0

Stefan's law - class-XI

Description: stefan's law
Number of Questions: 89
Created by:
Tags: physics heat transfer thermal properties
Attempted 0/88 Correct 0 Score 0

The value of solar constant is approximately  :

  1. $ 1340\ watt/m^{2}$

  2. $ 430\ watt/m^{2}$

  3. $ 340\ watt/m^{2}$

  4. $ 1388\ watt/m^{2}$


Correct Option: D
Explanation:

The solar constant is defined as the amount of heat energy received per second per unit area by a perfect black body placed at the surface of the Earth with its surface being held perpendicular to the direction of the sun's rays.

The value of solar constant is $1388$($\dfrac{watt}{{meter}^{2}}$) or $2$($\dfrac{cal}{{cm}{\times} {min}}$)

A heated body emits radiation which has maximum intensity at frequency $v _m$. If the temperature of the body is doubled

  1. the maximum intensity radiation will be at frequency $2v _m$

  2. the maximum intensity radiation will be at frequency $\displaystyle\dfrac{1}{2}v _m$

  3. the total emitted energy will increase by a factor of $16$

  4. the total emitted energy will increase by a factor of $2$


Correct Option: A,C
Explanation:

Wien's displacement law states maximum intensity wavength $ \lambda _{m}\propto \dfrac{1}{T}$
Also for any photon,$ \lambda \propto \dfrac{1}{\nu}$
Hence, frequency $\nu _m \propto T$
Doubling of temperature leads to doubling of frequency from $\nu _m$ to $ 2\nu _m$
From Stefan's law, power is directly proportional to $T^4$
Hence $ T \rightarrow 2T \Rightarrow E \rightarrow (\dfrac {2T}{T})^4E=16E$

The amount of radiations emitted by a black body depends on its

  1. size

  2. mass

  3. temperature

  4. density


Correct Option: C
Explanation:

The radiations emitted by the body only depend on the type of surface emitting and the temperature difference between the body and the surroundings.

The amplitudes of radiations from a cylindrical heat source is related to the distance are

  1. $ A \propto 1/{d}^2$

  2. $\displaystyle A \propto \frac{1}{ d} $

  3. $ A \propto d$

  4. $ A \propto d^2$


Correct Option: B
Explanation:

The intensity is inversely proportional to  square of distance and intensity is directly proportional to square of amplitude. So, amplitude is inversely proportional to distance. 

Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is

  1. $(\cfrac{65}{2})^{\frac{1}{4}}T$

  2. $(\cfrac{97}{4})^{\frac{1}{4}}T$

  3. $(\cfrac{97}{2})^{\frac{1}{4}}T$

  4. $(97)^{\frac{1}{4}}T$


Correct Option: C

The energy emitted by a black body at $727^oC$ is E. If the temperature of the body is increased by $227^oC$, the emitted energy will become

  1. 13 times

  2. 2.27 times

  3. 1.9 times

  4. 3.9 times


Correct Option: B
Explanation:

Here we know that energy emitted by any body is given by ${E}=\sigma{T^{4}}$

So, at temperature ${T}={727}^{o}C$ energy emitted will be ${E}$
at temperature ${T} _{1}={727+227}={954}^{o}C$ energy emitted will be ${E} _{1}={\sigma}{T} _{1}^{4}$
$\dfrac { E }{ { E } _{ 1 } } =\dfrac { { T }^{ 4 } }{ { T } _{ 1 }^{ 4 } }$
${ E } _{ 1 }=\dfrac { E\times { T } _{ 1 }^{ 4 } }{ { T }^{ 4 } } =\dfrac { E\times { 954 }^{ 4 } }{ { 727 }^{ 4 } } =2.96E$

The radiation emitted by a star $A$ is $10000$ times that of the sun. If the surface temperature of the sun and star $A$ are $6000:K$ and $2000:K$, respectively, the ratio of the radii of the star $A$ and the sun is

  1. $300:1$

  2. $600:1$

  3. $900:1$

  4. $1200:1$


Correct Option: C
Explanation:
Stars can be approximated as black bodies.
Hence by stefan's law, power emitted=P=$\sigma AT^4$
Thus,$ \dfrac{P _A}{P _{sun}}=\dfrac{r _A^2T _A^4}{r _{sun}^2T _{sun}^4}$
$\Rightarrow (\dfrac{r _A}{r _{sun}})^2=10000\times (\dfrac{6000}{2000})^4 \rightarrow \dfrac{r _A}{r _{sun}}=900$
So required ratio is 900:1

In pyrometer , temperature measured is proportional to $\underline{\hspace{0.5in}}$ energy emitted by the body 

  1. light

  2. electric

  3. radiation

  4. All the above


Correct Option: C
Explanation:

Stefan- Boltzann law, $j^{ \star  }=\varepsilon \sigma T^{ 4 }$ connects temperature T with thermal radiation or irradiance  $j^{ \star  }$.
Thus measuring the irradiance with pyrometer yields the temperature of the body.

Two bodies of same shape and having emissivities 0.1 and 0.9 respectively radiate same energy per second. The ratio of their temperature is :

  1. $\sqrt{3}:1$

  2. $1:\sqrt{3}$

  3. $3:1$

  4. $1:3$


Correct Option: A
Explanation:
$\dfrac{E}{t}=e \sigma A T^4$
From above equation which is Stefan's Law of radiation, it is clear that:
$\dfrac{E _1}{E _2} = \dfrac{{e} _{1}\sigma{T} _{1}^{4}}{{e} _{2}\sigma{T} _{2}^{4}}$

$1 = \dfrac{{0.1}{T} _{1}^{4}}{{0.9}{T} _{2}^{4}}$

$\dfrac{{T} _{1}}{{T} _{2}} = \dfrac{\sqrt{3}}{1} $

Two bodies A and B are kept in an evacuated chamber at $27^oC$. The temperature of A and B are $327^oC$ and $427^oC$ respectively. The ratio of rate of loss of heat from A and B will be

  1. 0.25

  2. 0.52

  3. 1.52

  4. 2.52


Correct Option: B
Explanation:

The power radiated is directly proportional to fourth power of absolute temperature.
i.e.
$P \propto T^{4}$
$\frac{P _{1}}{P _{2}} = (\frac{T _{1}}{T _{2}})^{4}$
$\frac{P _{1}}{P _{2}} = (\frac{327+273}{427+273})^{4} = 0.53$
Hence the ratio of rate of heat loss = 0.53
Hence option B is correct.

The radiation emitted by a perfectly black body is proportional to 

  1. temperature on ideal gas scale

  2. fourth root of temperature on ideal gas scale

  3. fourth power of temperature on ideal gas scale

  4. square of temperature on ideal gas scale


Correct Option: C
Explanation:

Stefan-Boltzmann law states that total power radiated by a perfectly black body is
$P=A\sigma { T }^{ 4 }$
so the radiation emitted by a perfectly black body is proportional to fourth power of temperature on ideal gas scale.
option (C) is the correct answer.

The amount of heat energy radiated per second by a surface depends upon:

  1. Area of the surface

  2. Difference of temperature between the surface and its surroundings

  3. Nature of the surface

  4. All the above


Correct Option: D
Explanation:

Refer Stefan's Law of Radiation: $Q = \eta \sigma A \delta{T}^{4}$
where, $\sigma = conductivity\ A$  = area $T$ = difference of the temperature 

The thermal radiation emitted by a body is proportional to $T^{n}$ where $T$ is its absolute temperature. The value of $n$ is exactly $4$ for

  1. a blackbody

  2. all bodies

  3. bodies painted balck only

  4. polished bodies only


Correct Option: B
Explanation:

By Stefan's Law, rate of thermal radiation is directly proportional to fourth power of temperature of the body.
$Q = \sigma {T}^{4}$

A black body radiates energy at the rate of $E\ watt/m$$^{2}$ at a high temperature $T^{o}K$ when the temperature is reduced to $\left [ \dfrac{T}{2} \right ]^{o}K$ Then radiant energy is

  1. $4E$

  2. $16E$

  3. $\dfrac{E}{4}$

  4. $\dfrac{E}{16}$


Correct Option: D
Explanation:

We know that from stefans-boltzman law: $E\propto { T }^{ 4 }$
if temperature will be reduces half form the initial value, then
${E} _{1}\propto ({ \dfrac { T }{ 2 } ) }^{ 4 }$
${E} _{1}\propto\dfrac{E}{16}$

All bodies emit heat energy from their surfaces by virtue of their temperature. This heat energy is called radiant energy of thermal radiation. The heat that we receive from the sun is transferred to us by a process which, unlike conduction orconvection, does not require the help of a medium in the intervening space which is almost free of particles. Radiant energy travels in space as electromagnetic spectrum. Thermal radiations travel through vacuum with the speed oflight. Thermal radiations obey the same laws of reflection and refraction as light does. They exhibit the phenomena of interference, diffraction and polarization as light does.
The emission of radiation from a hot body is expressed in terms of that emitted from a reference body (called the black body) at the same temperature. A black body absorbs and hence emits radiations of all wavelengts. The total energy E emitted by a unit area of a black bodyper second is given by $E =\sigma T^{4}$ where T is the absolute temperature of the body and $\sigma $ is a constant known as Stefans constant. If the body is not a perfect black body, then $E =\varepsilon \sigma  T^{4}$where $\varepsilon $ is the emissivity of the body.

