149442
Temperature of two stars are in ratio $3: 2$. If wavelength of maximum intensity of first body is $4000 \AA$, what is corresponding wavelength of second body?
1 $9000 \AA$
2 $6000 \AA$
3 $2000 \AA$
4 $8000 \AA$
Explanation:
B Given, $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{3}{2}$ $\left(\lambda_{\mathrm{m}}\right)_{1}=4000 \AA$ According to Wien's displacement law $\lambda_{\mathrm{m}} \mathrm{T}=\text { constant }$ $\frac{\left(\lambda_{\mathrm{m}}\right)_{1}}{\left(\lambda_{\mathrm{m}}\right)_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{4000}{\left(\lambda_{\mathrm{m}}\right)_{2}}=\frac{2}{3}$ $\left(\lambda_{\mathrm{m}}\right)_{2}=6000 \AA$
UPSEE - 2010
Heat Transfer
149444
If the emission rate of blackbody at $0^{\circ} \mathrm{C}$ is $R$, then the rate of emission at $273^{\circ} \mathrm{c}$ is
1 $2 \mathrm{R}$
2 $4 \mathrm{R}$
3 $8 \mathrm{R}$
4 $16 \mathrm{R}$
5 $32 \mathrm{R}$
Explanation:
D Given, $\mathrm{T}_{1}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ $\mathrm{T}_{2}=273^{\circ} \mathrm{C}+273=546 \mathrm{~K}$ According to Stefan's law, Emission rate of a ideal blackbody, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} \propto\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{273}{546}\right)^{4}$ $\frac{\mathrm{R}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{R}$
Kerala CEE -2018
Heat Transfer
149445
Two perfectly black spheres $A$ and $B$ having radii $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ are maintained at temperatures $127^{\circ} \mathrm{C}$ and $527^{\circ} \mathrm{C}$, respectively. The ratio of the energy radiated by $A$ to that by $B$ is
1 $1: 2$
2 $1: 1$
3 $2: 1$
4 $1: 4$
5 $1: 16$
Explanation:
B We know, radiation of energy $\mathrm{E}=\sigma \mathrm{AT}^{4}$ Where $\sigma=$ Stefan's Boltsmann constant $\mathrm{A}=$ surface area $\mathrm{T}=$ Temperature at which radiation occur. Given, $\mathrm{R}_{1}=8 \mathrm{~cm}=8 \times 10^{-2} \mathrm{~m}$ $\mathrm{R}_{2}=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}$ $\mathrm{T}_{1}=127+273=400 \mathrm{~K}$ $\mathrm{T}_{2}=527+273=800$ Then $\mathrm{A}_{1}=4 \pi \mathrm{r}_{1}{ }^{2}=4 \pi\left(8 \times 10^{-2}\right)^{2}$ $\mathrm{A}_{2}=4 \pi \mathrm{r}_{2}^{2}=4 \pi\left(2 \times 10^{-2}\right)^{2}$ $\mathrm{E}_{1}=\sigma \times 4 \pi\left(8 \times 10^{-2}\right)^{2} \times(400)^{4}$ $\mathrm{E}_{2}=\sigma \times 4 \pi\left(2 \times 10^{-2}\right)^{2} \times(800)^{4}$ Dividing eq ${ }^{\mathrm{n}}$ (i) / (ii) $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{64 \times 400^{4}}{4 \times 800^{4}}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=1: 1$
Kerala CEE - 2015
Heat Transfer
149446
The plots of intensity of radiation versus wavelength of three black bodies at temperatures $T_{1}, T_{2}$ and $T_{3}$ are shown. Then,
1 $\mathrm{T}_{3}>\mathrm{T}_{2}>\mathrm{T}_{1}$
