Tag: physics

Questions Related to physics

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 $k=1.38\times 10^{-23}J/K$]-

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

  1. $10^{7}K$

  2. $10^{5}K$

  3. $10^{3}K$

  4. $10^{9}K$

Show answer
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:

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

  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 $2901:K$

Show answer
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)

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

  1. $150\;k$

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

  3. $1/2\;k$

  4. None of these

Show answer
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

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

  1. 1.36 times

  2. 2 times

  3. 4 times

  4. 8 times

Show answer
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:

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

  1. 40

  2. 160

  3. 200

  4. 400

Show answer
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

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

  1. 5.67 J

  2. 56.7 J

  3. 567 J

  4. 5670 J

Show answer
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 $1.4kW{{m}^{-2}}$

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

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

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

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

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

Show answer
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

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

  1. 650K

  2. 450K

  3. 350K

  4. 250K

Show answer
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

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

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

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

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

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

Show answer
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

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

  1. 273 K

  2. 652 K

  3. 6520 K

  4. 68520 K

Show answer
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.