Stefan-Boltzmann & Wien's Law — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Stefan-Boltzmann & Wien's Law MCQs with step-by-step solutions (30 questions). Part of Kinetic Theory of Gases & Radiation. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Stefan-Boltzmann & Wien's Law · easy · theory
According to Stefan's law, the energy radiated per unit area per unit time by a perfect blackbody at absolute temperature $T$ is:
A. $\sigma T$
B. $\sigma T^4$ ✓ Correct
C. $\sigma T^2$
D. $\sigma T^3$
Solution: The Stefan-Boltzmann law states $E = \sigma T^4$, where $\sigma = 5.67 \times 10^{-8}\text{ W}\cdot\text{m}^{-2}\cdot\text{K}^{-4}$.
Q2 — Stefan-Boltzmann & Wien's Law · easy · numerical
If the absolute temperature of a blackbody is doubled, the rate of energy radiated per unit area increases by a factor of:
A. $2$
B. $8$
C. $16$ ✓ Correct
D. $4$
Solution: $E \propto T^4$, so doubling the temperature multiplies the emitted power by $2^4 = 16$.
Q3 — Stefan-Boltzmann & Wien's Law · medium · numerical
A blackbody at $27^\circ\text{C}$ radiates heat at $5\text{ W/cm}^2$. The rate of radiation by the same body at $327^\circ\text{C}$ is:
A. $10\text{ W/cm}^2$
B. $40\text{ W/cm}^2$
C. $80\text{ W/cm}^2$ ✓ Correct
D. $160\text{ W/cm}^2$
Solution: In kelvin, $T_1 = 300\text{ K}$ and $T_2 = 600\text{ K}$. So $\dfrac{E_2}{E_1} = \left(\dfrac{600}{300}\right)^4 = 16$, giving $E_2 = 16 \times 5 = 80\text{ W/cm}^2$.
Q4 — Stefan-Boltzmann & Wien's Law · medium · theory
The net rate of loss of radiant energy by a body of emissivity $e$ and surface area $A$ at temperature $T$, placed in surroundings at temperature $T_0$, is:
A. $e\sigma A T^4$
B. $e\sigma A(T - T_0)^4$
C. $e\sigma A(T^4 - T_0^4)$ ✓ Correct
D. $e\sigma A(T^4 + T_0^4)$
Solution: The body radiates $e\sigma A T^4$ and simultaneously absorbs $e\sigma A T_0^4$ from its surroundings, so the net loss is the difference.
Q5 — Stefan-Boltzmann & Wien's Law · medium · theory
The value of Stefan's constant $\sigma$ is:
A. $5.67 \times 10^{-8}\text{ W}\cdot\text{m}^{-2}\cdot\text{K}^{-1}$
B. $6.67 \times 10^{-11}\text{ W}\cdot\text{m}^{-2}\cdot\text{K}^{-4}$
C. $5.67 \times 10^{-8}\text{ W}\cdot\text{m}^{-2}\cdot\text{K}^{-4}$ ✓ Correct
D. $1.38 \times 10^{-23}\text{ W}\cdot\text{m}^{-2}\cdot\text{K}^{-4}$
Solution: $\sigma = 5.67 \times 10^{-8}\text{ W}\cdot\text{m}^{-2}\cdot\text{K}^{-4}$. The value $1.38 \times 10^{-23}$ is Boltzmann's constant and $6.67 \times 10^{-11}$ the gravitational constant.
Q6 — Stefan-Boltzmann & Wien's Law · medium · numerical
A perfect blackbody is maintained at $300\text{ K}$. The energy it radiates per unit area per second is approximately ($\sigma = 5.67 \times 10^{-8}$):
A. $115\text{ W/m}^2$
B. $1837\text{ W/m}^2$
C. $17\text{ W/m}^2$
D. $459\text{ W/m}^2$ ✓ Correct
Solution: $E = \sigma T^4 = 5.67 \times 10^{-8} \times (300)^4 = 5.67 \times 10^{-8} \times 8.1 \times 10^9 \approx 459\text{ W/m}^2$.
