149550
A star $A$ is 100 times brighter than star $B$. Then $m_{B}-m_{A}$ the difference in their apparent magnitudes is
1 100
2 0.01
3 5
4 0.2
Explanation:
C Given that, $\mathrm{I}_{\mathrm{B}}: \mathrm{I}_{\mathrm{A}}=1: 100$ As we know that, the difference in apparent magnitude is equal to $\log$ value of intensities. $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=-2.5 \log \left[\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}\right]$ $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=-2.5 \log \left(\frac{1}{100}\right)$ $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=-2.5 \log \left(10^{-2}\right)$ $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=5$
COMEDK 2012
Heat Transfer
149551
If the temperature of the Sun gets doubled, the rate of energy received on the Earth will increase by a factor of
1 2
2 4
3 8
4 16
Explanation:
D For hot objects, the power radiated per unit area given by Stefan-Boltzmann's law- $\mathrm{E}=\sigma \mathrm{AT}^{4}$ $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}}{2 \mathrm{~T}}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{E}$ This means that the rate of energy received on earth becomes 16 times when the temperature of sun get doubled.
WB JEE 2016
Heat Transfer
149553
A black body of surface area $10 \mathrm{~cm}^{2}$ is at $27^{\circ} \mathrm{C}$. The rate of energy radiated by it is $E$. If its temperature is raised to $627^{\circ} \mathrm{C}$, the rate of energy radiated will increase by
1 $16 \mathrm{E}$
2 $27 \mathrm{E}$
3 $80 \mathrm{E}$
4 $81 \mathrm{E}$
Explanation:
D Given that, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}$ $\mathrm{~T}_{2}=627^{\circ} \mathrm{C}+273=900 \mathrm{~K}$ $\mathrm{E}_{1}=\mathrm{E} \quad \mathrm{E}_{2}=?$ We know that Stefan Boltzman law $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{300}{900}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{3}{9}\right)^{4}$ $\mathrm{E}_{2}=81 \mathrm{E}$
UP CPMT-2001
Heat Transfer
149556
The wavelength of the radiation emitted by a black body is $6 \mathrm{~mm}$ and Wien's constant is $3 \times 10^{-3} \mathrm{mK}$. Then temperature of black body is:
1 $5 \mathrm{~K}$
2 $3 \mathrm{~K}$
3 $0.5 \mathrm{~K}$
4 $50 \mathrm{~K}$
Explanation:
C Given $\lambda=6 \mathrm{~mm}=6 \times 10^{-3} \mathrm{~m}$ $\mathrm{b}=3 \times 10^{-3} \mathrm{mK} \text {. }$ According to Wien's law, $\lambda \mathrm{T}=\mathrm{b}$ $\mathrm{T}=\frac{\mathrm{b}}{\lambda}$ $\mathrm{T}=\frac{3 \times 10^{-3}}{6 \times 10^{-3}}$ $\mathrm{~T}=\frac{0.5 \times 10^{-3}}{10^{-3}}$ $\mathrm{~T}=0.5 \mathrm{~K}$
149550
A star $A$ is 100 times brighter than star $B$. Then $m_{B}-m_{A}$ the difference in their apparent magnitudes is
1 100
2 0.01
3 5
4 0.2
Explanation:
C Given that, $\mathrm{I}_{\mathrm{B}}: \mathrm{I}_{\mathrm{A}}=1: 100$ As we know that, the difference in apparent magnitude is equal to $\log$ value of intensities. $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=-2.5 \log \left[\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}\right]$ $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=-2.5 \log \left(\frac{1}{100}\right)$ $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=-2.5 \log \left(10^{-2}\right)$ $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=5$
COMEDK 2012
Heat Transfer
149551
If the temperature of the Sun gets doubled, the rate of energy received on the Earth will increase by a factor of
1 2
2 4
3 8
4 16
Explanation:
D For hot objects, the power radiated per unit area given by Stefan-Boltzmann's law- $\mathrm{E}=\sigma \mathrm{AT}^{4}$ $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}}{2 \mathrm{~T}}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{E}$ This means that the rate of energy received on earth becomes 16 times when the temperature of sun get doubled.
