Differential Equations jee mains questions | JEE Main Maths Differential Equations Previous Year Question

(2)

Given differential equation is

$\frac{d T}{d t} = - K (T - 80)$

Now, $\frac{d T}{T - 80} = - K d t$

On integrating, we get

$\int \frac{d T}{T - 80} = - K \int d t$

$\Rightarrow \log | T - 80 | = - K t + C$ ... (i)

When t = 0, T = 160º F

$∴ \log | 160 - 80 | = - K \times 0 + C$

$\Rightarrow C = \log (80)$

From (i),

log |T – 80| = – Kt + log (80) ...(ii)

Now, when t = 15, T = 120º F

$∴$ From (ii),

$\log (120 - 80) = - K \times 15 + \log (80)$

$\Rightarrow 15 K = \log (\frac{80}{40}) = \log (2) ∴ k = \frac{\log (2)}{15}$

When t = 45, then from (ii),

$\log (T - 80) = \frac{- \log (2)}{15} \times 45 + \log (80)$

$\Rightarrow \log (T - 80) = - 3 \log (2) + \log (80) = \log (\frac{80}{8}) = \log (10)$

$\Rightarrow \log (T - 80) = 1 \Rightarrow T - 80 = 10^{1}$

$\Rightarrow T = 90 ° F$

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