Which of the following devices is used to detect thermal radiations?

  1. Constant volume air thermometer

  2. Platinum resistance thermometer

  3. Thermostat

  4. Thermopile


Correct Option: D
Explanation:

Thermopile is a very sensitive device which converts thermal energy to electrical energy.
It is used to detect thermal radiations.

The rate of radiation from a black body at $0$$^{o}$C is $E$. The rate of radiation from this black body at $273$$^{o}$C is :

  1. $2E$

  2. $E/2$

  3. $16E$

  4. $E/16$


Correct Option: C
Explanation:

As we know that: ${E} \ {\propto} \ {T}^{4}$
So, $\dfrac{E}{{T} _{1}^{4}}=\dfrac{{E} _{2}}{{T} _{2}^{4}}$
${E} _{2}=\dfrac{{T} _{2}^{4}\times{E}}{{T} _{1}^{4}}=\dfrac{{546}^{4}}{{273}^{4}}$
${E} _{2}={16E}$

Two spherical black bodies of radii $r _{1} $ and $  r _{2}$ are with surface temperatures $T _{1} $ and $ T _{2}$ respectively radiate the same power. $r _{1} / r _{2}$ must be equal to

  1. $(T _{1}/T _{2})^{2}$

  2. $(T _{2}/T _{1})^{2}$

  3. $(T _{1}/T _{2})^{4}$

  4. $(T _{2}/T _{1})^{4}$


Correct Option: B
Explanation:

$E=\varepsilon \sigma A{ T }^{ 4 }$

Given that both spherical body radiate with same power.
So equating ${E} _{1}={E} _{2}$
${A}=4{\pi}{r}^{2}$
${ r } _{ 1 }^{ 2 }{ T } _{ 1 }^{ 2 }={ r } _{ 2 }^{ 2 }{ T } _{ 2 }^{ 4 }$

$\dfrac { { r } _{ 1 } }{ { r } _{ 2 } } ={ (\dfrac { { T } _{ 2 } }{ { T } _{ 1 } } ) }^{ 2 }$

The temperature of the sun is doubled, the rate of energy received on earth will be increased by a factor of :

  1. 2

  2. 4

  3. 8

  4. 16


Correct Option: D
Explanation:

We know that from Stefan's Boltzmann relation: $E\propto { T }^{ 4 }$
if the temperature of the sun will be doubled, then : $E\propto ({ 2T })^{ 4 }$
Hence, $E\  will\  increase \ by \ the \ factor \ of \  { 16}$.

A black body is at temperature $300K$. It emits energy at a rate, which is proportional to 

  1. ${(300)}^{4}$

  2. ${(300)}^{3}$

  3. ${(300)}^{2}$

  4. $300$


Correct Option: A
Explanation:

For black body radiation
$E=\sigma{T}^{4}$ or $E\propto {T}^{4}$
Rate of emission of energy $\propto {(300)}^{4}$

If the absolute temperature of a blackbody is doubled, then the maximum energy density

  1. Increases to 16 times

  2. Increases to 32 times

  3. Decreases to 16 times

  4. Decreases to 32 times


Correct Option: A
Explanation:

The power with which a body(in this case black body) radiates is directly proportional to the fourth power of absolute temperature:
$P = kT^{4}$
i.e.
$P _{1} = kT _{1}^{4}$
If the absolute temperature is doubled,
$P _{2} = k(2T _{1})^{4} = 16kT _{1}^{4}$
Now $\dfrac{P _{2}}{P _{1}} = \dfrac{16}{1}$ 

Hence energy density is increased 16 times.

Intensity of heat radiation emitted by body is believed to be proportional to fourth power of absolute temperature of the body. The proportionality constant also known as Boltzmann's constant may have possible value of :

  1. $5.67\times 10^{-8} watt/K^4 $

  2. $5.67\times 10^{-8} watt/m^2 K^4 $

  3. $5.67\times 10^{-8} J/K^4 $

  4. $5.67\times 10^{-8} Js/K^4 $


Correct Option: B
Explanation:

From the given question,

$I\propto T^4$

$I=k T^4$

where $k=$ Stefan-Boltzmann's constant

$k=5.67\times10^{-8} W/m^2K^4 $

The correct option is B.

A black body at a temperature of $227^oC$ radiates heat energy at the rate 5 cal/cm$^{2}-s$. At a temperature of $727^oC$, the rate of heat radiated per unit area in cal/cm$^2$ will be

  1. 80

  2. 160

  3. 250

  4. 500


Correct Option: A
Explanation:

According to Stefen's Law, the rate of heat radiation from body is proportional to the fourth power of body's temperature.

Thus $P\propto T^4$
$\implies \dfrac{P _2}{P _1}=\dfrac{T _2^4}{T _1^4}$
$=16$
$\implies P _2=80cal/cm^2-s$

For a block body temperature $727^{o}C,$ its rate of energy loss is $20\ watt$ and temperature of surrounding is $227^{o}C.$ If temperature of black body is changed to $1227^{o}C$ then its rate of energy loss will be:

  1. $320\ W$

  2. $\dfrac {304}{3}\ W$

  3. $240 W$

  4. $120 W$


Correct Option: A
Explanation:

It is given that,

Temperature of surrounding

  $ {{T} _{0}}={{227}^{0}}C=500\ K $

 $ {{T} _{1}}={{727}^{0}}C=1000\ K $

 $ {{T} _{2}}={{1227}^{0}}C=1500\ K $

 $ {{E} _{1}}=20\ Watt\,\, $

 $ {{E} _{2}}=? $

According to Stefn boltzmann law:

$ E=\sigma {{T}^{4}} $

Or

 $ {{E} _{1}}=\sigma ({{T} _{1}}-{{T} _{0}})^4 $

 $ {{E} _{2}}=\sigma ({{T} _{2}}-{{T} _{0}})^4 $

Taking ratios of above equations:

For $ {{E} _{1}}=20\ Watt $

 $ \dfrac{20}{{{E} _{2}}}={{\left( \dfrac{500}{1000} \right)}^{4}} $

 $ \dfrac{20}{{{E} _{2}}}=\left( \dfrac{1}{16} \right) $

 $ {{E} _{2}}=320\ Watt $

The power received at distance $d$ from a small metallic sphere of radius $r(<<d)$ and at absolute temperature $T$ is $P$. If the temperature is doubled and distance reduced to half of the initial value, then the power received at that point will be:

  1. $4p$

  2. $8p$

  3. $32p$

  4. $64p$


Correct Option: D
Explanation:
Energy received per second i.e., power $P\alpha \dfrac{T^4}{d^2}=k\dfrac{T^4}{d^2}$
if temperature is double than T become 2T and distance become half than d become $\dfrac{d}{2}$
than power $ p _{1}=k\dfrac{(2T)^4}{(\dfrac{d}{2})^2}=64k\dfrac{T^4}{d^2}=64P$
Hence D option is correct.

What is the value of solar constant if the energy received by $ 12$ m$^2$ area in $2$ minutes is $2016$ kJ?

