149515
If temperature of black body increases from $17^{\circ} \mathrm{C}$ to $307^{\circ} \mathrm{C}$, then the rate of radiation increases by
1 16
2 2
3 4
4 $\left(\frac{307}{17}\right)^{4}$
Explanation:
A Given that, $\mathrm{T}_{1}=17^{\circ} \mathrm{C}=273+17=290^{\circ} \mathrm{K}$ $\mathrm{T}_{2}=307^{\circ} \mathrm{C}=307+273=580 \mathrm{~K}$ Energy radiation per second is given by Stefan's law, $\mathrm{E}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\sigma \mathrm{eAT}^{4}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}=\frac{\sigma \mathrm{eAT}_{1}^{4}}{\sigma \mathrm{eAT}_{2}^{4}}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}\left(\frac{290}{580}\right)^{4}=\left(\frac{1}{2}\right)^{4}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}=\frac{1}{16}$ $\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}=16 \times\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}=\frac{16}{1}$
MHT-CET 2020
Heat Transfer
149517
Two spherical black bodies of radius ' $r{ }_{1}$ ' and ' $r_{2}$ ' with surface temperature ' $T_{1}$ ' and ' $T_{2}$ ' respectively, radiate same power, then $r_{1}: r_{2}$ is
A Given that, Both spherical bodies radiate same power i.e. $\mathrm{E}_{1}=\mathrm{E}_{2}$ Power radiated by the black body $\mathrm{E}=\mathrm{e} \sigma \mathrm{AT}^{4}$ $\therefore \quad \sigma \mathrm{e}_{1} \mathrm{~T}_{1}^{4}=\sigma \mathrm{eA}_{2} \mathrm{~T}_{2}^{4}$ $4 \pi \mathrm{r}_{1}^{2} \times \mathrm{T}_{1}^{4}=4 \pi \mathrm{r}_{2} \times \mathrm{T}_{2}^{4}$ $\frac{\mathrm{r}_{1}^{2}}{\mathrm{r}_{2}^{2}}=\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}$ $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}$
MHT-CET 2020
Heat Transfer
149518
The energy spectrum of a black body exhibits a maximum around a wavelength ' $\lambda$ '. The temperature of a black body is now changed such that the energy is maximum around a wavelength $3 \lambda / 4$. The power radiated by a black body will now increase by a factor of
1 $\frac{128}{27}$
2 $\frac{128}{81}$
3 $\frac{256}{27}$
4 $\frac{256}{81}$
Explanation:
D Given that, $\lambda_{1}=\lambda$ $\lambda_{2}=3 \lambda / 4$ As we know that rate of radiated energy, $\mathrm{E}=\sigma \mathrm{eAT} \mathrm{T}^{4}$ $\therefore \mathrm{T} \propto \frac{1}{\lambda}$ $\mathrm{E}=\sigma \mathrm{eA} \frac{1}{\lambda^{4}}$ $\mathrm{E}_{1}=\sigma \mathrm{eA} \frac{1}{\lambda_{1}^{4}}$ $\mathrm{E}_{2}=\sigma \mathrm{eA} \frac{1}{\lambda_{2}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\lambda_{1}^{4}}{\lambda_{2}^{4}}=\frac{(\lambda)^{4}}{(3 \lambda / 4)^{4}}=\left(\frac{4}{3}\right)^{4}$ $\therefore \quad \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{256}{81}$
MHT-CET 2020
Heat Transfer
149519
Rate of radiation by a black body is ' $R$ ' at temperature ' $T$ '. Another body has same area but emissivity is 0.2 and temperature ' $3 T$ ' Its rate of radiation is
1 $(24.3) \mathrm{R}$
2 $(8.1) \mathrm{R}$
3 $(16.2) \mathrm{R}$
4 $(32.4) \mathrm{R}$
Explanation:
C Given that, For another body area $=\mathrm{A}_{1}=\mathrm{A}_{2}=\mathrm{A}$ $\mathrm{e}_{1}=0.2 \mathrm{e}$ $\mathrm{T}_{2}=3 \mathrm{~T}$ Rate of radiation $=\sigma e \mathrm{AT}_{2}^{4}$ $=\sigma(0.2 \mathrm{e}) \mathrm{A}(3 \mathrm{~T})^{4}$ $=0.2 \times 81 \sigma \mathrm{eAT}^{4}$ $=(16.2) \sigma \mathrm{eAT}^{4}$ Rate of radiation $=(16.2) \mathrm{R}$
149515
