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MHT CET 2026 (15 Apr, Shift 1) · Newton's Law of Cooling

Newton's Law of Cooling question from MHT CET 2026 (15 Apr, Shift 1)

Q4MHT CET 2026 · 15 APR · SHIFT 1Medium

Newton's Law of Cooling

An ordinary body cools from 4θ4\theta to 3θ3\theta in 't' minutes. The temperature of the body after next 't' minutes is (Assume Newton's law of cooling and room temperature as θ\theta)

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Answer: (b)

  1. Newton's law: T1−T2t=k(T1+T22−θ)\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - \theta\right).
  2. First tt: θt=k(3.5θ−θ)=2.5kθ\frac{\theta}{t} = k(3.5\theta - \theta) = 2.5k\theta
  3. Next tt: 3θ−Tt=k(3θ+T2−θ)=kθ+T2\frac{3\theta - T}{t} = k\left(\frac{3\theta + T}{2} - \theta\right) = k\frac{\theta + T}{2}
  4. Dividing: 3θ−Tθ=θ+T5θ⇒15θ−5T=θ+T⇒T=7θ3\frac{3\theta - T}{\theta} = \frac{\theta + T}{5\theta} \Rightarrow 15\theta - 5T = \theta + T \Rightarrow T = \frac{7\theta}{3}
  5. So the answer is (b) 7θ3\frac{7\theta}{3}.
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