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Differential Equations question

2024 · 31 Jan · Shift 2 · Q42
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  5. /2024 · 31 Jan · Shift 2 · Q42

Differential Equations question

2024 · 31 Jan · Shift 2 · Q42

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
The temperature T(t)T(t)T(t) of a body at time t=0t=0t=0 is 160∘F160^{\circ} \mathrm{F}160∘F and it decreases continuously as per the differential equation dTdt=−K(T−80)\frac{d T}{d t}=-K(T-80)dtdT​=−K(T−80), where KKK is a positive constant. If T(15)=120∘FT(15)=120^{\circ} \mathrm{F}T(15)=120∘F, then T(45)T(45)T(45) is equal to
  1. A
    90 ∘^\circ∘ F
  2. B
    85 ∘^\circ∘ F
  3. C
    80 ∘^\circ∘ F
  4. D
    95 ∘^\circ∘ F
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