Q7 — Stefan-Boltzmann & Wien's Law · easy · numerical
If the absolute temperature of a blackbody is tripled, the energy radiated per unit area per second becomes:
A. $81$ times ✓ Correct
B. $3$ times
C. $9$ times
D. $27$ times
Solution: $E \propto T^4$, so tripling the temperature multiplies the emission by $3^4 = 81$.
Q8 — Stefan-Boltzmann & Wien's Law · easy · theory
Wien's displacement law relates the wavelength $\lambda_m$ of maximum emission to the absolute temperature $T$ by:
A. $\lambda_m T^4 = b$
B. $\lambda_m T = b$ ✓ Correct
C. $\lambda_m = b T$
D. $\dfrac{\lambda_m}{T} = b$
Solution: Wien's law states $\lambda_m T = b$, where $b = 2.9 \times 10^{-3}\text{ m}\cdot\text{K}$ is Wien's constant.
Q9 — Stefan-Boltzmann & Wien's Law · medium · numerical
Two stars radiate maximum energy at wavelengths $400\text{ nm}$ and $600\text{ nm}$ respectively. The ratio of their absolute temperatures $T_1 : T_2$ is:
A. $4 : 9$
B. $3 : 2$ ✓ Correct
C. $2 : 3$
D. $9 : 4$
Solution: By Wien's law $\lambda_m T = $ constant, so $\dfrac{T_1}{T_2} = \dfrac{\lambda_2}{\lambda_1} = \dfrac{600}{400} = \dfrac{3}{2}$.
Q10 — Stefan-Boltzmann & Wien's Law · medium · theory
The value of Wien's constant $b$ is approximately:
A. $5.67 \times 10^{-8}\text{ m}\cdot\text{K}$
B. $2.9 \times 10^{-3}\text{ m}\cdot\text{K}$ ✓ Correct
C. $6.63 \times 10^{-34}\text{ m}\cdot\text{K}$
D. $2.9 \times 10^{-3}\text{ m}/\text{K}$
Solution: Wien's constant is $b = \lambda_m T \approx 2.9 \times 10^{-3}\text{ m}\cdot\text{K}$ (often quoted as $2.898 \times 10^{-3}$).
Q11 — Stefan-Boltzmann & Wien's Law · medium · numerical
The Sun emits maximum energy at a wavelength of about $500\text{ nm}$. Its surface temperature is approximately:
A. $11600\text{ K}$
B. $5800\text{ K}$ ✓ Correct
C. $2900\text{ K}$
D. $1450\text{ K}$
Solution: $T = \dfrac{b}{\lambda_m} = \dfrac{2.9 \times 10^{-3}}{500 \times 10^{-9}} = \dfrac{2.9 \times 10^{-3}}{5 \times 10^{-7}} = 5800\text{ K}$.
Q12 — Stefan-Boltzmann & Wien's Law · easy · theory
As the temperature of a blackbody increases, the wavelength at which it emits maximum energy:
A. First increases and then decreases
B. Shifts towards longer wavelengths
C. Shifts towards shorter wavelengths ✓ Correct
D. Remains unchanged
Solution: Since $\lambda_m \propto \dfrac{1}{T}$, hotter bodies peak at shorter wavelengths. This is why an iron bar glows first dull red, then orange, then white as it is heated.
Q13 — Stefan-Boltzmann & Wien's Law · easy · theory
For a body that is not a perfect blackbody, the rate of radiation per unit area at temperature $T$ is:
A. $\dfrac{\sigma T^4}{e}$
B. $e\sigma T^4$ ✓ Correct
C. $e\sigma T$
D. $\sigma T^4$
Solution: The blackbody value $\sigma T^4$ is multiplied by the emissivity $e$ of the surface, which lies between $0$ and $1$.
Q14 — Stefan-Boltzmann & Wien's Law · medium · numerical
Two spheres of the same material have radii in the ratio $1 : 2$ and are maintained at the same temperature. The ratio of the rates at which they radiate energy is:
A. $1 : 4$ ✓ Correct
B. $1 : 8$
C. $1 : 16$
D. $1 : 2$
Solution: At equal temperature the total radiated power is proportional to the surface area, which varies as $r^2$. Hence the ratio is $1^2 : 2^2 = 1 : 4$.