If temperature of black body increases from $17^{\circ} \mathrm{C}$ to $307^{\circ} \mathrm{C}$, then the rate of radiation increases by
1 16
2 2
3 4
4 $\left(\frac{307}{17}\right)^{4}$
Explanation:
A Given that, $\mathrm{T}_{1}=17^{\circ} \mathrm{C}=273+17=290^{\circ} \mathrm{K}$ $\mathrm{T}_{2}=307^{\circ} \mathrm{C}=307+273=580 \mathrm{~K}$ Energy radiation per second is given by Stefan's law, $\mathrm{E}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\sigma \mathrm{eAT}^{4}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}=\frac{\sigma \mathrm{eAT}_{1}^{4}}{\sigma \mathrm{eAT}_{2}^{4}}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}\left(\frac{290}{580}\right)^{4}=\left(\frac{1}{2}\right)^{4}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}=\frac{1}{16}$ $\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}=16 \times\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}=\frac{16}{1}$
MHT-CET 2020
Heat Transfer
149517
Two spherical black bodies of radius ' $r{ }_{1}$ ' and ' $r_{2}$ ' with surface temperature ' $T_{1}$ ' and ' $T_{2}$ ' respectively, radiate same power, then $r_{1}: r_{2}$ is
A Given that, Both spherical bodies radiate same power i.e. $\mathrm{E}_{1}=\mathrm{E}_{2}$ Power radiated by the black body $\mathrm{E}=\mathrm{e} \sigma \mathrm{AT}^{4}$ $\therefore \quad \sigma \mathrm{e}_{1} \mathrm{~T}_{1}^{4}=\sigma \mathrm{eA}_{2} \mathrm{~T}_{2}^{4}$ $4 \pi \mathrm{r}_{1}^{2} \times \mathrm{T}_{1}^{4}=4 \pi \mathrm{r}_{2} \times \mathrm{T}_{2}^{4}$ $\frac{\mathrm{r}_{1}^{2}}{\mathrm{r}_{2}^{2}}=\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}$ $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}$
MHT-CET 2020
Heat Transfer
149518
The energy spectrum of a black body exhibits a maximum around a wavelength ' $\lambda$ '. The temperature of a black body is now changed such that the energy is maximum around a wavelength $3 \lambda / 4$. The power radiated by a black body will now increase by a factor of
1 $\frac{128}{27}$
2 $\frac{128}{81}$
3 $\frac{256}{27}$
4 $\frac{256}{81}$
Explanation:
D Given that, $\lambda_{1}=\lambda$ $\lambda_{2}=3 \lambda / 4$ As we know that rate of radiated energy, $\mathrm{E}=\sigma \mathrm{eAT} \mathrm{T}^{4}$ $\therefore \mathrm{T} \propto \frac{1}{\lambda}$ $\mathrm{E}=\sigma \mathrm{eA} \frac{1}{\lambda^{4}}$ $\mathrm{E}_{1}=\sigma \mathrm{eA} \frac{1}{\lambda_{1}^{4}}$ $\mathrm{E}_{2}=\sigma \mathrm{eA} \frac{1}{\lambda_{2}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\lambda_{1}^{4}}{\lambda_{2}^{4}}=\frac{(\lambda)^{4}}{(3 \lambda / 4)^{4}}=\left(\frac{4}{3}\right)^{4}$ $\therefore \quad \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{256}{81}$
MHT-CET 2020
Heat Transfer
149519
Rate of radiation by a black body is ' $R$ ' at temperature ' $T$ '. Another body has same area but emissivity is 0.2 and temperature ' $3 T$ ' Its rate of radiation is
1 $(24.3) \mathrm{R}$
2 $(8.1) \mathrm{R}$
3 $(16.2) \mathrm{R}$
4 $(32.4) \mathrm{R}$
Explanation:
C Given that, For another body area $=\mathrm{A}_{1}=\mathrm{A}_{2}=\mathrm{A}$ $\mathrm{e}_{1}=0.2 \mathrm{e}$ $\mathrm{T}_{2}=3 \mathrm{~T}$ Rate of radiation $=\sigma e \mathrm{AT}_{2}^{4}$ $=\sigma(0.2 \mathrm{e}) \mathrm{A}(3 \mathrm{~T})^{4}$ $=0.2 \times 81 \sigma \mathrm{eAT}^{4}$ $=(16.2) \sigma \mathrm{eAT}^{4}$ Rate of radiation $=(16.2) \mathrm{R}$
149515
