8} \mathrm{Wm}^{-2}$
149520
Temperatures of two stars are in the ratio 3:2. If wavelength of maximum intensity of first star is $4500 \AA$, the corresponding wavelength for second star is
1 $2250 \AA$
2 $3000 \AA$
3 $9000 \AA$
4 $6750 \AA$
Explanation:
D Given that, $\mathrm{T}_{1}: \mathrm{T}_{2}=3: 2$ $\lambda_{1}=4500 \AA, \lambda_{2}=$ ? According to Wein's displacement law, $\lambda \mathrm{T}=$ constant $\lambda_{1} T_{1}=\lambda_{2} T_{2}$ $\lambda_{2}=\lambda_{1} \times \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\lambda_{2}=\lambda_{1} \times \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\therefore \quad 4500 \times \frac{3}{2}$ $6750 \AA$
MHT-CET 2020
Heat Transfer
149521
An iron nail changes its colour from red to orange red and then to bluish white, when heated strongly in flames. This change of colour can be explained on the basis of
1 Newton's law of cooling
2 Kirchoff's law
3 Wein's displacement law
4 Stefin's law of radiation
Explanation:
C According to Wein's displacement law, $\lambda_{\max } \mathrm{T}=\mathrm{b}$ Where, $\lambda_{\text {max }}=$ wavelength $\mathrm{T}=\text { Temperature }$ $\lambda_{\max } \mathrm{T}=\mathrm{b}$ $\lambda=\frac{\mathrm{b}}{\mathrm{T}}$ $\lambda \propto \frac{1}{\mathrm{~T}} \quad \text { (Let, } \mathrm{b}=\text { constant) }$ It states that wavelength changes with temperature. That's why change of colour can be explained by Wein's displacement law.
MHT-CET 2019
Heat Transfer
149522
The original temperature of a black body is $727^{\circ} \mathrm{C}$. The temperature to which the black body must be raised so as to double the total radiant energy is
149523
The maximum wavelength of radiation emitted by a star is $289.8 \mathrm{~nm}$. Then intensity of radiation for the star is (Given : Stefan's constant $=5.67 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-4}$, Wien's constant, $b=\mathbf{2 8 9 8} \mu \mathrm{mK}$ )
1 $5.67 \times 10^{-12} \mathrm{Wm}^{-2}$
2 $5.67 \times 10^{8} \mathrm{Wm}^{-2}$
3 $10.67 \times 10^{14} \mathrm{Wm}^{-2}$
4 $10.67 \times 10^{7} \mathrm{Wm}^{-2}$
Explanation:
B Given that, $\lambda=289.8 \mathrm{~nm}$ Wein's constant $b=2898 \mu \mathrm{mK}$ Applying Wein's displacement law, $\lambda \times \mathrm{T}=\mathrm{b}$ $\mathrm{T}=\frac{\mathrm{b}}{\lambda}$ $\mathrm{T}=\frac{2898 \mu \mathrm{mk}}{289.8 \mathrm{~nm}}$ $\mathrm{~T}=10^{4} \mathrm{~K}$ According to Stefan's Boltzmann law Radiation intensity $\mathrm{R}=\sigma \mathrm{T}^{4}$ $=5.67 \times 10^{-8} \times\left(10^{4}\right)^{4}$ $=5.67 \times 10^{8} \mathrm{Wm}^{-2}$
149520
Temperatures of two stars are in the ratio 3:2. If wavelength of maximum intensity of first star is $4500 \AA$, the corresponding wavelength for second star is
1 $2250 \AA$
2 $3000 \AA$
3 $9000 \AA$
4 $6750 \AA$
Explanation:
D Given that, $\mathrm{T}_{1}: \mathrm{T}_{2}=3: 2$ $\lambda_{1}=4500 \AA, \lambda_{2}=$ ? According to Wein's displacement law, $\lambda \mathrm{T}=$ constant $\lambda_{1} T_{1}=\lambda_{2} T_{2}$ $\lambda_{2}=\lambda_{1} \times \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\lambda_{2}=\lambda_{1} \times \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}$ $\therefore \quad 4500 \times \frac{3}{2}$ $6750 \AA$
MHT-CET 2020
Heat Transfer
149521
An iron nail changes its colour from red to orange red and then to bluish white, when heated strongly in flames. This change of colour can be explained on the basis of
1 Newton's law of cooling
2 Kirchoff's law
3 Wein's displacement law
4 Stefin's law of radiation
Explanation:
C According to Wein's displacement law, $\lambda_{\max } \mathrm{T}=\mathrm{b}$ Where, $\lambda_{\text {max }}=$ wavelength $\mathrm{T}=\text { Temperature }$ $\lambda_{\max } \mathrm{T}=\mathrm{b}$ $\lambda=\frac{\mathrm{b}}{\mathrm{T}}$ $\lambda \propto \frac{1}{\mathrm{~T}} \quad \text { (Let, } \mathrm{b}=\text { constant) }$ It states that wavelength changes with temperature. That's why change of colour can be explained by Wein's displacement law.
MHT-CET 2019
Heat Transfer
149522
The original temperature of a black body is $727^{\circ} \mathrm{C}$. The temperature to which the black body must be raised so as to double the total radiant energy is
149523
The maximum wavelength of radiation emitted by a star is $289.8 \mathrm{~nm}$. Then intensity of radiation for the star is (Given : Stefan's constant $=5.67 \times 10^{-8} \mathrm{Wm}^{-2} \mathrm{~K}^{-4}$, Wien's constant, $b=\mathbf{2 8 9 8} \mu \mathrm{mK}$ )
1 $5.67 \times 10^{-12} \mathrm{Wm}^{-2}$
2 $5.67 \times 10^{8} \mathrm{Wm}^{-2}$
3 $10.67 \times 10^{14} \mathrm{Wm}^{-2}$
4 $10.67 \times 10^{7} \mathrm{Wm}^{-2}$
Explanation:
B Given that, $\lambda=289.8 \mathrm{~nm}$ Wein's constant $b=2898 \mu \mathrm{mK}$ Applying Wein's displacement law, $\lambda \times \mathrm{T}=\mathrm{b}$ $\mathrm{T}=\frac{\mathrm{b}}{\lambda}$ $\mathrm{T}=\frac{2898 \mu \mathrm{mk}}{289.8 \mathrm{~nm}}$ $\mathrm{~T}=10^{4} \mathrm{~K}$ According to Stefan's Boltzmann law Radiation intensity $\mathrm{R}=\sigma \mathrm{T}^{4}$ $=5.67 \times 10^{-8} \times\left(10^{4}\right)^{4}$ $=5.67 \times 10^{8} \mathrm{Wm}^{-2}$