Q15 — Stefan-Boltzmann & Wien's Law · hard · numerical
If both the radius and the absolute temperature of a spherical blackbody are doubled, the total power it radiates becomes:
A. $8$ times
B. $16$ times
C. $64$ times ✓ Correct
D. $32$ times
Solution: Total power is $P = \sigma A T^4 \propto r^2 T^4$. Doubling $r$ gives a factor $4$ and doubling $T$ a factor $16$, so together $4 \times 16 = 64$.
Q16 — Stefan-Boltzmann & Wien's Law · easy · theory
Stefan's law applied to the net exchange between a body at $T$ and surroundings at $T_0$ predicts zero net radiation loss when:
A. $T_0 = 0$
B. $T = 2T_0$
C. $T = T_0$ ✓ Correct
D. $T = 0$
Solution: The net rate is $e\sigma A(T^4 - T_0^4)$, which vanishes only when the body and its surroundings are at the same temperature — thermal equilibrium.
Q17 — Stefan-Boltzmann & Wien's Law · medium · numerical
The absolute temperature of a blackbody is raised from $300\text{ K}$ to $900\text{ K}$. The energy it radiates per unit area increases by a factor of:
A. $9$
B. $27$
C. $81$ ✓ Correct
D. $3$
Solution: $E \propto T^4$ and the temperature ratio is $3$, so the emission rises by $3^4 = 81$ times.
Q18 — Stefan-Boltzmann & Wien's Law · medium · numerical
A body radiates maximum energy at a wavelength of $1450\text{ nm}$. Its temperature is ($b = 2.9 \times 10^{-3}\text{ m}\cdot\text{K}$):
A. $2000\text{ K}$ ✓ Correct
B. $4000\text{ K}$
C. $500\text{ K}$
D. $1000\text{ K}$
Solution: By Wien's law, $T = \dfrac{b}{\lambda_m} = \dfrac{2.9 \times 10^{-3}}{1450 \times 10^{-9}} = 2000\text{ K}$.
Q19 — Stefan-Boltzmann & Wien's Law · medium · numerical
A star has a surface temperature of $7250\text{ K}$. The wavelength at which it radiates maximum energy is ($b = 2.9 \times 10^{-3}\text{ m}\cdot\text{K}$):
A. $725\text{ nm}$
B. $400\text{ nm}$ ✓ Correct
C. $580\text{ nm}$
D. $250\text{ nm}$
Solution: $\lambda_m = \dfrac{b}{T} = \dfrac{2.9 \times 10^{-3}}{7250} = 4 \times 10^{-7}\text{ m} = 400\text{ nm}$.
Q20 — Stefan-Boltzmann & Wien's Law · hard · numerical
The radius of a spherical blackbody is doubled while its absolute temperature is halved. The total power it radiates becomes:
A. Unchanged
B. $\dfrac{1}{16}$ of the original
C. $\dfrac{1}{4}$ of the original ✓ Correct
D. $4$ times the original
Solution: $P = \sigma AT^4 \propto r^2T^4$. Doubling $r$ gives a factor $4$, halving $T$ gives a factor $\dfrac{1}{16}$, so overall $\dfrac{4}{16} = \dfrac{1}{4}$.
Q21 — Stefan-Boltzmann & Wien's Law · medium · numerical
The absolute temperature of a blackbody rises from $200\text{ K}$ to $800\text{ K}$. Its emissive power increases by a factor of:
A. $256$ ✓ Correct
B. $4$
C. $64$
D. $16$
Solution: The temperature ratio is $4$, and $E \propto T^4$, so the emission rises by $4^4 = 256$ times.
Q22 — Stefan-Boltzmann & Wien's Law · medium · numerical
A body radiates maximum energy at $580\text{ nm}$. Its temperature is ($b = 2.9 \times 10^{-3}\text{ m}\cdot\text{K}$):
A. $2000\text{ K}$
B. $4000\text{ K}$
C. $5800\text{ K}$
D. $5000\text{ K}$ ✓ Correct
Solution: $T = \dfrac{b}{\lambda_m} = \dfrac{2.9 \times 10^{-3}}{5.8 \times 10^{-7}} = 5000\text{ K}$.