If temperature of black body increases from $17^{\circ} \mathrm{C}$ to $307^{\circ} \mathrm{C}$, then the rate of radiation increases by
1 16
2 2
3 4
4 $\left(\frac{307}{17}\right)^{4}$
Explanation:
A Given that, $\mathrm{T}_{1}=17^{\circ} \mathrm{C}=273+17=290^{\circ} \mathrm{K}$ $\mathrm{T}_{2}=307^{\circ} \mathrm{C}=307+273=580 \mathrm{~K}$ Energy radiation per second is given by Stefan's law, $\mathrm{E}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\sigma \mathrm{eAT}^{4}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}=\frac{\sigma \mathrm{eAT}_{1}^{4}}{\sigma \mathrm{eAT}_{2}^{4}}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}\left(\frac{290}{580}\right)^{4}=\left(\frac{1}{2}\right)^{4}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}=\frac{1}{16}$ $\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}=16 \times\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}=\frac{16}{1}$
MHT-CET 2020
Heat Transfer
149517
Two spherical black bodies of radius ' $r{ }_{1}$ ' and ' $r_{2}$ ' with surface temperature ' $T_{1}$ ' and ' $T_{2}$ ' respectively, radiate same power, then $r_{1}: r_{2}$ is
A Given that, Both spherical bodies radiate same power i.e. $\mathrm{E}_{1}=\mathrm{E}_{2}$ Power radiated by the black body $\mathrm{E}=\mathrm{e} \sigma \mathrm{AT}^{4}$ $\therefore \quad \sigma \mathrm{e}_{1} \mathrm{~T}_{1}^{4}=\sigma \mathrm{eA}_{2} \mathrm{~T}_{2}^{4}$ $4 \pi \mathrm{r}_{1}^{2} \times \mathrm{T}_{1}^{4}=4 \pi \mathrm{r}_{2} \times \mathrm{T}_{2}^{4}$ $\frac{\mathrm{r}_{1}^{2}}{\mathrm{r}_{2}^{2}}=\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}$ $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}$
MHT-CET 2020
Heat Transfer
149518
The energy spectrum of a black body exhibits a maximum around a wavelength ' $\lambda$ '. The temperature of a black body is now changed such that the energy is maximum around a wavelength $3 \lambda / 4$. The power radiated by a black body will now increase by a factor of
1 $\frac{128}{27}$
2 $\frac{128}{81}$
3 $\frac{256}{27}$
4 $\frac{256}{81}$
Explanation:
D Given that, $\lambda_{1}=\lambda$ $\lambda_{2}=3 \lambda / 4$ As we know that rate of radiated energy, $\mathrm{E}=\sigma \mathrm{eAT} \mathrm{T}^{4}$ $\therefore \mathrm{T} \propto \frac{1}{\lambda}$ $\mathrm{E}=\sigma \mathrm{eA} \frac{1}{\lambda^{4}}$ $\mathrm{E}_{1}=\sigma \mathrm{eA} \frac{1}{\lambda_{1}^{4}}$ $\mathrm{E}_{2}=\sigma \mathrm{eA} \frac{1}{\lambda_{2}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\lambda_{1}^{4}}{\lambda_{2}^{4}}=\frac{(\lambda)^{4}}{(3 \lambda / 4)^{4}}=\left(\frac{4}{3}\right)^{4}$ $\therefore \quad \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{256}{81}$
MHT-CET 2020
Heat Transfer
149519
Rate of radiation by a black body is ' $R$ ' at temperature ' $T$ '. Another body has same area but emissivity is 0.2 and temperature ' $3 T$ ' Its rate of radiation is
1 $(24.3) \mathrm{R}$
2 $(8.1) \mathrm{R}$
3 $(16.2) \mathrm{R}$
4 $(32.4) \mathrm{R}$
Explanation:
C Given that, For another body area $=\mathrm{A}_{1}=\mathrm{A}_{2}=\mathrm{A}$ $\mathrm{e}_{1}=0.2 \mathrm{e}$ $\mathrm{T}_{2}=3 \mathrm{~T}$ Rate of radiation $=\sigma e \mathrm{AT}_{2}^{4}$ $=\sigma(0.2 \mathrm{e}) \mathrm{A}(3 \mathrm{~T})^{4}$ $=0.2 \times 81 \sigma \mathrm{eAT}^{4}$ $=(16.2) \sigma \mathrm{eAT}^{4}$ Rate of radiation $=(16.2) \mathrm{R}$
149515
If temperature of black body increases from $17^{\circ} \mathrm{C}$ to $307^{\circ} \mathrm{C}$, then the rate of radiation increases by
