Q.Mark the correct options: (Note: more than one of the given options may be correct.)

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.When an iron rod is heated, the variation of color from dull red to white can be explained by (A) Boltzman law (B) Newton's law of cooling (C) Stefen's law of radiation (D) Wien's displacement law

   ›Reveal solutionSolution

   The colour shift from dull red to white as a hot object's temperature rises is a direct visual signature of Wien's displacement law — the peak emission wavelength moves to shorter (bluer) wavelengths as $T$ increases.

   Concept and Intuition

   A heated solid radiates a continuous (black-body-like) spectrum of electromagnetic waves. Wien's displacement law states that the wavelength at which the emitted intensity peaks is inversely proportional to the absolute temperature: $\lambda_{max} = b/T$. At modest temperatures the peak lies in the infrared, and only the weak, red-shifted tail of the curve reaches into the visible band — giving a dull red glow. As temperature rises, the peak moves toward and then through the visible band; once it moves further into blue/violet, the rod emits a broad, roughly-flat mix of visible wavelengths, which the eye perceives as white.

   Step-by-Step Solution

   1. A hot iron rod is (approximately) a black-body radiator, and its spectral emission follows Planck's radiation law with a single peak.
   2. Wien's displacement law: $\lambda_{max}T = b$, where $b \approx 2.898\times10^{-3}\ \text{m K}$ is Wien's constant.
   3. At relatively low temperature, $\lambda_{max}$ is large (infrared); the rod glows dull red because only the long-wavelength tail of the curve overlaps the red end of the visible spectrum.
   4. As $T$ increases, $\lambda_{max}$ decreases (shifts toward shorter wavelengths) per the inverse relationship in Wien's law. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A blackened platinum wire of surface area $10^{-5}m^2$ is maintained at temperature of 3000K. At what rate the wire is loosing energy [Stefen-Boltzman constant $\sigma = 5.67 \times 10^{-8}\ W\,m^{-2}K^{-4}$] (A) 50 W (B) 46 W (C) 76 W (D) 38 W

   ›Reveal solutionSolution

   Direct application of the Stefan–Boltzmann law for a blackbody radiator. Answer: 46 W.

   Concept and Intuition

   A blackened surface behaves as an ideal (black-body) radiator, so the power it emits per unit area is $\sigma T^4$ — an extremely steep (fourth-power) dependence on absolute temperature. Multiplying by the surface area gives the total rate of energy loss by radiation.

   Step-by-Step Solution

   1. Stefan-Boltzmann law: $P=\sigma A T^4$.
   2. Compute $T^4$: $3000^4=(3\times10^3)^4=81\times10^{12}=8.1\times10^{13}$.
   3. $P=5.67\times10^{-8}\times10^{-5}\times8.1\times10^{13}$. …

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3. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A metal ball of emissivity $\dfrac{4}{7}$ and surface area $100\ \text{cm}^2$ is at a temperature of $127^{\circ}\text{C}$. If the temperature of the surroundings is $27^{\circ}\text{C}$, then the rate of loss of heat of the ball is (Stefan-Boltzmann constant $= 5.67 \times 10^{-8}\ W\, m^{-2}\, K^{-4}$) (A) 2.835 W (B) 22.68 W (C) 5.67 W (D) 11.34 W

   ›Reveal solutionSolution

   Tests direct application of the Stefan-Boltzmann law for a body radiating into surroundings that are also at a finite temperature (net radiative loss).

   Concept and Intuition

   A real body doesn't just radiate energy outward in isolation — it also absorbs radiation from its surroundings. The net rate of heat loss is the difference between what it emits (at its own temperature $T$) and what it absorbs (radiation from the surroundings, at $T_0$, that it would emit if it were at $T_0$), scaled by its emissivity.

   Step-by-Step Solution

   1. Convert temperatures to Kelvin: $T = 127+273=400\,K$, $T_0 = 27+273=300\,K$.
   2. Net rate of heat loss: $\dfrac{dQ}{dt} = e\sigma A\left(T^4-T_0^4\right)$.
   3. Compute the fourth powers: $400^4 = 2.56\times10^{10}$, $300^4 = 8.1\times10^{9}=0.81\times10^{10}$.
   4. Difference: $T^4-T_0^4 = 1.75\times10^{10}\,K^4$.
   5. Substitute: $\dfrac{4}{7}\times(5.67\times10^{-8})\times(100\times10^{-4}\,m^2)\times(1.75\times10^{10})$. …

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4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.A spherical perfect black body of radius 10 cm is maintained at $727°C$. The total power radiated from it is (approximately) Stefan-Boltzmann constant, $\sigma = 5.67\times10^{-8}\ \mathrm{Wm^{-2}K^{-4}}$ (A) $7120\ W$ (B) $7270\ W$ (C) $1000\ W$ (D) $7000\ W$

   ›Reveal solutionSolution

   This tests the Stefan-Boltzmann law for total radiated power from a spherical black body, requiring correct conversion of temperature to kelvin and computing the surface area.