  1. $1.4 \times 10^2 J s^{-1}  m^{-2}$

  2. $1400 J s^{-1}m^{-2}$

  3. $84 kJ s^{-1} m^{-2}$

  4. $84 J s^{-1} m^{-2}$


Correct Option: B
Explanation:
Solar constant = Energy received by unit area in unit time from Sun
$=\cfrac{2016\times 10^3}{12\times (2\times 60)} \\=1400Js^{-1}m^2$

If a graph is plotted by taking spectral emissive power along $y-$axis and wavelength along x-axis is:

  1. Emissivity

  2. Total intensity of radiation

  3. Diffusivity

  4. Solar constant


Correct Option: A
Explanation:

Emissive power varies according to Stefan-Boltzmann law as :

$E=\sigma {{T}^{4}}$

According to Planck’s distribution law:

${{E} _{\lambda }}(\lambda ,T)=\dfrac{{{C} _{1}}}{{{\lambda }^{5}}\left[ \exp (\dfrac{{{C} _{2}}}{\lambda T})-1 \right]}$

The graph shows that the emitted radiation varies with wavelength and also it shows Emissivity.

 

A spherical body of area A and emissivity $0.6$ is kept inside a perfectly black body. Total heat radiated by the body at temperature T is?

  1. $0.4\sigma AT^4$

  2. $0.8\sigma AT^4$

  3. $0.6\sigma AT^4$

  4. $1.0\sigma AT^4$


Correct Option: D
Explanation:

When a non black body is placed inside a hollow enclosure the total radiation from the body is the sum of what it would emit in the open ( with e<1 ) and the part (1-a) of the incident radiation from the walls reflected by it.

The two add up to a black body radiation . Hence the total radiation emitted by the body is $1.0\sigma AT^4$
1.0σAT4.

The rate of emission of radiation of ablack body at temperature $27^oC $ is $ E _1 $ . If its temperature is increased to $ 327^oC $ the rate of emission of radiation is $ E _2 . $ The relation between $ E _1 $ and $ E _2 $ is:

  1. $ E _2 = 24 E _1 $

  2. $ E _2 =16 E _1 $

  3. $ E _2 = 8 E _1 $

  4. $ E _2 = 4 E _1 $


Correct Option: B
Explanation:

In black body radiation 

$\dfrac{d\theta}{dt}=(4\pi r^{2})\sigma T^{4}$
If at $T=27^{o}C=300\ K, \dfrac{d\theta}{dt}=E _{1}$
Then, 
$E _{1}=(4\pi R^{2})\sigma(300)^{4}$
If at $T=327^{o}C=600\ k, \dfrac{d\theta}{dt}=E _{2}$
$E _{2}=(4\pi R^{2})\sigma (2)^{4}(300)^{4}$
So, $\dfrac{E _{2}}{E _{1}}=(2)^{4}\dfrac{(4\pi R^{2}\sigma (300)^{4}}{4\pi R^{2}\sigma(300)^{4}}$
$E _{2}=(2)^{4}E _{1}$
$\Rightarrow E _{2}10 E _{1}$
Option $B$ is correct






Two identical objects $A$ and $B$ are at temperatures $T _A$ and $T _B$. respectively. Both objects are placed in a room with perfectly absorbing walls maintained at a temperature $T$ ($T _A$ > $T$> $T _B$). The objects $A$ and $B$ attain the temperature $T$ eventually. Select the correct statements from the following

  1. $A$ only emits radiation, while $B$ only absorbs it until both attain the temperature $T$

  2. $A$ loses more heat by radiation than it absorbs, while $B$ absorbs more radiation than it emits until they attain the temperature $T$

  3. Both $A$ and $B$ only absorb radiation, but do not emit it, until they attain the temperature $T$

  4. Each object continuous to emit and absorb radiation even after attaining the temperature $T$


Correct Option: B
Explanation:

Since the temperature of $A$ is higher than the temperature of the surrounding hence $A$ radiates heat much larger than it absorbs heat. Since the temperature of $B$ is lower than the temperature of the surrounding hence $B$ absorbs heat much larger than it radiates.
This process goes on until both $A$ and $B$ reach the temperature $T$.
Even after reaching thermal equilibrium, both bodies keep radiating and absorbing.
Hence options $B$, $D$. 

A planet is at an average distance $d$ from the sun and its average surface temperature is $T$. Assume that the planet receives energy only from the sun and loses energy only through radiation from the surface. Neglect atmospheric effects. If $T$ $\propto d^{-n}$, the value of $n$ is :

  1. $2$

  2. $1$

  3. $\displaystyle \frac{1}{2}$

  4. $\displaystyle \frac{1}{4}$


Correct Option: C
Explanation:
Let P=power radiated by Sun
R=Radius of planet
E=energy received by planet=$\cfrac{P}{4 \pi d^2}\times \pi R^2$
Energy radiated by planet=$(4 \pi R^2)\sigma T^4 $
For thermal equilibriums:
$\Rightarrow \cfrac { P }{ 4\pi d^{ 2 } } \pi R^{ 2 }=4\pi R^{ 2 }\sigma T^{ 4 }\\ \Rightarrow T^{ 4 }\alpha \cfrac { 1 }{ d^{ 2 } } \\ \Rightarrow T\alpha \cfrac { 1 }{ d^{ { 1 }/{ 2 } } } \\ \Rightarrow T\alpha { d }^{ -\cfrac { 1 }{ 2 }  }$
So n=$\cfrac{1}{2}$

A planet radiates heat at a rate proportional to the fourth power of its surface temperature $T$. If such a steady temperature of the planet is due to an exactly equal amount of heat received from the sun then which of the following statements is true?

  1. The planet's surface temperature varies inversely as the distance of the sun

  2. The planet's surface temperature varies directly as the square of its distance from the sun

  3. The planet's surface temperature varies inversely as the square root of its distance from the sun

  4. The planet's surface temperature is proportional to the fourth power of distance from the sun


Correct Option: C
Explanation:
Planet's surface temperature varies inversely as square root of its distance from the Sun.
${ T }^{ 4 }\alpha \cfrac { 1 }{ { d }^{ 2 } } \Rightarrow T\alpha \cfrac { 1 }{ \sqrt { d }  } $

The radiation emitted by a star $A$ is $1000$ times that of the sun. If the surface temperatures of the sun and star $A$ are $6000 K$ and $2000 K$, respectively, the ratio of the radii of the star $A$ and the Sun is:

  1. 300:1

  2. 600:1

  3. 900:1

  4. 1200:1


Correct Option: C
Explanation:

$E\propto A{ T }^{ 4 }$


$\displaystyle \frac { { E } _{ 1 } }{ { E } _{ 2 } } =\frac { 1000 }{ 1 } =\frac { \pi { { r } _{ 1 } }^{ 2 }\times { T }^{ 4 } }{ \pi { { r } _{ 2 } }^{ 2 }\times { T }^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 }\times { T }^{ 4 } }{ { { r } _{ 2 } }^{ 2 }\times { T }^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 }\times { 2000 }^{ 4 } }{ { { r } _{ 2 } }^{ 2 }\times 6000^{ 4 } } =\frac { { { r } _{ 1 } }^{ 2 } }{ { { r } _{ 2 } }^{ 2 }\times 81 } $

$\displaystyle \frac { { r } _{ 1 } }{ { r } _{ 2 } } =\sqrt { \frac { 1000 }{ 1 } \times \frac { 81 }{ 1 }  } =284.6:1$

The number of oxygen molecules in a cylinder of volume $1 \mathrm { m } ^ { 3 }$ at a temperature of $27 ^ { \circ } C$ and pressure $13.8 Pa$ is
 (Boltzmaan's constant $k = 1.38 \times 10 ^ { - 23 } \mathrm { JK } ^ { - 1 }$)

  1. $6.23 \times 10 ^ { 26 }$

  2. $0.33 \times 10 ^ { 28 }$

  3. $3.3 \times 10 ^ { 21 }$

  4. none of these


Correct Option: A

A solid sphere of mass m and radius $R$ is painted black and placed inside a vacuum chamber. The walls of the chamber are maintained at temperature $T 0$ the initial temperature of the sphere is $3T _0$. The specific heat capacity of the sphere material varies with its temperature $T$ as $\alpha T^3$ where $\alpha$ is a constant. Then the sphere will cool down to temperature $2T _0$ in time ________ ($\sigma$ = Stefan Boltzmann constant)

  1. $\dfrac{m\alpha}{16\pi R^2\sigma}\ell n\left(\dfrac{16}{3}\right)$

  2. $\dfrac{m\alpha}{8\pi R^2\sigma}\ell n\left(\dfrac{4}{3}\right)$

  3. $\dfrac{m\alpha}{8\pi R^2\sigma}\ell n\left(\dfrac{3}{2}\right)$

  4. $\dfrac{m\alpha}{4\pi R^2\sigma}\ell n\left(\dfrac{8}{3}\right)$


Correct Option: A

In the nuclear fusion, $ _{1}^{2}{H}+ _{1}^{3}{H}\rightarrow _{2}^{4}{He}+n$ given that the repulsive potential energy between the two nuclie is $7.7\times 10^{-14}J$, the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann's constant $k=1.38\times 10^{-23}J/K$]-

  1. $10^{7}K$

  2. $10^{5}K$

  3. $10^{3}K$

  4. $10^{9}K$


Correct Option: D
Explanation:

Energy    $E \approx kT$

So,     $7.7\times 10^{-14} \approx 1.38\times 10^{-23}\times T$
$\implies \ T\approx  5.6\times 10^9 \ K$
Correct answer is option D.

