JEE Main 2021 · 27 July · morning · Single correct · +4 / −1

A body takes 4 min. to cool from 61° C to 59° C. If the temperature of the surroundings is 30° C, the time taken by the body to cool from 51° C to 49° C is :

  1. (a) 4 min.
  2. (b) 3 min.
  3. (c) 8 min.
  4. (d) 6 min.
Answer and solution

Answer (d) 6 min.

Solution

Physics sir · 4 steps

Same 2 degree drop both times, so compare how far each average temperature sits above the 30 °C room.

  1. Newton's law. For a small drop, take the rate of cooling as the drop divided by the time. Set it proportional to the average temperature minus the surroundings: (Δ T)/t = k(T_(avg) - T₀).
  2. First interval. From 61 °C to 59 °C the average is 60 °C, which is 30 °C above the room. So 2/4 = k × 30.
  3. Second interval. From 51 °C to 49 °C the average is 50 °C, only 20 °C above the room. So 2/t = k × 20.

    (2/4)/(2/t) = 30/20 ⇒ t = 4 × 30/20 = 6 min

  4. So. The body is closer to room temperature, so it cools more slowly and needs 6 min, option (d).

The idea to keep. For equal small drops, the time taken is inversely proportional to the average excess temperature over the surroundings. A body nearer room temperature cools more slowly.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Heat Transfer

Asked about 30 times in JEE Main since 2020. This question is from 2021.