2 $\mathrm{T}_{1}>\mathrm{T}_{2}>\mathrm{T}_{3}$
3 $T_{2}>T_{3}>T_{1}$
4 $T_{1}>T_{3}>T_{2}$
5 $\mathrm{T}_{3}>\mathrm{T}_{1}>\mathrm{T}_{2}$
Explanation:
D According to Wien's law $\lambda_{\mathrm{m}} \propto \frac{1}{\mathrm{~T}}$ And from the figure $\left(\lambda_{\mathrm{m}}\right)_{1} \lt \left(\lambda_{\mathrm{m}}\right)_{3} \lt \left(\lambda_{\mathrm{m}}\right)_{2}$ Therefore, $\mathrm{T}_{1}>\mathrm{T}_{3}>\mathrm{T}_{2}$
Kerala CEE - 2008
Heat Transfer
149447
A black body has maximum wavelength $\lambda_{\mathrm{m}}$ at $2000 \mathrm{~K}$. Its corresponding wavelength at $\mathbf{3 0 0 0}$ $\mathrm{K}$ will be:
1 $\frac{3}{2} \lambda_{m}$
2 $\frac{2}{3} \lambda_{m}$
3 $\frac{16}{81} \lambda_{m}$
4 $\frac{81}{16} \lambda_{m}$
5 $\frac{4}{3} \lambda_{\mathrm{m}}$
Explanation:
B Given, $\lambda_{\mathrm{m}_{1}}=\lambda_{\mathrm{m}}$ $\mathrm{T}_{1}=2000 \mathrm{~K}$ $\mathrm{~T}_{2}=3000 \mathrm{~K}$ $\lambda_{\mathrm{m}_{2}}=?$ From Wein's displacement law $\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{C}$ $\lambda_{\mathrm{m}_{2}} \mathrm{~T}_{2}=\lambda_{\mathrm{m}_{1}} \times \mathrm{T}_{1}$ $\lambda_{\mathrm{m}_{2}}=\lambda_{\mathrm{m}_{1}} \times \mathrm{T}_{1} / \mathrm{T}_{2}$ $\lambda_{\mathrm{m}_{2}}=\lambda_{\mathrm{m}} \times 2000 / 3000$ $\lambda_{\mathrm{m}_{2}}=\frac{2}{3} \lambda_{\mathrm{m}}$
149442
Temperature of two stars are in ratio $3: 2$. If wavelength of maximum intensity of first body is $4000 \AA$, what is corresponding wavelength of second body?
1 $9000 \AA$
2 $6000 \AA$
3 $2000 \AA$
4 $8000 \AA$
Explanation:
B Given, $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{3}{2}$ $\left(\lambda_{\mathrm{m}}\right)_{1}=4000 \AA$ According to Wien's displacement law $\lambda_{\mathrm{m}} \mathrm{T}=\text { constant }$ $\frac{\left(\lambda_{\mathrm{m}}\right)_{1}}{\left(\lambda_{\mathrm{m}}\right)_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{4000}{\left(\lambda_{\mathrm{m}}\right)_{2}}=\frac{2}{3}$ $\left(\lambda_{\mathrm{m}}\right)_{2}=6000 \AA$
UPSEE - 2010
Heat Transfer
149444
If the emission rate of blackbody at $0^{\circ} \mathrm{C}$ is $R$, then the rate of emission at $273^{\circ} \mathrm{c}$ is
1 $2 \mathrm{R}$
2 $4 \mathrm{R}$
3 $8 \mathrm{R}$
4 $16 \mathrm{R}$
5 $32 \mathrm{R}$
Explanation:
D Given, $\mathrm{T}_{1}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ $\mathrm{T}_{2}=273^{\circ} \mathrm{C}+273=546 \mathrm{~K}$ According to Stefan's law, Emission rate of a ideal blackbody, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} \propto\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{273}{546}\right)^{4}$ $\frac{\mathrm{R}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{R}$
Kerala CEE -2018
Heat Transfer
149445
