149600
The surface temperature is the sun which has maximum energy emission at $500 \mathrm{~nm}$ is $6000 \mathrm{~K}$. The temperature of a star which has maximum energy emission at $400 \mathrm{~nm}$ will be
1 $8500 \mathrm{~K}$
2 $4500 \mathrm{~K}$
3 $7500 \mathrm{~K}$
4 $6500 \mathrm{~K}$
Explanation:
C Given data, $\lambda_{1}=500 \mathrm{~nm}$ $\lambda_{2}=400 \mathrm{~nm}$ $\mathrm{T}_{1}=6000 \mathrm{~K}$ $\mathrm{T}_{2}=$ ? According to Wein's Displacement law- $\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2}$ Where, $\lambda_{1}=$ wavelength at $T_{1}$ and $\mathrm{T}_{2}$ $500 \times 6000=400 \times \mathrm{T}_{2}$ $\mathrm{~T}_{2}=\frac{500 \times 6000}{400}$ $\mathrm{~T}_{2}=\frac{30,00000}{400}$ $\mathrm{~T}_{2}=7500 \mathrm{~K}$
JIPMER-2012
Heat Transfer
149601
The temperature of the black body increases from $T$ to $2 T$. the factor by which the rate of emission will increase, is ?
1 4
2 2
3 16
4 8
Explanation:
C Given, $\mathrm{T}_{1}=\mathrm{T}, \mathrm{T}_{2}=2 \mathrm{~T}$ According to Stefan's law- $\therefore \mathrm{E}=\sigma \mathrm{T}^{4}$ Where $\sigma$ is Stefan's constant $\therefore \quad \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}^{4}}{(2 \mathrm{~T})^{4}}$ $\frac{E_{1}}{E_{2}}=\frac{1}{2^{4}}=\frac{1}{16}$ $\mathrm{E}_{2}=16 \mathrm{E}_{1}$ Hence, rate of emission increases sixteen time.
JIPMER-2005
Heat Transfer
149602
A metal rod at a temperature of $145^{\circ} \mathrm{C}$, radiates energy at a rate of $17 \mathrm{~W}$. If its temperature is increased to $273^{\circ} \mathrm{C}$, then it will radiate at the rate of
1 $49.6 \mathrm{~W}$
2 $17.5 \mathrm{~W}$
3 $5.3 \mathrm{~W}$
4 $67.5 \mathrm{~W}$
Explanation:
A Given that, $\mathrm{T}_{1}=145^{\circ} \mathrm{C}=273+145=418 \mathrm{~K}$ $\mathrm{E}_{1}=17 \mathrm{~W}$ $\mathrm{~T}_{2}=273^{\circ} \mathrm{C}=273+273=546 \mathrm{~K}$ According to Stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{418}{546}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=0.343$ Therefore final radiated energy, $\mathrm{E}_{2}=\frac{\mathrm{E}_{1}}{0.343}=\frac{17}{0.343}$ $\mathrm{E}_{2}=49.6 \mathrm{~W}$
JIPMER-2016
Heat Transfer
149605
When the temperature of a body is increased from $27^{\circ} \mathrm{C}$ to $127^{\circ} \mathrm{C}$. the radiation emitted by it increases by a factor of
1 $\frac{256}{81}$
2 $\frac{15}{19}$
3 $\frac{4}{5}$
4 $\frac{12}{27}$
Explanation:
A Given that, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{~T}_{2}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}$ According to stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\mathrm{E}=\sigma \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{T}_{2}^{4}}{\mathrm{~T}_{1}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\left(\frac{400}{300}\right)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{256}{81}$
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Heat Transfer
149600
The surface temperature is the sun which has maximum energy emission at $500 \mathrm{~nm}$ is $6000 \mathrm{~K}$. The temperature of a star which has maximum energy emission at $400 \mathrm{~nm}$ will be
1 $8500 \mathrm{~K}$
2 $4500 \mathrm{~K}$
3 $7500 \mathrm{~K}$
4 $6500 \mathrm{~K}$
Explanation:
C Given data, $\lambda_{1}=500 \mathrm{~nm}$ $\lambda_{2}=400 \mathrm{~nm}$ $\mathrm{T}_{1}=6000 \mathrm{~K}$ $\mathrm{T}_{2}=$ ? According to Wein's Displacement law- $\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2}$ Where, $\lambda_{1}=$ wavelength at $T_{1}$ and $\mathrm{T}_{2}$ $500 \times 6000=400 \times \mathrm{T}_{2}$ $\mathrm{~T}_{2}=\frac{500 \times 6000}{400}$ $\mathrm{~T}_{2}=\frac{30,00000}{400}$ $\mathrm{~T}_{2}=7500 \mathrm{~K}$
JIPMER-2012
Heat Transfer
149601
The temperature of the black body increases from $T$ to $2 T$. the factor by which the rate of emission will increase, is ?
