JEE Main 2024MathematicsDifferential EquationsMediumMCQ

JEE Main 2024 — Differential Equations Question with Solution

JEE Main 2024 (31 Jan Shift 2)

Question

The temperature Tt of a body at time t=0 is 160° F and it decreases continuously as per the differential equation dTdt=−KT−80, where K is positive constant. If T15=120° F, then T45 is equal to

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Show full solutionCorrect option: C
Correct answer
C90° F

Step-by-step explanation

Given: dTdt=−KT−80

⇒dTT−80=−Kdt

⇒∫160TdTT−80=∫0t−Kdt

⇒logT−80160T=−Kt

⇒logT−80−log80=−Kt

⇒logT−8080=−Kt

⇒T=80+80e−Kt

Now, using the value T15=120° we get,

⇒120=80+80e−K·15

⇒4080=e−15k

⇒e−15k=12

∴T45=80+80e−45k

⇒T45=80+80e−15k3

⇒T45=80+80×18

⇒T45=90° F

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About this question

This is a previous-year question from JEE Main 2024, covering the Differential Equations chapter of Mathematics. 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.