Two perfectly black spheres $A$ and $B$ having radii $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ are maintained at temperatures $127^{\circ} \mathrm{C}$ and $527^{\circ} \mathrm{C}$, respectively. The ratio of the energy radiated by $A$ to that by $B$ is
1 $1: 2$
2 $1: 1$
3 $2: 1$
4 $1: 4$
5 $1: 16$
Explanation:
B We know, radiation of energy $\mathrm{E}=\sigma \mathrm{AT}^{4}$ Where $\sigma=$ Stefan's Boltsmann constant $\mathrm{A}=$ surface area $\mathrm{T}=$ Temperature at which radiation occur. Given, $\mathrm{R}_{1}=8 \mathrm{~cm}=8 \times 10^{-2} \mathrm{~m}$ $\mathrm{R}_{2}=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}$ $\mathrm{T}_{1}=127+273=400 \mathrm{~K}$ $\mathrm{T}_{2}=527+273=800$ Then $\mathrm{A}_{1}=4 \pi \mathrm{r}_{1}{ }^{2}=4 \pi\left(8 \times 10^{-2}\right)^{2}$ $\mathrm{A}_{2}=4 \pi \mathrm{r}_{2}^{2}=4 \pi\left(2 \times 10^{-2}\right)^{2}$ $\mathrm{E}_{1}=\sigma \times 4 \pi\left(8 \times 10^{-2}\right)^{2} \times(400)^{4}$ $\mathrm{E}_{2}=\sigma \times 4 \pi\left(2 \times 10^{-2}\right)^{2} \times(800)^{4}$ Dividing eq ${ }^{\mathrm{n}}$ (i) / (ii) $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{64 \times 400^{4}}{4 \times 800^{4}}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=1: 1$
Kerala CEE - 2015
Heat Transfer
149446
The plots of intensity of radiation versus wavelength of three black bodies at temperatures $T_{1}, T_{2}$ and $T_{3}$ are shown. Then,
1 $\mathrm{T}_{3}>\mathrm{T}_{2}>\mathrm{T}_{1}$
2 $\mathrm{T}_{1}>\mathrm{T}_{2}>\mathrm{T}_{3}$
3 $T_{2}>T_{3}>T_{1}$
4 $T_{1}>T_{3}>T_{2}$
5 $\mathrm{T}_{3}>\mathrm{T}_{1}>\mathrm{T}_{2}$
Explanation:
D According to Wien's law $\lambda_{\mathrm{m}} \propto \frac{1}{\mathrm{~T}}$ And from the figure $\left(\lambda_{\mathrm{m}}\right)_{1} \lt \left(\lambda_{\mathrm{m}}\right)_{3} \lt \left(\lambda_{\mathrm{m}}\right)_{2}$ Therefore, $\mathrm{T}_{1}>\mathrm{T}_{3}>\mathrm{T}_{2}$
Kerala CEE - 2008
Heat Transfer
149447
A black body has maximum wavelength $\lambda_{\mathrm{m}}$ at $2000 \mathrm{~K}$. Its corresponding wavelength at $\mathbf{3 0 0 0}$ $\mathrm{K}$ will be:
1 $\frac{3}{2} \lambda_{m}$
2 $\frac{2}{3} \lambda_{m}$
3 $\frac{16}{81} \lambda_{m}$
4 $\frac{81}{16} \lambda_{m}$
5 $\frac{4}{3} \lambda_{\mathrm{m}}$
Explanation:
B Given, $\lambda_{\mathrm{m}_{1}}=\lambda_{\mathrm{m}}$ $\mathrm{T}_{1}=2000 \mathrm{~K}$ $\mathrm{~T}_{2}=3000 \mathrm{~K}$ $\lambda_{\mathrm{m}_{2}}=?$ From Wein's displacement law $\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{C}$ $\lambda_{\mathrm{m}_{2}} \mathrm{~T}_{2}=\lambda_{\mathrm{m}_{1}} \times \mathrm{T}_{1}$ $\lambda_{\mathrm{m}_{2}}=\lambda_{\mathrm{m}_{1}} \times \mathrm{T}_{1} / \mathrm{T}_{2}$ $\lambda_{\mathrm{m}_{2}}=\lambda_{\mathrm{m}} \times 2000 / 3000$ $\lambda_{\mathrm{m}_{2}}=\frac{2}{3} \lambda_{\mathrm{m}}$
149442
Temperature of two stars are in ratio $3: 2$. If wavelength of maximum intensity of first body is $4000 \AA$, what is corresponding wavelength of second body?
