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 a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is :

According to Newton's Law of Cooling, the rate of cooling of a body is directly proportional to the difference in temperature between the body and its surroundings. For a small temperature interval, we can use the average temperature approximation:

$\frac{T_1-T_2}{t}=K\left(\frac{T_1+T_2}{2}-T_0\right)$

where:
- $T_1$ is the initial temperature,
- $T_2$ is the final temperature,
- $T_0$ is the room (surroundings) temperature,
- $t$ is the time interval, and
- $K$ is a positive constant.

Step 1: Apply the formula to the first case.
The cup of coffee cools from 90°C to 80°C in $t$ minutes at a room temperature of 20°C.
Here, $T_1=90$, $T_2=80$, and $T_0=20$.
Substituting these values into the approximation equation:

$\frac{90-80}{t}=K\left(\frac{90+80}{2}-20\right)$

$\frac{10}{t}=K\left(85-20\right)$

$\frac{10}{t}=65K$ --- (Equation 1)

Step 2: Apply the formula to the second case.
Let the time taken by a similar cup of coffee to cool from 80°C to 60°C at the same room temperature of 20°C be $t\text{'}$.
Here, $T_1=80$, $T_2=60$, and $T_0=20$.
Substituting these values:

$\frac{80-60}{t\text{'}}=K\left(\frac{80+60}{2}-20\right)$

$\frac{20}{t\text{'}}=K\left(70-20\right)$

$\frac{20}{t\text{'}}=50K$ --- (Equation 2)

Step 3: Solve for $t\text{'}$ by dividing Equation 1 by Equation 2.

$\frac{10/t}{20/t\text{'}}=\frac{65K}{50K}$

$\frac{10}{t}\times \frac{t\text{'}}{20}=\frac{65}{50}$

$\frac{t\text{'}}{2t}=\frac{13}{10}$

Multiplying both sides by $2t$:

$t\text{'}=\frac{13}{10}\times 2t$

$t\text{'}=\frac{13}{5}t$

Thus, the correct option is (13/5)t.