JEE Main 2024 — Mathematics Calculus

Given: $\frac{dT}{dt}=-K(T-80)$dtdT​=−K(T−80)

$$\Rightarrow \frac{dT}{(T-80)}=-Kdt$$⇒(T−80)dT​=−Kdt

$$\Rightarrow \int _{160}^{T}\frac{dT}{(T-80)}=\int _{0}^{t}-Kdt$$⇒∫160T​(T−80)dT​=∫0t​−Kdt

$$\Rightarrow {[\mathrm{log}|T-80|]}_{160}^{T}=-Kt$$⇒[log∣T−80∣]160T​=−Kt

$$\Rightarrow \mathrm{log}|T-80|-\mathrm{log}80=-Kt$$⇒log∣T−80∣−log80=−Kt

$$\Rightarrow \mathrm{log}|\frac{T-80}{80}|=-Kt$$⇒log∣80T−80​∣=−Kt

$$\Rightarrow T=80+80{e}^{-Kt}$$⇒T=80+80e−Kt

Now, using the value $T(15)={120}^{^{\circ}}$T(15)=120∘ we get,

$$\Rightarrow 120=80+80{e}^{-K\cdot 15}$$⇒120=80+80e−K⋅15

$$\Rightarrow \frac{40}{80}={e}^{-15k}$$⇒8040​=e−15k

$$\Rightarrow {e}^{-15k}=\frac{1}{2}$$⇒e−15k=21​

$$\therefore T(45)=80+80{e}^{-45k}$$∴T(45)=80+80e−45k

$$\Rightarrow T(45)=80+80{({e}^{-15k})}^{3}$$⇒T(45)=80+80(e−15k)3

$$\Rightarrow T(45)=80+80\times \frac{1}{8}$$⇒T(45)=80+80×81​

$$\Rightarrow T(45)={90}^{^{\circ}}F$$⇒T(45)=90∘F