Easy MCQ +4 / -1 PYQ · JEE Mains 2023

A bowl filled with very hot soup cools from 98$^\circ$C to 86$^\circ$C in 2 minutes when the room temperature is 22$^\circ$C. How long it will take to cool from 75$^\circ$C to 69$^\circ$C?

  1. A 2 minutes
  2. B 0.5 minute
  3. C 1.4 minutes Correct answer
  4. D 1 minute

Solution

From Newton's law of cooling. <br/><br/> $\frac{d T}{d t}=-k\left(T-T_{s}\right)$ <br/><br/> Case $\mathrm{I}: d T=12^{\circ} \mathrm{C}, d t=2 \min$ <br/><br/> $\frac{12}{2}=-k\left[92-22^{\circ}\right]=-k 70$ <br/><br/> Case II : $d T=6^{\circ} \mathrm{C}$ <br/><br/> $\frac{6}{d t}=-k[72-22]=-k 50$ <br/><br/> From (1) and (2) <br/><br/> $d t=1.4 \mathrm{~min}$

About this question

Subject: Physics · Chapter: Properties of Solids and Liquids · Topic: Elasticity

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