Tag: physics

Questions Related to physics

There are two planets $A$ and $B$ at a large distance Planet $A$ is bigger and hotter than planet $B$. The angular diameter of planet $A$ is $40$ minute of arc as seen from planet $B$. The energy received by planet $B$ is $3cal-cm^{-2}$ per minute. Assuming the radiation to be black body in character. Given that stefan costant is $5.67\times 10^{-8}\ Wm^{-2}\ K^{-4}$. The temperature of planet $A$ is

stefan's law black body radiation heat transfer thermal properties physics

  1. $(10.93\times 10^{14})^{1/4}\ K$

  2. $(53.21\times 10^{14})^{1/4}\ K$

  3. $(63.63\times 10^{14})^{1/4}\ K$

  4. $(63.21\times 10^{14})^{1/4}\ K$

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Correct Option: C

A blackened steel plate is put in a dark room after being heated up to a high temperature. A white spot on the plate appears. 

stefan's law black body radiation heat transfer thermal properties physics

  1. brighter than the plate

  2. as bright as the plate

  3. dull as compared to the plate

  4. appears to be yellow

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Correct Option: A

Boltzmann's constant$ K = 1.38 \times 10^{-23} J/k $ The energy associated with helium atom the surface of sun, where surface temperature is 6000 K is

stefan's law black body radiation heat transfer thermal properties physics

  1. $ 1.242 \times 10^{-19} J $

  2. $ 2.484 \times 10^{-19} J $

  3. $ 207 \times 10^{-19} J $

  4. $ 0.621 \times 10^{-19} J $

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Correct Option: A

If in an ideal gas $r$ is radius of molecule, $P$ is pressure, $T$ is absolute temperature and $k$ is Boltzmann's constant, then mean free path $\overline { \lambda  } $ of gas molecules is given as

stefan's law black body radiation heat transfer thermal properties physics

  1. $\dfrac { 4\pi \sqrt { 2 } PT }{ k{ r }^{ 2 } } $

  2. $\dfrac { 4\pi \sqrt { 2 } kT }{ P{ r }^{ 2 } } $

  3. $\dfrac { kP }{ 4\pi \sqrt { 2 } { r }^{ 2 }T } $

  4. $\dfrac { kT }{ 4\pi \sqrt { 2 } { r }^{ 2 }P } $

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Correct Option: D

Solar constant for earth is $2 \mathrm { cal } / \mathrm { min } \mathrm { cm } ^ { 2 } ,$ if distance ofmerary from sun is 0.4 times than distance of earthfrom sun then solar constant for mercury will be? 

stefan's law black body radiation heat transfer thermal properties physics

  1. 12.5$\mathrm { cal } / \mathrm { min } \mathrm { cm } ^ { 2 }$

  2. 25$\mathrm { cal } / \mathrm { min } \mathrm { cm } ^ { 2 }$

  3. 0.32$\mathrm { cal } / \mathrm { min } \mathrm { cm } ^ { 2 }$

  4. 2$\mathrm { cal } / \mathrm { min } \mathrm { cm } ^ { 2 }$

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Correct Option: C

The solar energy incident on the roof in 1 hour of dimension $ 8m \times 20m$ will be

stefan's law black body radiation heat transfer thermal properties physics

  1. $5.76\times { 10 }^{ 8 }J$

  2. $5.76\times { 10 }^{ 7 }J$

  3. $5.76\times { 10 }^{ 6 }J$

  4. $5.76\times { 10 }^{ 5 }J$

Show answer
Correct Option: A
Explanation:

Here the power per unit area is given, $I=10^3  W/m^2$

So, the total power $=I\times $ area of roof $=10^3\times (8\times 20)=1.6\times 10^5 W$ 
Since power is the energy divided by time so, energy, $E=Pt=1.6\times 10^5 \times (3600)=5.76\times 10^8  J$

The Sun delivers ${{10}^{3}}W/{{m}^{2}}$ of electromagnetic flux to the Earth's surface.The total power that is incident on a roof of dimensions $8m\times 20m$, will be

stefan's law black body radiation heat transfer thermal properties physics

  1. $6.4\times { 10 }^{ 3 }W$

  2. $3.4\times { 10 }^{ 4 }W$

  3. $1.6\times { 10 }^{ 5 }W$

  4. none of these

Show answer
Correct Option: C
Explanation:

Here the power per unit area is given, $I=10^3  W/m^2$

So, the total power $=I\times $ area of roof $=10^3\times (8\times 20)=1.6\times 10^5 W$ 

Choose the correct relation, when the temperature of an isolated black body falls from $T _{1}$ to $T _{2}$ in time $'t'$, and assume $'c'$ to be a constant.

stefan's law black body radiation heat transfer thermal properties physics

  1. $t - c \left (\dfrac {1}{T _{2}} - \dfrac {1}{T _{1}}\right )$

  2. $t = c \left (\dfrac {1}{T _{2}^{2}} - \dfrac {1}{T _{1}^{2}}\right )$

  3. $t = c \left (\dfrac {1}{T _{2}^{3}} - \dfrac {1}{T _{1}^{3}}\right )$

  4. $t = c \left (\dfrac {1}{T _{2}^{4}} - \dfrac {1}{T _{1}^{4}}\right )$

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Correct Option: C

Calculate the surface temperature of the planet, if the energy radiated by unit area in unit time is $5.67 \times 10^4$ watt.

stefan's law black body radiation heat transfer thermal properties physics

  1. $1273^{\circ}C$

  2. $1000^{\circ}C$

  3. $727^{\circ}C$

  4. 727K

Show answer
Correct Option: C
Explanation:

According to stefan's Boltzmann law, the energy radiated per unit time:
$E=\sigma A{ T }^{ 4 }$
It is given that: ${E}={5.67}\times{10}^{4}$
Therefore, ${5.67}\times{10}^{4}={5.67}\times{10}^{-8}\times1\times{T}^{4}$
So, ${T}={1000}K$
${T}={1000-273}={727} \  ^oC$

A hot liquid is kept in a big room . the logarithm of the numerical value of the temperature difference between the liquid and the room is plotted against time. the plot will be very nearly

stefan's law black body radiation heat transfer thermal properties physics

  1. a straight line

  2. a circular arc

  3. a parabola

  4. an ellipse

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Correct Option: A