Q65 / 75 · JEE Main 2025 (29 Jan Shift 2) · Physics

A cup of coffee cools from $90^{\circ} \mathrm{C}$ to $80^{\circ} \mathrm{C}$ in t minutes when the room temperature is $20^{\circ} \mathrm{C}$. The time taken by the similar cup of coffee to cool from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ at the same room temperature is :

  • A
    $\frac{13}{10} \mathrm{t}$
  • B
    $\frac{10}{13} \mathrm{t}$
  • C
    $\frac{5}{13} \mathrm{t}$
  • Correct answer
    D
    $\frac{13}{5} \mathrm{t}$

Solution

By using average form of Newton's law of cooling $\begin{aligned} & \frac{90-80}{\mathrm{t}}=\mathrm{k}\left(\frac{90+80}{2}-20\right) \\ & \frac{80-60}{\mathrm{t}^{\prime}}=\mathrm{k}\left(\frac{80+60}{2}-20\right) \end{aligned}$ $\begin{aligned} & \text { (i)/(ii) } \\ & \frac{10 \times \mathrm{t}^{\prime}}{\mathrm{t} \times 20}=\frac{65}{50} \\ & \mathrm{t}^{\prime}=\frac{65}{50} \times 2 \mathrm{t}=\frac{65}{25} \mathrm{t}=\frac{13}{5} \mathrm{t} \end{aligned}$

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

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