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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
View written solutionFree

Correct answer: A

  1. Given differential equation

The temperature satisfies

dTdt=−K(T−80),K>0\frac{dT}{dt}=-K(T-80), \qquad K>0dtdT​=−K(T−80),K>0

with initial condition

T(0)=160.T(0)=160.T(0)=160.

This is Newton's law of cooling.

  1. Solve the differential equation

Let

y=T−80.y=T-80.y=T−80.

Then

dydt=−Ky.\frac{dy}{dt}=-Ky.dtdy​=−Ky.

This is a separable differential equation:

dyy=−K dt.\frac{dy}{y}=-K\,dt.ydy​=−Kdt.

Integrating,

ln⁡∣y∣=−Kt+C.\ln|y|=-Kt+C.ln∣y∣=−Kt+C.

So,

y=Ae−Kty=Ae^{-Kt}y=Ae−Kt

for some constant AAA.

Hence,

T−80=Ae−KtT-80=Ae^{-Kt}T−80=Ae−Kt

or

T(t)=80+Ae−Kt.T(t)=80+Ae^{-Kt}.T(t)=80+Ae−Kt.
  1. Use the initial condition

Since T(0)=160T(0)=160T(0)=160,

160=80+A.160=80+A.160=80+A.

Thus,

A=80.A=80.A=80.

So the temperature function is

T(t)=80+80e−Kt.T(t)=80+80e^{-Kt}.T(t)=80+80e−Kt.
  1. Use the condition T(15)=120T(15)=120T(15)=120

Substitute t=15t=15t=15:

120=80+80e−15K.120=80+80e^{-15K}.120=80+80e−15K.

Therefore,

40=80e−15K40=80e^{-15K}40=80e−15K e−15K=12.e^{-15K}=\frac{1}{2}.e−15K=21​.
  1. Find T(45)T(45)T(45)

Now,

T(45)=80+80e−45K.T(45)=80+80e^{-45K}.T(45)=80+80e−45K.

But

e−45K=(e−15K)3=(12)3=18.e^{-45K}=(e^{-15K})^3=\left(\frac12\right)^3=\frac18.e−45K=(e−15K)3=(21​)3=81​.

So,

T(45)=80+80⋅18=80+10=90.T(45)=80+80\cdot\frac18=80+10=90.T(45)=80+80⋅81​=80+10=90.
  1. Check options
  • A: 90∘F90^\circ F90∘F ✅
  • B: 85∘F85^\circ F85∘F ❌
  • C: 80∘F80^\circ F80∘F ❌
  • D: 95∘F95^\circ F95∘F ❌

Therefore, the correct answer is Option A.

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