A body cools from a temperature 3θ to 2θ in 10 minute. The room… | MHT CET 2026 (18 Apr, Shift 2)

Q2MHT CET 2026 · 18 APR · SHIFT 2Medium

Newton's Law of Cooling

A body cools from a temperature $3\theta$ to $2\theta$ in 10 minute. The room temperature is $\theta$ . The temperature of the body at the end of the next 10 minute is ' $x$ '. Assuming that Newton's law of cooling is applicable, the value of ' $x$ ' will be

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Answer: (c)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
First 10 min: $\frac{3\theta - 2\theta}{10} = k(2.5\theta - \theta) \Rightarrow k = \frac{1}{15}$ per min
Next 10 min: $\frac{2\theta - x}{10} = \frac{1}{15}\left(\frac{2\theta + x}{2} - \theta\right) = \frac{x}{30}$
$2\theta - x = \frac{x}{3} \Rightarrow x = \frac{3}{2}\theta$
So the answer is (c) $\frac{3}{2}\theta$ .

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