Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A cup of tea cools from 80 °C to 60 °C in 40 seconds. The ambient temperature is 30 °C. In cooling from 60 °C to 50 °C, It will take time :
PhysicsThermal Properties of MatterJEE Main
Options:
  • A 35 s
     
     
  • B 30 s
     
  • C 32 s
  • D 48 s
     
Solution:
2944 Upvotes Verified Answer
The correct answer is: 32 s
From θ 2 - θ 1 t = k θ 1 + θ 2 2 - θ 0

we have 8 0 - 6 0 4 0 = k 8 0 + 6 0 2 - 3 0

⇒    k = 1 8 0

Now 6 0 - 5 0 t = k 6 0 + 5 0 2 - 3 0

⇒    1 0 t = 1 8 0 2 5 t = 3 2   s

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.