Q66 / 75 · JEE Main 2025 (24 Jan Shift 2) · Physics

The temperature of a body in air falls from $40^{\circ} \mathrm{C}$ to $24^{\circ} \mathrm{C}$ in 4 minutes. The temperature of the air is $16^{\circ} \mathrm{C}$. The temperature of the body in the next 4 minutes will be:

  • A
    $\frac{14}{3}{ }^{\circ} \mathrm{C}$
  • B
    $\frac{42}{3}^{\circ} \mathrm{C}$
  • C
    $\frac{28}{3}{ }^{\circ} \mathrm{C}$
  • Correct answer
    D
    $\frac{56}{3}{ }^{\circ} \mathrm{C}$

Solution

$\begin{aligned} & \Delta T=T_2-T_1=16^{\circ} \mathrm{C} \\ & \text { And } T_0=16^{\circ} \mathrm{C} \\ & \frac{\Delta T}{t_1}=-k\left(32-16^{\circ}\right)...(i) \\ & \frac{\left(24-T_3\right)}{4}=-k\left(\frac{24+T_3}{2}-16\right) ....(ii) \\ & \frac{16}{4}=-k(16) \\ & \Rightarrow \frac{\left(24-T_3\right)}{4}=-k\left(12+\frac{T_3}{2}-16\right) \\ & \Rightarrow \frac{16}{24-T_3}=\frac{16}{T_3} T_2 \\ & \Rightarrow \frac{T_3}{2}-4=24-T_3 \\ & \Rightarrow \frac{3 T_3}{2}=28 \\ & \Rightarrow T_3=\frac{56}{3}{ }^{\circ} \mathrm{C} \end{aligned}$

Hence, the correct answer is option: 'D'.

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