   Concept and Intuition

   A perfect black body radiates total power according to the Stefan-Boltzmann law, $P = \sigma A T^4$, where $T$ must be the absolute temperature (kelvin), and $A$ is the total surface area emitting radiation — for a sphere, $A = 4\pi r^2$. The key steps are converting $727°C$ correctly to kelvin ($1000\ K$, a deliberately round number for easy computation) and carefully computing $T^4 = 10^{12}$.

   Step-by-Step Solution

   1. Convert temperature: $T = 727 + 273 = 1000\ K$.
   2. Surface area of the sphere: $A = 4\pi r^2 = 4\pi(0.1)^2 = 4\pi(0.01) = 0.12566\ m^2$.
   3. Apply Stefan-Boltzmann law: $P = \sigma A T^4 = 5.67\times10^{-8}\times0.12566\times(1000)^4$. …

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5. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If the wavelengths of maximum intensity of radiation emitted by two black bodies A and B are $0.5\ \mu m$ and 0.1 mm respectively, then ratio of the temperatures of the bodies A and B is (A) 5 (B) 25 (C) 100 (D) 200

   ›Reveal solutionSolution

   This tests Wien's displacement law — the peak-emission wavelength of a black body is inversely proportional to its absolute temperature. The ratio $T_A:T_B = 200$.

   Concept and Intuition

   A hotter black body radiates more strongly at shorter wavelengths — its emission spectrum peak shifts toward blue/UV as temperature rises. Wien's law quantifies this: $\lambda_{max} T = b$, where $b \approx 2.898\times10^{-3}\ \mathrm{m\,K}$ is Wien's constant. Since $b$ is fixed, a body with a smaller peak wavelength must have a higher temperature — they are inversely proportional.

   Step-by-Step Solution

   1. Write Wien's law for each body: $\lambda_A T_A = b$ and $\lambda_B T_B = b$.
   2. Since both equal the same constant $b$: $\lambda_A T_A = \lambda_B T_B$.
   3. Rearranging: $\dfrac{T_A}{T_B} = \dfrac{\lambda_B}{\lambda_A}$. …

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6. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.A spherical black body of radius 12 cm at a temperature of T K radiates a power of 400 W. If the radius of the sphere is doubled and absolute temperature is halved, then the power radiated by the body is (A) 100 W (B) 200 W (C) 400 W (D) 1600 W

   ›Reveal solutionSolution

   Stefan-Boltzmann power scales as (radius)$^2\times$(temperature)$^4$; doubling the radius but halving the temperature drops the radiated power to a quarter, i.e. 100 W.

   Concept and Intuition

   A black body radiates power $P=\sigma A T^4 = \sigma(4\pi r^2)T^4$, so $P\propto r^2T^4$. Because temperature enters to the fourth power, even a modest change in $T$ dominates the effect of a change in size.

   Step-by-Step Solution

   1. Original: radius $r$, temperature $T$, power $P_1=400$ W.
   2. New: radius $2r$, temperature $T/2$. …

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7. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.A solid sphere at a temperature of $400\ \mathrm{K}$ radiates a power P. If the radius of the sphere is halved and its absolute temperature is doubled, then the power radiated by it is (A) $\dfrac{P}{4}$ (B) $\dfrac{P}{2}$ (C) 2P (D) 4P

   ›Reveal solutionSolution

   Since radiated power scales as $r^2T^4$, halving $r$ (factor $1/4$) and doubling $T$ (factor $16$) together multiply the power by 4.

   Concept and Intuition

   The Stefan–Boltzmann law states that the power radiated by a blackbody (or any body of fixed emissivity) is $P = \varepsilon\sigma A T^4$, where $A$ is the surface area. For a sphere, $A = 4\pi r^2$, so $P\propto r^2 T^4$. This means radiated power is very sensitive to temperature (fourth power) but only quadratically sensitive to size.

   Step-by-Step Solution

   1. Original power: $P \propto r^2 T^4$ (with $T = 400\ \text{K}$).
   2. New radius: $r' = r/2 \Rightarrow (r')^2 = r^2/4$.
   3. New temperature: $T' = 2T \Rightarrow (T')^4 = 16T^4$. …

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8. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.The rate of emission of radiation of a black body at $27°\mathrm{C}$ is $E_1$. If the temperature increased to $327°\mathrm{C}$ the emission is $E_2$ then $E_2 =$ (A) $4E_1$ (B) $8E_1$ (C) $16E_1$ (D) $24E_1$

   ›Reveal solutionSolution

   This tests the Stefan-Boltzmann law ($E\propto T^4$) and the crucial step of converting given Celsius temperatures to absolute (Kelvin) temperatures before taking the ratio.

   Concept and Intuition

   The Stefan-Boltzmann law states the total radiant power emitted per unit area by a black body scales as the fourth power of absolute temperature. Because of this strong (4th-power) dependence, even a modest-looking temperature change (like doubling from 300 K to 600 K) produces a dramatic increase in emitted power. It's essential to convert to Kelvin first — using Celsius values directly would be a serious error since the law only holds for absolute temperature.