1 $9000 \AA$
2 $6000 \AA$
3 $2000 \AA$
4 $8000 \AA$
Explanation:
B Given, $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{3}{2}$ $\left(\lambda_{\mathrm{m}}\right)_{1}=4000 \AA$ According to Wien's displacement law $\lambda_{\mathrm{m}} \mathrm{T}=\text { constant }$ $\frac{\left(\lambda_{\mathrm{m}}\right)_{1}}{\left(\lambda_{\mathrm{m}}\right)_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{4000}{\left(\lambda_{\mathrm{m}}\right)_{2}}=\frac{2}{3}$ $\left(\lambda_{\mathrm{m}}\right)_{2}=6000 \AA$
UPSEE - 2010
Heat Transfer
149444
If the emission rate of blackbody at $0^{\circ} \mathrm{C}$ is $R$, then the rate of emission at $273^{\circ} \mathrm{c}$ is
1 $2 \mathrm{R}$
2 $4 \mathrm{R}$
3 $8 \mathrm{R}$
4 $16 \mathrm{R}$
5 $32 \mathrm{R}$
Explanation:
D Given, $\mathrm{T}_{1}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ $\mathrm{T}_{2}=273^{\circ} \mathrm{C}+273=546 \mathrm{~K}$ According to Stefan's law, Emission rate of a ideal blackbody, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} \propto\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{273}{546}\right)^{4}$ $\frac{\mathrm{R}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{R}$
Kerala CEE -2018
Heat Transfer
149445
Two perfectly black spheres $A$ and $B$ having radii $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ are maintained at temperatures $127^{\circ} \mathrm{C}$ and $527^{\circ} \mathrm{C}$, respectively. The ratio of the energy radiated by $A$ to that by $B$ is
1 $1: 2$
2 $1: 1$
3 $2: 1$
4 $1: 4$
5 $1: 16$
Explanation:
B We know, radiation of energy $\mathrm{E}=\sigma \mathrm{AT}^{4}$ Where $\sigma=$ Stefan's Boltsmann constant $\mathrm{A}=$ surface area $\mathrm{T}=$ Temperature at which radiation occur. Given, $\mathrm{R}_{1}=8 \mathrm{~cm}=8 \times 10^{-2} \mathrm{~m}$ $\mathrm{R}_{2}=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}$ $\mathrm{T}_{1}=127+273=400 \mathrm{~K}$ $\mathrm{T}_{2}=527+273=800$ Then $\mathrm{A}_{1}=4 \pi \mathrm{r}_{1}{ }^{2}=4 \pi\left(8 \times 10^{-2}\right)^{2}$ $\mathrm{A}_{2}=4 \pi \mathrm{r}_{2}^{2}=4 \pi\left(2 \times 10^{-2}\right)^{2}$ $\mathrm{E}_{1}=\sigma \times 4 \pi\left(8 \times 10^{-2}\right)^{2} \times(400)^{4}$ $\mathrm{E}_{2}=\sigma \times 4 \pi\left(2 \times 10^{-2}\right)^{2} \times(800)^{4}$ Dividing eq ${ }^{\mathrm{n}}$ (i) / (ii) $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{64 \times 400^{4}}{4 \times 800^{4}}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=1: 1$
Kerala CEE - 2015
Heat Transfer
149446
The plots of intensity of radiation versus wavelength of three black bodies at temperatures $T_{1}, T_{2}$ and $T_{3}$ are shown. Then,
1 $\mathrm{T}_{3}>\mathrm{T}_{2}>\mathrm{T}_{1}$
2 $\mathrm{T}_{1}>\mathrm{T}_{2}>\mathrm{T}_{3}$
3 $T_{2}>T_{3}>T_{1}$
4 $T_{1}>T_{3}>T_{2}$
5 $\mathrm{T}_{3}>\mathrm{T}_{1}>\mathrm{T}_{2}$
Explanation:
D According to Wien's law $\lambda_{\mathrm{m}} \propto \frac{1}{\mathrm{~T}}$ And from the figure $\left(\lambda_{\mathrm{m}}\right)_{1} \lt \left(\lambda_{\mathrm{m}}\right)_{3} \lt \left(\lambda_{\mathrm{m}}\right)_{2}$ Therefore, $\mathrm{T}_{1}>\mathrm{T}_{3}>\mathrm{T}_{2}$
Kerala CEE - 2008
Heat Transfer
149447
A black body has maximum wavelength $\lambda_{\mathrm{m}}$ at $2000 \mathrm{~K}$. Its corresponding wavelength at $\mathbf{3 0 0 0}$ $\mathrm{K}$ will be:
1 $\frac{3}{2} \lambda_{m}$
2 $\frac{2}{3} \lambda_{m}$
3 $\frac{16}{81} \lambda_{m}$
4 $\frac{81}{16} \lambda_{m}$
5 $\frac{4}{3} \lambda_{\mathrm{m}}$
Explanation:
B Given, $\lambda_{\mathrm{m}_{1}}=\lambda_{\mathrm{m}}$ $\mathrm{T}_{1}=2000 \mathrm{~K}$ $\mathrm{~T}_{2}=3000 \mathrm{~K}$ $\lambda_{\mathrm{m}_{2}}=?$ From Wein's displacement law $\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{C}$ $\lambda_{\mathrm{m}_{2}} \mathrm{~T}_{2}=\lambda_{\mathrm{m}_{1}} \times \mathrm{T}_{1}$ $\lambda_{\mathrm{m}_{2}}=\lambda_{\mathrm{m}_{1}} \times \mathrm{T}_{1} / \mathrm{T}_{2}$ $\lambda_{\mathrm{m}_{2}}=\lambda_{\mathrm{m}} \times 2000 / 3000$ $\lambda_{\mathrm{m}_{2}}=\frac{2}{3} \lambda_{\mathrm{m}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Heat Transfer
149442
Temperature of two stars are in ratio $3: 2$. If wavelength of maximum intensity of first body is $4000 \AA$, what is corresponding wavelength of second body?