WB JEE 2016
Heat Transfer
149553
A black body of surface area $10 \mathrm{~cm}^{2}$ is at $27^{\circ} \mathrm{C}$. The rate of energy radiated by it is $E$. If its temperature is raised to $627^{\circ} \mathrm{C}$, the rate of energy radiated will increase by
1 $16 \mathrm{E}$
2 $27 \mathrm{E}$
3 $80 \mathrm{E}$
4 $81 \mathrm{E}$
Explanation:
D Given that, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}$ $\mathrm{~T}_{2}=627^{\circ} \mathrm{C}+273=900 \mathrm{~K}$ $\mathrm{E}_{1}=\mathrm{E} \quad \mathrm{E}_{2}=?$ We know that Stefan Boltzman law $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{300}{900}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{3}{9}\right)^{4}$ $\mathrm{E}_{2}=81 \mathrm{E}$
UP CPMT-2001
Heat Transfer
149556
The wavelength of the radiation emitted by a black body is $6 \mathrm{~mm}$ and Wien's constant is $3 \times 10^{-3} \mathrm{mK}$. Then temperature of black body is:
1 $5 \mathrm{~K}$
2 $3 \mathrm{~K}$
3 $0.5 \mathrm{~K}$
4 $50 \mathrm{~K}$
Explanation:
C Given $\lambda=6 \mathrm{~mm}=6 \times 10^{-3} \mathrm{~m}$ $\mathrm{b}=3 \times 10^{-3} \mathrm{mK} \text {. }$ According to Wien's law, $\lambda \mathrm{T}=\mathrm{b}$ $\mathrm{T}=\frac{\mathrm{b}}{\lambda}$ $\mathrm{T}=\frac{3 \times 10^{-3}}{6 \times 10^{-3}}$ $\mathrm{~T}=\frac{0.5 \times 10^{-3}}{10^{-3}}$ $\mathrm{~T}=0.5 \mathrm{~K}$
149550
A star $A$ is 100 times brighter than star $B$. Then $m_{B}-m_{A}$ the difference in their apparent magnitudes is
1 100
2 0.01
3 5
4 0.2
Explanation:
C Given that, $\mathrm{I}_{\mathrm{B}}: \mathrm{I}_{\mathrm{A}}=1: 100$ As we know that, the difference in apparent magnitude is equal to $\log$ value of intensities. $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=-2.5 \log \left[\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}\right]$ $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=-2.5 \log \left(\frac{1}{100}\right)$ $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=-2.5 \log \left(10^{-2}\right)$ $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=5$
COMEDK 2012
Heat Transfer
149551
If the temperature of the Sun gets doubled, the rate of energy received on the Earth will increase by a factor of
1 2
2 4
3 8
4 16
Explanation:
D For hot objects, the power radiated per unit area given by Stefan-Boltzmann's law- $\mathrm{E}=\sigma \mathrm{AT}^{4}$ $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}}{2 \mathrm{~T}}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{E}$ This means that the rate of energy received on earth becomes 16 times when the temperature of sun get doubled.
WB JEE 2016
Heat Transfer
149553
A black body of surface area $10 \mathrm{~cm}^{2}$ is at $27^{\circ} \mathrm{C}$. The rate of energy radiated by it is $E$. If its temperature is raised to $627^{\circ} \mathrm{C}$, the rate of energy radiated will increase by
1 $16 \mathrm{E}$
2 $27 \mathrm{E}$
3 $80 \mathrm{E}$
4 $81 \mathrm{E}$
Explanation:
D Given that, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}$ $\mathrm{~T}_{2}=627^{\circ} \mathrm{C}+273=900 \mathrm{~K}$ $\mathrm{E}_{1}=\mathrm{E} \quad \mathrm{E}_{2}=?$ We know that Stefan Boltzman law $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{300}{900}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{3}{9}\right)^{4}$ $\mathrm{E}_{2}=81 \mathrm{E}$