   Step-by-Step Solution

   1. Convert given Celsius temperatures to Kelvin: $T_1 = 27+273=300$ K, $T_2=327+273=600$ K. …

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9. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.The absorption coefficient value of a perfect black body is (A) zero (B) $< 1$ (C) $> 1$ (D) $1$

   ›Reveal solutionSolution

   By definition, a perfect black body absorbs all incident radiation, so its absorption coefficient is exactly $1$.

   Concept and Intuition

   The absorption coefficient (or absorptive power) of a surface is the ratio $\dfrac{\text{energy absorbed}}{\text{energy incident}}$. A perfect black body is precisely defined as the idealised surface that absorbs the entirety of any radiation falling on it, regardless of wavelength or angle — nothing is reflected, nothing is transmitted.

   Step-by-Step Solution

   1. Absorption coefficient $a = \dfrac{Q_{absorbed}}{Q_{incident}}$.
   2. For a perfect black body, $Q_{absorbed} = Q_{incident}$ (total absorption, by definition).
   3. Hence $a = 1$. …

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10. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
    Q.A 567 W bulb has a tungsten filament of length 40 cm and radius $\dfrac{2}{\pi}$ mm. If the radiation of filament is 81% of that of a perfect black body, then the temperature of filament is (Stefan's constant, $\sigma = 5.67\times10^{-8}\ \mathrm{Wm^{-2}K^{-4}}$) (A) $2500\ \mathrm{K}$ (B) $1666.7\ \mathrm{K}$ (C) $1333.3\ \mathrm{K}$ (D) $999.6\ \mathrm{K}$

    ›Reveal solutionSolution

    This applies the Stefan–Boltzmann law to a cylindrical filament that radiates like a grey body (emissivity 0.81) to find its operating temperature.

    Concept and Intuition

    A hot filament radiates power according to $P=e\,\sigma A T^4$, where $e$ is the emissivity (here 0.81, since it radiates 81% of what an ideal black body at the same temperature and area would). The relevant radiating area of a thin wire/filament is its curved lateral surface, $A=2\pi r L$, not its cross-section.

    Step-by-Step Solution

    1. Compute the lateral surface area: $r=\dfrac{2}{\pi}\times10^{-3}\ \text m$, $L=0.40\ \text m$.

    $A=2\pi r L = 2\pi\left(\frac{2}{\pi}\times10^{-3}\right)(0.4)=1.6\times10^{-3}\ \mathrm{m^2}$

    (the $\pi$'s cancel neatly, which is why the radius was given as $2/\pi$ mm.)

    2. Apply Stefan's law with emissivity 0.81:

    $P = 0.81\,\sigma A T^{4}$

    $567 = 0.81\times(5.67\times10^{-8})\times(1.6\times10^{-3})\times T^{4}$ …

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11. AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
    Q.The temperature of the surface of sun is 6000 K. The wavelength corresponding to the maximum energy is nearly (Wien's constant, $b = 2.89 \times 10^{-3} \, \mathrm{mK}$) (A) $1.82 \times 10^{-7} \, \mathrm{m}$ (B) $6.82 \times 10^{-7} \, \mathrm{m}$ (C) $5.32 \times 10^{-7} \, \mathrm{m}$ (D) $4.82 \times 10^{-7} \, \mathrm{m}$

    ›Reveal solutionSolution

    Wien's displacement law directly gives the peak-emission wavelength as $b/T$, which evaluates to about $4.82\times10^{-7}$ m for the Sun's surface temperature.

    Concept and Intuition

    Wien's displacement law states that a blackbody's peak emission wavelength is inversely proportional to its absolute temperature: $\lambda_{max} T = b$, where $b$ is Wien's constant. Hotter bodies emit peak radiation at shorter wavelengths; this explains why hotter stars appear bluer and cooler stars appear redder.

    Step-by-Step Solution

    1. Given: $T = 6000$ K, $b = 2.89\times10^{-3}$ m·K.
    2. $\lambda_{max} = \dfrac{b}{T} = \dfrac{2.89\times10^{-3}}{6000}$.
    3. Compute: $\dfrac{2.89\times10^{-3}}{6\times10^3} = \dfrac{2.89}{6}\times10^{-6} \approx 0.4817\times10^{-6} = 4.817\times10^{-7}$ m. …

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12. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.The emissivity of a perfect black body is increased to 16 times by increasing its temperature. If the initial temperature is T, then final temperature of that black body is (A) 4 T (B) 8 T (C) 2 T (D) 16 T

    ›Reveal solutionSolution

    The Stefan-Boltzmann law says radiated power scales as $T^4$; a 16-fold increase means the temperature only doubles.

    Concept and Intuition

    For a black body, total emissive power $E=\sigma T^4$. "Emissivity increased 16 times" here means the total radiated power (emissive power) went up 16-fold due to the temperature rise — since a perfect black body's emissivity is always 1 and can't itself change, the intended meaning is the $T^4$-scaling power output.

    Step-by-Step Solution

    1. $E=\sigma T^4$ (with emissivity $=1$ for a perfect black body). …

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