JEE Main 2019PhysicsThermal Properties of MatterHardMCQ

JEE Main 2019Thermal Properties of Matter Question with Solution

JEE Main 2019 (08 Apr Shift 1)

Question

Two identical beakers A and B contain equal volumes of two different liquids at 60°C each and left to cool down. Liquid in A has density of 8×102 kg m-3 and specific heat of 2000 J kg-1K-1 while the liquid in B has density 103 kg m-3 and specific heat of 4000 J kg-1K-1. Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)

Choose an option

Show full solutionCorrect option: B
Correct answer
B

Step-by-step explanation

From Newton's law of cooling,
Rate of cooling: -dTdt=4σeAT03T-T0ms 1ms  
ρA<ρB
mA<mB
And sA<sB
mAsA<mBsB
At  t=0,-dTdtA>-dTdtB
Initially, the rate of cooling for A will be higher than B and temperature (T)  vs time (t) graph will be a curve.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Thermal Properties of Matter chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2019, covering the Thermal Properties of Matter chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.