A cup of coffee cools from 908C to 808C in t minutes, when the room temperature is 208C. The time taken by a similar cup of coffee to cool from 808C to 608C at a room temperature same at 208C is :

The correct option is 13/5 t.

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.

Using the average temperature approximation for Newton's Law of Cooling, we have:
$\frac{T_1-T_2}{t}=K(\frac{T_1+T_2}{2}-T_s)$
where:
- $T_1$ is the initial temperature,
- $T_2$ is the final temperature,
- $T_s$ is the surrounding room temperature,
- $t$ is the time taken, and
- $K$ is a constant.

Case 1: The cup of coffee cools from 90°C to 80°C in time $t$ minutes at room temperature 20°C.
Substituting $T_1=90$, $T_2=80$, and $T_s=20$ into the equation:
$\frac{90-80}{t}=K(\frac{90+80}{2}-20)$
Simplifying this:
$\frac{10}{t}=K(85-20)$
$\frac{10}{t}=65K$ — (Equation 1)

Case 2: The similar cup of coffee cools from 80°C to 60°C in time $t^{\prime }$ minutes at the same room temperature 20°C.
Substituting $T_1=80$, $T_2=60$, and $T_s=20$ into the equation:
$\frac{80-60}{t^{\prime }}=K(\frac{80+60}{2}-20)$
Simplifying this:
$\frac{20}{t^{\prime }}=K(70-20)$
$\frac{20}{t^{\prime }}=50K$ — (Equation 2)

Now, divide Equation 1 by Equation 2 to eliminate $K$:
$\frac{10/t}{20/t^{\prime }}=\frac{65K}{50K}$
$\frac{10}{20}\times \frac{t^{\prime }}{t}=\frac{65}{50}$
$\frac{1}{2}\times \frac{t^{\prime }}{t}=\frac{13}{10}$
Multiplying both sides by 2:
$\frac{t^{\prime }}{t}=\frac{26}{10}=\frac{13}{5}$
$t^{\prime }=\frac{13}{5}t$

Thus, the time taken to cool from 80°C to 60°C is 13/5 t.