1 $9000 \AA$
2 $6000 \AA$
3 $2000 \AA$
4 $8000 \AA$
Explanation:
B Given, $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{3}{2}$ $\left(\lambda_{\mathrm{m}}\right)_{1}=4000 \AA$ According to Wien's displacement law $\lambda_{\mathrm{m}} \mathrm{T}=\text { constant }$ $\frac{\left(\lambda_{\mathrm{m}}\right)_{1}}{\left(\lambda_{\mathrm{m}}\right)_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{4000}{\left(\lambda_{\mathrm{m}}\right)_{2}}=\frac{2}{3}$ $\left(\lambda_{\mathrm{m}}\right)_{2}=6000 \AA$
UPSEE - 2010
Heat Transfer
149444
If the emission rate of blackbody at $0^{\circ} \mathrm{C}$ is $R$, then the rate of emission at $273^{\circ} \mathrm{c}$ is
1 $2 \mathrm{R}$
2 $4 \mathrm{R}$
3 $8 \mathrm{R}$
4 $16 \mathrm{R}$
5 $32 \mathrm{R}$
Explanation:
D Given, $\mathrm{T}_{1}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ $\mathrm{T}_{2}=273^{\circ} \mathrm{C}+273=546 \mathrm{~K}$ According to Stefan's law, Emission rate of a ideal blackbody, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} \propto\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{273}{546}\right)^{4}$ $\frac{\mathrm{R}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{R}$
Kerala CEE -2018
Heat Transfer
149445
Two perfectly black spheres $A$ and $B$ having radii $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ are maintained at temperatures $127^{\circ} \mathrm{C}$ and $527^{\circ} \mathrm{C}$, respectively. The ratio of the energy radiated by $A$ to that by $B$ is
1 $1: 2$
2 $1: 1$
3 $2: 1$
4 $1: 4$
5 $1: 16$
Explanation:
B We know, radiation of energy $\mathrm{E}=\sigma \mathrm{AT}^{4}$ Where $\sigma=$ Stefan's Boltsmann constant $\mathrm{A}=$ surface area $\mathrm{T}=$ Temperature at which radiation occur. Given, $\mathrm{R}_{1}=8 \mathrm{~cm}=8 \times 10^{-2} \mathrm{~m}$ $\mathrm{R}_{2}=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}$ $\mathrm{T}_{1}=127+273=400 \mathrm{~K}$ $\mathrm{T}_{2}=527+273=800$ Then $\mathrm{A}_{1}=4 \pi \mathrm{r}_{1}{ }^{2}=4 \pi\left(8 \times 10^{-2}\right)^{2}$ $\mathrm{A}_{2}=4 \pi \mathrm{r}_{2}^{2}=4 \pi\left(2 \times 10^{-2}\right)^{2}$ $\mathrm{E}_{1}=\sigma \times 4 \pi\left(8 \times 10^{-2}\right)^{2} \times(400)^{4}$ $\mathrm{E}_{2}=\sigma \times 4 \pi\left(2 \times 10^{-2}\right)^{2} \times(800)^{4}$ Dividing eq ${ }^{\mathrm{n}}$ (i) / (ii) $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{64 \times 400^{4}}{4 \times 800^{4}}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=1: 1$
Kerala CEE - 2015
Heat Transfer
149446
The plots of intensity of radiation versus wavelength of three black bodies at temperatures $T_{1}, T_{2}$ and $T_{3}$ are shown. Then,
1 $\mathrm{T}_{3}>\mathrm{T}_{2}>\mathrm{T}_{1}$
2 $\mathrm{T}_{1}>\mathrm{T}_{2}>\mathrm{T}_{3}$