Two bodies $A$ and $B$ have thermal emissivities of $0.01$ and $0.81$ respectively. The outer surface area of the two bodies are the same. The two bodies radiate energy at the same rate. The wavelength $\lambda _{B}$, corresponding to the maximum spectral radiancy in the radiation from $B$, is shifted from the wavelength corresponding to the maximum spectral radiancy in the radiation from $A$ by $1.00 :\mu m$. If the temperature of $A$ is $5802 :K$, then:

  1. the temperature of $B$ is $1934:K$

  2. $\lambda _{B}=1.5:\mu m$

  3. the temperature of $B$ is $11604:K$

  4. the temperature of $B$ is $2901:K$


Correct Option: A,B
Explanation:

From Stefan's Law:
$\sigma A\epsilon _AT _A^4=\sigma A\epsilon _BT _B^4$  ....(1)
where, $T _A=5802:K$ is temp of A and $T _B$ is temp of B,

$\epsilon _A=0.01$ is emissivity of A,
$\epsilon _B=0.81$ is emissivity of B,
$\sigma$ is Stefan's constant,
$A$ is the surface area of the bodies A and B

Substituting the values in (1)

$0.01 \times 5802^4 = 0.81 T _B^4$

or, $\left (\dfrac{T _B}{5802}\right )^4 = \dfrac{0.01}{0.81}=\left ( \dfrac{1}{3} \right )^4$

$\therefore T _B= \dfrac{5802}{3}=1934:K$

From Wien's displacement Law
$(\lambda _A) _mT _A=(\lambda _B) _mT _B$    ......(2)

Given, $(\lambda _B) _m = (\lambda _A) _m + 1\times 10^{-6}$  ....(3)

Substituting $(\lambda _B) _m$ from (3) in (2)
$(\lambda _A) _mT _A=( (\lambda _A) _m + 1\times 10^{-6}) T _B$
$ \therefore (\lambda _A) _m \times  3 = (\lambda _A) _m + 1\times 10^{-6}$ since $\dfrac{T _A}{T _B}=3$
$\therefore 2 (\lambda _A) _m = 10^{-6}$
$\therefore (\lambda _A) _m= 0.5\times 10^{-6}$
$\therefore (\lambda _B) _m = 0.5\times 10^{-6} + 1\times 10^{-6} =1.5 \times  10^{-6}=1.5 \mu m$

Energy associated with each molecule per degree of freedom o a system at room temperature $(27^{\circ}C)$ will be ($k$ is Boltzmann's constant)

  1. $150\;k$

  2. $(27/2)\;k$

  3. $1/2\;k$

  4. None of these


Correct Option: C

The temperature of a piece of metal is raised from $27^oC$ to $51.2^oC$. The rate at which the metal radiates energy increases nearly

  1. 1.36 times

  2. 2 times

  3. 4 times

  4. 8 times


Correct Option: A
Explanation:

The rate at which a substance radiates is directly proportional to the fourth power of the absolute temperature.
The temperature increases from $300K$ to $324.2K$ which is an increase by $1.080$
Hence the rate at which the metal radiates would increase by $(1.08)^{4}$ = $1.36$

A black body at a temperature $77^oC$ radiates heat at a rate of $10 calcm^{-2}s^{-1}$. The rate at which this body would radiate heat in units of $cal \ cm^{-2} \ s^{-1}$ at $427^oC$ is closest to:

  1. 40

  2. 160

  3. 200

  4. 400


Correct Option: B
Explanation:

Energy radiated $P=\sigma AT^4$
$ \displaystyle \frac{P _1}{P _2} = \cfrac{T _1^4}{T _2^4}= { \bigg ( \frac {350}{700} \bigg ) }^4 = \frac{10}{P _2} \space or P _2 = 160$

The amount of thermal radiations emitted from one square centimeter area of a black body in a second when at a temperature of 1000K

  1. 5.67 J

  2. 56.7 J

  3. 567 J

  4. 5670 J


Correct Option: A
Explanation:
Stefan's law $\Delta Q = \sigma ST^4\Delta t$
$\sigma=5.67E- 8W/m^2K^4;\, S=1E-4m^2;\, T=1000K; \, \Delta t=1s;$
subsutituting value in Stefan's law $\Delta Q=5.67J$
The value can be directly calculated by the Stefan's equation. After substituting the parameters sigma $= 5.67 E-8  W/m^2/K^4, A=10^-4 m^4,$ $T=1000K$ the value comes $5.67$ J
Thus, A is correct answer.

Find the radiation pressure of solar radiation on the surface of earth. Solar constant is $1.4kW{{m}^{-2}}$

  1. $4.7\times { 10 }^{ -5 }Pa$

  2. $4.7\times { 10 }^{ -6 }Pa$

  3. $2.37\times { 10 }^{ -6 }Pa$

  4. $9.4\times { 10 }^{ -6 }Pa$


Correct Option: B
Explanation:
METHOD-1:
${Pressure} _{absorbed}=\dfrac{E _{f}}{c}$
${E} _{f}=$ energy flux, $C=$ Speed of light

${ Pressure } _{ absorbed }=\dfrac { 1.4\times 1000 }{ 3\times { 10 }^{ 8 } } =4.66\times { 10 }^{ -6 }Pa$

METHOD-2: (checking the unit,if formula is not remembered)

We know that:
${Power}={force}\times{velocity}$-----(1)
${Pressure}=\dfrac{force}{area}$------(2)

${Force}=\dfrac{power}{velocity}=\dfrac { 1.4\times 1000 }{ 3\times { 10 }^{ 8 } } =4.66\times { 10 }^{ -6 } newton$

Put in equation (2)
${ Pressure } _{ absorbed }=\dfrac { 4.66\times { 10 }^{ -6 }N }{ { m }^{ 2 } } =4.66\times { 10 }^{ -6 }Pa$

The temperature of a black body corresponding to which it will emit energy at the rate of $1 watt/cm^2$ will be

  1. 650K

  2. 450K

  3. 350K

  4. 250K


Correct Option: A
Explanation:

$E\propto { T }^{ 4 }\quad \Longrightarrow \quad E=\sigma { T }^{ 4 }$
$\sigma =5.67*{ 10 }^{ 8 }W{ m }^{ 2 }{ k }^{ 4 }$
$1*{ 10 }^{ -4 }=5.67*{ 10 }^{ 8 }*{ T }^{ 4 }$
$T=648k\cong 650k$



The solar constant for the earth is $\Sigma$. The surface temperature of the sun is $T$ K. The sun subtends an angle $\theta$ at the earth

  1. $\Sigma \space \propto \space T^4$

  2. $\Sigma \space \propto \space T^2$

  3. $\Sigma \space \propto \space \theta^4$

  4. $\Sigma \space \propto \space \theta$


Correct Option: A
Explanation:

By Stephan Boltzman law,


$P=\sigma (4\pi { R }^{ 2 }){ T }^{ 4 }$

$\theta =\dfrac { 2r }{ R } $ where R is distance between earth and sun and r is radius of earth.

Hence, $\sum {  } =\dfrac { P }{ 4\pi { r }^{ 2 } } =C{ T }^{ 4 }{ \left( \dfrac { R }{ r }  \right)  }^{ 2 }=K{ T }^{ 4 }{ \theta  }^{ 2 }$

Hence $\sum {  } \alpha { T }^{ 4 }$ and $\sum {  } \alpha { \theta }^{ 2 }$

Answer is option A.