UP CPMT-2001
Heat Transfer
149556
The wavelength of the radiation emitted by a black body is $6 \mathrm{~mm}$ and Wien's constant is $3 \times 10^{-3} \mathrm{mK}$. Then temperature of black body is:
1 $5 \mathrm{~K}$
2 $3 \mathrm{~K}$
3 $0.5 \mathrm{~K}$
4 $50 \mathrm{~K}$
Explanation:
C Given $\lambda=6 \mathrm{~mm}=6 \times 10^{-3} \mathrm{~m}$ $\mathrm{b}=3 \times 10^{-3} \mathrm{mK} \text {. }$ According to Wien's law, $\lambda \mathrm{T}=\mathrm{b}$ $\mathrm{T}=\frac{\mathrm{b}}{\lambda}$ $\mathrm{T}=\frac{3 \times 10^{-3}}{6 \times 10^{-3}}$ $\mathrm{~T}=\frac{0.5 \times 10^{-3}}{10^{-3}}$ $\mathrm{~T}=0.5 \mathrm{~K}$
149550
A star $A$ is 100 times brighter than star $B$. Then $m_{B}-m_{A}$ the difference in their apparent magnitudes is
1 100
2 0.01
3 5
4 0.2
Explanation:
C Given that, $\mathrm{I}_{\mathrm{B}}: \mathrm{I}_{\mathrm{A}}=1: 100$ As we know that, the difference in apparent magnitude is equal to $\log$ value of intensities. $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=-2.5 \log \left[\frac{\mathrm{I}_{\mathrm{B}}}{\mathrm{I}_{\mathrm{A}}}\right]$ $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=-2.5 \log \left(\frac{1}{100}\right)$ $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=-2.5 \log \left(10^{-2}\right)$ $\mathrm{m}_{\mathrm{B}}-\mathrm{m}_{\mathrm{A}}=5$
COMEDK 2012
Heat Transfer
149551
If the temperature of the Sun gets doubled, the rate of energy received on the Earth will increase by a factor of
1 2
2 4
3 8
4 16
Explanation:
D For hot objects, the power radiated per unit area given by Stefan-Boltzmann's law- $\mathrm{E}=\sigma \mathrm{AT}^{4}$ $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}}{2 \mathrm{~T}}\right)^{4}$ $\mathrm{E}_{2}=16 \mathrm{E}$ This means that the rate of energy received on earth becomes 16 times when the temperature of sun get doubled.
WB JEE 2016
Heat Transfer
149553
A black body of surface area $10 \mathrm{~cm}^{2}$ is at $27^{\circ} \mathrm{C}$. The rate of energy radiated by it is $E$. If its temperature is raised to $627^{\circ} \mathrm{C}$, the rate of energy radiated will increase by
1 $16 \mathrm{E}$
2 $27 \mathrm{E}$
3 $80 \mathrm{E}$
4 $81 \mathrm{E}$
Explanation:
D Given that, $\mathrm{T}_{1}=27^{\circ} \mathrm{C}+273=300 \mathrm{~K}$ $\mathrm{~T}_{2}=627^{\circ} \mathrm{C}+273=900 \mathrm{~K}$ $\mathrm{E}_{1}=\mathrm{E} \quad \mathrm{E}_{2}=?$ We know that Stefan Boltzman law $\mathrm{E} \propto \mathrm{T}^{4}$ $\frac{\mathrm{E}_{1}}{\mathrm{E}_{2}}=\left(\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{300}{900}\right)^{4}$ $\frac{\mathrm{E}}{\mathrm{E}_{2}}=\left(\frac{3}{9}\right)^{4}$ $\mathrm{E}_{2}=81 \mathrm{E}$
UP CPMT-2001
Heat Transfer
149556
The wavelength of the radiation emitted by a black body is $6 \mathrm{~mm}$ and Wien's constant is $3 \times 10^{-3} \mathrm{mK}$. Then temperature of black body is:
1 $5 \mathrm{~K}$
2 $3 \mathrm{~K}$
3 $0.5 \mathrm{~K}$
4 $50 \mathrm{~K}$
Explanation:
C Given $\lambda=6 \mathrm{~mm}=6 \times 10^{-3} \mathrm{~m}$ $\mathrm{b}=3 \times 10^{-3} \mathrm{mK} \text {. }$ According to Wien's law, $\lambda \mathrm{T}=\mathrm{b}$ $\mathrm{T}=\frac{\mathrm{b}}{\lambda}$ $\mathrm{T}=\frac{3 \times 10^{-3}}{6 \times 10^{-3}}$ $\mathrm{~T}=\frac{0.5 \times 10^{-3}}{10^{-3}}$ $\mathrm{~T}=0.5 \mathrm{~K}$