3 $T_{2}>T_{3}>T_{1}$
4 $T_{1}>T_{3}>T_{2}$
5 $\mathrm{T}_{3}>\mathrm{T}_{1}>\mathrm{T}_{2}$
Explanation:
D According to Wien's law $\lambda_{\mathrm{m}} \propto \frac{1}{\mathrm{~T}}$ And from the figure $\left(\lambda_{\mathrm{m}}\right)_{1} \lt \left(\lambda_{\mathrm{m}}\right)_{3} \lt \left(\lambda_{\mathrm{m}}\right)_{2}$ Therefore, $\mathrm{T}_{1}>\mathrm{T}_{3}>\mathrm{T}_{2}$
Kerala CEE - 2008
Heat Transfer
149447
A black body has maximum wavelength $\lambda_{\mathrm{m}}$ at $2000 \mathrm{~K}$. Its corresponding wavelength at $\mathbf{3 0 0 0}$ $\mathrm{K}$ will be:
1 $\frac{3}{2} \lambda_{m}$
2 $\frac{2}{3} \lambda_{m}$
3 $\frac{16}{81} \lambda_{m}$
4 $\frac{81}{16} \lambda_{m}$
5 $\frac{4}{3} \lambda_{\mathrm{m}}$
Explanation:
B Given, $\lambda_{\mathrm{m}_{1}}=\lambda_{\mathrm{m}}$ $\mathrm{T}_{1}=2000 \mathrm{~K}$ $\mathrm{~T}_{2}=3000 \mathrm{~K}$ $\lambda_{\mathrm{m}_{2}}=?$ From Wein's displacement law $\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{C}$ $\lambda_{\mathrm{m}_{2}} \mathrm{~T}_{2}=\lambda_{\mathrm{m}_{1}} \times \mathrm{T}_{1}$ $\lambda_{\mathrm{m}_{2}}=\lambda_{\mathrm{m}_{1}} \times \mathrm{T}_{1} / \mathrm{T}_{2}$ $\lambda_{\mathrm{m}_{2}}=\lambda_{\mathrm{m}} \times 2000 / 3000$ $\lambda_{\mathrm{m}_{2}}=\frac{2}{3} \lambda_{\mathrm{m}}$
149442
Temperature of two stars are in ratio $3: 2$. If wavelength of maximum intensity of first body is $4000 \AA$, what is corresponding wavelength of second body?
1 $9000 \AA$
2 $6000 \AA$
3 $2000 \AA$
4 $8000 \AA$
Explanation:
B Given, $\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\frac{3}{2}$ $\left(\lambda_{\mathrm{m}}\right)_{1}=4000 \AA$ According to Wien's displacement law $\lambda_{\mathrm{m}} \mathrm{T}=\text { constant }$ $\frac{\left(\lambda_{\mathrm{m}}\right)_{1}}{\left(\lambda_{\mathrm{m}}\right)_{2}}=\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$ $\frac{4000}{\left(\lambda_{\mathrm{m}}\right)_{2}}=\frac{2}{3}$ $\left(\lambda_{\mathrm{m}}\right)_{2}=6000 \AA$
UPSEE - 2010
Heat Transfer
149444
If the emission rate of blackbody at $0^{\circ} \mathrm{C}$ is $R$, then the rate of emission at $273^{\circ} \mathrm{c}$ is
1 $2 \mathrm{R}$
2 $4 \mathrm{R}$
3 $8 \mathrm{R}$
4 $16 \mathrm{R}$
5 $32 \mathrm{R}$
Explanation:
D Given, $\mathrm{T}_{1}=0^{\circ} \mathrm{C}=273 \mathrm{~K}$ $\mathrm{T}_{2}=273^{\circ} \mathrm{C}+273=546 \mathrm{~K}$ According to Stefan's law, Emission rate of a ideal blackbody, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}} \propto\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{273}{546}\right)^{4}$ $\frac{\mathrm{R}}{\mathrm{E}_{2}}=\left(\frac{1}{2}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{R}$
Kerala CEE -2018
Heat Transfer
149445
