NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Heat Transfer
149469
Let there be four identical cubes having colours blue, red, black and white. When they are heated together and allowed to cool under identical conditions, which cube will cool at the earliest?
1 Blue
2 Red
3 Black
4 White
Explanation:
C The black colour absorbs and emits the heat at a faster rate compared to all other colour. Body having higher absorptivity will have a higher emissivity.
CG PET- 2017
Heat Transfer
149477
Two bodies $A$ and $B$ having temperatures $327^{\circ} \mathrm{C}$ and $427^{\circ} \mathrm{C}$ are radiating heat to the surrounding. The surrounding temperature is $27^{\circ} \mathrm{C}$. The ratio of rates of heat radiation of $A$ to that of $B$ is
1 0.52
2 0.31
3 0.81
4 0.42
Explanation:
A If temperature of surrounding is considered, then net loss of energy of a body by radiation $\quad \mathrm{Q} =\mathrm{A} \sigma\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}\right) \Rightarrow \mathrm{Q} \propto\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}\right)$ $\therefore \quad \frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}} =\frac{\mathrm{T}_{1}^{4}-\mathrm{T}_{0}^{4}}{\mathrm{~T}_{2}^{4}-\mathrm{T}_{0}^{4}}$ $=\frac{(273+327)^{4}-(273+27)^{4}}{(273+427)^{4}-(273+27)^{4}}$ $=\frac{(600)^{4}-(300)^{4}}{(700)^{4}-(300)^{4}}=0.52$
Manipal UGET-2010
Heat Transfer
149482
If $\lambda$ denotes the wavelength at which the radiative emission from a black body at a temperature $T$ is maximum, then
1 $\lambda \propto \mathrm{T}^{-1}$
2 $\lambda \propto T^{4}$
3 $\lambda$ is independent of $\mathrm{T}$
4 $\lambda \propto \mathrm{T}$
Explanation:
A According to Wien's displacement law, $\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{b}$ Where $\lambda_{\mathrm{m}}=$ wavelength corresponding to maximum radiation of energy from black body, $b=$ Wien's constant and $\mathrm{T}=$ temperature. $\therefore$ From Eq. (i), we get $\lambda_{\mathrm{m}} \propto \frac{1}{\mathrm{~T}} \Rightarrow \lambda_{\mathrm{m}} \propto \mathrm{T}^{-1}$
TS- EAMCET.14.09.2020
Heat Transfer
149486
The surface temperature is maximum for :
1 blue star
2 yellow star
3 green star
4 red star
Explanation:
A According to Wien's displacement law- $\lambda_{\mathrm{m}} \mathrm{T}=\text { constant }$ Where, $\lambda_{\mathrm{m}}=$ maximum wavelength $\mathrm{T}=$ absolute temperature. Since, wavelength of blue star is maximum. Note:- Surface temperature will be minimum for red star.
149469
Let there be four identical cubes having colours blue, red, black and white. When they are heated together and allowed to cool under identical conditions, which cube will cool at the earliest?
1 Blue
2 Red
3 Black
4 White
Explanation:
C The black colour absorbs and emits the heat at a faster rate compared to all other colour. Body having higher absorptivity will have a higher emissivity.
CG PET- 2017
Heat Transfer
149477
Two bodies $A$ and $B$ having temperatures $327^{\circ} \mathrm{C}$ and $427^{\circ} \mathrm{C}$ are radiating heat to the surrounding. The surrounding temperature is $27^{\circ} \mathrm{C}$. The ratio of rates of heat radiation of $A$ to that of $B$ is
1 0.52
2 0.31
3 0.81
4 0.42
Explanation:
A If temperature of surrounding is considered, then net loss of energy of a body by radiation $\quad \mathrm{Q} =\mathrm{A} \sigma\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}\right) \Rightarrow \mathrm{Q} \propto\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}\right)$ $\therefore \quad \frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}} =\frac{\mathrm{T}_{1}^{4}-\mathrm{T}_{0}^{4}}{\mathrm{~T}_{2}^{4}-\mathrm{T}_{0}^{4}}$ $=\frac{(273+327)^{4}-(273+27)^{4}}{(273+427)^{4}-(273+27)^{4}}$ $=\frac{(600)^{4}-(300)^{4}}{(700)^{4}-(300)^{4}}=0.52$
Manipal UGET-2010
Heat Transfer
149482
If $\lambda$ denotes the wavelength at which the radiative emission from a black body at a temperature $T$ is maximum, then
1 $\lambda \propto \mathrm{T}^{-1}$
2 $\lambda \propto T^{4}$
3 $\lambda$ is independent of $\mathrm{T}$
4 $\lambda \propto \mathrm{T}$
Explanation:
A According to Wien's displacement law, $\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{b}$ Where $\lambda_{\mathrm{m}}=$ wavelength corresponding to maximum radiation of energy from black body, $b=$ Wien's constant and $\mathrm{T}=$ temperature. $\therefore$ From Eq. (i), we get $\lambda_{\mathrm{m}} \propto \frac{1}{\mathrm{~T}} \Rightarrow \lambda_{\mathrm{m}} \propto \mathrm{T}^{-1}$
TS- EAMCET.14.09.2020
Heat Transfer
149486
The surface temperature is maximum for :
1 blue star
2 yellow star
3 green star
4 red star
Explanation:
A According to Wien's displacement law- $\lambda_{\mathrm{m}} \mathrm{T}=\text { constant }$ Where, $\lambda_{\mathrm{m}}=$ maximum wavelength $\mathrm{T}=$ absolute temperature. Since, wavelength of blue star is maximum. Note:- Surface temperature will be minimum for red star.
