StreakPeaked· Practice

ExamsJEE AdvancedPhysics

A solid metallic sphere of diameter 0.1 m is at a temperature of 427 deg C and is placed in an enclosure at 27 deg C. The emissivity of the surface is 0.84. Given: Stefan constant = (20/3) * 10⁻⁸ W/(m² K⁴), specific heat capacity = 0.1 kcal/(kg K), density = 9280 kg/m³, and 1 kcal = 4200 J. If the initial rate of decrease of temperature of the sphere is N * 10⁻³ deg C/s, find N/40.

  1. 1
  2. 2
  3. 3
  4. 5

Correct answer: 5

Solution

Radius r = 0.05 m. A = 4*pi*(0.05)² = 0.01*pi m². V = (4/3)*pi*(0.05)³ = 5.236*10⁻⁴ m³. Mass m = 9280 * 5.236*10⁻⁴ = 4.859 kg. c = 0.1*4200 = 420 J/(kg K). T = 700 K, T0 = 300 K. T⁴ - T0⁴ = 2401*10⁸ - 81*10⁸ = 2320*10⁸ K⁴. P = 0.84 * (20/3)*10⁻⁸ * 0.01*pi * 2320*10⁸ = 0.84*(20/3)*0.01*pi*2320 = 0.84*6.667*72.89 = 408 W. dT/dt = 408/(4.859*420) = 408/2041 = 0.200 deg C/s = 200*10⁻³ deg C/s. So N = 200, N/40 = 5.

Related JEE Advanced Physics questions

⚔️ Practice JEE Advanced Physics free + battle 1v1 →