Q23 — Stefan-Boltzmann & Wien's Law · medium · numerical
A blackbody at $1000\text{ K}$ radiates maximum energy at a wavelength of ($b = 2.9 \times 10^{-3}\text{ m}\cdot\text{K}$):
A. $29000\text{ nm}$
B. $290\text{ nm}$
C. $2900\text{ nm}$ ✓ Correct
D. $580\text{ nm}$
Solution: $\lambda_m = \dfrac{b}{T} = \dfrac{2.9 \times 10^{-3}}{1000} = 2.9 \times 10^{-6}\text{ m} = 2900\text{ nm}$.
Q24 — Stefan-Boltzmann & Wien's Law · medium · numerical
A body at $290\text{ K}$ radiates maximum energy at a wavelength of ($b = 2.9 \times 10^{-3}\text{ m}\cdot\text{K}$):
A. $10\,\mu\text{m}$ ✓ Correct
B. $1\,\mu\text{m}$
C. $0.1\,\mu\text{m}$
D. $100\,\mu\text{m}$
Solution: $\lambda_m = \dfrac{2.9 \times 10^{-3}}{290} = 10^{-5}\text{ m} = 10\,\mu\text{m}$ — in the infrared, as expected for a body near room temperature.
Q25 — Stefan-Boltzmann & Wien's Law · easy · numerical
Two stars have absolute temperatures in the ratio $2 : 1$. The ratio of the wavelengths at which they radiate maximum energy is:
A. $1 : 2$ ✓ Correct
B. $1 : 4$
C. $2 : 1$
D. $4 : 1$
Solution: By Wien's law $\lambda_m \propto \dfrac{1}{T}$, so the hotter star peaks at half the wavelength.
Q26 — Stefan-Boltzmann & Wien's Law · hard · numerical
The radius of a spherical blackbody is halved while its absolute temperature is doubled. Its total radiated power becomes:
A. $4$ times ✓ Correct
B. $16$ times
C. $\dfrac{1}{4}$ times
D. $64$ times
Solution: $P \propto r^2T^4$. Halving $r$ gives $\dfrac{1}{4}$ and doubling $T$ gives $16$, so overall $\dfrac{16}{4} = 4$ times.
Q27 — Stefan-Boltzmann & Wien's Law · medium · numerical
A body radiates maximum energy at $725\text{ nm}$. Its temperature is ($b = 2.9 \times 10^{-3}\text{ m}\cdot\text{K}$):
A. $5000\text{ K}$
B. $8000\text{ K}$
C. $4000\text{ K}$ ✓ Correct
D. $2000\text{ K}$
Solution: $T = \dfrac{2.9 \times 10^{-3}}{7.25 \times 10^{-7}} = 4000\text{ K}$.
Q28 — Stefan-Boltzmann & Wien's Law · medium · numerical
A perfect blackbody at $500\text{ K}$ radiates energy at the rate of ($\sigma = 5.67 \times 10^{-8}$):
A. $459\text{ W/m}^2$
B. $1452\text{ W/m}^2$
C. $3544\text{ W/m}^2$ ✓ Correct
D. $7088\text{ W/m}^2$
Solution: $E = \sigma T^4 = 5.67 \times 10^{-8} \times 6.25 \times 10^{10} \approx 3544\text{ W/m}^2$.
Q29 — Stefan-Boltzmann & Wien's Law · easy · numerical
If the absolute temperature of a blackbody is made four times as large, its emissive power becomes:
A. $256$ times ✓ Correct
B. $16$ times
C. $4$ times
D. $64$ times
Solution: $E \propto T^4$, so multiplying the temperature by $4$ multiplies the emission by $4^4 = 256$.
Q30 — Stefan-Boltzmann & Wien's Law · easy · numerical
Two blackbodies at the same temperature have surface areas in the ratio $1 : 3$. The ratio of the total powers they radiate is:
A. $1 : 3$ ✓ Correct
B. $3 : 1$
C. $1 : 9$
D. $1 : 81$
Solution: At equal temperature $P = \sigma AT^4 \propto A$, so the powers follow the area ratio $1 : 3$.