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

Correct answer: D. (13)/(5) t By using average form of Newton's law of cooling aligned & 90-80 t =k((90+80)/(2)-20) & 80-60 t^ =k((80+60)/(2)-20) aligned aligned & (i)/(ii) & 10 × t^ t × 20 =(65)/(50) & t^ =(65)/(50) × 2 t=(65)/(25) t=(13)/(5) t aligned

JEE Main 202529 JanuaryEvening Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Evening Shift
Subject: Physics
Chapter: Thermal Properties of Matter

JEE Main 2025 · 29 January · Evening Shift · Physics · Thermal Properties of Matter

A cup of coffee cools from 90^ C to 80^ C in t minutes when the...

A cup of coffee cools from

90^{\circ} \mathrm{C}

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}

60^{\circ} \mathrm{C}

at the same room temperature is :

Options

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

Correct answer

D. $\frac{13}{5} \mathrm{t}$

Step-by-step 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}

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