149469
Let there be four identical cubes having colours blue, red, black and white. When they are heated together and allowed to cool under identical conditions, which cube will cool at the earliest?
1 Blue
2 Red
3 Black
4 White
Explanation:
C The black colour absorbs and emits the heat at a faster rate compared to all other colour. Body having higher absorptivity will have a higher emissivity.
CG PET- 2017
Heat Transfer
149477
Two bodies $A$ and $B$ having temperatures $327^{\circ} \mathrm{C}$ and $427^{\circ} \mathrm{C}$ are radiating heat to the surrounding. The surrounding temperature is $27^{\circ} \mathrm{C}$. The ratio of rates of heat radiation of $A$ to that of $B$ is
1 0.52
2 0.31
3 0.81
4 0.42
Explanation:
A If temperature of surrounding is considered, then net loss of energy of a body by radiation $\quad \mathrm{Q} =\mathrm{A} \sigma\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}\right) \Rightarrow \mathrm{Q} \propto\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}\right)$ $\therefore \quad \frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}} =\frac{\mathrm{T}_{1}^{4}-\mathrm{T}_{0}^{4}}{\mathrm{~T}_{2}^{4}-\mathrm{T}_{0}^{4}}$ $=\frac{(273+327)^{4}-(273+27)^{4}}{(273+427)^{4}-(273+27)^{4}}$ $=\frac{(600)^{4}-(300)^{4}}{(700)^{4}-(300)^{4}}=0.52$
Manipal UGET-2010
Heat Transfer
149482
If $\lambda$ denotes the wavelength at which the radiative emission from a black body at a temperature $T$ is maximum, then
1 $\lambda \propto \mathrm{T}^{-1}$
2 $\lambda \propto T^{4}$
3 $\lambda$ is independent of $\mathrm{T}$
4 $\lambda \propto \mathrm{T}$
Explanation:
A According to Wien's displacement law, $\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{b}$ Where $\lambda_{\mathrm{m}}=$ wavelength corresponding to maximum radiation of energy from black body, $b=$ Wien's constant and $\mathrm{T}=$ temperature. $\therefore$ From Eq. (i), we get $\lambda_{\mathrm{m}} \propto \frac{1}{\mathrm{~T}} \Rightarrow \lambda_{\mathrm{m}} \propto \mathrm{T}^{-1}$
TS- EAMCET.14.09.2020
Heat Transfer
149486
The surface temperature is maximum for :
1 blue star
2 yellow star
3 green star
4 red star
Explanation:
A According to Wien's displacement law- $\lambda_{\mathrm{m}} \mathrm{T}=\text { constant }$ Where, $\lambda_{\mathrm{m}}=$ maximum wavelength $\mathrm{T}=$ absolute temperature. Since, wavelength of blue star is maximum. Note:- Surface temperature will be minimum for red star.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Heat Transfer
149469
Let there be four identical cubes having colours blue, red, black and white. When they are heated together and allowed to cool under identical conditions, which cube will cool at the earliest?
1 Blue
2 Red
3 Black
4 White
Explanation:
C The black colour absorbs and emits the heat at a faster rate compared to all other colour. Body having higher absorptivity will have a higher emissivity.
CG PET- 2017
Heat Transfer
149477
Two bodies $A$ and $B$ having temperatures $327^{\circ} \mathrm{C}$ and $427^{\circ} \mathrm{C}$ are radiating heat to the surrounding. The surrounding temperature is $27^{\circ} \mathrm{C}$. The ratio of rates of heat radiation of $A$ to that of $B$ is
1 0.52
2 0.31
3 0.81
4 0.42
Explanation:
A If temperature of surrounding is considered, then net loss of energy of a body by radiation $\quad \mathrm{Q} =\mathrm{A} \sigma\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}\right) \Rightarrow \mathrm{Q} \propto\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}\right)$ $\therefore \quad \frac{\mathrm{Q}_{1}}{\mathrm{Q}_{2}} =\frac{\mathrm{T}_{1}^{4}-\mathrm{T}_{0}^{4}}{\mathrm{~T}_{2}^{4}-\mathrm{T}_{0}^{4}}$ $=\frac{(273+327)^{4}-(273+27)^{4}}{(273+427)^{4}-(273+27)^{4}}$ $=\frac{(600)^{4}-(300)^{4}}{(700)^{4}-(300)^{4}}=0.52$
Manipal UGET-2010
Heat Transfer
149482
If $\lambda$ denotes the wavelength at which the radiative emission from a black body at a temperature $T$ is maximum, then
1 $\lambda \propto \mathrm{T}^{-1}$
2 $\lambda \propto T^{4}$
3 $\lambda$ is independent of $\mathrm{T}$
4 $\lambda \propto \mathrm{T}$
Explanation:
A According to Wien's displacement law, $\lambda_{\mathrm{m}} \mathrm{T}=\mathrm{b}$ Where $\lambda_{\mathrm{m}}=$ wavelength corresponding to maximum radiation of energy from black body, $b=$ Wien's constant and $\mathrm{T}=$ temperature. $\therefore$ From Eq. (i), we get $\lambda_{\mathrm{m}} \propto \frac{1}{\mathrm{~T}} \Rightarrow \lambda_{\mathrm{m}} \propto \mathrm{T}^{-1}$
TS- EAMCET.14.09.2020
Heat Transfer
149486
The surface temperature is maximum for :
1 blue star
2 yellow star
3 green star
4 red star
Explanation:
A According to Wien's displacement law- $\lambda_{\mathrm{m}} \mathrm{T}=\text { constant }$ Where, $\lambda_{\mathrm{m}}=$ maximum wavelength $\mathrm{T}=$ absolute temperature. Since, wavelength of blue star is maximum. Note:- Surface temperature will be minimum for red star.