1 16
2 2
3 4
4 $\left(\frac{307}{17}\right)^{4}$
Explanation:
A Given that, $\mathrm{T}_{1}=17^{\circ} \mathrm{C}=273+17=290^{\circ} \mathrm{K}$ $\mathrm{T}_{2}=307^{\circ} \mathrm{C}=307+273=580 \mathrm{~K}$ Energy radiation per second is given by Stefan's law, $\mathrm{E}=\frac{\mathrm{dQ}}{\mathrm{dt}}=\sigma \mathrm{eAT}^{4}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}=\frac{\sigma \mathrm{eAT}_{1}^{4}}{\sigma \mathrm{eAT}_{2}^{4}}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}\left(\frac{290}{580}\right)^{4}=\left(\frac{1}{2}\right)^{4}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}=\frac{1}{16}$ $\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}=16 \times\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}$ $\frac{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{2}}{\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{1}}=\frac{16}{1}$
MHT-CET 2020
Heat Transfer
149517
Two spherical black bodies of radius ' $r{ }_{1}$ ' and ' $r_{2}$ ' with surface temperature ' $T_{1}$ ' and ' $T_{2}$ ' respectively, radiate same power, then $r_{1}: r_{2}$ is
A Given that, Both spherical bodies radiate same power i.e. $\mathrm{E}_{1}=\mathrm{E}_{2}$ Power radiated by the black body $\mathrm{E}=\mathrm{e} \sigma \mathrm{AT}^{4}$ $\therefore \quad \sigma \mathrm{e}_{1} \mathrm{~T}_{1}^{4}=\sigma \mathrm{eA}_{2} \mathrm{~T}_{2}^{4}$ $4 \pi \mathrm{r}_{1}^{2} \times \mathrm{T}_{1}^{4}=4 \pi \mathrm{r}_{2} \times \mathrm{T}_{2}^{4}$ $\frac{\mathrm{r}_{1}^{2}}{\mathrm{r}_{2}^{2}}=\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{4}$ $\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right)^{2}$
MHT-CET 2020
Heat Transfer
149518
The energy spectrum of a black body exhibits a maximum around a wavelength ' $\lambda$ '. The temperature of a black body is now changed such that the energy is maximum around a wavelength $3 \lambda / 4$. The power radiated by a black body will now increase by a factor of
1 $\frac{128}{27}$
2 $\frac{128}{81}$
3 $\frac{256}{27}$
4 $\frac{256}{81}$
Explanation:
D Given that, $\lambda_{1}=\lambda$ $\lambda_{2}=3 \lambda / 4$ As we know that rate of radiated energy, $\mathrm{E}=\sigma \mathrm{eAT} \mathrm{T}^{4}$ $\therefore \mathrm{T} \propto \frac{1}{\lambda}$ $\mathrm{E}=\sigma \mathrm{eA} \frac{1}{\lambda^{4}}$ $\mathrm{E}_{1}=\sigma \mathrm{eA} \frac{1}{\lambda_{1}^{4}}$ $\mathrm{E}_{2}=\sigma \mathrm{eA} \frac{1}{\lambda_{2}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\lambda_{1}^{4}}{\lambda_{2}^{4}}=\frac{(\lambda)^{4}}{(3 \lambda / 4)^{4}}=\left(\frac{4}{3}\right)^{4}$ $\therefore \quad \frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{256}{81}$
MHT-CET 2020
Heat Transfer
149519
Rate of radiation by a black body is ' $R$ ' at temperature ' $T$ '. Another body has same area but emissivity is 0.2 and temperature ' $3 T$ ' Its rate of radiation is
1 $(24.3) \mathrm{R}$
2 $(8.1) \mathrm{R}$
3 $(16.2) \mathrm{R}$
4 $(32.4) \mathrm{R}$
Explanation:
C Given that, For another body area $=\mathrm{A}_{1}=\mathrm{A}_{2}=\mathrm{A}$ $\mathrm{e}_{1}=0.2 \mathrm{e}$ $\mathrm{T}_{2}=3 \mathrm{~T}$ Rate of radiation $=\sigma e \mathrm{AT}_{2}^{4}$ $=\sigma(0.2 \mathrm{e}) \mathrm{A}(3 \mathrm{~T})^{4}$ $=0.2 \times 81 \sigma \mathrm{eAT}^{4}$ $=(16.2) \sigma \mathrm{eAT}^{4}$ Rate of radiation $=(16.2) \mathrm{R}$
