The temperature {\T}( {\t}) of a body at time {\t}=0 is 160^{\circ} F and it decreases continuously as per the differential equation { {\dT}}/{ {\dt}}=- {\K}( {\T}-80), where {\K} is positive constant. If {\T}(15)=120^{\circ} {\F}, then {\T}(45) is equal to [31-Jan-2024 Shift 2]

JEE Main 31-Jan-2024 Shift 2 Solved Paper

Section: Mathematics

Question : 8 of 90

Marks: +1, -0

The temperature T(t) of a body at time t=0 is 160∘ F and it decreases continuously as per the differential equation ‌

=−K(T−80), where K is positive constant. If T(15)=120∘F, then T(45) is equal to

[31-Jan-2024 Shift 2]

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