In the Orion stellar system the shining of a star is $17\space \times 10^3$ times that of the sun. If the temperature of the surface of the sun $6 \times 10^3 K$ then the temperature of this star will be

  1. 273 K

  2. 652 K

  3. 6520 K

  4. 68520 K


Correct Option: D
Explanation:

The power radiated is directly proportional to the fourth power of the absolute temperature.
$P \propto T^{4}$
$P = kT^{4}$ 

Let P = Power radiated by the sun
Power radiated by star $ = 17 \times  10^{3}P $
$\dfrac{T _{star}^{4}}{T _{sun}^{4}} = 17 \times  10^{3}$

$\dfrac{T _{star}}{T _{sun}} = 11.418$

$T _{star} = 6000 \times  11.418 = 68,511K$

Hence, the answer is option D.

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

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


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. 

  1. brighter than the plate

  2. as bright as the plate

  3. dull as compared to the plate

  4. appears to be yellow


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

  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 $


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

  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 } $


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? 

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


Correct Option: C

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

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

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

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

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


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

  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


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.

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


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.

  1. $1273^{\circ}C$

  2. $1000^{\circ}C$

  3. $727^{\circ}C$

  4. 727K


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

  1. a straight line

  2. a circular arc

  3. a parabola

  4. an ellipse


Correct Option: A

A solid at temperature $ T _1 $ is kept in an evacuated chamber at Temperature $ T _2 > T _1 $ . the rate of increase of temperature of the body is proportional to

  1. $ T _2- T _1 $

  2. $ T^2 _2 - T^2 _1 $

  3. $ T^3 _2 -T^3 _1 $

  4. $ T^4 _1 - T^4 _1 $


Correct Option: D

A black body radiates energy at the rate of $E$ watt per metr$e^2$ at a high ternperature $T$ K. when the temperature is reduced to $(T/2)$ K, the radiant energy will be

  1. $E/16$

  2. $E/4$

  3. $E/2$

  4. $2E$


Correct Option: A
Explanation:

By Stefan-Boltzmann law black body radiation of energy $E$ is directly proportional to fourth power of $T$ temperature of the black body.
$E\quad \propto \quad { T }^{ 4 }$
If the temperature of the body is reduced to $\dfrac{T}{2}$, the energy of radiation will be $\dfrac{E}{16}$
option (A) is the correct answer.

The rate of radiation of a black body at $0^{\circ}C$ is $E$ J/s. Then the rate of radiation of this black body at $273^{\circ}C$ will be

  1. 16 E

  2. 8 E

  3. 4 E

  4. E


Correct Option: A
Explanation:

From the stefan's law: ${E}={\sigma}{A}{T}^{4}$
So, ${E}={\sigma}{A}{273}^{4}$------(1)
Now, for second one ${E} _{1}={\sigma}{A}{546}^{4}$-----(2)
On dividing equation(2) by (1), we get
$\dfrac{{E} _{1}}{E}=\dfrac{{546}^{4}}{{273}^{4}}={16}$

Or, we can say ${E} _{1}={16}{E}$

The temperature of a spherical planet is related to the distance from sun as :

  1. $T \propto 1/d^{2}$

  2. $T\propto \dfrac{1}{\sqrt{d}}$

  3. $T\propto d$

  4. $T\propto d^{2}$


Correct Option: B
Explanation:
A planet reaches its equilibrium temperature when the amount of heat it absorbs from the sun is equal to the amount that it radiates back to space.
The energy absorbed is equal to the insolation at planet's distance (which is Sun's bolometric output divided by $4\pi$ times the distance squared) times the planet's profile area times the planet's Bond aldedo. The amount radiated (assuming that the thermal radiation is approximately black body radiation) is proportional to the planet's surface area times its emmisivity times the forth power of its temperature.

Now, throwing away 2, $\pi$ and other constants that do not vary with temperature or distance.
${ T }^{ 4 }\alpha \cfrac { 1 }{ { d }^{ 2 } } \\ \Rightarrow T\alpha \cfrac { 1 }{ \sqrt { d }  } $

Assertion (A): The radiation from the sun surface varies as the fourth power of its absolute temperature.
Reason (R): Sun is not a black body

  1. Both A and R are true, R is correct explanation of A

  2. Both A and R are true, R is not correct explanation of A

  3. A is true but R is false

  4. Both A and R are false


Correct Option: B
Explanation:

According to Stefan's law, energy emitted is proportional to fourth power of absolute temperature.
So, the same will hold for the sun.
Also, Sun can absorb/emit radiations of all wavelengths - so it is a black body.
So, assertion and reason are both correct.
But reason is not a correct explanation of assertion.
Because all bodies emit energy which is proportional to fourth power of absolute temperature, not only sun.

Three bodies A, B, C are at $-27^{o}$C, $0^{o}$C, $100^{o}$C respectively. The body which does not radiate heat is:

  1. A

  2. B

  3. none as all the bodies radiate heat

  4. C


Correct Option: C
Explanation:

All bodies radiate heat irrespective of temperature.
Heat radiated per unit time is given by Stefan's law.

A solid shpere and a hollow sphere of the same material and of equal radii are heated to the same temperature

  1. both will emit equal amount of radiation per unit time in the beginning.

  2. both will absorbs equal amount of radiation per second from the surrounding in the beginning.

  3. the initial rate of cooling will be the same for both the spheres

  4. the two spheres will have equal temperature at any instant


Correct Option: A,B
Explanation:

According to many radiation laws like Stefan Boltzmann we know that radiation emission and absorption are a purely surface phenomenon. Since the two bodies are of same material, same radii, and same temperature they will at that instant radiate and absorb at the same rates.
But however since the hollow sphere has lesser mass, the rate at which it's temperature will rise will be different from that of the solid sphere. Hence their rates of cooling would be varied and they would have different temperatures at different times.

A black body at 127$^{o}$C emits the energy at the rate of 10$^{6}$ J/m$^{2}$ s. The temperature of a black body at which the rate of energy emission is 16x10$^{6}$ J/m$^{2}$ s is :

  1. $508^{o}C$

  2. $273^{o}C$

  3. $400^{o}C$

  4. $527^{o}C$


Correct Option: D
Explanation:

Using Stefan's Law:
$\dfrac { { E } _{ 2 } }{ { E } _{ 1 } } ={ \left( \dfrac { { T } _{ 2 } }{ { T } _{ 1 } }  \right)  }^{ 4 }\ { T } _{ 2 }={ T } _{ 1 }\sqrt [ 4 ]{ \dfrac { { E } _{ 2 } }{ { E } _{ 1 } }  } \quad \ { T } _{ 2 }={ (127+273) }\sqrt [ 4 ]{ \dfrac { 16\times { 10 }^{ 6 } }{ { 10 }^{ 6 } }  } =\quad 800K\ { T } _{ 2 }={ 527 }^{ \circ  }C$

Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures of 2T and 3T respectively. The temperatures of the middle (i.e. second) plate under steady state condition is then

  1. $\left ( \dfrac{64}{2} \right )^{\dfrac{1}{4}}T$

  2. $\left ( \dfrac{97}{4} \right )^{\dfrac{1}{4}}T$

  3. $\left ( \dfrac{97}{2} \right )^{\dfrac{1}{4}}T$

  4. $\left ( 97\right )^{\dfrac{1}{4}}T$


Correct Option: C

The rectangular surface of area $8cm$ $\times$ $4 cm$ of a black body at temperature $127^{\circ}C$ emits energy $E$ per second. If the length and breadth are reduced to half of the initial value and the temperature is raised to $327^{\circ}C$, the rate of emission of energy becomes

  1. $\displaystyle \frac{3}{8}E$

  2. $\displaystyle \frac{81}{16}E$

  3. $\displaystyle \frac{9}{16}E$

  4. $\displaystyle \frac{81}{64}E$


Correct Option: D
Explanation:

By $Stefan's$ Law

$Power=\sigma A T^4$
$A=length\times breadth$
$\dfrac{P _2}{E}=\dfrac{l _2b _2T _2^4}{l _1b _1T _1^4}$
Here$\dfrac{l _2}{l _1}=\dfrac{1}{2}$       $\dfrac{b _2}{b _1}=\dfrac{1}{2}$       $\dfrac{T _2}{T _1}=\dfrac{3}{2}$

$\implies P _2=\dfrac{81}{64}E$

If the temperature of a hot body is raised by $0.5\%$, then the heat energy radiated would increase by :