1 4
2 2
3 16
4 8
Explanation:
C Given, $\mathrm{T}_{1}=\mathrm{T}, \mathrm{T}_{2}=2 \mathrm{~T}$ According to Stefan's law- $\therefore \mathrm{E}=\sigma \mathrm{T}^{4}$ Where $\sigma$ is Stefan's constant $\therefore \quad \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}^{4}}{(2 \mathrm{~T})^{4}}$ $\frac{E_{1}}{E_{2}}=\frac{1}{2^{4}}=\frac{1}{16}$ $\mathrm{E}_{2}=16 \mathrm{E}_{1}$ Hence, rate of emission increases sixteen time.
JIPMER-2005
Heat Transfer
149602
A metal rod at a temperature of $145^{\circ} \mathrm{C}$, radiates energy at a rate of $17 \mathrm{~W}$. If its temperature is increased to $273^{\circ} \mathrm{C}$, then it will radiate at the rate of
1 $49.6 \mathrm{~W}$
2 $17.5 \mathrm{~W}$
3 $5.3 \mathrm{~W}$
4 $67.5 \mathrm{~W}$
Explanation:
A Given that, $\mathrm{T}_{1}=145^{\circ} \mathrm{C}=273+145=418 \mathrm{~K}$ $\mathrm{E}_{1}=17 \mathrm{~W}$ $\mathrm{~T}_{2}=273^{\circ} \mathrm{C}=273+273=546 \mathrm{~K}$ According to Stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{418}{546}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=0.343$ Therefore final radiated energy, $\mathrm{E}_{2}=\frac{\mathrm{E}_{1}}{0.343}=\frac{17}{0.343}$ $\mathrm{E}_{2}=49.6 \mathrm{~W}$
JIPMER-2016
Heat Transfer
149605
When the temperature of a body is increased from $27^{\circ} \mathrm{C}$ to $127^{\circ} \mathrm{C}$. the radiation emitted by it increases by a factor of
1 $\frac{256}{81}$
2 $\frac{15}{19}$
3 $\frac{4}{5}$
4 $\frac{12}{27}$
Explanation:
A Given that, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{~T}_{2}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}$ According to stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\mathrm{E}=\sigma \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{T}_{2}^{4}}{\mathrm{~T}_{1}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\left(\frac{400}{300}\right)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{256}{81}$
149600
The surface temperature is the sun which has maximum energy emission at $500 \mathrm{~nm}$ is $6000 \mathrm{~K}$. The temperature of a star which has maximum energy emission at $400 \mathrm{~nm}$ will be
1 $8500 \mathrm{~K}$
2 $4500 \mathrm{~K}$
3 $7500 \mathrm{~K}$
4 $6500 \mathrm{~K}$
Explanation:
C Given data, $\lambda_{1}=500 \mathrm{~nm}$ $\lambda_{2}=400 \mathrm{~nm}$ $\mathrm{T}_{1}=6000 \mathrm{~K}$ $\mathrm{T}_{2}=$ ? According to Wein's Displacement law- $\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2}$ Where, $\lambda_{1}=$ wavelength at $T_{1}$ and $\mathrm{T}_{2}$ $500 \times 6000=400 \times \mathrm{T}_{2}$ $\mathrm{~T}_{2}=\frac{500 \times 6000}{400}$ $\mathrm{~T}_{2}=\frac{30,00000}{400}$ $\mathrm{~T}_{2}=7500 \mathrm{~K}$
JIPMER-2012
Heat Transfer
149601
The temperature of the black body increases from $T$ to $2 T$. the factor by which the rate of emission will increase, is ?
1 4
2 2
3 16
4 8
Explanation:
C Given, $\mathrm{T}_{1}=\mathrm{T}, \mathrm{T}_{2}=2 \mathrm{~T}$ According to Stefan's law- $\therefore \mathrm{E}=\sigma \mathrm{T}^{4}$ Where $\sigma$ is Stefan's constant $\therefore \quad \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}^{4}}{(2 \mathrm{~T})^{4}}$ $\frac{E_{1}}{E_{2}}=\frac{1}{2^{4}}=\frac{1}{16}$ $\mathrm{E}_{2}=16 \mathrm{E}_{1}$ Hence, rate of emission increases sixteen time.