Two perfectly black spheres $A$ and $B$ having radii $8 \mathrm{~cm}$ and $2 \mathrm{~cm}$ are maintained at temperatures $127^{\circ} \mathrm{C}$ and $527^{\circ} \mathrm{C}$, respectively. The ratio of the energy radiated by $A$ to that by $B$ is
1 $1: 2$
2 $1: 1$
3 $2: 1$
4 $1: 4$
5 $1: 16$
Explanation:
B We know, radiation of energy $\mathrm{E}=\sigma \mathrm{AT}^{4}$ Where $\sigma=$ Stefan's Boltsmann constant $\mathrm{A}=$ surface area $\mathrm{T}=$ Temperature at which radiation occur. Given, $\mathrm{R}_{1}=8 \mathrm{~cm}=8 \times 10^{-2} \mathrm{~m}$ $\mathrm{R}_{2}=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}$ $\mathrm{T}_{1}=127+273=400 \mathrm{~K}$ $\mathrm{T}_{2}=527+273=800$ Then $\mathrm{A}_{1}=4 \pi \mathrm{r}_{1}{ }^{2}=4 \pi\left(8 \times 10^{-2}\right)^{2}$ $\mathrm{A}_{2}=4 \pi \mathrm{r}_{2}^{2}=4 \pi\left(2 \times 10^{-2}\right)^{2}$ $\mathrm{E}_{1}=\sigma \times 4 \pi\left(8 \times 10^{-2}\right)^{2} \times(400)^{4}$ $\mathrm{E}_{2}=\sigma \times 4 \pi\left(2 \times 10^{-2}\right)^{2} \times(800)^{4}$ Dividing eq ${ }^{\mathrm{n}}$ (i) / (ii) $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{64 \times 400^{4}}{4 \times 800^{4}}$ $\mathrm{E}_{1}: \mathrm{E}_{2}=1: 1$
Kerala CEE - 2015
Heat Transfer
149446
The plots of intensity of radiation versus wavelength of three black bodies at temperatures $T_{1}, T_{2}$ and $T_{3}$ are shown. Then,
1 $\mathrm{T}_{3}>\mathrm{T}_{2}>\mathrm{T}_{1}$
2 $\mathrm{T}_{1}>\mathrm{T}_{2}>\mathrm{T}_{3}$
3 $T_{2}>T_{3}>T_{1}$
4 $T_{1}>T_{3}>T_{2}$
5 $\mathrm{T}_{3}>\mathrm{T}_{1}>\mathrm{T}_{2}$
Explanation:
D According to Wien's law $\lambda_{\mathrm{m}} \propto \frac{1}{\mathrm{~T}}$ And from the figure $\left(\lambda_{\mathrm{m}}\right)_{1} \lt \left(\lambda_{\mathrm{m}}\right)_{3} \lt \left(\lambda_{\mathrm{m}}\right)_{2}$ Therefore, $\mathrm{T}_{1}>\mathrm{T}_{3}>\mathrm{T}_{2}$
Kerala CEE - 2008
Heat Transfer
149447
A black body has maximum wavelength $\lambda_{\mathrm{m}}$ at $2000 \mathrm{~K}$. Its corresponding wavelength at $\mathbf{3 0 0 0}$ $\mathrm{K}$ will be:
1 $\frac{3}{2} \lambda_{m}$
2 $\frac{2}{3} \lambda_{m}$
3 $\frac{16}{81} \lambda_{m}$
4 $\frac{81}{16} \lambda_{m}$
5 $\frac{4}{3} \lambda_{\mathrm{m}}$
Explanation:
B Given, $\lambda_{\mathrm{m}_{1}}=\lambda_{\mathrm{m}}$ $\mathrm{T}_{1}=2000 \mathrm{~K}$ $\mathrm{~T}_{2}=3000 \mathrm{~K}$ $\lambda_{\mathrm{m}_{2}}=?$ From Wein's displacement law $\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{C}$ $\lambda_{\mathrm{m}_{2}} \mathrm{~T}_{2}=\lambda_{\mathrm{m}_{1}} \times \mathrm{T}_{1}$ $\lambda_{\mathrm{m}_{2}}=\lambda_{\mathrm{m}_{1}} \times \mathrm{T}_{1} / \mathrm{T}_{2}$ $\lambda_{\mathrm{m}_{2}}=\lambda_{\mathrm{m}} \times 2000 / 3000$ $\lambda_{\mathrm{m}_{2}}=\frac{2}{3} \lambda_{\mathrm{m}}$