  1. 0.5% 

  2. 1.0%

  3. 1.5%

  4. 2.0%


Correct Option: D
Explanation:
The rate of heat energy radiated by black body is given as:
$Q=\sigma T^4A$ where $\sigma$ is the stefen-boltzmann constant, $A$ is the area of the radiating body.
So $Q\propto T^4=kT^4$, where $k$ is the propotionality constant that we have assumed here.
Taking log of above equation, $logQ=log(kT^4)=logk+4logT$
Taking differential of above equation, $\dfrac{dQ}{Q}=0+\dfrac{4dT}{T}$ (because logk is constant).
Here $dQ$ and $dT$ indicate very small change in the $Q$ and $T$ respectively.
Multiplying by $100$,
$\dfrac{dQ}{Q}\times100=4\dfrac{dT}{T}\times100$
$\Rightarrow$ Percentage change in Heat rate $=4\times $ percentage change in temperature
So here, percentage change in rate of heat energy radiated $=4\times0.5=2\%$

A black body is at a temperature of $500$K. It emits its energy at a rate which is proportional to :

  1. $500$

  2. $(500)^{2}$

  3. $(500)^{3}$

  4. $(500)^{4}$


Correct Option: D
Explanation:

From stefan's boltzmann relation we know that: $E\propto { T }^{ 4 }$
So, $E\propto { 500}^{ 4 }$

The rate of emission of a black body at temperature $27$$^{o}$C is $E _{1}$. If its temperature is increased to $327$$^{o}$C, the rate of emission of radiation is $E _{2}$. The relation between $E _{1} $ and $  E _{2}$ is :

  1. $E _{2}=24E _{1}$

  2. $E _{2}=16E _{1}$

  3. $E _{2}=8E _{1}$

  4. $E _{2}=4E _{1}$


Correct Option: B
Explanation:

We know the relation for emissive power of body: $\dfrac { { E } _{ 1 } }{ { E } _{ 2 } } =\dfrac { { T } _{ 1 }^{ 4 } }{ { T } _{ 2 }^{ 4 } } $
${T} _{1}=27+273=300\ K$
${T} _{2}=327+273=600\ K$
So, putting all these data in above formulas, we get
${E} _{2}=16{E} _{1}$

The temperature of a black body is increased by $50\%$ . Then the percentage of increase of radiation is approximately

  1. 100%

  2. 25%

  3. 400%

  4. 500%


Correct Option: C
Explanation:

By $Stefan's$ $Law$,

Power = ${\sigma A {T}^{4}}$ for a black body
$\sigma$ is known as Stefan's constant
$\dfrac{P _1}{P _2}$ = $\dfrac{ T _1^4}{ T _2^4}$
${T _2}$ = $\dfrac{3 T _1}{2}$
$\dfrac{T _1^4}{T _2^4}$= $\dfrac{16}{81}$
By the above equations we get
$P _2$=$\dfrac{81 P _1}{16}$
Percentage increase in radiation = $\dfrac{P _2 - P _1}{P _1}$ x $100$ = $406.25$%

The wave length corresponding to maximum intensity of radiation emitted by a star is $289.8$nm. The intensity of radiation for the star is :

(Stefans constant $=$ 5.6x10$^{-8}Wm^{-2}K^{-4}$, Wien's displacement constant = $2898 \times 10^{-6} mK$ )

  1. 5.67 x 10$^{8}Wm^{-2}$

  2. 5.67 x 10$^{4}Wm^{-2}$

  3. 10.67 x 10$^{7}Wm^{-2}$

  4. 10.67 x 10$^{4}Wm^{-2}$


Correct Option: A
Explanation:
Given, Wavelength corresponding to maximum radiation $=289.8nm$
Stefans constant$=5.6\times 10^{-8}Wm^{-2}K^{-4}$
From Wien's law,
$\lambda _{max} T$= constant (b)
$b=2898\times 10^{-6}$
$\lambda _{max}=289.8$
$\therefore$ $T=\frac{b}{\lambda _{max}}=\frac{2898\times 10^{-6}}{289.8\times 10^{-9}}=10^4 K$

Intensity of radiation E
$E=\sigma T^4=5.6\times 10^{-8}\times 10^{16}= 5.6\times 10^8 Wm^{-2}$

All bodies emit heat energy from their surfaces by virtue of their temperature. This heat energy is called radiant energy of thermal radiation. The heat that we receive from the sun is transferred to us by a process which, unlike conduction or convection, does not require the help of a medium in the intervening space which is almost free of particles. Radiant energy travels in space as electromagnetic spectrum. Thermal radiations travel through vacuum with the speed of light. Thermal radiations obey the same laws of reflection and refraction as light does. They exhibit the phenomena of interference, diffraction and polarization as light does.
The emission of radiation from a hot body is expressed in terms of that emitted from a reference body (called the black body) at the same temperature. A black body absorbs and hence emits radiations of all wavelengths. The total energy E emitted by a unit area of a black body per second is given by $E =\sigma T^{4}$ where T is the absolute temperature of the body and $\sigma $ is a constant known as Stefans constant. If the body is not a perfect black body, then $E =\varepsilon \sigma  T^{4}$where $\varepsilon $ is the emissivity of the body.

From stefan-Boltzmann law, the dimensions of Stefans constant $\sigma $ are :

  1. $ML^{-2}T^{-2}K^{-4}$

  2. $ML^{-1}T^{-2}K^{-4}$

  3. $MLT^{-3}K^{-4}$

  4. $ML^{0}T^{-3}K^{-4}$


Correct Option: D
Explanation:

Stefan's law:
$E = \sigma T^{4}$ 
$\sigma = E/ T^{4}$
$E = energy/(area*time) = ( M L^{2} T^{-2} )/ ( L^{2} T)$
$\sigma =  M T^{-3} K^{-4}$

The power radiated by a black body is $P$ and it radiates maximum energy around the wavelength $\lambda  _{o}$ . If the temperature of the black body is now changed so that it radiates maximum energy around a wavelength $3\lambda  _{o}/4$ , the power radiated by it will increase by a factor of :

  1. $4/3$

  2. $16/9$

  3. $64/27$

  4. $256/81$


Correct Option: D
Explanation:
From Wein's displacement law, we know that ${\lambda} _{max}{T}={CONSTANT}$

So, let us assume that initial temperature of the body be ${T} _{1}$
and by changing the temperature from ${T} _{1}   \ to \  {T}$, wavelength will change from $\lambda$ to $\dfrac{3{\lambda}}{4}$,so applying the above relation,

${\lambda} _{0}{T} _{1}=\dfrac{3{\lambda} _{0}}{4}{T}$

Hence, ${T}=\dfrac{4}{3}T _1$

and we know that ${P}={e}{{T}^{4}}$

So, ${P'}=\dfrac{256}{81}P$

The rays of sun are focussed on a piece of ice through a lens of diameter $5$ cm, as a result of which $10$ grams of ice melts in $10$ min. The amount of heat received from Sun is (per unit area per min)

  1. 4 $cal/cm^{2} : min$

  2. 40 $cal/cm^{2} : min$

  3. 4 $J/cm^{2} : min$

  4. 400 $J/cm^{2} : min$


Correct Option: A
Explanation:

The latent heat of fusion is $80cal/g$.

So heat received per unit times is $10g\times 80cal g^{-1}/ 10min=80cal/min$
The are is $\pi r^2=\pi\times(2.5)^2cm^2\approx 20cm^2$
So, amountof heat per unit area per unit time is $80/20=4cal/cm^2\ min$

The emissive power of a sphere of radius $5$cm coated with lamp black is $1500$Wm$^{-2}$. The amount of energy radiated per second is.