JIPMER-2005
Heat Transfer
149602
A metal rod at a temperature of $145^{\circ} \mathrm{C}$, radiates energy at a rate of $17 \mathrm{~W}$. If its temperature is increased to $273^{\circ} \mathrm{C}$, then it will radiate at the rate of
1 $49.6 \mathrm{~W}$
2 $17.5 \mathrm{~W}$
3 $5.3 \mathrm{~W}$
4 $67.5 \mathrm{~W}$
Explanation:
A Given that, $\mathrm{T}_{1}=145^{\circ} \mathrm{C}=273+145=418 \mathrm{~K}$ $\mathrm{E}_{1}=17 \mathrm{~W}$ $\mathrm{~T}_{2}=273^{\circ} \mathrm{C}=273+273=546 \mathrm{~K}$ According to Stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{418}{546}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=0.343$ Therefore final radiated energy, $\mathrm{E}_{2}=\frac{\mathrm{E}_{1}}{0.343}=\frac{17}{0.343}$ $\mathrm{E}_{2}=49.6 \mathrm{~W}$
JIPMER-2016
Heat Transfer
149605
When the temperature of a body is increased from $27^{\circ} \mathrm{C}$ to $127^{\circ} \mathrm{C}$. the radiation emitted by it increases by a factor of
1 $\frac{256}{81}$
2 $\frac{15}{19}$
3 $\frac{4}{5}$
4 $\frac{12}{27}$
Explanation:
A Given that, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{~T}_{2}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}$ According to stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\mathrm{E}=\sigma \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{T}_{2}^{4}}{\mathrm{~T}_{1}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\left(\frac{400}{300}\right)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{256}{81}$
149600
The surface temperature is the sun which has maximum energy emission at $500 \mathrm{~nm}$ is $6000 \mathrm{~K}$. The temperature of a star which has maximum energy emission at $400 \mathrm{~nm}$ will be
1 $8500 \mathrm{~K}$
2 $4500 \mathrm{~K}$
3 $7500 \mathrm{~K}$
4 $6500 \mathrm{~K}$
Explanation:
C Given data, $\lambda_{1}=500 \mathrm{~nm}$ $\lambda_{2}=400 \mathrm{~nm}$ $\mathrm{T}_{1}=6000 \mathrm{~K}$ $\mathrm{T}_{2}=$ ? According to Wein's Displacement law- $\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2}$ Where, $\lambda_{1}=$ wavelength at $T_{1}$ and $\mathrm{T}_{2}$ $500 \times 6000=400 \times \mathrm{T}_{2}$ $\mathrm{~T}_{2}=\frac{500 \times 6000}{400}$ $\mathrm{~T}_{2}=\frac{30,00000}{400}$ $\mathrm{~T}_{2}=7500 \mathrm{~K}$
JIPMER-2012
Heat Transfer
149601
The temperature of the black body increases from $T$ to $2 T$. the factor by which the rate of emission will increase, is ?
1 4
2 2
3 16
4 8
Explanation:
C Given, $\mathrm{T}_{1}=\mathrm{T}, \mathrm{T}_{2}=2 \mathrm{~T}$ According to Stefan's law- $\therefore \mathrm{E}=\sigma \mathrm{T}^{4}$ Where $\sigma$ is Stefan's constant $\therefore \quad \frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}^{4}}{(2 \mathrm{~T})^{4}}$ $\frac{E_{1}}{E_{2}}=\frac{1}{2^{4}}=\frac{1}{16}$ $\mathrm{E}_{2}=16 \mathrm{E}_{1}$ Hence, rate of emission increases sixteen time.
JIPMER-2005
Heat Transfer
149602
A metal rod at a temperature of $145^{\circ} \mathrm{C}$, radiates energy at a rate of $17 \mathrm{~W}$. If its temperature is increased to $273^{\circ} \mathrm{C}$, then it will radiate at the rate of
1 $49.6 \mathrm{~W}$
2 $17.5 \mathrm{~W}$
3 $5.3 \mathrm{~W}$
4 $67.5 \mathrm{~W}$
Explanation:
A Given that, $\mathrm{T}_{1}=145^{\circ} \mathrm{C}=273+145=418 \mathrm{~K}$ $\mathrm{E}_{1}=17 \mathrm{~W}$ $\mathrm{~T}_{2}=273^{\circ} \mathrm{C}=273+273=546 \mathrm{~K}$ According to Stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{418}{546}\right)^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=0.343$ Therefore final radiated energy, $\mathrm{E}_{2}=\frac{\mathrm{E}_{1}}{0.343}=\frac{17}{0.343}$ $\mathrm{E}_{2}=49.6 \mathrm{~W}$
JIPMER-2016
Heat Transfer
149605
When the temperature of a body is increased from $27^{\circ} \mathrm{C}$ to $127^{\circ} \mathrm{C}$. the radiation emitted by it increases by a factor of
1 $\frac{256}{81}$
2 $\frac{15}{19}$
3 $\frac{4}{5}$
4 $\frac{12}{27}$
Explanation:
A Given that, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$ $\mathrm{~T}_{2}=127^{\circ} \mathrm{C}=127+273=400 \mathrm{~K}$ According to stefan's law, $\mathrm{E} \propto \mathrm{T}^{4}$ $\mathrm{E}=\sigma \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\frac{\mathrm{T}_{1}^{4}}{\mathrm{~T}_{2}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{\mathrm{T}_{2}^{4}}{\mathrm{~T}_{1}^{4}}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\left(\frac{400}{300}\right)^{4}$ $\frac{\mathrm{E}_{2}}{\mathrm{E}_{1}}=\frac{256}{81}$