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NTA Abhyas JEE Main2020PhysicsThermal Properties of MatterPractice

The graph shown in the adjacent diagram represents the variation of temperature ( T ) of two bodies x and y having the same surface area, with time ( t ). Both of these bodies lose heat only due to the emission of radiation. Find the correct relation between the emissive and absorptive power of the two bodies.

Options

  1. AE x > E y &#xA0; and &#xA0; a x < a y
  2. BE x < E y &#xA0; and &#xA0; a x > a y
  3. CE x > E y &#xA0; and &#xA0; a x > a y
  4. DE x < E y &#xA0; and &#xA0; a x < a y

Correct answer

C. E x > E y &#xA0; and &#xA0; a x > a y

Step-by-step solution

Rate of cooling - dT dt &#8733; emissivity e From the graph, - dT dt x &#62; - dT dt y &#8756; e x &#62; e y Further emissivity ( e ) &#8733; absorptive power ( a ) (good absorbers are good emitters also) &#8756; a x &#62; a y

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