  1. 15.7 J

  2. 3.14 J

  3. 47.10 J

  4. 4.71 J


Correct Option: C
Explanation:

We know that emissive power is given as: ${E}={\sigma}{T}^{4}$
${T}^{4}=\dfrac{E}{\sigma}$
${T}=\dfrac{1500}{5.67\times{10}^{-8}}=403.29 K$
Now, we know that radiation from the surface is given as $E=\sigma \epsilon A{ T }^{ 4 }$
${E}={5.67\times{10}^{-8}\times4\times{\pi}\times{0.05}^{2}\times{403.29}^{4}}=47.15 J $

Match the physical quantities given in Column I with their dimensional formula given in ColumnII

Column-I Column-II
(a) Thermal conductivity (p) is a dimensionless quantity
(b) Stefans constant (q) $ML^{o}T^{o}K$
(c) Wiens constant (r) $ML^2T^{-3}K^{-1}$
(d) Emissivity (s) $ML^{o}T^{-3}K^{-4}$
  1. a-s, b-r, c-p, d-q

  2. a-r, b-s, c-q, d-p

  3. a-s, b-q, c-r, d-p

  4. a-p, b-q, c-r, d-s


Correct Option: B
Explanation:

Thermal resistance $=\dfrac{kelvin}{watt}$
Unit of power is watt so power = force$\times$ velocity
Force = mass$\times$acceleration = M$\times$$\dfrac{m}{{sec}^{2}}$
Force = $ML{T}^{-2}$
Power = $M{L}^{2}{T}^{-3}$
So, Thermal resistance $={M}^{-1}{L}^{-2}{T}^{3}{\theta}$
and we know that relation between conductivity and resistevity is that they are reciprocal to each other,so thermal conductivity =${M}^{1}{L}^{2}{T}^{-3}{\theta}^{-1}$

Stefan's constant $=5.64\times$${10}^{-8}$$\dfrac { w }{ { m }^{ 2 }{ k }^{ 4 } } $
$\sigma$=${ M{ L }^{ 0 }{ T }^{ -3 }{ \theta  }^{ 4 } }$

Wein's constant: ${\lambda}{\theta}$ = constant = [$L{\theta}$]

Emissivity is define as the ration of two same quantity so it is dimensionless.

A black body emits maximum radiation of wavelength $\displaystyle \lambda _{1}=2000A $ at a certain temperature $\displaystyle T _{1} $ On increasing the temperature the total energy of radiation emitted is increased $16$ times at temperature $\displaystyle T _{2} $ If $\displaystyle \lambda _{2} $ is the wavelength corresponding to which maximum radiation emitted at temperature  $\displaystyle T _{2} $ Calculate the value of $\displaystyle \left ( \frac{\lambda _{1}}{\lambda _{2}} \right ) $

  1. $2:1$

  2. $1:2$

  3. $3:4$

  4. $4:3$


Correct Option: A
Explanation:

The rate of radiation, emitted by per unit area of a body at temperature $T$ , is given by Stefan-Boltzmann's law as,

   $E\propto T^{4}$ ,
hence , $E _{1}/E _{2}=(T _{1}/T _{2})^{4}$ ,
given , $E _{2}=16E _{1}$ ,
therefore ,
            $E _{1}/16E _{1}=(T _{1}/T _{2})^{4}$ ,
or         $T _{2}/T _{1}=2$ ,
now by Wein's law ,
            $\lambda _{m}\propto 1/T$ ,
hence  $\lambda _{m1}/\lambda _{m2}=T _{2}/T _{1}$ ,
therefore ,  $\lambda _{m1}/\lambda _{m2}=T _{2}/T _{1}=2$ ,
               

All bodies emit heat energy from their surfaces by virtue of their temperature. This heat energy is called radiant energy of thermal radiation. The heat that we receive from the sun is transferred to us by a process which, unlike conduction or convection, does not require the help of a medium in the intervening space which is almost free of particles. Radiant energy travels in space as electromagnetic spectrum. Thermal radiations travel through vacuum with the speed of light. Thermal radiations obey the same laws of reflection and refraction as light does. They exhibit the phenomena of interference, diffraction and polarisation as light does.
The emission of radiation from a hot body is expressed in terms of that emitted from a reference body (called the black body) at the same temperature. A black body absorbs and hence emits radiations of all wavelengths. The total energy $E$ emitted by a unit area of a black body per second is given by $E =\sigma T^{4}$ where $T$ is the absolute temperature of the body and $\sigma $ is a constant known as Stefan's constant. If the body is not a perfect black body, then $E =\varepsilon \sigma  T^{4}$where $\varepsilon $ is the emissivity of the body.

In which region of the electromagnetic spectrum do thermal radiations lie?

  1. Visible region

  2. Infrared region

  3. Ultraviolet region

  4. Microwave region


Correct Option: B
Explanation:

Stefan's law states:
$E = \sigma T^{4}$ 
Normally temperature of a body never exceeds more than 1000K. 
As the temperature is quite less compared to the sun's temperature. Photons of this radiation have less energy and hence greater wavelength and fall into infrared region.
Alternately by Wein's displacement law
$ \lambda _{max} T = constant$
By applying above law on sun and normal body, wavelength of normal thermal radiation falls in infrared region.

All bodies emit heat energy from their surfaces by virtue of their temperature. This heat energy is called radiant energy of thermal radiation. The heat that we receive from the sun is transferred to us by a process which, unlike conduction or convection, does not require the help of a medium in the intervening space which is almost free of particles. Radiant energy travels in space as electromagnetic spectrum. Thermal radiations travel through vacuum with the speed of light. Thermal radiations obey the same laws of reflection and refraction as light does. They exhibit the phenomena of interference, diffraction and polarization as light does.
The emission of radiation from a hot body is expressed in terms of that emitted from a reference body (called the black body) at the same temperature. A black body absorbs and hence emits radiations of all wavelengths. The total energy E emitted by a unit area of a black body per second is given by $E =\sigma T^{4}$ where T is the absolute temperature of the body and $\sigma $ is a constant known as Stefan's constant. If the body is not a perfect black body, then $E =\varepsilon \sigma  T^{4}$where $\varepsilon $ is the emissivity of the body.

What is the SI unit of Stefan's constant?

  1. $Js^{-1}K^{-4}$

  2. $Wm^{-1}K^{-4}$

  3. $Wm^{-2}K^{-4}$

  4. $Jm^{-2}K^{-4}$


Correct Option: C
Explanation:

Stefan's law:
$E = \sigma T^{4}$ 
$\sigma = E/ T^{4}$
$E = energy/(area*time) = W m^{-2}$
$\sigma =  W m^{-2} K^{-4}$

Match the physical quantities given in Column I with their SI units given in Cloumn II :

Column-I Column-II
(a) Thermal conductivity (p) Wm$^{-2}$K$^{-4}$
(b) Stefans constant (q) m-K
(c) Wiens constant (r) J kg$^{-1}$K$^{-1}$
(d) Specific heat (s)Wm$^{-1}$K$^{-1}$
  1. a-s, b-p, c-q, d-r

  2. a-s, b-p, c-r, d-q

  3. a-s, b-r, c-p, d-q

  4. a-r, b-s, c-p, d-q


Correct Option: A
Explanation:

Thermal conductivity =$\dfrac{Watt}{Metre.Kelvin}$

Stefan's constant $=5.64\times$ ${10}^{-8}$$\dfrac { w }{ { m }^{ 2 }{ K }^{4}}$

Wien's constant $={\lambda}{\theta}={constant}={Metre}  {Kelvin}$

Specific heat $=\dfrac{Joule} {{Kg}  {Kelvin}}$

Which of the following statements is true/correct?

  1. During clear nights, the temperature rises steadily upward near the ground level

  2. Newton's law of cooling, and approximate form of Stefan's law, is valid only for natural convection

  3. The total energy emitted by a black body per unit time per unit area is proportional to the square of its temperature in the Kelvin scale

  4. Two spheres of the same material have radii $1 m$ and $4 m$ and temperatures $4000 K$ and $2000 K$ respectively. The energy radiated per second by the first sphere is greater than that radiated per second by the second sphere


Correct Option: B
Explanation:

During clear nights object on surface of earth radiate out heat and temperature falls. Hence option (a) is wrong.
The total energy radiated by a body per unit time per unit area $E \propto {T}^{4}$. Hence option (c) is wrong.
Energy radiated per second is given by
$\dfrac { Q }{ t } =PA\varepsilon \sigma { T }^{ 4 }$
$\Rightarrow \dfrac { { P } _{ 1 } }{ { P } _{ 2 } } =\dfrac { { A } _{ 1 } }{ { A } _{ 2 } } { \left( \dfrac { { T } _{ 1 } }{ { T } _{ 2 } }  \right)  }^{ 4 }={ \left( \dfrac { { r } _{ 1 } }{ { r } _{ 2 } }  \right)  }^{ 2 }\cdot { \left( \dfrac { { T } _{ 1 } }{ { T } _{ 2 } }  \right)  }^{ 2 }$
$={ \left( \dfrac { 1 }{ 4 }  \right)  }^{ 2 }\left( \dfrac { 4000 }{ 200 }  \right) =\dfrac { 1 }{ 1 } $
$\because    {P} _{1} = {P} _{2}$ hence option (d) is wrong.
Newton's law is an approximate from of Stefan's law of radiation and works well for natural convection. Hence option (b) is correct.

