Q137 / 150 · MHT CET 2026 (15 April Shift 1) · Physics
An ordinary body cools from $4\theta$ to $3\theta$ in 't' minutes. The temperature of the body after next 't' minutes is (Assume Newton's law of cooling and room temperature as $\theta$)
- A$\dfrac{2\theta}{3}$
- Correct answerB$\dfrac{7\theta}{3}$
- C$\dfrac{5\theta}{3}$
- D$\dfrac{8\theta}{3}$
Solution
According to Newton's law of cooling, the temperature $T$ at time $t$ is given by:
$T - T_s = (T_i - T_s)e^{-kt}$
where $T_s$ is the surrounding temperature and $T_i$ is the initial temperature.
For the first $t$ minutes, $T_i = 4\theta$, $T = 3\theta$, and $T_s = \theta$:
$3\theta - \theta = (4\theta - \theta)e^{-kt}$
$2\theta = 3\theta e^{-kt} \Rightarrow e^{-kt} = \dfrac{2}{3}$
For the next $t$ minutes, the initial temperature is $3\theta$. Let the final temperature be $T'$:
$T' - \theta = (3\theta - \theta)e^{-kt}$
$T' - \theta = 2\theta \left(\dfrac{2}{3}\right) = \dfrac{4\theta}{3}$
$T' = \theta + \dfrac{4\theta}{3} = \dfrac{7\theta}{3}$
Answer: $\dfrac{7\theta}{3}$
$T - T_s = (T_i - T_s)e^{-kt}$
where $T_s$ is the surrounding temperature and $T_i$ is the initial temperature.
For the first $t$ minutes, $T_i = 4\theta$, $T = 3\theta$, and $T_s = \theta$:
$3\theta - \theta = (4\theta - \theta)e^{-kt}$
$2\theta = 3\theta e^{-kt} \Rightarrow e^{-kt} = \dfrac{2}{3}$
For the next $t$ minutes, the initial temperature is $3\theta$. Let the final temperature be $T'$:
$T' - \theta = (3\theta - \theta)e^{-kt}$
$T' - \theta = 2\theta \left(\dfrac{2}{3}\right) = \dfrac{4\theta}{3}$
$T' = \theta + \dfrac{4\theta}{3} = \dfrac{7\theta}{3}$
Answer: $\dfrac{7\theta}{3}$
Hence, the correct answer is option: 'B'.