Exams › JEE Main › Physics

A black sphere of diameter L is enclosed (without touching) inside a black cube of side L, with vacuum in between. The sphere is at temperature T1 and the cube at T2. What is the net rate of radiative heat exchange between them?

1. sigma*pi*L²*(T1⁴ - T2⁴)
2. sigma*L²*(T1⁴ - T2⁴)
3. 6*sigma*L²*(T1⁴ - T2⁴)
4. 4*sigma*pi*L²*(T1⁴ - T2⁴)

Correct answer: sigma*pi*L²*(T1⁴ - T2⁴)

Solution

Net exchange = sigma * A_sphere * (T1⁴ - T2⁴), and the sphere's area is pi*L², giving sigma*pi*L²*(T1⁴ - T2⁴).

Related JEE Main Physics questions

• A cube initially at 0°C is subjected to the same external pressure P on all its faces, causing it to be compressed. How much must its temperature be increased so that the cube regains its original size? The bulk modulus of the cube material is B and its coefficient of linear expansion is α.
• If λm represents the wavelength at which a black body emits radiation most strongly at temperature T K, then which relation is true?
• Which of the following statements about heat transfer is/are incorrect?
• A composite wall consists of two slabs, A and B, of equal thickness but made of different materials. Their thermal conductivities satisfy K_A = 3K_B. If the total temperature difference across the wall is 20°C, then in steady thermal equilibrium:
• An evacuated bulb contains a filament that is 10 cm long and 0.2 mm in diameter. If its emissivity is 0.2, determine the radiant power emitted by the filament at 2000 K. Take σ = 5.67 × 10⁻⁸ W/m² K⁴.
• A rectangular solid is raised in temperature from 0°C to 100°C. If its length increases by 0.10% over this interval, what is the corresponding percentage increase in its volume?

⚔️ Practice JEE Main Physics free + battle 1v1 →