STATEMENT-1 : Animals curl into a ball, when they feel very cold.
STATEMENT-2 : Animals by curling their body reduces the surface area.

  1. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

  2. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

  3. STATEMENT-1 is True, STATEMENT-2 is False

  4. STATEMENT-1 is False, STATEMENT-2 is True


Correct Option: A
Explanation:

Both statements are true as the animals curl their body in very cold environment so they can reduce their surface area , and reduce the heat leaving their body in form of radiation as heat emitted is directly proportional to surface area.

The dimensions of Stefan's constant are

  1. $\left[ { M }^{ 0 }{ L }^{ 1 }{ T }^{ -3 }{ K }^{ -4 } \right] $

  2. $\left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -3 }{ K }^{ -3 } \right] $

  3. $\left[ { M }^{ 1 }{ L }^{ 2 }{ T }^{ -3 }{ K }^{ -4 } \right] $

  4. $\left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -3 }{ K }^{ -4 } \right] $


Correct Option: D
Explanation:

Power radiated by a body $P = \sigma AeT^4$

where $\sigma$ is the Stefan's constant, $e$ is the emmissivity of the body, $A$ is the surface area of the body and $T$ is its temperature.
Dimensions of power $[P] = [ML^2T^{-3}]$
Dimensions of area $[A] = [L^2]$
Dimensions of temperature $[t] = [K]$
Emmissivity $e$ is a dimensionless quantity.
$\therefore$ Dimensions of Stefan's constant $[\sigma] = \dfrac{[ML^2T^{-3}]}{[L^2] [K^4]}$
$\implies$ $[\sigma] = [M^1 L^0 T^{-3} K^{-4}]$

A black body is heated from $27^oC  $ to $927^oC  $. The ratio of radiation emitted will be:

  1. $1 : 4$

  2. $1 : 8$

  3. $1 : 16$

  4. $1 : 256$


Correct Option: D
Explanation:

Energy radiated depends on the temperature of the body.
Stefan's law states that the total amount of energy radiated per second per unit area of a perfect black body is directly proportional to the fourth power of the absolute temperature of the surface of the body,ie,
$E\propto { T }^{ 4 }$
or $E=\sigma { T }^{ 4 }$
where $\sigma $ is Stefan's constant. It's value is $5.67\times { 10 }^{ -8 }W{ m }^{ -2 }{ K }^{ -4 }$
Here, ${ T } _{ 1 }=27+273=300K$
          ${ T } _{ 2 }=927+273=1200K$
$\therefore       \dfrac { { E } _{ 1 } }{ { E } _{ 2 } } ={ \left( \dfrac { 300 }{ 1200 }  \right)  }^{ 4 } = 1 : 256$

Two bodies A and B of equal surface area have thermal emissivities of $0.01$ and $0.81$ respectively. The two bodies are radiating energy at the same rate. Maximum energy is radiated from the two bodies A and B at wavelengths $\lambda _A$, and $\lambda _B$ respectively. Difference in these two wavelengths is 1 $\mu$. If the temperature of the body A is $5802\  K$, then value of $\lambda _B$ is :

  1. $\dfrac{3}{2}\mu m$

  2. $1\mu m$

  3. $2 \mu m$

  4. $\dfrac{3}{4} \mu m$


Correct Option: A
Explanation:

We know that as per stephan's boltzman radiation law, $P\ \alpha\  \sigma A{ T }^{ 4 }$.
Since surface area of two bodies is same,

Therefore, ${ \sigma  } _{ A }{ T } _{ A }^{ 4 }={ \sigma  } _{ B }{ T } _{ B }^{ 4 }$. Hence, ${ T } _{ A }=3{ T } _{ B }$.

Now, as per Wein's displacement law, $\lambda T=constant=k$, $\lambda =\dfrac { k }{ T } $.

${ \lambda  } _{ B }-{ \lambda  } _{ A }=k(\dfrac { 1 }{ { T } _{ B } } -\dfrac { 1 }{ { T } _{ A } } )=k(\dfrac { 1 }{ { T } _{ B } } -\dfrac { 1 }{ 3{ T } _{ B } } )=\dfrac { 2k }{ 3{ T } _{ B } } =1\mu $


${ T } _{ B }=\dfrac { { T } _{ A } }{ 3 } =1934\ K$

Putting in the above equation to calculate $k$ and the solving for ${ \lambda  } _{ B }=\dfrac { k }{ { T } _{ B } } $, we get ${ \lambda  } _{ B }=1.5\mu m$.


A black body at a high temperature $T$ radiates energy at the rate of $U\left( in\quad W/{ m }^{ 2 } \right) $. When the temperature falls to half (i.e $T/2$), the radiated energy $\left( in\quad W/{ m }^{ 2 } \right) $ will be

  1. $U/8$

  2. $U/16$

  3. $U/4$

  4. $U/2$


Correct Option: B
Explanation:

According to Stefan's law, rate of energy radiated by a black per unit area (inW/m2)(inW/m2) at temperature TT is given by
U=σT4...(i)U=σT4...(i)
when the temperature falls to half (i.e., T/2T/2
radiated energy (inW/m2)...(ii)(inW/m2)...(ii)
From (i) and (ii) we get
UU=(12)4=116U′U=(12)4=116
or U=U16

 If the radius of a star is R and it acts as a black body, what would be the temperature of the star, in which the rate of energy production is 0? (a stands for Stefan's constant.)

  1. $

    \left(\frac{4 \pi R^{2} Q}{\sigma}\right)^{1 / 4}

    $

  2. $

    \left(\frac{Q}{4 \pi R^{2} \sigma}\right)^{1 / 4}

    $

  3. $

    \frac{Q}{4 \pi R^{2} \sigma}

    $

  4. $

    \left(\frac{Q}{4 \pi R^{2} \sigma}\right)^{-1 / 2}

    $


Correct Option: C

$\dfrac {watt} {kelvin}$ is the unit of 

  1. Stefan's constant

  2. Wien's constant

  3. Cooling's constant

  4. Thermal constant


Correct Option: A

Assuming the Sun to be a spherical body of radius $R$ at a temperature of $T\ K$. Evaluate the intensity of radiant power, incident on Earth, at a distance $r$ from the Sun where $r _{0}$ is the radius of the Earth and $\sigma$ is Stefan's constant :

  1. $\dfrac{R^{2}\sigma T^{4} }{r^{2}}$

  2. $\dfrac{4\pi ^{2}R^{2}\sigma T^{4}}{r^{2}}$

  3. $\dfrac{\pi ^{2}R^{2}\sigma T^{4}}{r^{2}}$

  4. $\dfrac{\pi ^{2}R^{2}\sigma T^{4}}{4\pi r^{2}}$


Correct Option: A
Explanation:
Total power radiated by the sun

 $=\sigma { T }^{ 4 }\times 4\pi { R }^{ 2 }$

The intensity of power at earth surface

$=\cfrac{\sigma { T }^{ 4 }\times 4\pi { R }^{ 2 }}{4\pi { r }^{ 2 }} \\=\cfrac{\sigma { T }^{ 4 } { R }^{ 2 }}{{ r }^{ 2 }}$

The rectangular surface of area $8 cm \times 4 cm$ of a black body at a temperature of $127^0C$ emits energy at the rate of $E$ per second. If both length and breadth of the surface are reduced to half of its initial value, and the temperature is raised to $327^0C$, then the rate of emission of energy will become :

  1. $\dfrac{3}{8}E$

  2. $\dfrac{81}{16}E$

  3. $\dfrac{9}{16}E$

  4. $\dfrac{81}{64}E$


Correct Option: D
Explanation:
Let $A _1=32$ as given.
Let $A _2$ be the area when length and breadth are reduced by half. Thus the area will be $\dfrac {1}{4}$th of $A _1$
$ \therefore A _2=\dfrac {1}{4} *32=8$
Given $T _1={127}^0C={400}^0K$
Given $T _2={327}^0C={600}^0K$
From Stefan's law $E=\sigma AT^4$
$ \therefore\dfrac{E _1}{E _2}= \dfrac {A _1{T _1}^4}{A _2{T _2}^4}=\dfrac {32*(400)^4}{8*(600)^4}=\dfrac{64}{81}$
$ \therefore \dfrac{E _2}{E _1}= \dfrac{81}{64}$
- Hide questions